- This Solutions Manual for Active Calculus is organized by section.
- Preview Activity solutions are not provided,
- but each activity within a section has a solution provided.
- The solution to each activity starts at the top of a new page,
- so instructors can easily extract the solution for a single activity if they wish to print it for students or post on their course's learning management system site.
- After the activity solutions come the exercise solutions.
- Solutions for
- exercises are not provided,
- since those solutions are automatically available in this book's HTML version.
- Each exercise also starts at the top of its own page to make it easy to extract only the solutions to certain exercises.
-
-
-
- Please do not post this solutions manual publicly on the internet nor in any electronic form where it is available in full to students.
- As much as possible,
- we aspire to keep this solutions manual as a resource for instructors only so that students get the full benefit of activities and exercises by having to struggle with them without looking at solutions.
-
-
-
- By extracting individual pages from the PDF,
- it is fine to share solutions to individual activities or exercises via course management system software.
- Please do not post these individual pages publicly on the internet.
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- This book was authored in .
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+ \newcommand{\dollar}{\$}
+ \DeclareMathOperator{\erf}{erf}
+ \DeclareMathOperator{\arctanh}{arctanh}
+ \DeclareMathOperator{\arcsec}{arcsec}
+
+ % This macro should only be used in ancillary production!
+ \raggedbottom
+
+
+
+ Preview Activity
+ Motivating Questions
+
+
+
+ <line>Solutions Manual</line>
+ <line>for</line>
+ <line>Active Calculus Single Variable</line>
+
+
+
+
+
+
+ Matthew Boelkins
+ Department of Mathematics
+ Grand Valley State University
+ boelkinm@gvsu.edu
+
+
+
+ Contributing Authors
+
+ Steven Schlicker
+ Department of Mathematics
+ Grand Valley State University
+ schlicks@gvsu.edu
+
+
+
+ Ted Sundstrom
+ Department of Mathematics
+ Grand Valley State University
+ sundstrt@gvsu.edu
+
+
+
+
+
+ Production Editor
+
+ Mitchel T. Keller
+ Department of Mathematics
+ University of Wisconsin-Madison
+ mitch@rellek.net
+
+
+
+
+
+
+
+
+
+
+ This Solutions Manual for Active Calculus is organized by section.
+ Preview Activity solutions are not provided,
+ but each activity within a section has a solution provided.
+ The solution to each activity starts at the top of a new page,
+ so instructors can easily extract the solution for a single activity if they wish to print it for students or post on their course's learning management system site.
+ After the activity solutions come the exercise solutions.
+ Solutions for
+ exercises are not provided,
+ since those solutions are automatically available in this book's HTML version.
+ Each exercise also starts at the top of its own page to make it easy to extract only the solutions to certain exercises.
+
+
+
+ Please do not post this solutions manual publicly on the internet nor in any electronic form where it is available in full to students.
+ As much as possible,
+ we aspire to keep this solutions manual as a resource for instructors only so that students get the full benefit of activities and exercises by having to struggle with them without looking at solutions.
+
+
+
+ By extracting individual pages from the PDF,
+ it is fine to share solutions to individual activities or exercises via course management system software.
+ Please do not post these individual pages publicly on the internet.
+
- Sketch at least two different possible graphs that satisfy the criteria for the function stated in each part. Make your graphs as significantly different as you can. If it is impossible for a graph to satisfy the criteria, explain why.
-
-
-
-
-
-
f is a function defined on [-1,7] such that f(1) = 4 and AV_{[1,3]} = -2.
-
-
-
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-
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-
-
-
-
-
-
- Since AV_{[1,3]} = -2 and f(1) = 4, we need
- \frac{f(3)-4}{3-1} = -2, so f(3) = 0. Any graph passing through
- (1,4) and (3,0) satisfies the conditions. Two examples:
- the line f(x) = -2x+6, or a curve such as f(x) = -(x-1)^2 + 4
- (which also passes through both required points).
-
-
-
-
-
-
-
-
-
-
-
g is a function defined on [-1,7] such that g(4) = 3, AV_{[0,4]} = 0.5, and g is not always increasing on (0,4).
-
-
-
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-
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-
-
-
-
-
-
- Since AV_{[0,4]} = 0.5 and g(4) = 3, we need
- \frac{3 - g(0)}{4} = 0.5, so g(0) = 1. Any graph passing through
- (0,1) and (4,3) that is not always increasing on (0,4)
- satisfies the conditions. For example, a curve that dips below the secant line
- between x=0 and x=4 before returning to (4,3).
-
-
-
-
-
-
-
-
-
-
-
h is a function defined on [-1,7] such that h(2) = 5, h(4) = 3 and AV_{[2,4]} = -2.
-
-
-
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-
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-
-
-
-
-
-
- These conditions are impossible to satisfy simultaneously. Given h(2) = 5
- and h(4) = 3, the average rate of change is determined:
- AV_{[2,4]} = \frac{h(4) - h(2)}{4-2} = \frac{3-5}{2} = -1 \ne -2.
- No function can have both the specified output values and the specified average
- rate of change on [2,4].
-
+ Sketch at least two different possible graphs that satisfy the criteria for the function stated in each part. Make your graphs as significantly different as you can. If it is impossible for a graph to satisfy the criteria, explain why.
+
+
+
+
+
+
f is a function defined on [-1,7] such that f(1) = 4 and AV_{[1,3]} = -2.
+
+
+
ADD ALT TEXT TO THIS IMAGE
+
ADD ALT TEXT TO THIS IMAGE
+
+
+
+
+ Any graph passing through (1,4) and (3,0) satisfies the conditions.
+
+
+
+ Since AV_{[1,3]} = -2 and f(1) = 4, we need
+ \frac{f(3)-4}{3-1} = -2, so f(3) = 0. Any graph passing through
+ (1,4) and (3,0) satisfies the conditions. Two examples:
+ the line f(x) = -2x+6, or a curve such as f(x) = -(x-1)^2 + 4
+ (which also passes through both required points).
+
+
+
+
+
+
+
g is a function defined on [-1,7] such that g(4) = 3, AV_{[0,4]} = 0.5, and g is not always increasing on (0,4).
+
+
+
ADD ALT TEXT TO THIS IMAGE
+
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+
+
+
+
+ Any graph passing through (0,1) and (4,3) that is not always increasing on (0,4) satisfies the conditions.
+
+
+
+ Since AV_{[0,4]} = 0.5 and g(4) = 3, we need
+ \frac{3 - g(0)}{4} = 0.5, so g(0) = 1. Any graph passing through
+ (0,1) and (4,3) that is not always increasing on (0,4)
+ satisfies the conditions. For example, a curve that dips below the secant line
+ between x=0 and x=4 before returning to (4,3).
+
+
+
+
+
+
+
+
+
+
h is a function defined on [-1,7] such that h(2) = 5, h(4) = 3 and AV_{[2,4]} = -2.
+
+
+
ADD ALT TEXT TO THIS IMAGE
+
ADD ALT TEXT TO THIS IMAGE
+
+
+
+
+ These conditions are impossible to satisfy simultaneously.
+
+
+
+ These conditions are impossible to satisfy simultaneously. Given h(2) = 5
+ and h(4) = 3, the average rate of change is determined:
+ AV_{[2,4]} = \frac{h(4) - h(2)}{4-2} = \frac{3-5}{2} = -1 \ne -2.
+ No function can have both the specified output values and the specified average
+ rate of change on [2,4].
+
- According to the US census,
- the populations of Kent and Ottawa Counties in Michigan where GVSU is located
- (Grand Rapids is in Kent, Allendale in Ottawa)
- from 1960 to 2010 measured in 10-year intervals are given in the following tables.
-
- Let K(Y) represent the population of Kent County in year Y and W(Y) the population of Ottawa County in year Y.
-
-
-
-
-
-
- Compute AV_{[1990,2010]} for both K and W.
-
-
-
-
-
-
- For Kent County, the population in 1990 was 500,631 and in 2010 was 602,622, so
-
- AV_{[1990,2010]} &= \frac{602{,}622 - 500{,}631}{2010 - 1990}
- &= \frac{101{,}991}{20}
- &\approx 5099.55 \text{ people per year.}
-
-
-
- For Ottawa County, the population in 1990 was 187,768 and in 2010 was 263,801, so
-
- AV_{[1990,2010]} &= \frac{263{,}801 - 187{,}768}{2010 - 1990}
- &= \frac{76{,}033}{20}
- &\approx 3801.65 \text{ people per year.}
-
-
-
-
-
-
-
- What are the units on each of the quantities you computed in (a.)?
-
-
-
-
-
-
- The units on AV_{[1990,2010]} for both counties are people per year.
-
-
-
-
-
-
- Write a careful sentence that explains the meaning of the average rate of change of the Ottawa county population on the time interval [1990,2010]. Your sentence should begin something like
- In an average year between 1990 and 2010, the population of Ottawa County was \ldots
-
-
-
-
-
- Between 1990 and 2010, the population of Ottawa County grew by an average of
- approximately 3801.65 people per year.
-
-
-
-
-
-
- Which county had a greater average rate of change during the time interval [2000,2010]?
- Were there any intervals in which one of the counties had a negative average rate of change?
-
-
-
-
-
-
- For Kent County on [2000,2010]:
- AV_{[2000,2010]} = \frac{602{,}622 - 574{,}336}{10} = \frac{28{,}286}{10} \approx 2828.6 \text{ people per year.}
- For Ottawa County on [2000,2010]:
- AV_{[2000,2010]} = \frac{263{,}801 - 238{,}313}{10} = \frac{25{,}488}{10} = 2548.8 \text{ people per year.}
- Kent County had a greater average rate of change during [2000,2010].
- In both counties, the population increased in every decade from 1960 to 2010,
- so the average rate of change was positive on every decade interval for both counties.
-
-
-
-
-
-
-
-
- Using the given data, what do you predict will be the population of Ottawa County in 2018? Why?
-
-
-
-
-
-
- Using the most recent decade's rate of change as a guide, Ottawa County grew at
- 2548.8 people per year from 2000 to 2010. Projecting this rate forward
- eight years from 2010:
- 263{,}801 + 8 \cdot 2548.8 \approx 284{,}191 \text{ people.}
-
+ According to the US census,
+ the populations of Kent and Ottawa Counties in Michigan where GVSU is located
+ (Grand Rapids is in Kent, Allendale in Ottawa)
+ from 1960 to 2010 measured in 10-year intervals are given in the following tables.
+
+ Let K(Y) represent the population of Kent County in year Y and W(Y) the population of Ottawa County in year Y.
+
+
+
+
+
+
+ Compute AV_{[1990,2010]} for both K and W.
+
+
+
+
+ For Kent County, AV_{[1990,2010]} \approx 5099.55 people per year.
+ For Ottawa County, AV_{[1990,2010]} \approx 3801.65 people per year.
+
+
+
+ For Kent County, the population in 1990 was 500,631 and in 2010 was 602,622, so
+
+ AV_{[1990,2010]} &= \frac{602{,}622 - 500{,}631}{2010 - 1990}
+ &= \frac{101{,}991}{20}
+ &\approx 5099.55 \text{ people per year.}
+
+
+
+ For Ottawa County, the population in 1990 was 187,768 and in 2010 was 263,801, so
+
+ AV_{[1990,2010]} &= \frac{263{,}801 - 187{,}768}{2010 - 1990}
+ &= \frac{76{,}033}{20}
+ &\approx 3801.65 \text{ people per year.}
+
+
+
+
+
+
+
+ What are the units on each of the quantities you computed in (a.)?
+
+
+
+
+ The units are people per year for both.
+
+
+
+ The units on AV_{[1990,2010]} for both counties are people per year.
+
+
+
+
+
+
+ Write a careful sentence that explains the meaning of the average rate of change of the Ottawa county population on the time interval [1990,2010]. Your sentence should begin something like
+ In an average year between 1990 and 2010, the population of Ottawa County was \ldots
+
+
+
+ In an average year between 1990 and 2010, the population of Ottawa County was increasing by approximately 3801.65 people per year.
+
+
+
+ In an average year between 1990 and 2010, the population of Ottawa County was increasing by approximately 3801.65 people per year.
+
+
+
+
+
+
+ Which county had a greater average rate of change during the time interval [2000,2010]?
+ Were there any intervals in which one of the counties had a negative average rate of change?
+
+
+
+
+ Kent County had a greater average rate of change during the time interval [2000,2010].
+ There were no intervals in which either county had a negative average rate of change.
+
+
+
+ For Kent County on [2000,2010]:
+ AV_{[2000,2010]} = \frac{602{,}622 - 574{,}336}{10} = \frac{28{,}286}{10} \approx 2828.6 \text{ people per year.}
+ For Ottawa County on [2000,2010]:
+ AV_{[2000,2010]} = \frac{263{,}801 - 238{,}313}{10} = \frac{25{,}488}{10} = 2548.8 \text{ people per year.}
+ Kent County had a greater average rate of change during [2000,2010].
+ In both counties, the population increased in every decade from 1960 to 2010,
+ so the average rate of change was positive on every decade interval for both counties.
+
+
+
+
+
+
+
+
+ Using the given data, what do you predict will be the population of Ottawa County in 2018? Why?
+
+
+
+
\approx 284{,}191 \text{ people.}
+
+
+ Using the most recent decade's rate of change as a guide, Ottawa County grew at
+ 2548.8 people per year from 2000 to 2010. Projecting this rate forward
+ eight years from 2010:
+ 263{,}801 + 8 \cdot 2548.8 \approx 284{,}191 \text{ people.}
+
- Let's consider two different functions and see how different computations of their average rate of change tells us about their respective behavior. Plots of q and h are shown in the figures in part (c).
-
-
-
-
-
-
- Consider the function q(x) = 4-(x-2)^2.
- Compute AV_{[0,1]},
- AV_{[1,2]}, AV_{[2,3]}, and AV_{[3,4]}.
- What do your last two computations tell you about the behavior of the function q on [2,4]?
-
-
-
-
-
-
- Computing the function values: q(0)=0, q(1)=3, q(2)=4,
- q(3)=3, q(4)=0. Thus:
- AV_{[0,1]} = \frac{3-0}{1} = 3, \quad AV_{[1,2]} = \frac{4-3}{1} = 1,
- AV_{[2,3]} = \frac{3-4}{1} = -1, \quad AV_{[3,4]} = \frac{0-3}{1} = -3.
- Since AV_{[2,3]} and AV_{[3,4]} are both negative, q is
- decreasing on [2,4].
-
-
-
-
-
-
- Consider the function h(t) = 3 - 2(0.5)^t.
- Compute AV_{[-1,1]},
- AV_{[1,3]}, and AV_{[3,5]}.
- What do your computations tell you about the behavior of the function h on [-1,5]?
-
-
-
-
-
-
- Computing function values: h(-1) = 3 - 2(0.5)^{-1} = 3 - 4 = -1,
- h(1) = 3 - 2(0.5) = 2, h(3) = 3 - 2(0.5)^3 = \frac{11}{4} \approx 2.75,
- h(5) = 3 - 2(0.5)^5 = \frac{47}{16} \approx 2.9375. Thus:
- AV_{[-1,1]} = \frac{2-(-1)}{2} = \frac{3}{2} = 1.5,
- AV_{[1,3]} = \frac{\frac{11}{4} - 2}{2} = \frac{3}{8} \approx 0.375,
- AV_{[3,5]} = \frac{\frac{47}{16} - \frac{11}{4}}{2} = \frac{3}{32} \approx 0.094.
- All three average rates of change are positive but decreasing, so h is
- increasing on [-1,5], but at a decreasing rate.
-
-
-
-
-
-
- On the graphs that follow (q at left, h at right), plot the line segments whose respective slopes are the average rates of change you computed in (a) and (b).
-
-
-
-
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-
-
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-
-
-
-
-
-
-
On the graph of q, draw four line segments: one connecting (0,0) and (1,3) (slope 3), one connecting (1,3) and (2,4) (slope 1), one connecting (2,4) and (3,3) (slope -1), and one connecting (3,3) and (4,0) (slope -3). On the graph of h, draw three line segments: one connecting (-1,-1) and (1,2) (slope 1.5), one connecting (1,2) and (3,2.75) (slope 0.375), and one connecting (3,2.75) and (5,2.9375) (slope 0.094). The decreasing slopes on the left graph reflect that q increases then decreases, while the positive but decreasing slopes on the right reflect that h is increasing at a decreasing rate.
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
- True or false: Since AV_{[0,3]} = 1, the function q is increasing on the interval (0,3). Justify your decision.
-
-
-
-
-
-
- False. Although AV_{[0,3]} = \frac{q(3)-q(0)}{3} = \frac{3-0}{3} = 1 > 0,
- a positive average rate of change over an interval does not mean the function is
- increasing everywhere on that interval. From part (a), AV_{[2,3]} = -1 < 0,
- so q is actually decreasing on [2,3].
-
-
-
-
-
-
- Give an example of a function that has the same average rate of change no matter what interval you choose.
- You can provide your example through a table, a graph,
- or a formula;
- regardless of your choice, write a sentence to explain.
-
-
-
-
-
-
- Any linear function has constant average rate of change. For example,
- f(x) = x has AV_{[a,b]} = \frac{b-a}{b-a} = 1 for every interval
- [a,b].
-
-
-
-
-
-
-
-
See solutions to individual tasks above.
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ Let's consider two different functions and see how different computations of their average rate of change tells us about their respective behavior. Plots of q and h are shown in the figures in part (c).
+
+
+
+
+
+
+ Consider the function q(x) = 4-(x-2)^2.
+ Compute AV_{[0,1]},
+ AV_{[1,2]}, AV_{[2,3]}, and AV_{[3,4]}.
+ What do your last two computations tell you about the behavior of the function q on [2,4]?
+
+
+
+
+ AV_{[0,1]} = 3, AV_{[1,2]} = 1, AV_{[2,3]} = -1, and AV_{[3,4]} = -3.
+ The function q is decreasing on [2,4].
+
+
+
+ Computing the function values: q(0)=0, q(1)=3, q(2)=4,
+ q(3)=3, q(4)=0. Thus:
+ AV_{[0,1]} = \frac{3-0}{1} = 3, \quad AV_{[1,2]} = \frac{4-3}{1} = 1,
+ AV_{[2,3]} = \frac{3-4}{1} = -1, \quad AV_{[3,4]} = \frac{0-3}{1} = -3.
+ Since AV_{[2,3]} and AV_{[3,4]} are both negative, q is
+ decreasing on [2,4].
+
+
+
+
+
+
+ Consider the function h(t) = 3 - 2(0.5)^t.
+ Compute AV_{[-1,1]},
+ AV_{[1,3]}, and AV_{[3,5]}.
+ What do your computations tell you about the behavior of the function h on [-1,5]?
+
+
+
+
+ AV_{[-1,1]} = 1.5, AV_{[1,3]} \approx 0.375, and AV_{[3,5]} \approx 0.094.
+ On [-1,5], the function h is increasing but at a decreasing rate.
+
+
+
+ Computing function values: h(-1) = 3 - 2(0.5)^{-1} = 3 - 4 = -1,
+ h(1) = 3 - 2(0.5) = 2, h(3) = 3 - 2(0.5)^3 = \frac{11}{4} \approx 2.75,
+ h(5) = 3 - 2(0.5)^5 = \frac{47}{16} \approx 2.9375. Thus:
+ AV_{[-1,1]} = \frac{2-(-1)}{2} = \frac{3}{2} = 1.5,
+ AV_{[1,3]} = \frac{\frac{11}{4} - 2}{2} = \frac{3}{8} \approx 0.375,
+ AV_{[3,5]} = \frac{\frac{47}{16} - \frac{11}{4}}{2} = \frac{3}{32} \approx 0.094.
+ All three average rates of change are positive but decreasing, so h is
+ increasing on [-1,5], but at a decreasing rate.
+
+
+
+
+
+
+ On the graphs that follow (q at left, h at right), plot the line segments whose respective slopes are the average rates of change you computed in (a) and (b).
+
+
+
+
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+
+
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+
+
+
+
+
+
+ (a)
+
+
+ plot of average rate of change on the graph of q
+ plot of average rate of change on the graph of q
+ plot of average rate of change on the graph of q
+
+
+ (b)
+
+
+ plot of average rate of change on the graph of h
+ plot of average rate of change on the graph of h
+ plot of average rate of change on the graph of h
+
+
+
+
On the graph of q, draw four line segments: one connecting (0,0) and (1,3) (slope 3), one connecting (1,3) and (2,4) (slope 1), one connecting (2,4) and (3,3) (slope -1), and one connecting (3,3) and (4,0) (slope -3). On the graph of h, draw three line segments: one connecting (-1,-1) and (1,2) (slope 1.5), one connecting (1,2) and (3,2.75) (slope 0.375), and one connecting (3,2.75) and (5,2.9375) (slope 0.094). The decreasing slopes on the left graph reflect that q increases then decreases, while the positive but decreasing slopes on the right reflect that h is increasing at a decreasing rate.
+
+
+ plot of average rate of change on the graph of q
+ plot of average rate of change on the graph of q
+ plot of average rate of change on the graph of q
+
+
+ plot of average rate of change on the graph of h
+ plot of average rate of change on the graph of h
+ plot of average rate of change on the graph of h
+
+
+
+
+
+
+ True or false: Since AV_{[0,3]} = 1, the function q is increasing on the interval (0,3). Justify your decision.
+
+
+
+
+
+ False.
+
+
+
+
+ False. Although AV_{[0,3]} = \frac{q(3)-q(0)}{3} = \frac{3-0}{3} = 1 > 0,
+ a positive average rate of change over an interval does not mean the function is
+ increasing everywhere on that interval. From part (a), AV_{[2,3]} = -1 < 0,
+ so q is actually decreasing on [2,3].
+
+
+
+
+
+
+ Give an example of a function that has the same average rate of change no matter what interval you choose.
+ You can provide your example through a table, a graph,
+ or a formula;
+ regardless of your choice, write a sentence to explain.
+
+
+
+
Answers will vary but will be linear.
+
+
+ Any linear function has constant average rate of change. For example,
+ f(x) = x has AV_{[a,b]} = \frac{b-a}{b-a} = 1 for every interval
+ [a,b].
+
- Consider the functions f and g defined by the following figures. Assume that the given lines and curves pass through intersection points on the grid when it looks plausible. For instance, (0,2.5) and (3,-0.5) lie on the graph of f, and (-1,3) and (1.5, 1.5) lie on the graph of g.
-
-
-
-
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-
-
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-
-
-
-
-
-
-
- Determine the exact value of (f+g)(0).
-
-
-
-
-
-
- From the graphs, f(0) = \frac{5}{2} and g(0) = 1, so
- (f+g)(0) = \frac{5}{2} + 1 = \frac{7}{2}.
-
-
-
-
-
-
- Determine the exact value of (g-f)(1).
-
-
-
-
-
-
- From the graphs, f(1) = 4 and g(1) = 3, so
- (g-f)(1) = 3 - 4 = -1.
-
-
-
-
-
-
- Determine the exact value of (f \cdot g)(-1).
-
-
-
-
-
-
- From the graphs, f(-1) = 3 and g(-1) = 3, so
- (f \cdot g)(-1) = 3 \cdot 3 = 9.
-
-
-
-
-
-
- Are there any values of x for which \left( \frac{f}{g} \right)(x) is undefined? If not, explain why. If so, determine the values and justify your answer.
-
-
-
-
-
-
- \left(\frac{f}{g}\right)(x) is undefined wherever g(x) = 0.
- From the graph, g(x) = 0 at x = -2 and x = 3.
-
-
-
-
-
-
- For what values of x is (f \cdot g)(x) = 0? Why?
-
-
-
-
-
-
- (f \cdot g)(x) = 0 wherever f(x) = 0 or g(x) = 0.
- From the graphs, f(x) = 0 at x = \frac{7}{3}, and
- g(x) = 0 at x = -2 and x = 3.
- So (f \cdot g)(x) = 0 at x \in \left\{-2,\, \frac{7}{3},\, 3\right\}.
-
-
-
-
-
-
- Are there any values of x for which (f-g)(x) = 0? Why or why not?
-
-
-
-
-
-
- Yes: (f-g)(x) = 0 when f(x) = g(x).
- One intersection is visible from the graphs at x = -1.
- On the interval -1 \lt x \le 1, the formulas are f(x) = \frac{5}{2} + x
- and g(x) = -x^2 + 4; setting them equal gives:
-
- x^2 + x - \frac{3}{2} &= 0
- x &= \frac{-1 + \sqrt{7}}{2}.
-
- So (f-g)(x) = 0 at x = -1 and x = \dfrac{\sqrt{7}-1}{2}.
-
+ Consider the functions f and g defined by the following figures. Assume that the given lines and curves pass through intersection points on the grid when it looks plausible. For instance, (0,2.5) and (3,-0.5) lie on the graph of f, and (-1,3) and (1.5, 1.5) lie on the graph of g.
+
+
+
+
+
+
+
A piecewise linear function f(x). The first piece goes through (-2,3.5) and ends at the closed circle (-1,3), while the second piece starts at an open circle at (-1,1) and ends at a closed circle at (1,4), and the third piece starts at an open circle at (1,1) passes through an open circle at (2,1/4) and continues to the edge of the graph.
+
+
+
+
+
+
+
+
A piecewise function g(x). The first piece is a parabola opening downward, with an open circle at the vertex (0,4) and a closed circle at the point (0,1). The first piece ends at a closed circle at (1,3). The second piece is a line starting at an open circle at (1,2) passing through an open circle at (2,1), and continuing to the edge of the graph.
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ Determine the exact value of (f+g)(0).
+
+
+
+
(f+g)(0) = \frac{7}{2}
+
+
+ From the graphs, f(0) = \frac{5}{2} and g(0) = 1, so
+ (f+g)(0) = \frac{5}{2} + 1 = \frac{7}{2}.
+
+
+
+
+
+
+ Determine the exact value of (g-f)(1).
+
+
+
+
(g-f)(1) = -1
+
+
+ From the graphs, f(1) = 4 and g(1) = 3, so
+ (g-f)(1) = 3 - 4 = -1.
+
+
+
+
+
+
+ Determine the exact value of (f \cdot g)(-1).
+
+
+
+
(f \cdot g)(-1) = 3
+
+
+ From the graphs, f(-1) = 1 and g(-1) = 3, so
+ (f \cdot g)(-1) = 1 \cdot 3 = 3.
+
+
+
+
+
+
+ Are there any values of x for which \left( \frac{f}{g} \right)(x) is undefined? If not, explain why. If so, determine the values and justify your answer.
+
+
+
+
x = -2 and x = 3
+
+
+ \left(\frac{f}{g}\right)(x) is undefined wherever g(x) = 0.
+ From the graph, g(x) = 0 at x = -2 and x = 3.
+
+
+
+
+
+
+ For what values of x is (f \cdot g)(x) = 0? Why?
+
+
+
+
(f \cdot g)(x) = 0 at x= -2,\, \frac{7}{3},\, 3
+
+
+ (f \cdot g)(x) = 0 wherever f(x) = 0 or g(x) = 0.
+ From the graphs, f(x) = 0 at x = \frac{7}{3}, and
+ g(x) = 0 at x = -2 and x = 3.
+ So (f \cdot g)(x) = 0 at x= -2,\, \frac{7}{3},\, 3.
+
+
+
+
+
+
+ Are there any values of x for which (f-g)(x) = 0? Why or why not?
+
+
+
+
(f-g)(x) = 0 at one value between x=0 and x=1. Using algebraic formulas for the functions, x = \dfrac{\sqrt{7}-1}{2}
+
+
+ Yes: (f-g)(x) = 0 when f(x) = g(x).
+ On the interval 0 \lt x \le 1, the function f starts out below the function g and ends up above the function g with no jumps or holes, so there must have been a place where the two functions crossed. The formulas are f(x) = \frac{5}{2} + x
+ and g(x) = -x^2 + 4; setting them equal gives:
+
+ x^2 + x - \frac{3}{2} &= 0
+ x &= \frac{-1 + \sqrt{7}}{2}.
+
+ So (f-g)(x) = 0 at x = \dfrac{\sqrt{7}-1}{2}, and it looks like they'll cross again off the right side of the graph as well if both functions continue.
+
- Let f be a function that measures a car's fuel economy in the following way.
- Given an input velocity v in miles per hour,
- f(v) is the number of gallons of fuel that the car consumes per mile (i.e., gallons per mile).
- We know that f(60) = 0.04.
-
-
-
-
-
-
- What is the meaning of the statement
- f(60) = 0.04
- in the context of the problem?
- That is, what does this say about the car's fuel economy?
- Write a complete sentence to explain.
-
-
-
-
-
-
- At a speed of 60 miles per hour, the car consumes 0.04 gallons of fuel
- for each mile traveled.
-
-
-
-
-
-
- Consider the function g(v) = \frac{1}{f(v)}.
- What is the value of g(60)?
- What are the units on g?
- What does g measure?
-
-
-
-
-
-
- g(60) = \frac{1}{f(60)} = \frac{1}{0.04} = 25 miles per gallon.
- The function g measures the car's fuel economy in miles per gallon
- at a given speed.
-
-
-
-
-
-
- Consider the function h(v) = v \cdot f(v).
- What is the value of h(60)?
- What are the units on h?
- What does h measure?
-
-
-
-
-
-
- h(60) = 60 \cdot f(60) = 60 \cdot 0.04 = 2.4 gallons per hour.
- The units follow from: miles/hour \times gallons/mile = gallons/hour.
- The function h measures the rate at which the car consumes fuel
- (in gallons per hour) at a given speed.
-
-
-
-
-
-
- Do f(60), g(60), and h(60) tell us fundamentally different information, or are they all essentially saying the same thing? Explain.
-
-
-
-
-
-
- All three convey information about fuel consumption at 60 mph, but with
- different units and perspectives: f(60) = 0.04 gives gallons per mile,
- g(60) = 25 gives miles per gallon, and h(60) = 2.4 gives
- gallons per hour. They describe the same phenomenon from different angles.
-
-
-
-
-
-
- Suppose we also know that f(70) = 0.045.
- Find the average rate of change of f on the interval [60,70].
- What are the units on the average rate of change of f?
- What does this quantity measure?
- Write a complete sentence to explain.
-
-
-
-
-
-
- AV_{[60,70]} = \frac{f(70)-f(60)}{70-60} = \frac{0.045 - 0.04}{10} = 0.0005
- \ \frac{\text{gal/mi}}{\text{mph}}.
- This measures how much the car's fuel consumption rate (in gallons per mile)
- increases per additional mile per hour of speed.
-
+ Let f be a function that measures a car's fuel economy in the following way.
+ Given an input velocity v in miles per hour,
+ f(v) is the number of gallons of fuel that the car consumes per mile (i.e., gallons per mile).
+ We know that f(60) = 0.04.
+
+
+
+
+
+
+ What is the meaning of the statement
+ f(60) = 0.04
+ in the context of the problem?
+ That is, what does this say about the car's fuel economy?
+ Write a complete sentence to explain.
+
+
+
+
+ At a speed of 60 miles per hour, the car consumes 0.04 gallons of fuel
+ for each mile traveled.
+
+
+
+
+ At a speed of 60 miles per hour, the car consumes 0.04 gallons of fuel
+ for each mile traveled.
+
+
+
+
+
+
+ Consider the function g(v) = \frac{1}{f(v)}.
+ What is the value of g(60)?
+ What are the units on g?
+ What does g measure?
+
+
+
+
+ g(60) = 25 miles per gallon.
+ The function g measures the car's fuel economy in miles per gallon
+ at a given input speed.
+
+
+
+
+ g(60) = \frac{1}{f(60)} = \frac{1}{0.04} = 25 miles per gallon.
+ The function g measures the car's fuel economy in miles per gallon
+ at a given speed.
+
+
+
+
+
+
+ Consider the function h(v) = v \cdot f(v).
+ What is the value of h(60)?
+ What are the units on h?
+ What does h measure?
+
+
+
+
+ h(60) = 2.4 gallons per hour.
+ The function h measures the rate at which the car consumes fuel
+ (in gallons per hour) at a given speed.
+
+
+
+
+ h(60) = 60 \cdot f(60) = 60 \cdot 0.04 = 2.4 gallons per hour.
+ The units follow from: miles/hour \times gallons/mile = gallons/hour.
+ The function h measures the rate at which the car consumes fuel
+ (in gallons per hour) at a given speed.
+
+
+
+
+
+
+ Do f(60), g(60), and h(60) tell us fundamentally different information, or are they all essentially saying the same thing? Explain.
+
+
+
+
+ All three convey information about the fuel consumption at 60 mph, but with
+ different units and perspectives.
+
+
+
+
+ All three convey information about fuel consumption at 60 mph, but with
+ different units and perspectives: f(60) = 0.04 gives gallons per mile,
+ g(60) = 25 gives miles per gallon, and h(60) = 2.4 gives
+ gallons per hour. They describe the same phenomenon from different angles.
+
+
+
+
+
+
+ Suppose we also know that f(70) = 0.045.
+ Find the average rate of change of f on the interval [60,70].
+ What are the units on the average rate of change of f?
+ What does this quantity measure?
+ Write a complete sentence to explain.
+
+
+
+
+ AV_{[60,70]} = 0.0005 \ \frac{\text{gal/mi}}{\text{mph}}.
+ This measures how much the car's fuel consumption rate (in gallons per mile)
+ increases per additional mile per hour of speed.
+
+
+
+
+ AV_{[60,70]} = \frac{f(70)-f(60)}{70-60} = \frac{0.045 - 0.04}{10} = 0.0005
+ \ \frac{\text{gal/mi}}{\text{mph}}.
+ This measures how much the car's fuel consumption rate (in gallons per mile)
+ increases per additional mile per hour of speed.
+
- In what follows,
- we work to understand two different piecewise functions entirely by hand based on familiar properties of linear and quadratic functions.
-
-
-
-
-
-
- Consider the function p defined by the following rule:
-
- p(x) =
- \begin{cases}
- -(x+2)^2 + 2, \amp x \lt 0 \\
- \frac{1}{2}(x-2)^2 + 1, \amp x \ge 0
- \end{cases}
-
-
- What are the values of p(-4), p(-2), p(0), p(2), and p(4)?
-
- What point is the vertex of the quadratic part of p that is valid for x \lt 0? What point is the vertex of the quadratic part of p that is valid for x \ge 0?
-
-
-
-
-
-
- The left-side parabola -(x+2)^2 + 2 (valid for x \lt 0) opens
- downward and has vertex (-2, 2).
- The right-side parabola \frac{1}{2}(x-2)^2 + 1 (valid for x \ge 0)
- opens upward and has vertex (2, 1).
-
-
-
-
-
-
- For what values of x is p(x) = 0? In addition, what is the y-intercept of p?
-
-
-
-
-
-
- For x \lt 0, setting -(x+2)^2 + 2 = 0 gives (x+2)^2 = 2,
- so x = -2 \pm \sqrt{2}. Both values x = -2 + \sqrt{2} \approx -0.586
- and x = -2 - \sqrt{2} \approx -3.414 lie in x \lt 0, so both are
- zeros of p. The right-side parabola \frac{1}{2}(x-2)^2+1 \ge 1 \gt 0
- for all x \ge 0, so it contributes no zeros.
- The y-intercept is p(0) = 3.
-
-
-
-
-
-
- Sketch an accurate, labeled graph of y = p(x) on the axes provided at left in the following figure.
-
-
-
-
ADD ALT TEXT TO THIS IMAGE
-
-
ADD ALT TEXT TO THIS IMAGE
-
-
-
-
-
-
-
The graph of y = p(x) consists of two parabolic pieces. For x \lt 0, the piece -(x+2)^2+2 is a downward-opening parabola with vertex (-2,2) that passes through (-4,-2), (-2+\sqrt{2},0), (-2-\sqrt{2},0), and approaches (0,3) from the left (open endpoint). For x \ge 0, the piece \frac{1}{2}(x-2)^2+1 is an upward-opening parabola with vertex (2,1) that starts at (0,3) (closed endpoint) and passes through (4,3). The graph has a jump at x = 0 but the values approach 3 from both sides, so the graph is actually continuous there.
-
-
-
-
-
-
-
- For the function f defined by the righthand figure in (d), determine a piecewise-defined formula for f that is expressed in bracket notation similar to the definition of y = p(x) above.
-
-
-
-
-
-
Reading from the graph of f, the function appears piecewise linear on three intervals. For -2.5 \lt x \le -1, the line passes through (-2,3.5) and (-1,3), giving slope -\frac{1}{2} and formula f(x) = 3 - \frac{1}{2}(x+1). For -1 \lt x \le 1, the line passes through (0,2.5) and (1,4), giving slope \frac{3}{2} and formula f(x) = 4 + \frac{3}{2}(x-1). For 1 \lt x \lt 3.5 (excluding x = 2), the line passes through (1,1) and (3,-0.5), giving slope -\frac{3}{4} and formula f(x) = 1 - \frac{3}{4}(x-1). Thus f(x) = \begin{cases} 3 - \frac{1}{2}(x+1), \amp -2.5 \lt x \le -1 \\ 4 + \frac{3}{2}(x-1), \amp -1 \lt x \le 1 \\ 1 - \frac{3}{4}(x-1), \amp 1 \lt x \lt 3.5, \ x \ne 2 \end{cases}
-
-
-
-
-
-
-
See solutions to individual tasks above.
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ In what follows,
+ we work to understand two different piecewise functions entirely by hand based on familiar properties of linear and quadratic functions.
+
+
+
+
+
+
+ Consider the function p defined by the following rule:
+
+ p(x) =
+ \begin{cases}
+ -(x+2)^2 + 2, \amp x \lt 0 \\
+ \frac{1}{2}(x-2)^2 + 1, \amp x \ge 0
+ \end{cases}
+
+
+ What are the values of p(-4), p(-2), p(0), p(2), and p(4)?
+
+ What point is the vertex of the quadratic part of p that is valid for x \lt 0? What point is the vertex of the quadratic part of p that is valid for x \ge 0?
+
+
+
+
The parabola that is valid for x \lt 0 has vertex (-2, 2). The parabola that is valid for x \ge 0 has vertex (2, 1).
+
+
+
+ The left-side parabola -(x+2)^2 + 2 (valid for x \lt 0) opens
+ downward and has vertex (-2, 2).
+ The right-side parabola \frac{1}{2}(x-2)^2 + 1 (valid for x \ge 0)
+ opens upward and has vertex (2, 1).
+
+
+
+
+
+
+ For what values of x is p(x) = 0? In addition, what is the y-intercept of p?
+
+
+
+
p(x)=0 when x = -2 + \sqrt{2} \approx -0.586
+ and x = -2 - \sqrt{2} \approx -3.414. The y-intercept is p(0) = 3.
+
+
+
+ For x \lt 0, setting -(x+2)^2 + 2 = 0 gives (x+2)^2 = 2,
+ so x = -2 \pm \sqrt{2}. Both values x = -2 + \sqrt{2} \approx -0.586
+ and x = -2 - \sqrt{2} \approx -3.414 lie in x \lt 0, so both are
+ zeros of p. The right-side parabola \frac{1}{2}(x-2)^2+1 \ge 1 \gt 0
+ for all x \ge 0, so it contributes no zeros.
+ The y-intercept is p(0) = 3.
+
+
+
+
+
+
+ Sketch an accurate, labeled graph of y = p(x) on the axes provided at left in the following figure.
+
+
+
+
+
+
+
+
Blank axes for graphing y = p(x)
+
+
+
+
+
+
+
A piecewise linear function f(x). The first piece goes through (-2,3.5) and ends at the closed circle (-1,3), while the second piece starts at an open circle at (-1,1) and ends at a closed circle at (1,4), and the third piece starts at an open circle at (1,1) passes through an open circle at (2,1/4) and continues to the edge of the graph.
+
+
+
+
+
+
+
+
+
+
+
+
+
A piecewise function p(x). The first piece is an downward-opening parabola with vertex (-2,2) ending in an open circle at (0,-2), while the second piece is an upward-opening parabola starting at a closed circle at (0,3) with vertex (2,1).
+
+
+
+
+
+
The graph of y = p(x) consists of two parabolic pieces. For x \lt 0, the piece -(x+2)^2+2 is a downward-opening parabola with vertex (-2,2) that passes through (-4,-2), (-2+\sqrt{2},0), (-2-\sqrt{2},0), and approaches (0,-2) from the left (open endpoint). For x \ge 0, the piece \frac{1}{2}(x-2)^2+1 is an upward-opening parabola with vertex (2,1) that starts at (0,3) (closed endpoint) and passes through (4,3). The graph has a jump at x = 0.
+
+
+
+
+
+
A piecewise function p(x). The first piece is an downward-opening parabola with vertex (-2,2) ending in an open circle at (0,-2), while the second piece is an upward-opening parabola starting at a closed circle at (0,3) with vertex (2,1).
+
+
+
+
+
+
+
+
+
+
+ For the function f defined by the righthand figure in (d), determine a piecewise-defined formula for f that is expressed in bracket notation similar to the definition of y = p(x) above.
+
+
+
+
f(x) = \begin{cases} 3 - \frac{1}{2}(x+1), \amp -2.5 \lt x \le -1 \\ 4 + \frac{3}{2}(x-1), \amp -1 \lt x \le 1 \\ 1 - \frac{3}{4}(x-1), \amp 1 \lt x \lt 3.5, \ x \ne 2 \end{cases}
+
+
+
Reading from the graph of f, the function appears piecewise linear on three intervals. For -2.5 \lt x \le -1, the line passes through (-2,3.5) and (-1,3), giving slope -\frac{1}{2} and formula f(x) = 3 - \frac{1}{2}(x+1). For -1 \lt x \le 1, the line passes through (0,2.5) and (1,4), giving slope \frac{3}{2} and formula f(x) = 4 + \frac{3}{2}(x-1). For 1 \lt x \lt 3.5 (excluding x = 2), the line passes through (1,1) and (3,-0.5), giving slope -\frac{3}{4} and formula f(x) = 1 - \frac{3}{4}(x-1). Thus f(x) = \begin{cases} 3 - \frac{1}{2}(x+1), \amp -2.5 \lt x \le -1 \\ 4 + \frac{3}{2}(x-1), \amp -1 \lt x \le 1 \\ 1 - \frac{3}{4}(x-1), \amp 1 \lt x \lt 3.5, \ x \ne 2 \end{cases}
- Determine the most simplified expression you can for the average rate of change of f on the interval [1,1+h].
- That is, determine AV_{[1,1+h]} for f and simplify the result as much as possible.
-
- Compute g(1+h).
- Is there any valid algebra you can do to write g(1+h) more simply?
-
-
-
-
-
-
- g(1+h) = \frac{5}{1+h}.
- This expression cannot be simplified further.
-
-
-
-
-
-
- Determine the most simplified expression you can for the average rate of change of g on the interval [1,1+h].
- That is, determine AV_{[1,1+h]} for g and simplify the result.
-
+ Determine the most simplified expression you can for the average rate of change of f on the interval [1,1+h].
+ That is, determine AV_{[1,1+h]} for f and simplify the result as much as possible.
+
+ Compute g(1+h).
+ Is there any valid algebra you can do to write g(1+h) more simply?
+
+
+
+
g(1+h) = \frac{5}{1+h}
+
+
+ g(1+h) = \frac{5}{1+h}.
+ This expression cannot be simplified further.
+
+
+
+
+
+
+ Determine the most simplified expression you can for the average rate of change of g on the interval [1,1+h].
+ That is, determine AV_{[1,1+h]} for g and simplify the result.
+
- Let F = D(N) = 40 + 0.25N be Dolbear's function that converts an input of number of chirps per minute to degrees Fahrenheit, and let C = G(F) = \frac{5}{9}(F-32) be the function that converts an input of degrees Fahrenheit to an output of degrees Celsius.
-
-
-
-
-
-
- Determine a formula for the new function H = (G \circ D) that depends only on the variable N.
-
- What is the meaning of the function you found in (a)?
-
-
-
-
-
-
- The function H converts the number of cricket chirps per minute directly
- into a temperature in degrees Celsius.
-
-
-
-
-
-
- How does a plot of the function H = (G \circ D) compare to that of Dolbear's function? Sketch a plot of y = H(N) = (G \circ D)(N) on the blank axes to the right of the plot of Dolbear's function, and discuss the similarities and differences between them. Be sure to label the vertical scale on your axes.
-
-
-
-
ADD ALT TEXT TO THIS IMAGE
-
-
ADD ALT TEXT TO THIS IMAGE
-
-
-
-
-
-
-
Both Dolbear's function D and H = (G \circ D) are linear functions of N, so their graphs have the same shape. Dolbear's function has slope \frac{1}{4} (degrees Fahrenheit per chirp per minute), while H(N) = \frac{5}{9}(0.25N+8) has a smaller slope of \frac{5}{36} (degrees Celsius per chirp per minute). The vertical scale on the plot of H should reflect the Celsius range: H(40) = 10 (corresponding to 50^\circ F) and H(180) = \frac{265}{9} \approx 29.4 (corresponding to 85^\circ F). The plot of H is an increasing line on [40, 180], similar to Dolbear's function but with a shallower slope and Celsius units on the vertical axis.
-
-
-
-
-
-
-
-
-
-
- What is the domain of the function H = G \circ D? What is its range?
-
-
-
-
-
-
- As an abstract mathematical function, both the domain and range are all real
- numbers. In the context of Dolbear's model (domain [40, 180] chirps per
- minute), the domain of H is [40, 180] and the corresponding range
- is approximately [10, 29.4] degrees Celsius.
-
-
-
-
-
-
-
-
See solutions to individual tasks above.
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ Let F = D(N) = 40 + 0.25N be Dolbear's function that converts an input of number of chirps per minute to degrees Fahrenheit, and let C = G(F) = \frac{5}{9}(F-32) be the function that converts an input of degrees Fahrenheit to an output of degrees Celsius.
+
+
+
+
+
+
+ Determine a formula for the new function H = (G \circ D) that depends only on the variable N.
+
+ What is the meaning of the function you found in (a)?
+
+
+
+
+ The function H converts the number of cricket chirps per minute directly
+ into a temperature in degrees Celsius.
+
+
+
+ The function H converts the number of cricket chirps per minute directly
+ into a temperature in degrees Celsius.
+
+
+
+
+
+
+ How does a plot of the function H = (G \circ D) compare to that of Dolbear's function? Sketch a plot of y = H(N) = (G \circ D)(N) on the blank axes to the right of the plot of Dolbear's function, and discuss the similarities and differences between them. Be sure to label the vertical scale on your axes.
+
Plot of D(N) with horizontal axis labeled N (chirps per minute) and vertical axis in degrees Fahrenheit.
+
+ The graph is a line segment from (40,50) to (180,85), also going through the points (80, 60), (120, 70), and (160, 80).
+
+
+
+
+
+
+ h(x)=5/9*(8+0.25*x)
+
+
+
+
+
+
+
+
Plot of the function H(N) with horizontal axis labeled N (chirps per minute) and vertical axis labeled T (degrees Celsius).
+
+ The graph is a line segment from (40,10) to (180, \frac{265}{9}).
+
+
+
+
+
+
+
Both Dolbear's function D and H = (G \circ D) are linear functions of N, so their graphs have the same shape. Dolbear's function has slope \frac{1}{4} (degrees Fahrenheit per chirp per minute), while H(N) = \frac{5}{9}(0.25N+8) has a smaller slope of \frac{5}{36} (degrees Celsius per chirp per minute). The vertical scale on the plot of H should reflect the Celsius range: H(40) = 10 (corresponding to 50^\circ F) and H(180) = \frac{265}{9} \approx 29.4 (corresponding to 85^\circ F). The plot of H is an increasing line on [40, 180], similar to Dolbear's function but with a shallower slope and Celsius units on the vertical axis.
Plot of D(N) with horizontal axis labeled N (chirps per minute) and vertical axis in degrees Fahrenheit.
+
+ The graph is a line segment from (40,50) to (180,85), also going through the points (80, 60), (120, 70), and (160, 80).
+
+
+
+
+
+
+ h(x)=5/9*(8+0.25*x)
+
+
+
+
+
+
+
+
Plot of the function H(N) with horizontal axis labeled N (chirps per minute) and vertical axis labeled T (degrees Celsius).
+
+ The graph is a line segment from (40,10) to (180, \frac{265}{9}).
+
+
+
+
+
+
+
+
+
+
+
+ What is the domain of the function H = G \circ D? What is its range?
+
+
+
+
The domain of H is [40, 180] and the range is approximately [10, 29.4].
+
+
+ As an abstract mathematical function, both the domain and range are all real
+ numbers. In the context of Dolbear's model (domain [40, 180] chirps per
+ minute), the domain of H is [40, 180] and the corresponding range
+ is approximately [10, 29.4] degrees Celsius.
+
- Let functions p and q be given by the graphs in the figure below, which are each piecewise linear - that is, parts that look like straight lines are straight lines. In addition, let f and g be given by the tables below.
-
- Compute each of the following quantities or explain why they are not defined.
-
-
-
-
-
-
- p(q(0))
-
-
-
-
-
-
- From the graph, q(0) = 2 and p(2) = 1, so p(q(0)) = 1.
-
-
-
-
-
-
- q(p(0))
-
-
-
-
-
-
- From the graph, p(0) = -\frac{1}{2} and q\!\left(-\frac{1}{2}\right) = 2,
- so q(p(0)) = 2.
-
-
-
-
-
- (p \circ p)(-1)
-
-
-
-
-
- From the graph, p(-1) = -1, so (p \circ p)(-1) = p(p(-1)) = p(-1) = -1.
-
-
-
-
-
-
- (f \circ g)(2)
-
-
-
-
-
-
- From the table, g(2) = 0 and f(0) = 6, so (f \circ g)(2) = 6.
-
-
-
-
-
-
- (g \circ f)(3)
-
-
-
-
-
-
- From the table, f(3) = 4 and g(4) = 2, so (g \circ f)(3) = 2.
-
-
-
-
-
-
- g(f(0))
-
-
-
-
-
-
- From the table, f(0) = 6, but g(6) is not defined in the table,
- so g(f(0)) is undefined.
-
-
-
-
-
-
- For what value(s) of x is f(g(x)) = 4?
-
-
-
-
-
-
- We need f(g(x)) = 4, which requires g(x) = 1 or g(x) = 3
- (since f(1)=4 and f(3)=4 from the table). From the table,
- g(0) = 1 and g(1) = 3, so f(g(x)) = 4 for x = 0
- and x = 1.
-
-
-
-
-
-
- For what value(s) of x is q(p(x)) = 1?
-
-
-
-
-
-
- We need q(p(x)) = 1, which (from the graph of q) requires
- p(x) = -\frac{4}{3} or p(x) = 2. From the graph of p,
- p(x) = 2 at x = 2 and at x \approx -2.5; there are no
- x-values for which p(x) = -\frac{4}{3}. So q(p(x)) = 1
- for x = 2 and x \approx -2.5.
-
+ Let functions p and q be given by the graphs in the figure below, which are each piecewise linear - that is, parts that look like straight lines are straight lines. In addition, let f and g be given by the tables below.
+
graph of two piecewise linear functions p(x) and q(x) with domain at least [-3.5,3.5]. Each function consists of three linear pieces.
+
The first piece of p(x) goes through the point (-3, 3) and ends at the point (-1, -1). The second piece of p(x) starts at the point (-1, -1) and ends at the point (1, 0). The third piece of p(x) starts at the point (1, 0) and goes through the point (2, 2).
+
The first piece of q(x) goes through the point (-2, -1) and ends at the point (-1, 2). The second piece of q(x) starts at the point (-1, 2) and ends at the point (1, 2). The third piece of q(x) starts at the point (1, 2) and goes through the point (3, 0).
+
+
+
+
+
+
+ Compute each of the following quantities or explain why they are not defined.
+
+
+
+
+
+
+ p(q(0))
+
+
+
+
p(q(0)) = 1
+
+
+ From the graph, q(0) = 2 and p(2) = 1, so p(q(0)) = 1.
+
+
+
+
+
+
+ q(p(0))
+
+
+
+
q(p(0)) = 2
+
+
+ From the graph, p(0) = -\frac{1}{2} and q\!\left(-\frac{1}{2}\right) = 2,
+ so q(p(0)) = 2.
+
+
+
+
+
+ (p \circ p)(-1)
+
+
+
(p \circ p)(-1) = -1
+
+
+ From the graph, p(-1) = -1, so (p \circ p)(-1) = p(p(-1)) = p(-1) = -1.
+
+
+
+
+
+
+ (f \circ g)(2)
+
+
+
+
(f \circ g)(2) = 6
+
+
+ From the table, g(2) = 0 and f(0) = 6, so (f \circ g)(2) = 6.
+
+
+
+
+
+
+ (g \circ f)(3)
+
+
+
+
(g \circ f)(3) = 2
+
+
+ From the table, f(3) = 4 and g(4) = 2, so (g \circ f)(3) = 2.
+
+
+
+
+
+
+ g(f(0))
+
+
+
+
g(f(0)) = \text{undefined}
+
+
+ From the table, f(0) = 6, but g(6) is not defined in the table,
+ so g(f(0)) is undefined.
+
+
+
+
+
+
+ For what value(s) of x is f(g(x)) = 4?
+
+
+
+
f(g(x)) = 4 when x = 0 or x = 1
+
+
+ We need f(g(x)) = 4, which requires g(x) = 1 or g(x) = 3
+ (since f(1)=4 and f(3)=4 from the table). From the table,
+ g(0) = 1 and g(1) = 3, so f(g(x)) = 4 for x = 0
+ and x = 1.
+
+
+
+
+
+
+ For what value(s) of x is q(p(x)) = 1?
+
+
+
+
q(p(x)) = 1 when x = 2 or x \approx -2.5
+
+
+ We need q(p(x)) = 1, which (from the graph of q) requires
+ p(x) = -\frac{4}{3} or p(x) = 2. From the graph of p,
+ p(x) = 2 at x = 2 and at x \approx -2.5; there are no
+ x-values for which p(x) = -\frac{4}{3}. So q(p(x)) = 1
+ for x = 2 and x \approx -2.5.
+
- Each of the following prompts describes a relationship between two quantities. For each, your task is to decide whether or not the relationship can be thought of as a function. If not, explain why. If so, state the domain and codomain of the function and write at least one sentence to explain the process that leads from the collection of inputs to the collection of outputs.
-
-
-
-
-
-
- The relationship between x and y in each of the graphs below (address each graph separately as a potential situation where y is a function of x). In the lefthand figure, any point on the circle relates x and y. For instance, the y-value \sqrt{7} is related to the x-value -3. In the righthand figure, any point on the blue curve relates x and y. For instance, when x = -1, the corresponding y-value is y = 3. An unfilled circle indicates that there is not a point on the graph at that specific location.
-
-
-
-
ADD ALT TEXT TO THIS IMAGE
-
-
ADD ALT TEXT TO THIS IMAGE
-
-
-
-
-
-
-
- The circle is not a function of x, because some values of x
- correspond to more than one value of y. For example, when x = 0,
- both y = 4 and y = -4 lie on the circle.
-
-
- The curve in the righthand figure is a function of x, because
- each value of x in the domain corresponds to exactly one value of y.
-
-
-
-
-
-
- The relationship between the day of the year and the value of the S&P500 stock index (at the close of trading on a given day), where we attempt to consider the index's value (at the close of trading) as a function of the day of the year.
-
-
-
-
-
-
- The closing value of the S&P500 can be expressed as a function of the day of
- the year, provided the domain is restricted to trading days (excluding weekends and
- market holidays). With this restriction, each day in the domain corresponds to
- exactly one closing value.
-
-
-
-
-
-
- The relationship between a car's velocity and its odometer, where we attempt to view the car's odometer reading as a function of its velocity.
-
-
-
-
-
-
- The odometer reading cannot be viewed as a function of the car's velocity.
- The same speed can occur at many different odometer readings throughout a trip,
- so a single velocity value can correspond to multiple odometer values.
-
-
-
-
-
-
- The relationship between x and y that is given in the following table where we attempt to view y as depending on x.
-
+ Each of the following prompts describes a relationship between two quantities. For each, your task is to decide whether or not the relationship can be thought of as a function. If not, explain why. If so, state the domain and codomain of the function and write at least one sentence to explain the process that leads from the collection of inputs to the collection of outputs.
+
+
+
+
+
+
+ The relationship between x and y in each of the graphs below (address each graph separately as a potential situation where y is a function of x). In the lefthand figure, any point on the circle relates x and y. For instance, the y-value \sqrt{7} is related to the x-value -3. In the righthand figure, any point on the blue curve relates x and y. For instance, when x = -1, the corresponding y-value is y = 3. An unfilled circle indicates that there is not a point on the graph at that specific location.
+
+
+
+
ADD ALT TEXT TO THIS IMAGE
+
+
ADD ALT TEXT TO THIS IMAGE
+
+
+
+
+
The circle is not a function of x, but the curve in the right-hand figure is a function of x.
+
+
+ The circle is not a function of x, because some values of x
+ correspond to more than one value of y. For example, when x = 0,
+ both y = 4 and y = -4 lie on the circle.
+
+
+ The curve in the right-hand figure is a function of x, because
+ each value of x in the domain corresponds to exactly one value of y.
+
+
+
+
+
+
+ The relationship between the day of the year and the value of the S&P500 stock index (at the close of trading on a given day), where we attempt to consider the index's value (at the close of trading) as a function of the day of the year.
+
+
+
+
+ The relationship between the day of the year and the S&P500 stock index is a function.
+
+
+
+ The closing value of the S&P500 can be expressed as a function of the day of
+ the year, provided the domain is restricted to trading days (excluding weekends and
+ market holidays). With this restriction, each day in the domain corresponds to
+ exactly one closing value.
+
+
+
+
+
+
+ The relationship between a car's velocity and its odometer, where we attempt to view the car's odometer reading as a function of its velocity.
+
+
+
+
+ The odometer reading cannot be viewed as a function of the car's velocity.
+
+
+
+ The odometer reading cannot be viewed as a function of the car's velocity.
+ The same speed can occur at many different odometer readings throughout a trip,
+ so a single velocity value can correspond to multiple odometer values.
+
+
+
+
+
+
+ The relationship between x and y that is given in the following table where we attempt to view y as depending on x.
+
- Consider a spherical tank of radius 4 m that is completely full of water. Suppose that the tank is being drained by regulating an exit valve in such a way that the height of the water in the tank is always decreasing at a rate of 0.5 meters per minute. Let
- V be the volume of water in the tank (in cubic meters) at a given time t (in minutes), and
- h the depth of the water (in meters) at the same time. It can be shown using calculus
- that V is a function of t according to the model
-
- V = p(t) = \frac{256\pi}{3} - \frac{\pi}{24} t^2(24-t)
- .
- In addition, let h = q(t) be the function whose output is the depth of the water in the tank at time t.
-
-
-
-
-
-
- What is the height of the water when t = 0? When t = 1? When t = 2? How long will it take the tank to completely drain? Why?
-
-
-
-
-
-
- At t = 0, the tank is full and the height of the water is 8 m.
- At t = 1 minute, 0.5 m of water has drained and the height is 7.5 m.
- At t = 2 minutes, 1 m of water has drained and the height is 7 m.
-
-
- The tank has 8 m of water (depth), which is 16 half-meters, with one
- half-meter draining each minute. Thus it will take 16 minutes for the tank
- to drain completely. The linear height model
- h = q(t) = 8 - 0.5t
- gives q(t) = 0 when t = 16.
-
-
-
-
-
-
- What is the domain of the model h = q(t)? What is the domain of the model V = p(t)?
-
-
-
-
-
-
- The domain of each model is determined by the time it takes the tank to drain.
- Assuming the tank begins draining at t = 0 minutes and is empty at
- t = 16 minutes, the domain of both models is 0 \le t \le 16,
- or equivalently [0, 16].
-
-
-
-
-
-
- How much water is in the tank when the tank is full? What is the range of the model h = q(t)? What is the range of the model V = p(t)?
-
-
-
-
-
-
- When the tank is full, the depth of the water equals the diameter of the tank,
- 2 \times 4 = 8 m. The volume of water in the full tank equals the volume
- of the sphere:
- \frac{4}{3}\pi (4)^3 = \frac{256\pi}{3} \approx 268 \text{ m}^3.
-
-
- The range of the volume model is 0 \le V \le \frac{256\pi}{3}, going from
- completely empty to completely full. The range of the height model is likewise
- 0 \le h \le 8.
-
-
-
-
-
-
- We will frequently use a graphing utility to help us understand function behavior, and strongly recommend Desmos because it is intuitive, online, and free. To learn more about Desmos, see their outstanding online tutorials.
-
- In this prepared Desmos worksheet, you can see how we enter the (abstract) function V = p(t) = \frac{256\pi}{3} - \frac{\pi}{24} t^2(24-t), as well as the corresponding graph the program generates. Make as many observations as you can about the model V = p(t). You should discuss its shape and overall behavior, its domain, its range, and more.
-
-
-
-
-
-
The graph of V = p(t) = \frac{256\pi}{3} - \frac{\pi}{24}t^2(24-t) on [0,16] is a curve that decreases from V(0) = \frac{256\pi}{3} \approx 268 cubic meters down to V(16) = 0. The curve is concave down for small t (volume decreasing rapidly as the full, wide sphere loses water) and concave up near t = 16 (volume decreasing more slowly as the narrow bottom empties). The domain is [0, 16] minutes and the range is \left[0, \frac{256\pi}{3}\right] cubic meters.
-
-
-
-
-
-
-
- How does the model V = p(t) = \frac{256\pi}{3} - \frac{\pi}{24} t^2(24-t) differ from the abstract function y = r(x) = \frac{256\pi}{3} - \frac{\pi}{24} x^2(24-x)? In particular, how do the domain and range of the model differ from those of the abstract function, if at all?
-
-
-
-
-
-
- The model V = p(t) differs from the abstract function y = r(x)
- in its domain and range. The model's domain is restricted to [0, 16]
- and its range to \left[0, \frac{256\pi}{3}\right] by the physical context.
- The abstract function has an unrestricted domain and an unrestricted range.
-
-
-
-
-
-
- How should the graph of the height function h = q(t) appear? Can you determine a formula for q? Explain your thinking.
-
-
-
-
-
-
- The height function should be linear and decreasing, because the height decreases
- by the same amount (0.5 m) every minute. The formula is
- q(t) = 8 - 0.5t.
-
+ Consider a spherical tank of radius 4 m that is completely full of water. Suppose that the tank is being drained by regulating an exit valve in such a way that the height of the water in the tank is always decreasing at a rate of 0.5 meters per minute. Let
+ V be the volume of water in the tank (in cubic meters) at a given time t (in minutes), and
+ h the depth of the water (in meters) at the same time. It can be shown using calculus
+ that V is a function of t according to the model
+
+ V = p(t) = \frac{256\pi}{3} - \frac{\pi}{24} t^2(24-t)
+ .
+ In addition, let h = q(t) be the function whose output is the depth of the water in the tank at time t.
+
+
+
+
+
+
+ What is the height of the water when t = 0? When t = 1? When t = 2? How long will it take the tank to completely drain? Why?
+
+
+
+
+ At t = 0, the height of the water is 8 m.
+ At t = 1 minute, the height is 7.5 m.
+ At t = 2 minutes, the height is 7 m.
+ It will take 16 minutes for the tank
+ to drain completely.
+
+
+
+
+ At t = 0, the tank is full and the height of the water is 8 m.
+ At t = 1 minute, 0.5 m of water has drained and the height is 7.5 m.
+ At t = 2 minutes, 1 m of water has drained and the height is 7 m.
+
+
+ The tank has 8 m of water (depth), which is 16 half-meters, with one
+ half-meter draining each minute. Thus it will take 16 minutes for the tank
+ to drain completely. The linear height model
+ h = q(t) = 8 - 0.5t
+ gives q(t) = 0 when t = 16.
+
+
+
+
+
+
+ What is the domain of the model h = q(t)? What is the domain of the model V = p(t)?
+
+
+
+
The domain of each model is 0 \le t \le 16,
+ or equivalently [0, 16].
+
+
+ The domain of each model is determined by the time it takes the tank to drain.
+ Assuming the tank begins draining at t = 0 minutes and is empty at
+ t = 16 minutes, the domain of both models is 0 \le t \le 16,
+ or equivalently [0, 16].
+
+
+
+
+
+
+ How much water is in the tank when the tank is full? What is the range of the model h = q(t)? What is the range of the model V = p(t)?
+
+
+
+
+ When the tank is full, there are \frac{256\pi}{3} \approx 268.1 cubic meters of water in the tank.
+ The range of the model h = q(t) is 0 \le h \le 8, or equivalently [0, 8].
+ The range of the model V = p(t) is 0 \le V \le \frac{256\pi}{3}, or equivalently \left[0, \frac{256\pi}{3}\right].
+
+
+
+
+ When the tank is full, the depth of the water equals the diameter of the tank,
+ 2 \times 4 = 8 m. The volume of water in the full tank equals the volume
+ of the sphere:
+ \frac{4}{3}\pi (4)^3 = \frac{256\pi}{3} \approx 268 \text{ m}^3.
+
+
+ The range of the volume model is 0 \le V \le \frac{256\pi}{3}, going from
+ completely empty to completely full. The range of the height model is likewise
+ 0 \le h \le 8.
+
+
+
+
+
+
+ We will frequently use a graphing utility to help us understand function behavior, and strongly recommend Desmos because it is intuitive, online, and free. To learn more about Desmos, see their outstanding online tutorials.
+
+ In this prepared Desmos worksheet, you can see how we enter the (abstract) function V = p(t) = \frac{256\pi}{3} - \frac{\pi}{24} t^2(24-t), as well as the corresponding graph the program generates. Make as many observations as you can about the model V = p(t). You should discuss its shape and overall behavior, its domain, its range, and more.
+
+
+
+
+ Descriptions will vary. The graph of V = p(t) = \frac{256\pi}{3} - \frac{\pi}{24}t^2(24-t) on [0,16] is a curve that decreases from V(0) = \frac{256\pi}{3} \approx 268 cubic meters down to V(16) = 0. The domain is [0, 16] minutes and the range is \left[0, \frac{256\pi}{3}\right] cubic meters.
+
+
+
+
The graph of V = p(t) = \frac{256\pi}{3} - \frac{\pi}{24}t^2(24-t) on [0,16] is a curve that decreases from V(0) = \frac{256\pi}{3} \approx 268 cubic meters down to V(16) = 0. The curve is concave down for small t (volume decreasing rapidly as the full, wide sphere loses water) and concave up near t = 16 (volume decreasing more slowly as the narrow bottom empties). The domain is [0, 16] minutes and the range is \left[0, \frac{256\pi}{3}\right] cubic meters.
+
+
+
+
+
+
+
+ How does the model V = p(t) = \frac{256\pi}{3} - \frac{\pi}{24} t^2(24-t) differ from the abstract function y = r(x) = \frac{256\pi}{3} - \frac{\pi}{24} x^2(24-x)? In particular, how do the domain and range of the model differ from those of the abstract function, if at all?
+
+
+
+
+ The model's domain is restricted to [0, 16]
+ and its range to \left[0, \frac{256\pi}{3}\right] by the physical context.
+ The abstract function has an unrestricted domain and an unrestricted range.
+
+
+
+
+ The model V = p(t) differs from the abstract function y = r(x)
+ in its domain and range. The model's domain is restricted to [0, 16]
+ and its range to \left[0, \frac{256\pi}{3}\right] by the physical context.
+ The abstract function has an unrestricted domain and an unrestricted range.
+
+
+
+
+
+
+ How should the graph of the height function h = q(t) appear? Can you determine a formula for q? Explain your thinking.
+
+
+
+
+ The height function should be linear and decreasing. The formula is
+ q(t) = 8 - 0.5t.
+
+
+
+
+ The height function should be linear and decreasing, because the height decreases
+ by the same amount (0.5 m) every minute. The formula is
+ q(t) = 8 - 0.5t.
+
- Consider a spherical tank of radius 4 m that is filling with water. Let
- V be the volume of water in the tank (in cubic meters) at a given time, and
- h the depth of the water (in meters) at the same time. It can be shown using calculus
- that V is a function of h according to the rule
-
- V = f(h) = \frac{\pi}{3} h^2(12-h)
- .
-
-
-
-
-
-
- What values of h make sense to consider in the context of this function? What values of V make sense in the same context?
-
-
-
-
-
-
- Since depth can't be negative and can't exceed the tank's diameter of 8 m,
- the values of h that make sense are 0 \le h \le 8.
- The corresponding values of V range from 0 (empty tank) to
- f(8) = \frac{256\pi}{3} (full tank), so 0 \le V \le \frac{256\pi}{3}.
-
-
-
-
-
-
- What is the domain of the function f in the context of the spherical tank? Why? What is the corresponding codomain? Why?
-
-
-
-
-
-
- The domain of f in context is [0, 8], since the water depth
- can range from 0 m (empty) to 8 m (full diameter).
- The range is \left[0, \frac{256\pi}{3}\right].
- Since volumes are non-negative, the codomain can be taken as [0, \infty),
- or more broadly as all real numbers if we consider the abstract formula
- without physical constraints.
-
-
-
-
-
-
- Determine and interpret (with appropriate units) the values
- f(2), f(4), and f(8). What is important about the value of f(8)?
-
-
-
-
-
-
-
- f(2) &= \frac{\pi}{3}(4)(10) = \frac{40\pi}{3} \approx 41.9 \text{ m}^3
- f(4) &= \frac{\pi}{3}(16)(8) = \frac{128\pi}{3} \approx 134.0 \text{ m}^3
- f(8) &= \frac{\pi}{3}(64)(4) = \frac{256\pi}{3} \approx 268.1 \text{ m}^3
-
- These give the volume of water when the depth is 2, 4, and 8
- meters respectively. The value f(8) = \frac{256\pi}{3} equals the volume
- of a sphere of radius 4, confirming that the tank is completely full when
- h = 8.
-
-
-
-
-
-
- Consider the claim: since f(9) = \frac{\pi}{3} 9^2(12-9) = 81\pi \approx 254.47, when the water is 9 meters deep, there is about 254.47 cubic meters of water in the tank. Is this claim valid? Why or why not? Further, does it make sense to observe that f(13) = -\frac{169\pi}{3}? Why or why not?
-
-
-
-
-
-
- Neither claim is valid. The domain of f is [0, 8], so
- h = 9 is outside the domain — the water cannot be 9 m deep in a
- tank whose diameter is only 8 m. Similarly, f(13) falls outside the
- domain, and the negative value it produces is further evidence that the formula
- should not be applied beyond h = 8.
-
-
-
-
-
-
- Can you determine a value of h for which f(h) = 300 cubic meters? Why or why not?
-
-
-
-
-
-
- No. The maximum value of f on its domain [0, 8] is
- f(8) = \frac{256\pi}{3} \approx 268.1 m^3, which is less than
- 300. The tank simply cannot hold 300 cubic meters of water.
-
+ Consider a spherical tank of radius 4 m that is filling with water. Let
+ V be the volume of water in the tank (in cubic meters) at a given time, and
+ h the depth of the water (in meters) at the same time. It can be shown using calculus
+ that V is a function of h according to the rule
+
+ V = f(h) = \frac{\pi}{3} h^2(12-h)
+ .
+
+
+
+
+
+
+ What values of h make sense to consider in the context of this function? What values of V make sense in the same context?
+
+
+
+
0 \le h \le 8 and 0 \le V \le \frac{256\pi}{3}
+
+
+ Since depth can't be negative and can't exceed the tank's diameter of 8 m,
+ the values of h that make sense are 0 \le h \le 8.
+ The corresponding values of V range from 0 (empty tank) to
+ f(8) = \frac{256\pi}{3} (full tank), so 0 \le V \le \frac{256\pi}{3}.
+
+
+
+
+
+
+ What is the domain of the function f in the context of the spherical tank? Why? What is the corresponding codomain? Why?
+
+
+
+
The domain is [0, 8] and the codomain is \left[0, \frac{256\pi}{3}\right].
+
+
+ The domain of f in context is [0, 8], since the water depth
+ can range from 0 m (empty) to 8 m (full diameter).
+ The range is \left[0, \frac{256\pi}{3}\right].
+ Since volumes are non-negative, the codomain can be taken as [0, \infty),
+ or more broadly as all real numbers if we consider the abstract formula
+ without physical constraints.
+
+
+
+
+
+
+ Determine and interpret (with appropriate units) the values
+ f(2), f(4), and f(8). What is important about the value of f(8)?
+
+
+ f(2) &= \frac{\pi}{3}(4)(10) = \frac{40\pi}{3} \approx 41.9 \text{ m}^3
+ f(4) &= \frac{\pi}{3}(16)(8) = \frac{128\pi}{3} \approx 134.0 \text{ m}^3
+ f(8) &= \frac{\pi}{3}(64)(4) = \frac{256\pi}{3} \approx 268.1 \text{ m}^3
+
+ These give the volume of water when the depth is 2, 4, and 8
+ meters respectively. The value f(8) = \frac{256\pi}{3} equals the volume
+ of a sphere of radius 4, confirming that the tank is completely full when
+ h = 8.
+
+
+
+
+
+
+ Consider the claim: since f(9) = \frac{\pi}{3} 9^2(12-9) = 81\pi \approx 254.47, when the water is 9 meters deep, there is about 254.47 cubic meters of water in the tank. Is this claim valid? Why or why not? Further, does it make sense to observe that f(13) = -\frac{169\pi}{3}? Why or why not?
+
+
+
+
Neither claim is valid.
+
+
+ Neither claim is valid. The domain of f is [0, 8], so
+ h = 9 is outside the domain — the water cannot be 9 m deep in a
+ tank whose diameter is only 8 m. Similarly, f(13) falls outside the
+ domain, and the negative value it produces is further evidence that the formula
+ should not be applied beyond h = 8.
+
+
+
+
+
+
+ Can you determine a value of h for which f(h) = 300 cubic meters? Why or why not?
+
+
+
+
No, the tank cannot hold 300 cubic meters of water.
+
+
+ No. The maximum value of f on its domain [0, 8] is
+ f(8) = \frac{256\pi}{3} \approx 268.1 m^3, which is less than
+ 300. The tank simply cannot hold 300 cubic meters of water.
+
- Recall Dolbear's function F = D(N) = 40 + \frac{1}{4}N that converts the number, N, of snowy tree cricket chirps per minute to a corresponding Fahrenheit temperature. We have earlier established that the domain of D is [40,180] and the range of D is [50,85].
-
-
-
-
-
-
- Solve the equation F = 40 + \frac{1}{4}N for N in terms of F. Call the resulting function N = E(F).
-
-
-
-
-
-
- Subtracting 40 and multiplying by 4:
- E(F) = N = 4(F - 40).
-
-
-
-
-
-
- Explain in words the process or effect of the function N = E(F). What does it take as input? What does it generate as output?
-
-
-
-
-
-
- The function E takes a temperature in degrees Fahrenheit as its input and
- outputs the corresponding number of snowy tree cricket chirps per minute.
-
-
-
-
-
-
- Use the function E that you found in (a.) to compute j(N) = E(D(N)). Simplify your result as much as possible. Do likewise for k(F) = D(E(F)). What do you notice about these two composite functions j and k?
-
-
-
-
-
-
- j(N) = E(D(N)) = 4\!\left(\left(40 + \frac{1}{4}N\right) - 40\right) = 4 \cdot \frac{N}{4} = N.
- k(F) = D(E(F)) = 40 + \frac{1}{4}(4(F-40)) = 40 + (F-40) = F.
- Both j and k are the identity: composing D and E
- in either order returns the original input, confirming that E and D
- are inverse functions.
-
-
-
-
-
-
- Consider the equations F = 40 + \frac{1}{4}N and N = 4(F-40). Do these equations express different relationships between F and N, or do they express the same relationship in two different ways? Explain.
-
-
-
-
-
-
- They express the same relationship between F and N, just solved
- for different variables. Each equation can be obtained from the other by
- algebraic manipulation, so they describe the same connection between temperature
- and chirp rate.
-
+ Recall Dolbear's function F = D(N) = 40 + \frac{1}{4}N that converts the number, N, of snowy tree cricket chirps per minute to a corresponding Fahrenheit temperature. We have earlier established that the domain of D is [40,180] and the range of D is [50,85].
+
+
+
+
+
+
+ Solve the equation F = 40 + \frac{1}{4}N for N in terms of F. Call the resulting function N = E(F).
+
+
+
+
N = E(F) = 4(F-40)
+
+
+ Subtracting 40 and multiplying by 4:
+ E(F) = N = 4(F - 40).
+
+
+
+
+
+
+ Explain in words the process or effect of the function N = E(F). What does it take as input? What does it generate as output?
+
+
+
+
The function E takes a temperature in degrees Fahrenheit as its input and outputs the corresponding number of snowy tree cricket chirps per minute.
+
+
+ The function E takes a temperature in degrees Fahrenheit as its input and
+ outputs the corresponding number of snowy tree cricket chirps per minute.
+
+
+
+
+
+
+ Use the function E that you found in (a.) to compute j(N) = E(D(N)). Simplify your result as much as possible. Do likewise for k(F) = D(E(F)). What do you notice about these two composite functions j and k?
+
+
+
+
j(N) = N and k(F) = F
+
+
+ j(N) = E(D(N)) = 4\!\left(\left(40 + \frac{1}{4}N\right) - 40\right) = 4 \cdot \frac{N}{4} = N.
+ k(F) = D(E(F)) = 40 + \frac{1}{4}(4(F-40)) = 40 + (F-40) = F.
+ Both j and k are the identity: composing D and E
+ in either order returns the original input, confirming that E and D
+ are inverse functions.
+
+
+
+
+
+
+ Consider the equations F = 40 + \frac{1}{4}N and N = 4(F-40). Do these equations express different relationships between F and N, or do they express the same relationship in two different ways? Explain.
+
+
+
+
They express the same relationship between F and N, just solved for different variables.
+
+
+ They express the same relationship between F and N, just solved
+ for different variables. Each equation can be obtained from the other by
+ algebraic manipulation, so they describe the same connection between temperature
+ and chirp rate.
+
- Determine, with justification, whether each of the following functions has an inverse function. For each function that has an inverse function, give two examples of values of the inverse function by writing statements such as s^{-1}(3) = 1.
-
-
-
-
-
-
- The function f : S \to S given by the table of values below, where S = \{0, 1, 2, 3, 4 \}.
-
- The function gdoes have an inverse, since all five output values
- are distinct. For example, g^{-1}(4) = 0 and g^{-1}(0) = 1.
-
-
-
-
-
-
- The function p given by p(t) = 7 - \frac{3}{5}t. Assume that the domain and codomain of p are both all real numbers.
-
-
-
-
-
-
- The function p(t) = 7 - \frac{3}{5}t is linear and one-to-one, so it
- does have an inverse. Solving y = 7 - \frac{3}{5}t for t:
- p^{-1}(y) = \frac{5}{3}(7-y).
- For example, p^{-1}(7) = 0 and p^{-1}(4) = 5.
-
-
-
-
-
-
- The function q given by q(t) = 7 - \frac{3}{5}t^4. Assume that the domain and codomain of q are both all real numbers.
-
-
-
-
-
-
- The function q(t) = 7 - \frac{3}{5}t^4 does not have an inverse.
- Since q(1) = q(-1) = 7 - \frac{3}{5}, the value q^{-1}\!\left(\frac{32}{5}\right)
- could be either 1 or -1.
-
-
-
-
-
-
- The functions r and s given by the graphs in the figures below. Assume that the graphs show all of the important behavior of the functions and that the apparent trends continue beyond what is pictured.
-
-
-
-
ADD ALT TEXT TO THIS IMAGE
-
-
ADD ALT TEXT TO THIS IMAGE
-
-
-
-
-
-
-
- The function r(t) passes the horizontal line test, so it does
- have an inverse. For example, r^{-1}(2) = -1 and r^{-1}(0) = 0.
-
-
- The function s(t) does not have an inverse, because it fails the
- horizontal line test: a horizontal line between y=1 and y=2 crosses
- the graph in more than one place.
-
+ Determine, with justification, whether each of the following functions has an inverse function. For each function that has an inverse function, give two examples of values of the inverse function by writing statements such as s^{-1}(3) = 1.
+
+
+
+
+
+
+ The function f : S \to S given by the table of values below, where S = \{0, 1, 2, 3, 4 \}.
+
The function gdoes have an inverse. For example, g^{-1}(4) = 0 and g^{-1}(0) = 1.
+
+
+ The function gdoes have an inverse, since all five output values
+ are distinct. For example, g^{-1}(4) = 0 and g^{-1}(0) = 1.
+
+
+
+
+
+
+ The function p given by p(t) = 7 - \frac{3}{5}t. Assume that the domain and codomain of p are both all real numbers.
+
+
+
+
The function pdoes have an inverse. For example, p^{-1}(7) = 0 and p^{-1}(4) = 5.
+
+
+ The function p(t) = 7 - \frac{3}{5}t is linear and one-to-one, so it
+ does have an inverse. Solving y = 7 - \frac{3}{5}t for t:
+ p^{-1}(y) = \frac{5}{3}(7-y).
+ For example, p^{-1}(7) = 0 and p^{-1}(4) = 5.
+
+
+
+
+
+
+ The function q given by q(t) = 7 - \frac{3}{5}t^4. Assume that the domain and codomain of q are both all real numbers.
+
+
+
+
The function q does not have an inverse, because q(1) = q(-1).
+
+
+ The function q(t) = 7 - \frac{3}{5}t^4 does not have an inverse.
+ Since q(1) = q(-1) = 7 - \frac{3}{5}, the value q^{-1}\!\left(\frac{32}{5}\right)
+ could be either 1 or -1.
+
+
+
+
+
+
+ The functions r and s given by the graphs in the figures below. Assume that the graphs show all of the important behavior of the functions and that the apparent trends continue beyond what is pictured.
+
+
+
+
+
+
+
+
+ f(x)=-2.5*atan(x)
+
+
+
+
+
+
Graph of a decreasing function r(t)that is first curving downward, goes through the origin, and starts curving upward.
Graph of a piecewise linear function s(t) whose first piece goes from (-5,5) to an open circle at (-3,1), and whose second piece goes from a closed circle at (-3,-2) ending with an open circle at (3,2), and whose third piece goes from a closed circle at (3,-3) to the edge of the graph at (5,-5).
+
+
+
+
+
The function r has an inverse, for example r^{-1}(2) = -1 and r^{-1}(0) = 0. The function s does not have an inverse.
+
+
+ The function r(t) passes the horizontal line test, so it does
+ have an inverse. For example, r^{-1}(2) = -1 and r^{-1}(0) = 0.
+
+
+ The function s(t) does not have an inverse, because it fails the
+ horizontal line test: a horizontal line between y=1 and y=2 crosses
+ the graph in more than one place.
+
- During a major rainstorm, the rainfall at Gerald R. Ford Airport is measured on a frequent basis for a 10-hour period of time. The following function g models the rate, R, at which the rain falls (in cm/hr) on the time interval t = 0 to t = 10:
-
- R = g(t) = \frac{4}{t+2} + 1
- .
-
-
-
-
-
-
- Compute g(3) and write a complete sentence to explain its meaning in the given context, including units.
-
-
-
-
-
-
- g(3) = \frac{4}{3+2} + 1 = \frac{4}{5} + 1 = \frac{9}{5} = 1.8 \text{ cm/hr.}
- At t = 3 hours into the storm, the rain is falling at a rate of 1.8
- centimeters per hour.
-
-
-
-
-
-
- Compute the average rate of change of g on the time interval [3,5] and write two careful complete sentences to explain the meaning of this value in the context of the problem, including units. Explicitly address what the value you compute tells you about how rain is falling over a certain time interval, and what you should expect as time goes on.
-
-
-
-
-
-
- Using g(3) = \frac{9}{5} and g(5) = \frac{4}{7} + 1 = \frac{11}{7}:
- AV_{[3,5]} = \frac{\frac{11}{7} - \frac{9}{5}}{2} = \frac{\frac{55-63}{35}}{2} = \frac{-8/35}{2} = -\frac{4}{35} \approx -0.114 \text{ cm/hr}^2.
- Between hours 3 and 5, the rate of rainfall decreased on average by about
- 0.114 cm/hr per hour. Since g is a decreasing function, we expect
- the rainfall rate to continue decreasing throughout the storm.
-
-
-
-
-
-
- Plot the function y = g(t) using a computational device. On the domain [0,10], what is the corresponding range of g? Why does the function g have an inverse function?
-
-
-
-
-
-
- At the endpoints: g(0) = \frac{4}{2}+1 = 3 and
- g(10) = \frac{4}{12}+1 = \frac{4}{3}. Since g is strictly
- decreasing on [0,10], the range is \left[\frac{4}{3}, 3\right].
- The function has an inverse because it is strictly decreasing (one-to-one) on
- this domain — it passes the horizontal line test.
-
-
-
-
-
-
-
-
- Determine g^{-1} \left( \frac{9}{5} \right) and write a complete sentence to explain its meaning in the given context.
-
-
-
-
-
-
- Setting \frac{9}{5} = \frac{4}{t+2} + 1:
-
- \frac{4}{5} &= \frac{4}{t+2}
- t + 2 &= 5 \implies t = 3.
-
- So g^{-1}\!\left(\frac{9}{5}\right) = 3: the rainfall rate equals
- 1.8 cm/hr at exactly 3 hours into the storm.
-
-
-
-
-
-
- According to the model g, is there ever a time during the storm that the rain falls at a rate of exactly 1 centimeter per hour? Why or why not? Provide an algebraic justification for your answer.
-
-
-
-
-
-
- No. Setting g(t) = 1 gives \frac{4}{t+2} + 1 = 1, so
- \frac{4}{t+2} = 0, which has no solution. Since \frac{4}{t+2} > 0
- for all t in [0,10], we have g(t) > 1 throughout the storm.
- (The range of g is \left[\frac{4}{3}, 3\right], and
- 1 \notin \left[\frac{4}{3}, 3\right].)
-
+ During a major rainstorm, the rainfall at Gerald R. Ford Airport is measured on a frequent basis for a 10-hour period of time. The following function g models the rate, R, at which the rain falls (in cm/hr) on the time interval t = 0 to t = 10:
+
+ R = g(t) = \frac{4}{t+2} + 1
+ .
+
+
+
+
+
+
+ Compute g(3) and write a complete sentence to explain its meaning in the given context, including units.
+
+
+
+
g(3) = 1.8 cm/hr. At t = 3 hours into the storm, the rain is falling at a rate of 1.8 centimeters per hour.
+
+
+ g(3) = \frac{4}{3+2} + 1 = \frac{4}{5} + 1 = \frac{9}{5} = 1.8 \text{ cm/hr.}
+ At t = 3 hours into the storm, the rain is falling at a rate of 1.8
+ centimeters per hour.
+
+
+
+
+
+
+ Compute the average rate of change of g on the time interval [3,5] and write two careful complete sentences to explain the meaning of this value in the context of the problem, including units. Explicitly address what the value you compute tells you about how rain is falling over a certain time interval, and what you should expect as time goes on.
+
+
+
+
AV_{[3,5]} = -\frac{4}{35}\text{cm/hr}^2. This means that between hours 3 and 5, the rate of rainfall decreased on average by about 0.114 cm/hr per hour. We expect the rainfall rate to continue decreasing.
+
+
+ Using g(3) = \frac{9}{5} and g(5) = \frac{4}{7} + 1 = \frac{11}{7}:
+ AV_{[3,5]} = \frac{\frac{11}{7} - \frac{9}{5}}{2} = \frac{\frac{55-63}{35}}{2} = \frac{-8/35}{2} = -\frac{4}{35} \approx -0.114 \text{ cm/hr}^2.
+ Between hours 3 and 5, the rate of rainfall decreased on average by about
+ 0.114 cm/hr per hour. Since g is a decreasing function, we expect
+ the rainfall rate to continue decreasing throughout the storm.
+
+
+
+
+
+
+ Plot the function y = g(t) using a computational device. On the domain [0,10], what is the corresponding range of g? Why does the function g have an inverse function?
+
+
+
+
+
+
+
+
+
+ g(x)=4/(x+2)+1
+
+
+
+
+
Graph of R=g(t), which goes through (0,3) and decreases curving upward until stopping at (10,\frac{4}{3}).
+
The range of g on [0,10] is \left[\frac{4}{3}, 3\right]. The function g has an inverse because it is strictly decreasing and passes the horizontal line test.
+
+
+ At the endpoints: g(0) = \frac{4}{2}+1 = 3 and
+ g(10) = \frac{4}{12}+1 = \frac{4}{3}. Since g is strictly
+ decreasing on [0,10], the range is \left[\frac{4}{3}, 3\right].
+ The function has an inverse because it is strictly decreasing (one-to-one) on
+ this domain — it passes the horizontal line test.
+
+
+
+
+
+
+
+ g(x)=4/(x+2)+1
+
+
+
+
+
Graph of R=g(t), which goes through (0,3) and decreases curving upward until stopping at (10,\frac{4}{3}).
+
+
+
+
+
+
+
+ Determine g^{-1} \left( \frac{9}{5} \right) and write a complete sentence to explain its meaning in the given context.
+
+
+
+
g^{-1}\!\left(\frac{9}{5}\right) = 3. The rainfall rate equals
+ 1.8 cm/hr at exactly 3 hours into the storm.
+
+
+ Setting \frac{9}{5} = \frac{4}{t+2} + 1:
+
+ \frac{4}{5} &= \frac{4}{t+2}
+ t + 2 &= 5 \implies t = 3.
+
+ So g^{-1}\!\left(\frac{9}{5}\right) = 3: the rainfall rate equals
+ 1.8 cm/hr at exactly 3 hours into the storm.
+
+
+
+
+
+
+ According to the model g, is there ever a time during the storm that the rain falls at a rate of exactly 1 centimeter per hour? Why or why not? Provide an algebraic justification for your answer.
+
+
+
+
No. Setting g(t) = 1 gives \frac{4}{t+2} + 1 = 1, but \frac{4}{t+2} \gt 0 for all t in [0,10], so there is no solution.
+
+
+ No. Setting g(t) = 1 gives \frac{4}{t+2} + 1 = 1, so
+ \frac{4}{t+2} = 0, which has no solution. Since \frac{4}{t+2} > 0
+ for all t in [0,10], we have g(t) > 1 throughout the storm.
+ (The range of g is \left[\frac{4}{3}, 3\right], and
+ 1 \notin \left[\frac{4}{3}, 3\right].)
+
- The summit of Africa's largest peak, Mt.
- Kilimanjaro, has two main ice fields and a glacier at its peak.
- Geologists measured the ice cover in the year 2000 (t = 0) to be approximately 1951 m^2;
- in the year 2007, the ice cover measured 1555 m^2.
-
-
-
-
-
-
- Suppose that the amount of ice cover at the peak of Mt.
- Kilimanjaro is changing at a constant average rate from year to year.
- Find a linear model A = f(t) whose output is the area of the ice cover,
- A,
- in square meters in year t
- (where t is the number of years after 2000).
-
-
-
-
-
-
- Using the points (0, 1951) and (7, 1555), the slope is
- \frac{1555-1951}{7-0} = \frac{-396}{7} \approx -56.6 m^2/year.
- Since t=0 gives A = 1951, the model is:
- f(t) = 1951 - \frac{396}{7} t \approx 1951 - 56.6t.
-
-
-
-
-
-
- What do the slope and A-intercept mean in the model you found in (a)? In particular, what are the units on the slope?
-
-
-
-
-
-
- The slope \approx -56.6 m^2/year means the ice cover decreases by
- about 56.6 square meters per year. The A-intercept of 1951 m^2
- represents the ice cover in the year 2000 (t=0).
-
-
-
-
-
-
- Compute f(17).
- What does this quantity measure?
- Write a complete sentence to explain.
-
-
-
-
-
-
- f(17) \approx 1951 - 56.6(17) \approx 989 \text{ m}^2.
- In the year 2017, the model predicts that the ice field near the summit of
- Mt. Kilimanjaro will have an area of approximately 989 square meters.
-
-
-
-
-
-
- If the model holds further into the future,
- when do we predict the ice cover will vanish?
-
-
-
-
-
-
- Setting f(t) = 0:
-
- 0 &\approx 1951 - 56.6t
- t &\approx \frac{1951}{56.6} \approx 34.5.
-
- The model predicts the ice will vanish around the year 2034.
-
-
-
-
-
-
- In light of your work above,
- what is a reasonable domain to use for the model A = f(t)?
- What is the corresponding range?
-
-
-
-
-
-
- A reasonable domain is 0 \le t \le 34.5 (from the year 2000 until the ice
- disappears). The corresponding range is 0 \le A \le 1951.
-
-
-
-
-
-
-
-
-
-
-
The main context of the sequence of questions in this activity comes from Exercise 30 on p.27 of Connally's
- Functions Modeling Change, 5th ed.
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ The summit of Africa's largest peak, Mt.
+ Kilimanjaro, has two main ice fields and a glacier at its peak.
+ Geologists measured the ice cover in the year 2000 (t = 0) to be approximately 1951 m^2;
+ in the year 2007, the ice cover measured 1555 m^2.
+
+
+
+
+
+
+ Suppose that the amount of ice cover at the peak of Mt.
+ Kilimanjaro is changing at a constant average rate from year to year.
+ Find a linear model A = f(t) whose output is the area of the ice cover,
+ A,
+ in square meters in year t
+ (where t is the number of years after 2000).
+
+
+
+
f(t) = 1951 - \frac{396}{7} t \approx 1951 - 56.6t.
+
+
+ Using the points (0, 1951) and (7, 1555), the slope is
+ \frac{1555-1951}{7-0} = \frac{-396}{7} \approx -56.6 m^2/year.
+ Since t=0 gives A = 1951, the model is:
+ f(t) = 1951 - \frac{396}{7} t \approx 1951 - 56.6t.
+
+
+
+
+
+
+ What do the slope and A-intercept mean in the model you found in (a)? In particular, what are the units on the slope?
+
+
+
+
+ The slope has units of \frac{\text{m}^2}{\text{year}}, or square meters per year, and represents how much the ice cover is changing each year. The A-intercept has units \text{m}^2 and represents the amount of ice cover in the year 2000.
+
+
+
+ The slope \approx -56.6 m^2/year means the ice cover decreases by
+ about 56.6 square meters per year. The A-intercept of 1951 m^2
+ represents the ice cover in the year 2000 (t=0).
+
+
+
+
+
+
+ Compute f(17).
+ What does this quantity measure?
+ Write a complete sentence to explain.
+
+
+
+
f(17) \approx 989 m^2. This is the predicted area of the ice cover in the year 2017.
+
+
+ f(17) \approx 1951 - 56.6(17) \approx 989 \text{ m}^2.
+ In the year 2017, the model predicts that the ice field near the summit of
+ Mt. Kilimanjaro will have an area of approximately 989 square meters.
+
+
+
+
+
+
+ If the model holds further into the future,
+ when do we predict the ice cover will vanish?
+
+
+
+
t \approx 34.5, which corresponds to the middle of the year 2034.
+
+
+ Setting f(t) = 0:
+
+ 0 &\approx 1951 - 56.6t
+ t &\approx \frac{1951}{56.6} \approx 34.5.
+
+ The model predicts the ice will vanish around the year 2034.
+
+
+
+
+
+
+ In light of your work above,
+ what is a reasonable domain to use for the model A = f(t)?
+ What is the corresponding range?
+
+
+
+
A reasonable domain is 0 \le t \le 34.5, and the corresponding range is 0 \le A \le 1951.
+
+
+ A reasonable domain is 0 \le t \le 34.5 (from the year 2000 until the ice
+ disappears). The corresponding range is 0 \le A \le 1951.
+
+
+
+
+
The main context of the sequence of questions in this activity comes from Exercise 30 on p.27 of Connally's
+ Functions Modeling Change, 5th ed.
- Find an equation for the line that is determined by the following conditions;
- write your answer in point-slope form wherever possible.
-
-
-
-
-
-
- The line with slope \frac{3}{7} that passes through (-11, -17).
-
-
-
-
-
-
- y - (-17) = \frac{3}{7}(x - (-11)), \quad \text{i.e.,} \quad y + 17 = \frac{3}{7}(x+11).
-
-
-
-
-
-
- The line passing through the points (-2,5) and (3,-1).
-
-
-
-
-
-
- The slope is m = \frac{-1-5}{3-(-2)} = -\frac{6}{5}. Using the point
- (3,-1):
- y + 1 = -\frac{6}{5}(x-3).
-
-
-
-
-
-
- The line passing through (4,9) that is parallel to the line 2x - 3y = 5.
-
-
-
-
-
-
- Solving 2x - 3y = 5 for y gives y = \frac{2}{3}x - \frac{5}{3},
- so the slope is m = \frac{2}{3}. Using the point (4,9):
- y - 9 = \frac{2}{3}(x-4).
-
-
-
-
-
-
- Explain why the function f given by the data in the following table appears to be linear and find a formula for f(x).
-
- The function appears to be linear because the rate of change is constant:
- for each increase of 1 in x, f(x) decreases by 2,
- giving slope m = -2. Using the point (1,7):
-
- y - 7 &= -2(x-1)
- f(x) &= 9 - 2x.
-
-
-
-
-
-
-
- Find a formula for the linear function shown in the following figure.
-
-
-
ADD ALT TEXT TO THIS IMAGE
-
-
-
-
-
-
- The graph has slope m = -\frac{1}{3} and passes through (4,1), so:
-
- y - 1 &= -\frac{1}{3}(x-4)
- h(x) &= \frac{7}{3} - \frac{1}{3}x.
-
-
+ Find an equation for the line that is determined by the following conditions;
+ write your answer in point-slope form wherever possible.
+
+
+
+
+
+
+ The line with slope \frac{3}{7} that passes through (-11, -17).
+
+
+
+
y + 17 = \frac{3}{7}(x+11)
+
+
+ y - (-17) = \frac{3}{7}(x - (-11)), \quad \text{i.e.,} \quad y + 17 = \frac{3}{7}(x+11).
+
+
+
+
+
+
+ The line passing through the points (-2,5) and (3,-1).
+
+
+
+
y + 1 = -\frac{6}{5}(x-3)
+
+
+ The slope is m = \frac{-1-5}{3-(-2)} = -\frac{6}{5}. Using the point
+ (3,-1):
+ y + 1 = -\frac{6}{5}(x-3).
+
+
+
+
+
+
+ The line passing through (4,9) that is parallel to the line 2x - 3y = 5.
+
+
+
+
y - 9 = \frac{2}{3}(x-4)
+
+
+ Solving 2x - 3y = 5 for y gives y = \frac{2}{3}x - \frac{5}{3},
+ so the slope is m = \frac{2}{3}. Using the point (4,9):
+ y - 9 = \frac{2}{3}(x-4).
+
+
+
+
+
+
+ Explain why the function f given by the data in the following table appears to be linear and find a formula for f(x).
+
The function appears to be linear because the rate of change is constant per 1 change in x. f(x) = 9 - 2x
+
+
+ The function appears to be linear because the rate of change is constant:
+ for each increase of 1 in x, f(x) decreases by 2,
+ giving slope m = -2. Using the point (1,7):
+
+ y - 7 &= -2(x-1)
+ f(x) &= 9 - 2x.
+
+
+
+
+
+
+
+ Find a formula for the linear function shown in the following figure.
+
+
+
Graph of a function y = h(x) which is a straight line going through the points (-2,3) and (4,1).
+
+
+
+
h(x) = \frac{7}{3} - \frac{1}{3}x
+
+
+ The graph has slope m = -\frac{1}{3} and passes through (4,1), so:
+
+ y - 1 &= -\frac{1}{3}(x-4)
+ h(x) &= \frac{7}{3} - \frac{1}{3}x.
+
+
- In each of the following prompts, we investigate linear functions in context.
-
-
-
-
-
-
- A town's population initially has 28750 people present and then grows at a constant rate of 825 people per year.
- Find a linear model P = f(t) for the number of people in the town in year t.
-
-
-
-
-
-
- f(t) = 28750 + 825t.
-
-
-
-
-
-
- A different town's population Q is given by the function Q = g(t) = 42505 - 465t. What is the slope of this function and what is its meaning in the model? Write a complete sentence to explain.
-
-
-
-
-
-
- The slope is -465 people per year. This means the town's population
- decreases by 465 people each year.
-
-
-
-
-
-
- A spherical tank is being drained with a pump. Initially the tank is full with \frac{32\pi}{3} cubic feet of water. Assume the tank is drained at a constant rate of 1.2 cubic feet per minute. Find a linear model V = p(t) for the total amount of water in the tank at time t. In addition, what is a reasonable approximate domain for the model?
-
-
-
-
-
-
- p(t) = \frac{32\pi}{3} - 1.2t.
- The tank empties when p(t) = 0, i.e., t = \frac{32\pi}{3 \cdot 1.2} \approx 27.9 minutes.
- A reasonable domain is [0, 27.9].
-
-
-
-
-
-
- A conical tank is being filled in such a way that the height of the water in the tank, h (in feet), at time t (in minutes) is given by the function h = q(t) = 0.65t. What can you say about how the water level is rising? Write at least one careful sentence to explain.
-
-
-
-
-
-
- Since q(t) = 0.65t is a linear function with slope 0.65, the water
- level rises at a constant rate of 0.65 feet per minute.
-
-
-
-
-
-
- Suppose we know that a 5-year old car's value is $10200, and that after 10 years its value is $4600. Assuming that the car's value depreciates linearly, find a function C = L(t) whose output is the value of the car in year t. What is a reasonable domain for the model? What is the value and meaning of the slope of the line? Write at least one careful sentence to explain.
-
-
-
-
-
-
- The slope is \frac{4600 - 10200}{10-5} = \frac{-5600}{5} = -1120 dollars per year.
- Using the point (5, 10200):
-
- C - 10200 &= -1120(t-5)
- L(t) &= 15800 - 1120t.
-
- The car reaches zero value when t \approx 14.1, so a reasonable domain is
- [0, 14.1]. The slope means the car loses \$1120 in value each year.
-
+ In each of the following prompts, we investigate linear functions in context.
+
+
+
+
+
+
+ A town's population initially has 28750 people present and then grows at a constant rate of 825 people per year.
+ Find a linear model P = f(t) for the number of people in the town in year t.
+
+
+
+
f(t) = 28750 + 825t
+
+
+ f(t) = 28750 + 825t
+
+
+
+
+
+
+ A different town's population Q is given by the function Q = g(t) = 42505 - 465t. What is the slope of this function and what is its meaning in the model? Write a complete sentence to explain.
+
+
+
+
+ The slope is -465. This means the town's population
+ decreases by 465 people each year.
+
+
+
+ The slope is -465 people per year. This means the town's population
+ decreases by 465 people each year.
+
+
+
+
+
+
+ A spherical tank is being drained with a pump. Initially the tank is full with \frac{32\pi}{3} cubic feet of water. Assume the tank is drained at a constant rate of 1.2 cubic feet per minute. Find a linear model V = p(t) for the total amount of water in the tank at time t. In addition, what is a reasonable approximate domain for the model?
+
+
+
+
p(t) = \frac{32\pi}{3} - 1.2t. A reasonable domain is [0, 27.9].
+
+
+ p(t) = \frac{32\pi}{3} - 1.2t.
+ The tank empties when p(t) = 0, i.e., t = \frac{32\pi}{3 \cdot 1.2} \approx 27.9 minutes.
+ A reasonable domain is [0, 27.9].
+
+
+
+
+
+
+ A conical tank is being filled in such a way that the height of the water in the tank, h (in feet), at time t (in minutes) is given by the function h = q(t) = 0.65t. What can you say about how the water level is rising? Write at least one careful sentence to explain.
+
+
+
+
+ The water level rises at a constant rate of 0.65 feet per minute.
+
+
+
+ Since q(t) = 0.65t is a linear function with slope 0.65, the water
+ level rises at a constant rate of 0.65 feet per minute.
+
+
+
+
+
+
+ Suppose we know that a 5-year old car's value is $10200, and that after 10 years its value is $4600. Assuming that the car's value depreciates linearly, find a function C = L(t) whose output is the value of the car in year t. What is a reasonable domain for the model? What is the value and meaning of the slope of the line? Write at least one careful sentence to explain.
+
+
+
+
C=L(t) = 15800 - 1120t. A reasonable domain is [0, 14.1]. The slope is -1120, meaning the car loses \$1120 in value each year.
+
+
+ The slope is \frac{4600 - 10200}{10-5} = \frac{-5600}{5} = -1120 dollars per year.
+ Using the point (5, 10200):
+
+ C - 10200 &= -1120(t-5)
+ L(t) &= 15800 - 1120t.
+
+ The car reaches zero value when t \approx 14.1, so a reasonable domain is
+ [0, 14.1]. The slope means the car loses \$1120 in value each year.
+
- A water balloon is tossed vertically from a window at an initial height of 37 feet and with an initial velocity of 41 feet per second.
-
-
-
-
-
-
- Determine a formula, s(t),
- for the function that models the height of the water balloon at time t.
-
-
-
-
-
-
- Using the standard model s(t) = -16t^2 + v_0 t + s_0 with initial height
- s_0 = 37 ft and initial velocity v_0 = 41 ft/s:
- s(t) = -16t^2 + 41t + 37.
-
-
-
-
-
-
- Plot the function in Desmos
- in an appropriate window. Sketch a copy of the graph here.
-
-
-
-
-
-
Plotting s(t) = -16t^2 + 41t + 37 in an appropriate window (for example, 0 \le t \le 4 and 0 \le s \le 70) shows an upside-down parabola that rises from s(0) = 37, reaches a maximum near t \approx 1.28 seconds and s \approx 63.3 feet, and then falls back to ground level near t \approx 3.3 seconds. Answers will vary on the exact sketch.
-
-
-
-
-
-
-
- Use the graph to estimate the time the water balloon lands.
-
-
-
-
-
-
- From the graph, the balloon appears to land at approximately t \approx 3.3 seconds.
-
-
-
-
-
-
- Use algebra to find the exact
- time the water balloon lands.
-
-
-
-
-
-
- Setting s(t) = 0 and applying the quadratic formula with a=-16,
- b=41, c=37:
-
- t &= \frac{-41 \pm \sqrt{41^2 - 4(-16)(37)}}{2(-16)}
- &= \frac{-41 \pm \sqrt{1681 + 2368}}{-32}
- &= \frac{-41 \pm \sqrt{4049}}{-32}.
-
- Taking the root that gives t > 0: t = \frac{-41 - \sqrt{4049}}{-32} \approx 3.27 seconds.
-
-
-
-
-
-
- Determine the exact time the water balloon reaches its highest point and its height at that time.
-
-
-
-
-
-
- The vertex occurs at t = -\frac{b}{2a} = -\frac{41}{2(-16)} = \frac{41}{32} \approx 1.28 seconds.
- The maximum height is
- s\!\left(\frac{41}{32}\right) = -16 \cdot \frac{1681}{1024} + 41 \cdot \frac{41}{32} + 37 = \frac{4049}{64} \approx 63.3 \text{ feet.}
-
-
-
-
-
-
- Compute the average rate of change of s on the intervals [1.5, 2],
- [2, 2.5], [2.5,3].
- Include units on your answers and write one sentence to explain the meaning of the values you found.
- Sketch appropriate lines on the graph of s whose respective slopes are the values of these average rates of change.
-
-
-
-
-
-
- Using s(1.5) = 62.5, s(2) = 55, s(2.5) = 39.5, s(3) = 16:
- AV_{[1.5,2]} = \frac{55-62.5}{0.5} = -15 \text{ ft/s},
- AV_{[2,2.5]} = \frac{39.5-55}{0.5} = -31 \text{ ft/s},
- AV_{[2.5,3]} = \frac{16-39.5}{0.5} = -47 \text{ ft/s.}
- These values represent the average downward velocity of the balloon over each
- half-second interval; the balloon is falling faster and faster as it descends.
-
+ A water balloon is tossed vertically from a window at an initial height of 37 feet and with an initial velocity of 41 feet per second.
+
+
+
+
+
+
+ Determine a formula, s(t),
+ for the function that models the height of the water balloon at time t.
+
+
+
+
s(t) = -16t^2 + 41t + 37
+
+
+ Using the standard model s(t) = -16t^2 + v_0 t + s_0 with initial height
+ s_0 = 37 ft and initial velocity v_0 = 41 ft/s:
+ s(t) = -16t^2 + 41t + 37.
+
+
+
+
+
+
+ Plot the function in Desmos
+ in an appropriate window. Sketch a copy of the graph here.
+
+
+
+
+
+
+
+ f(x)=-16*x^2+41*x+37
+
+
+
+
+
+
+
A graph of s(t).
+
+
+
Plotting s(t) = -16t^2 + 41t + 37 in an appropriate window (for example, 0 \le t \le 4 and 0 \le s \le 70) shows an upside-down parabola that rises from s(0) = 37, reaches a maximum near t \approx 1.28 seconds and s \approx 63.3 feet, and then falls back to ground level near t \approx 3.3 seconds. Answers will vary on the exact sketch.
+
+
+
+
+ f(x)=-16*x^2+41*x+37
+
+
+
+
+
+
+
A graph of s(t).
+
+
+
+
+
+
+
+
+ Use the graph to estimate the time the water balloon lands.
+
+
+
+
t \approx 3.25 seconds
+
+
+ From the graph, the balloon appears to land at approximately t \approx 3.25 seconds.
+
+
+
+
+
+
+ Use algebra to find the exact
+ time the water balloon lands.
+
+
+
+
t = \frac{-41 - \sqrt{4049}}{-32} \approx 3.27 seconds
+
+
+ Setting s(t) = 0 and applying the quadratic formula with a=-16,
+ b=41, c=37:
+
+ t &= \frac{-41 \pm \sqrt{41^2 - 4(-16)(37)}}{2(-16)}
+ &= \frac{-41 \pm \sqrt{1681 + 2368}}{-32}
+ &= \frac{-41 \pm \sqrt{4049}}{-32}.
+
+ Taking the root that gives t > 0: t = \frac{-41 - \sqrt{4049}}{-32} \approx 3.27 seconds.
+
+
+
+
+
+
+ Determine the exact time the water balloon reaches its highest point and its height at that time.
+
+ The vertex occurs at t = -\frac{b}{2a} = -\frac{41}{2(-16)} = \frac{41}{32} \approx 1.28 seconds.
+ The maximum height is
+ s\left(\frac{41}{32}\right) = -16 \cdot \frac{1681}{1024} + 41 \cdot \frac{41}{32} + 37 = \frac{4049}{64} \approx 63.3 \text{ feet.}
+
+
+
+
+
+
+ Compute the average rate of change of s on the intervals [1.5, 2],
+ [2, 2.5], [2.5,3].
+ Include units on your answers and write one sentence to explain the meaning of the values you found.
+ Sketch appropriate lines on the graph of s whose respective slopes are the values of these average rates of change.
+
+
+
+
AV_{[1.5,2]} = -15 \text{ ft/s}, AV_{[2,2.5]} = -31 \text{ ft/s}, AV_{[2.5,3]} = -47 \text{ ft/s}. Between t=1.5 and t=2, the water balloon is falling at an average rate of 15 feet per second; similarly for the other intervals. The balloon is falling faster and faster as it descends.
+
+
+
+ f(x)=-16*x^2+41*x+37
+
+
+
+
+
+
+
+
+
+
A graph of s(t) along with the lines representing the average rates of change on [1.5,2], [2,2.5], and [2.5,3].
+
+
+
+
+ Using s(1.5) = 62.5, s(2) = 55, s(2.5) = 39.5, s(3) = 16:
+ AV_{[1.5,2]} = \frac{55-62.5}{0.5} = -15 \text{ ft/s},
+ AV_{[2,2.5]} = \frac{39.5-55}{0.5} = -31 \text{ ft/s},
+ AV_{[2.5,3]} = \frac{16-39.5}{0.5} = -47 \text{ ft/s.}
+ These values represent the average velocity of the balloon over each
+ half-second interval; the balloon is falling faster and faster as it descends.
+
- Open a browser and point it to Desmos.
- In Desmos, enter q(x) = ax^2 + bx + c;
- you will be prompted to add sliders for a, b,
- and c. Do so.
- Then begin exploring with the sliders and respond to the following questions.
-
-
-
-
-
-
- Describe how changing the value of a affects the graph of q.
-
-
-
-
-
-
- If a > 0 the parabola opens upward; if a < 0 it opens downward.
- The larger |a| is, the narrower the parabola; the closer |a| is to
- zero, the wider the parabola.
-
-
-
-
-
-
- Describe how changing the value of b affects the graph of q.
-
-
-
-
-
-
- With a = 1 and c = 0, changing b shifts the vertex
- diagonally: making b > 0 moves the vertex left and down, while
- making b < 0 moves it right and down.
-
-
-
-
-
-
- Describe how changing the value of c affects the graph of q.
-
-
-
-
-
-
- With a = 1 and b = 0, changing c moves the vertex straight
- up or down along the y-axis. The vertex is at (0, c), and c
- is the y-intercept of the function.
-
-
-
-
-
-
- Which parameter seems to have the simplest effect?
- Which parameter seems to have the most complicated effect?
- Why?
-
-
-
-
-
-
- The parameter a has the most clearly understandable effect: it controls
- whether the parabola opens up or down and how wide it is. The parameter c
- has a simple effect as well (pure vertical translation). The parameter b
- has the most complicated effect because changing it moves the vertex diagonally
- rather than purely up, down, or in terms of width.
-
-
-
-
-
-
- Is it possible to find a formula for a quadratic function that passes through the points (0,8),
- (1,12), (2,12)? If yes, do so; if not, explain why not.
-
-
-
-
-
-
- Yes. Since (1,12) and (2,12) have the same output, the axis of
- symmetry is x = 1.5 and the vertex has x-coordinate 1.5.
- Since the parabola must pass through (0,8), which is below the values at
- x=1 and x=2, the parabola opens downward. With c = q(0) = 8
- and using vertex form or solving the system, the formula is
- q(x) = -2x^2 + 6x + 8.
-
+ Open a browser and point it to Desmos.
+ In Desmos, enter q(x) = ax^2 + bx + c;
+ you will be prompted to add sliders for a, b,
+ and c. Do so.
+ Then begin exploring with the sliders and respond to the following questions.
+
+
+
+
+
+
+ Describe how changing the value of a affects the graph of q.
+
+
+
+
If a > 0 the parabola opens upward; if a < 0 it opens downward.
+ The larger |a| is, the narrower the parabola; the closer |a| is to
+ zero, the wider the parabola.
+
+
+ If a > 0 the parabola opens upward; if a < 0 it opens downward.
+ The larger |a| is, the narrower the parabola; the closer |a| is to
+ zero, the wider the parabola.
+
+
+
+
+
+
+ Describe how changing the value of b affects the graph of q.
+
+
+
+
With a = 1 and c = 0, changing b shifts the vertex
+ diagonally: making b > 0 moves the vertex left and down, while
+ making b < 0 moves it right and down.
+
+
+ With a = 1 and c = 0, changing b shifts the vertex
+ diagonally: making b > 0 moves the vertex left and down, while
+ making b < 0 moves it right and down.
+
+
+
+
+
+
+ Describe how changing the value of c affects the graph of q.
+
+
+
+
With a = 1 and b = 0, changing c moves the vertex straight
+ up or down along the y-axis.
+
+
+ With a = 1 and b = 0, changing c moves the vertex straight
+ up or down along the y-axis. The vertex is at (0, c), and c
+ is the y-intercept of the function.
+
+
+
+
+
+
+ Which parameter seems to have the simplest effect?
+ Which parameter seems to have the most complicated effect?
+ Why?
+
+
+
+
The parameter c seems to have the simplest effect while b seems to have the most complicated effect.
+
+
+ The parameter a has the most clearly understandable effect: it controls
+ whether the parabola opens up or down and how wide it is. The parameter c
+ has a simple effect as well (pure vertical translation). The parameter b
+ has the most complicated effect because changing it moves the vertex diagonally
+ rather than purely up, down, or in terms of width.
+
+
+
+
+
+
+ Is it possible to find a formula for a quadratic function that passes through the points (0,8),
+ (1,12), (2,12)? If yes, do so; if not, explain why not.
+
+
+
+
q(x) = -2x^2 + 6x + 8
+
+
+ Yes. Since (1,12) and (2,12) have the same output, the axis of
+ symmetry is x = 1.5 and the vertex has x-coordinate 1.5.
+ Since the parabola must pass through (0,8), which is below the values at
+ x=1 and x=2, the parabola opens downward. With c = q(0) = 8
+ and using vertex form or solving the system, the formula is
+ q(x) = -2x^2 + 6x + 8.
+
- Reason algebraically using appropriate properties of quadratic functions to answer the following questions. Use Desmos to check your results graphically.
-
-
-
-
-
-
- How many quadratic functions have x-intercepts at (-5,0) and (10,0) and a y-intercept at (0,-1)? Can you determine an exact formula for such a function? If yes, do so. If not, explain why.
-
-
-
-
-
-
- Three points determine a unique quadratic, so exactly one such function exists.
- Using the factored form q(x) = a(x+5)(x-10) and the y-intercept
- q(0) = -1:
- a(5)(-10) = -50a = -1 \implies a = \frac{1}{50}.
- Thus q(x) = \frac{1}{50}x^2 - \frac{1}{10}x - 1.
-
-
-
-
-
-
- Suppose that a quadratic function q has vertex (-3,-4) and opens upward. How many x-intercepts can you guarantee the function has? Why?
-
-
-
-
-
-
- We can guarantee exactly two x-intercepts. Since the vertex (-3,-4)
- is below the x-axis and the parabola opens upward, both arms must cross the
- x-axis.
-
-
-
-
-
-
- In addition to the information in (b), suppose you know that q(-1) = -3. Can you determine an exact formula for q? If yes, do so. If not, explain why.
-
-
-
-
-
-
- Yes. Using vertex form q(x) = a(x+3)^2 - 4 and the point (-1,-3):
-
- -3 &= a(-1+3)^2 - 4 = 4a - 4
- a &= \frac{1}{4}.
-
- Thus q(x) = \frac{1}{4}(x+3)^2 - 4.
-
-
-
-
-
-
- Does the quadratic function p(x) = -3(x+1)^2 + 9 have 0, 1, or 2x-intercepts? Reason algebraically to determine the exact values of any such intercepts or explain why none exist.
-
-
-
-
-
-
- The vertex (-1,9) is above the x-axis and a=-3 < 0, so the
- parabola opens downward and has two x-intercepts. Setting p(x)=0:
-
- -3(x+1)^2 + 9 &= 0
- (x+1)^2 &= 3
- x &= -1 \pm \sqrt{3}.
-
-
-
-
-
-
-
- Does the quadratic function w(x) = -2x^2 + 10x - 20 have 0, 1, or 2x-intercepts? Reason algebraically to determine the exact values of any such intercepts or explain why none exist.
-
-
-
-
-
-
- The discriminant is b^2 - 4ac = 10^2 - 4(-2)(-20) = 100 - 160 = -60 < 0,
- so there are no real x-intercepts. Equivalently, the vertex is at
- x = \frac{-10}{2(-2)} = 2.5 with w(2.5) = -7.5 < 0, and since
- a = -2 < 0 the parabola opens downward and lies entirely below the
- x-axis.
-
+ Reason algebraically using appropriate properties of quadratic functions to answer the following questions. Use Desmos to check your results graphically.
+
+
+
+
+
+
+ How many quadratic functions have x-intercepts at (-5,0) and (10,0) and a y-intercept at (0,-1)? Can you determine an exact formula for such a function? If yes, do so. If not, explain why.
+
+
+
+
One, q(x) = \frac{1}{50}x^2 - \frac{1}{10}x - 1.
+
+
+ Three points determine a unique quadratic, so exactly one such function exists.
+ Using the factored form q(x) = a(x+5)(x-10) and the y-intercept
+ q(0) = -1:
+ a(5)(-10) = -50a = -1 \implies a = \frac{1}{50}.
+ Thus q(x) = \frac{1}{50}x^2 - \frac{1}{10}x - 1.
+
+
+
+
+
+
+ Suppose that a quadratic function q has vertex (-3,-4) and opens upward. How many x-intercepts can you guarantee the function has? Why?
+
+
+
+
Two, because the vertex is below the x-axis and the parabola opens upward.
+
+
+ We can guarantee exactly two x-intercepts. Since the vertex (-3,-4)
+ is below the x-axis and the parabola opens upward, both arms must cross the
+ x-axis.
+
+
+
+
+
+
+ In addition to the information in (b), suppose you know that q(-1) = -3. Can you determine an exact formula for q? If yes, do so. If not, explain why.
+
+
+
+
q(x) = \frac{1}{4}(x+3)^2 - 4
+
+
+ Yes. Using vertex form q(x) = a(x+3)^2 - 4 and the point (-1,-3):
+
+ -3 &= a(-1+3)^2 - 4 = 4a - 4
+ a &= \frac{1}{4}.
+
+ Thus q(x) = \frac{1}{4}(x+3)^2 - 4.
+
+
+
+
+
+
+ Does the quadratic function p(x) = -3(x+1)^2 + 9 have 0, 1, or 2x-intercepts? Reason algebraically to determine the exact values of any such intercepts or explain why none exist.
+
+
+
+
Two, because the vertex (-1,9) is above the x-axis and the parabola opens downward. x=-1 \pm \sqrt{3}
+
+
+ The vertex (-1,9) is above the x-axis and a=-3 < 0, so the
+ parabola opens downward and has two x-intercepts. Setting p(x)=0:
+
+ -3(x+1)^2 + 9 &= 0
+ (x+1)^2 &= 3
+ x &= -1 \pm \sqrt{3}.
+
+
+
+
+
+
+
+ Does the quadratic function w(x) = -2x^2 + 10x - 20 have 0, 1, or 2x-intercepts? Reason algebraically to determine the exact values of any such intercepts or explain why none exist.
+
+
+
+
Zero, because b^2 - 4ac = 10^2 - 4(-2)(-20) is negative.
+
+
+ The discriminant is b^2 - 4ac = 10^2 - 4(-2)(-20) = 100 - 160 = -60 < 0,
+ so there are no real x-intercepts. Equivalently, the vertex is at
+ x = \frac{-10}{2(-2)} = 2.5 with w(2.5) = -7.5 < 0, and since
+ a = -2 < 0 the parabola opens downward and lies entirely below the
+ x-axis.
+
- Consider a tank in the shape of an inverted circular cone (point down) where the tank's radius is 2 feet and its depth is 4 feet. Suppose that the tank is being filled with water that is entering at a constant rate of 0.75 cubic feet per minute.
-
-
-
-
-
-
- Sketch a labeled picture of the tank, including a snapshot of there being water in the tank prior to the tank being completely full.
-
-
-
-
-
-
Answers will vary. A sketch should show an inverted cone with labeled radius 2 ft and depth 4 ft, with water partially filling the bottom of the tank.
-
-
-
-
-
- What are some quantities that are changing in this scenario? What are some quantities that are not changing?
-
-
-
-
-
-
Quantities that are changing include the volume of water in the tank, the height of the water, the surface area of the top of the water, and time. Quantities that are not changing include the rate at which water enters, the height of the tank, and the radius of the tank.
-
-
-
-
-
- Fill in the following table of values to determine how much water, V, is in the tank at a given time in minutes, t, and thus generate a graph of the relationship between volume and time by plotting the data on the provided axes.
-
Since water enters at a constant rate of 0.75 cubic feet per minute, V = 0.75t. The table values are: t = 0, V = 0; t = 1, V = 0.75; t = 2, V = 1.5; t = 3, V = 2.25; t = 4, V = 3.0; t = 5, V = 3.75. The graph of V versus t is a straight line through the origin with slope 0.75.
-
-
-
-
-
- Finally, think about how the height, h, of the water changes in tandem with time. Without attempting to determine specific values of h at particular values of t, how would you expect the data for the relationship between h and t to appear? Use the provided axes to sketch at least two possibilities; write at least one sentence to explain how you think the graph should appear.
-
-
-
ADD ALT TEXT TO THIS IMAGE
-
ADD ALT TEXT TO THIS IMAGE
-
-
-
-
-
-
Since the radius of the cone increases as the water level rises, the height of the water increases rapidly at first but the rate of increase slows as the water level rises. Sketches should show a concave-down increasing curve approaching the maximum depth of 4 ft.
-
-
-
-
-
-
-
See solutions to individual tasks above.
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ Consider a tank in the shape of an inverted circular cone (point down) where the tank's radius is 2 feet and its depth is 4 feet. Suppose that the tank is being filled with water that is entering at a constant rate of 0.75 cubic feet per minute.
+
+
+
+
+
+
+ Sketch a labeled picture of the tank, including a snapshot of there being water in the tank prior to the tank being completely full.
+
+
+
+
Sketches should look something like the following image.
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
An inverted cone with labeled radius 2 ft and depth 4 ft, with water partially filling the bottom of the tank.
+
+
+
Answers will vary. A sketch should show an inverted cone with labeled radius 2 ft and depth 4 ft, with water partially filling the bottom of the tank.
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
An inverted cone with labeled radius 2 ft and depth 4 ft, with water partially filling the bottom of the tank.
+
+
+
+
+
+ What are some quantities that are changing in this scenario? What are some quantities that are not changing?
+
+
+
+
Changing: volume, height, surface area, and time. Not changing: rate of water entering, height of tank, and radius of tank.
+
+
Quantities that are changing include the volume of water in the tank, the height of the water, the surface area of the top of the water, and time. Quantities that are not changing include the rate at which water enters, the height of the tank, and the radius of the tank.
+
+
+
+
+
+ Fill in the following table of values to determine how much water, V, is in the tank at a given time in minutes, t, and thus generate a graph of the relationship between volume and time by plotting the data on the provided axes.
+
A graph of volume versus time, using the values from the table. The graph is a straight line through the origin with slope 0.75.
+
+
+
+
Since water enters at a constant rate of 0.75 cubic feet per minute, V = 0.75t. The table values are: t = 0, V = 0; t = 1, V = 0.75; t = 2, V = 1.5; t = 3, V = 2.25; t = 4, V = 3.0; t = 5, V = 3.75. The graph of V versus t is a straight line through the origin with slope 0.75.
A graph of volume versus time, using the values from the table. The graph is a straight line through the origin with slope 0.75.
+
+
+
+
+
+
+ Finally, think about how the height, h, of the water changes in tandem with time. Without attempting to determine specific values of h at particular values of t, how would you expect the data for the relationship between h and t to appear? Use the provided axes to sketch at least two possibilities; write at least one sentence to explain how you think the graph should appear.
+
+
+
+
+
+
+
+
+
+
+
Blank axes for making a graph of height versus time.
+
+
+
+
+
+
+
+
+
Blank axes for making a graph of height versus time.
+
+
+
+
+
Sketches should show a curve starting at the origin that is bending downwards as it approaches the maximum depth of 4 ft.
+
+
Since the radius of the cone increases as the water level rises, the height of the water increases rapidly at first but the rate of increase slows as the water level rises. Sketches should show a concave-down increasing curve approaching the maximum depth of 4 ft.
- Consider a tank in the shape of a sphere where the tank's radius is 3 feet. Suppose that the tank is initially completely full and that it is being drained by a pump at a constant rate of 1.2 cubic feet per minute.
-
-
-
-
-
-
- Sketch a labeled picture of the tank, including a snapshot of some water remaining in the tank prior to the tank being completely empty.
-
-
-
-
-
-
Answers will vary. A sketch should show a sphere with labeled radius 3 ft, with water partially filling the tank.
-
-
-
-
-
- What are some quantities that are changing in this scenario? What are some quantities that are not changing?
-
-
-
-
-
-
Quantities that are changing include the volume of water in the tank, the height of the water, the surface area of the top of the water, and time. Quantities that are not changing include the rate at which water is pumped out and the radius of the tank.
-
-
-
-
-
- Recall that the volume of a sphere of radius r is V = \frac{4}{3} \pi r^3. When the tank is completely full at time t = 0 right before it starts being drained, how much water is present?
-
-
-
-
-
-
Since the radius is 3 feet, the full tank contains \frac{4}{3}\pi(3)^3 = 36\pi \approx 113.1 cubic feet of water.
-
-
-
-
-
- How long will it take for the tank to drain completely?
-
-
-
-
-
-
Water is removed at 1.2 cubic feet per minute. The tank drains when 1.2t = 36\pi, so t = \frac{36\pi}{1.2} = 30\pi \approx 94.25 minutes.
-
-
-
-
-
- Fill in the following table of values to determine how much water, V, is in the tank at a given time in minutes, t, and thus generate a graph of the relationship between volume and time. Write a sentence to explain why the data's graph appears the way that it does.
-
Since water is removed at a constant rate, V = 36\pi - 1.2t. The approximate values are: t = 0, V \approx 113.1; t = 20, V \approx 89.1; t = 40, V \approx 65.1; t = 60, V \approx 41.1; t = 80, V \approx 17.1; t = 94.24, V \approx 0. The graph is linear because water is removed at a constant rate.
-
-
-
-
-
- Finally, think about how the height of the water changes in tandem with time. What is the height of the water when t = 0? What is the height when the tank is empty? How would you expect the data for the relationship between h and t to appear? Use the provided axes to sketch at least two possibilities; write at least one sentence to explain how you think the graph should appear.
-
-
-
ADD ALT TEXT TO THIS IMAGE
-
ADD ALT TEXT TO THIS IMAGE
-
-
-
-
-
-
When t = 0 the tank is full, so the height of the water is the diameter h = 6 ft. When the tank is empty, h = 0. The height decreases quickly at first, then the rate of decrease slows as the widest part of the sphere is reached, then speeds up again as the tank nears empty. Sketches should show an S-shaped decreasing curve from h = 6 to h = 0.
-
-
-
-
-
-
-
See solutions to individual tasks above.
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ Consider a tank in the shape of a sphere where the tank's radius is 3 feet. Suppose that the tank is initially completely full and that it is being drained by a pump at a constant rate of 1.2 cubic feet per minute.
+
+
+
+
+
+
+ Sketch a labeled picture of the tank, including a snapshot of some water remaining in the tank prior to the tank being completely empty.
+
+
+
+
Sketches should look something like the following image.
An sphere with labeled radius 3 ft, with water partially filling the bottom of the spherical tank.
+
+
+
+
+
+ What are some quantities that are changing in this scenario? What are some quantities that are not changing?
+
+
+
+
Changing: volume, height, surface area, and time. Not changing: rate of water being pumped out and radius of the tank.
+
+
Quantities that are changing include the volume of water in the tank, the height of the water, the surface area of the top of the water, and time. Quantities that are not changing include the rate at which water is pumped out and the radius of the tank.
+
+
+
+
+
+ Recall that the volume of a sphere of radius r is V = \frac{4}{3} \pi r^3. When the tank is completely full at time t = 0 right before it starts being drained, how much water is present?
+
+
+
+
\frac{4}{3}\pi(3)^3 = 36\pi \approx 113.1 cubic feet of water.
+
+
Since the radius is 3 feet, the full tank contains \frac{4}{3}\pi(3)^3 = 36\pi \approx 113.1 cubic feet of water.
+
+
+
+
+
+ How long will it take for the tank to drain completely?
+
+
+
+
Approximately 94.25 minutes
+
+
Water is removed at 1.2 cubic feet per minute. The tank drains when 1.2t = 36\pi, so t = \frac{36\pi}{1.2} = 30\pi \approx 94.25 minutes.
+
+
+
+
+
+ Fill in the following table of values to determine how much water, V, is in the tank at a given time in minutes, t, and thus generate a graph of the relationship between volume and time. Write a sentence to explain why the data's graph appears the way that it does.
+
A graph of volume versus time, using the values from the table. The graph is a straight line through (0, 36\pi) with slope -1.2.
+
+
+
+
+
+
Since water is removed at a constant rate, V = 36\pi - 1.2t. The approximate values are: t = 0, V \approx 113.1; t = 20, V \approx 89.1; t = 40, V \approx 65.1; t = 60, V \approx 41.1; t = 80, V \approx 17.1; t = 94.24, V \approx 0. The graph is linear because water is removed at a constant rate.
A graph of volume versus time, using the values from the table. The graph is a straight line through (0, 36\pi) with slope -1.2.
+
+
+
+
+
+
+
+
+ Finally, think about how the height of the water changes in tandem with time. What is the height of the water when t = 0? What is the height when the tank is empty? How would you expect the data for the relationship between h and t to appear? Use the provided axes to sketch at least two possibilities; write at least one sentence to explain how you think the graph should appear.
+
+
+
+
+
+
+
+
+
+
+
Blank axes for making a graph of height versus time.
+
+
+
+
+
+
+
+
+
Blank axes for making a graph of height versus time.
+
+
+
+
+
Sketches should show an S-shaped decreasing curve from h = 6 to h = 0.
+
+
When t = 0 the tank is full, so the height of the water is the diameter h = 6 ft. When the tank is empty, h = 0. The height decreases quickly at first, then the rate of decrease slows as the widest part of the sphere is reached, then speeds up again as the tank nears empty. Sketches should show an S-shaped decreasing curve from h = 6 to h = 0.
- Consider the functions f and g given in the following figures.
-
-
-
-
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-
-
ADD ALT TEXT TO THIS IMAGE
-
-
-
-
-
-
-
- Sketch an accurate graph of the transformation y = p(x) = -\frac{1}{2}f(x-1)+2. Write at least one sentence to explain how you developed the graph of p, and identify the point on p that corresponds to the original point (-2,2) on the graph of f.
-
-
-
-
-
-
- The function p(x) = -\frac{1}{2}f(x-1)+2 is obtained from f by
- shifting right 1 unit, applying a vertical compression by \frac{1}{2} and
- a reflection across the x-axis, then shifting up 2 units.
- The point (-2, 2) on f moves to (-1, 1) on p.
-
-
-
-
-
-
-
-
- Sketch an accurate graph of the transformation y = q(x) = 2g(x+0.5)-0.75. Write at least one sentence to explain how you developed the graph of q, and identify the point on q that corresponds to the original point (1.5,1.5) on the graph of g.
-
-
-
-
-
-
- The function q(x) = 2g(x+0.5) - 0.75 is obtained from g by
- shifting left 0.5 units, stretching vertically by a factor of 2, then
- shifting down 0.75 units.
- The point (1.5, 1.5) on g moves to (1, 2.25) on q.
-
-
-
-
-
-
-
-
- Is the function y = r(x) = \frac{1}{2}(-f(x-1) - 4) the same function as p in part (a) or different? Why? Explain in two different ways: discuss the algebraic similarities and differences between p and r, and also discuss how each is a transformation of f.
-
-
-
-
-
-
- Algebraically:
- r(x) = \frac{1}{2}(-f(x-1)-4) = -\frac{1}{2}f(x-1) - 2.
- Since p(x) = -\frac{1}{2}f(x-1)+2, we have r(x) \ne p(x)
- — they differ by 4 in output.
- As transformations, both shift f right 1 unit and apply a vertical
- compression and reflection by -\frac{1}{2}, but p then shifts
- the result up 2 units while r shifts it down 2 units.
-
-
-
-
-
-
- Find a formula for a function y = s(x) (in terms of g) that represents this transformation of g: a horizontal shift of 1.25 units left, followed by a reflection across the x-axis and a vertical stretch by a factor of 2.5 units, followed by a vertical shift of 1.75 units. Sketch an accurate, labeled graph of s on the following axes along with the given parent function g.
-
-
-
ADD ALT TEXT TO THIS IMAGE
-
-
-
-
-
-
- s(x) = -2.5\,g(x+1.25) + 1.75.
- The point (1.5, 1.5) on g moves to (0.25, -2) on s.
-
+ Consider the functions f and g given in the following figures.
+
+
+
+
+
+
+
Graph of a piecewise linear function f(x) whose first piece goes from (-4,2) to the point (-2,2), and whose second piece goes from (-2,2) to (2,2), and whose third piece goes from (2,2) to the edge of the graph at (4,2). The point (-2,2) is marked on the graph.
+
+
+
+
+
Graph of a function g(x) which is going up and down repeatedly, hitting peaks of 1.5 and valleys of -1.5. The point (1.5,1.5) is marked on the graph.
+
+
+
+
+
+
+
+ Sketch an accurate graph of the transformation y = p(x) = -\frac{1}{2}f(x-1)+2. Write at least one sentence to explain how you developed the graph of p, and identify the point on p that corresponds to the original point (-2,2) on the graph of f.
+
+
+
+
Shift right 1 unit, then compress vertically by a factor of \frac{1}{2}, then reflect across the x-axis, and lastly shift up 2 units. The point (-2, 2) on f moves to (-1, 1) on p.
+
+
+
+
+
Graph of the piecewise linear function f(x), along with a transformation of f.
The function p(x) is also piecewise linear, but has been flipped upside-down, shrunk vertically so that the slopes of each line are half what they are in g, and shifted to the right. The point (-1,1) is marked on the graph of p.
+
+
+
+
+ The function p(x) = -\frac{1}{2}f(x-1)+2 is obtained from f by
+ shifting right 1 unit, applying a vertical compression by \frac{1}{2} and
+ a reflection across the x-axis, then shifting up 2 units.
+ The point (-2, 2) on f moves to (-1, 1) on p.
+
+
+
+
+
+
+
Graph of the piecewise linear function f(x), along with a transformation of f.
The function p(x) is also piecewise linear, but has been flipped upside-down, shrunk vertically so that the slopes of each line are half the magnitude of what they are in g, and shifted to the right. The point (-1,1) is marked on the graph of p.
+
+
+
+
+
+
+
+
+ Sketch an accurate graph of the transformation y = q(x) = 2g(x+0.5)-0.75. Write at least one sentence to explain how you developed the graph of q, and identify the point on q that corresponds to the original point (1.5,1.5) on the graph of g.
+
+
+
+
Shift left 0.5 units, then stretch vertically by a factor of 2, and lastly
+ shift down 0.75 units. The point (1.5, 1.5) on g moves to (1, 2.25) on q.
+
+
+
+
+
Graph of the function g(x), along with a transformation of g.
The function g is going up and down repeatedly, hitting peaks of 1.5 and valleys of -1.5. The point (1.5,1.5) is marked on the graph of g.
The function q(x) is also going up and down repeatedly, but reaches higher peaks and valleys and has been shifted horizontally. The point (1,2.25) is marked on the graph of q.
+
+
+
+
+ The function q(x) = 2g(x+0.5) - 0.75 is obtained from g by
+ shifting left 0.5 units, stretching vertically by a factor of 2, then
+ shifting down 0.75 units.
+ The point (1.5, 1.5) on g moves to (1, 2.25) on q.
+
+
+
+
+
+
+
Graph of the function g(x), along with a transformation of g.
The function g is going up and down repeatedly, hitting peaks of 1.5 and valleys of -1.5. The point (1.5,1.5) is marked on the graph of g.
The function q(x) is also going up and down repeatedly, but reaches higher peaks and valleys and has been shifted horizontally. The point (1,2.25) is marked on the graph of q.
+
+
+
+
+
+
+
+ Is the function y = r(x) = \frac{1}{2}(-f(x-1) - 4) the same function as p in part (a) or different? Why? Explain in two different ways: discuss the algebraic similarities and differences between p and r, and also discuss how each is a transformation of f.
+
+
+
+
Both shift f right 1 unit and apply a vertical
+ compression and reflection by -\frac{1}{2}, but p then shifts
+ the result up 2 units while r shifts it down 2 units.
+
+
+ Algebraically:
+ r(x) = \frac{1}{2}(-f(x-1)-4) = -\frac{1}{2}f(x-1) - 2.
+ Since p(x) = -\frac{1}{2}f(x-1)+2, we have r(x) \ne p(x)
+ — they differ by 4 in output.
+ As transformations, both shift f right 1 unit and apply a vertical
+ compression and reflection by -\frac{1}{2}, but p then shifts
+ the result up 2 units while r shifts it down 2 units.
+
+
+
+
+
+
+ Find a formula for a function y = s(x) (in terms of g) that represents this transformation of g: a horizontal shift of 1.25 units left, followed by a reflection across the x-axis and a vertical stretch by a factor of 2.5 units, followed by a vertical shift of 1.75 units. Sketch an accurate, labeled graph of s on the following axes along with the given parent function g.
+
+
+
+
+
+
Graph of a function g(x) which is going up and down repeatedly, hitting peaks of 1.5 and valleys of -1.5. The point (1.5,1.5) is marked on the graph.
+
+
+
+
s(x) = -2.5\,g(x+1.25) + 1.75.
+
+
+
+
+
Graph of a function g(x) which is going up and down repeatedly, hitting peaks of 1.5 and valleys of -1.5. The point (1.5,1.5) is marked on the graph.
+
+
+
+ s(x) = -2.5\,g(x+1.25) + 1.75.
+ The point (1.5, 1.5) on g moves to (0.25, -2) on s.
+
+
+
+
+
+
Graph of a function g(x) which is going up and down repeatedly, hitting peaks of 1.5 and valleys of -1.5. The point (1.5,1.5) is marked on the graph.
- Consider the functions r and s given in the figures below.
-
-
-
-
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-
-
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-
-
-
-
-
-
-
- On the same axes as the plot of y = r(x), sketch the following graphs: y = g(x) = r(x) + 2, y = h(x) = r(x+1), and y = f(x) = r(x+1) + 2. Be sure to label the point on each of g, h, and f that corresponds to (-2,-1) on the original graph of r. In addition, write one sentence to explain the overall transformations that have resulted in g, h, and f.
-
-
-
-
-
-
- The graph of g(x) = r(x) + 2 is r shifted vertically up 2 units;
- the point (-2, -1) moves to (-2, 1).
- The graph of h(x) = r(x+1) is r shifted horizontally to the left
- 1 unit; the point (-2, -1) moves to (-3, -1).
- The graph of f(x) = r(x+1) + 2 combines both transformations — a shift
- left 1 unit and up 2 units — moving (-2, -1) to (-3, 1).
-
-
-
-
-
-
-
-
- On the same axes as the plot of y = s(x), sketch the following graphs: y = k(x) = s(x) - 1, y = j(x) = s(x-2), and y = m(x) = s(x-2) - 1. Be sure to label the point on each of k, j, and m that corresponds to (-2,-3) on the original graph of s. In addition, write one sentence to explain the overall transformations that have resulted in k, j, and m.
-
-
-
-
-
-
- The graph of k(x) = s(x) - 1 is s shifted down 1 unit;
- the point (-2, -3) moves to (-2, -4).
- The graph of j(x) = s(x-2) is s shifted right 2 units;
- the point (-2, -3) moves to (0, -3).
- The graph of m(x) = s(x-2) - 1 combines both — a shift right 2 units
- and down 1 unit — moving (-2, -3) to (0, -4).
-
-
-
-
-
-
- Now consider the function q(x) = x^2. Determine a formula for the function that is given by p(x) = q(x+3) - 4. How is p a transformation of q?
-
-
-
-
-
-
- Substituting q(x) = x^2:
-
- p(x) &= q(x+3) - 4 = (x+3)^2 - 4
- &= x^2 + 6x + 9 - 4 = x^2 + 6x + 5.
-
- The function p is a translation of q three units to the left
- and four units down.
-
+ Consider the functions r and s given in the figures below.
+
+
+
+
+
+
+
+
A parent function r(x).
+
+
+
+
+
+
+
+
+
+
A parent function s(x).
+
+
+
+
+
+
+
+
+
+
+
+ On the same axes as the plot of y = r(x), sketch the following graphs: y = g(x) = r(x) + 2, y = h(x) = r(x+1), and y = f(x) = r(x+1) + 2. Be sure to label the point on each of g, h, and f that corresponds to (2,-1) on the original graph of r. In addition, write one sentence to explain the overall transformations that have resulted in g, h, and f.
+
+
+
+
The function g(x) shifts r up 2 units, while the function h(x) shifts r left 1 unit, and the function f(x) shifts r both up 2 and left 1.
+
+
+
+
+
A parent function r(x) along with three transformations: h(x) which shifts to the left 1 unit, g(x) which shifts up 2 units, and f(x) which shifts both left 1 unit and up 2 units.
+
+
+
+
+
+
+ The graph of g(x) = r(x) + 2 is r shifted vertically up 2 units;
+ the point (2, -1) moves to (2, 1).
+ The graph of h(x) = r(x+1) is r shifted horizontally to the left
+ 1 unit; the point (2, -1) moves to (1, -1).
+ The graph of f(x) = r(x+1) + 2 combines both transformations — a shift
+ left 1 unit and up 2 units — moving (2, -1) to (1, 1).
+
+
+
+
+
+
+
A parent function r(x) along with three transformations: h(x) which shifts to the left 1 unit, g(x) which shifts up 2 units, and f(x) which shifts both left 1 unit and up 2 units.
+
+
+
+
+
+
+
+
+
+
+ On the same axes as the plot of y = s(x), sketch the following graphs: y = k(x) = s(x) - 1, y = j(x) = s(x-2), and y = m(x) = s(x-2) - 1. Be sure to label the point on each of k, j, and m that corresponds to (-2,-3) on the original graph of s. In addition, write one sentence to explain the overall transformations that have resulted in k, j, and m.
+
+
+
+
The function k(x) shifts s down 1 unit, while the function j(x) shifts s right 2 units, and the function m(x) shifts s both down 1 and right 2.
+
+
+
+
+
A parent function s(x) along with three transformations: j(x) which shifts to the right 2 units, k(x) which shifts down 1 units, and m(x) which shifts both right 2 units and down 1 unit.
+
+
+
+
+
+
+ The graph of k(x) = s(x) - 1 is s shifted down 1 unit;
+ the point (-2, -3) moves to (-2, -4).
+ The graph of j(x) = s(x-2) is s shifted right 2 units;
+ the point (-2, -3) moves to (0, -3).
+ The graph of m(x) = s(x-2) - 1 combines both — a shift right 2 units
+ and down 1 unit — moving (-2, -3) to (0, -4).
+
+
+
+
+
+
+
A parent function s(x) along with three transformations: j(x) which shifts to the right 2 units, k(x) which shifts down 1 units, and m(x) which shifts both right 2 units and down 1 unit.
+
+
+
+
+
+
+
+
+
+
+ Now consider the function q(x) = x^2. Determine a formula for the function that is given by p(x) = q(x+3) - 4. How is p a transformation of q?
+
+
+
+
+ The function p(x)=x^2 + 6x + 5, and is the result of translating q(x) three units to the left and four units down.
+
+
+
+
+ Substituting q(x) = x^2:
+
+ p(x) &= q(x+3) - 4 = (x+3)^2 - 4
+ &= x^2 + 6x + 9 - 4 = x^2 + 6x + 5.
+
+ The function p is a translation of q three units to the left
+ and four units down.
+
- Consider the functions r and s given in the figures below.
-
-
-
-
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-
-
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-
-
-
-
-
-
-
- On the same axes as the plot of y = r(x), sketch the following graphs: y = g(x) = 3r(x) and y = h(x) = \frac{1}{3}r(x). Be sure to label the point on g and h that corresponds to the point (-2,-1) on the original graph of r. In addition, write one sentence to explain the overall transformations that have resulted in g and h from r.
-
-
-
-
-
-
- The graph of g(x) = 3r(x) is a vertical stretch of r by a factor
- of 3; the point (-2, -1) moves to (-2, -3).
- The graph of h(x) = \frac{1}{3}r(x) is a vertical compression of r
- by a factor of \frac{1}{3}; the point (-2, -1) moves to
- (-2, -\frac{1}{3}).
-
-
-
-
-
-
-
-
- On the same axes as the plot of y = s(x), sketch the following graphs: y = k(x) = -s(x) and y = j(x) = -\frac{1}{2}s(x). Be sure to label the point on k and j that corresponds to the point (-2,-3) on the original graph of s. In addition, write one sentence to explain the overall transformations that have resulted in k and j from s.
-
-
-
-
-
-
- The graph of k(x) = -s(x) is a reflection of s across the
- x-axis; the point (-2, -3) moves to (-2, 3).
- The graph of j(x) = -\frac{1}{2}s(x) is a vertical compression by
- \frac{1}{2} combined with a reflection across the x-axis;
- the point (-2, -3) moves to (-2, \frac{3}{2}).
-
-
-
-
-
-
-
-
- On the additional copies of the two figures below, sketch the graphs of the following transformed functions: y = m(x) = 2r(x+1)-1 (at left) and y = n(x) = \frac{1}{2}s(x-2)+2. As above, be sure to label a key point on each graph that corresonds to the labeled point on the original parent function.
-
-
-
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-
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-
-
-
-
-
-
- For m(x) = 2r(x+1) - 1: the point (-2, -1) on r moves
- to (-3, -3) on m.
- For n(x) = \frac{1}{2}s(x-2) + 2: the point (-2, -3) on s
- moves to (0, \frac{1}{2}) on n.
-
-
-
-
-
-
-
-
-
-
-
- Describe in words how the function y = m(x) = 2r(x+1)-1 is the result of three elementary transformations of y = r(x). Does the order in which these transformations occur matter? Why or why not?
-
-
-
-
-
-
- The function m(x) = 2r(x+1) - 1 results from three elementary
- transformations of r: a horizontal shift left 1 unit (replacing x
- with x+1), a vertical stretch by a factor of 2 (multiplying by 2), and
- a vertical shift down 1 unit (subtracting 1).
- The vertical stretch and the horizontal shift may be applied in either order,
- but both must be applied before the vertical shift.
-
+ Consider the functions r and s given in the figures below.
+
+
+
+
+
+
+
A parent function r(x).
+
+
+
+
+
+
+
+
+
+
A parent function s(x).
+
+
+
+
+
+
+
+
+
+
+
+ On the same axes as the plot of y = r(x), sketch the following graphs: y = g(x) = 3r(x) and y = h(x) = \frac{1}{3}r(x). Be sure to label the point on g and h that corresponds to the point (-2,1) on the original graph of r. In addition, write one sentence to explain the overall transformations that have resulted in g and h from r.
+
+
+
+
+ The graph of g(x) = 3r(x) is a vertical stretch of r by a factor
+ of 3 while the graph of h(x) = \frac{1}{3}r(x) is a vertical compression of r
+ by a factor of \frac{1}{3}.
+
+
+
+
+
+
A parent function r(x) along with two transformations: g(x) which stretches vertical heights by a factor of 3, and h(x) which vertically compresses vertical heights by a factor of 1/3.
+
+
+
+
+
+
+ The graph of g(x) = 3r(x) is a vertical stretch of r by a factor
+ of 3; the point (-2, 1) moves to (-2, 3).
+ The graph of h(x) = \frac{1}{3}r(x) is a vertical compression of r
+ by a factor of \frac{1}{3}; the point (-2, 1) moves to
+ (-2, \frac{1}{3}).
+
+
+
+
+
+
+
A parent function r(x) along with two transformations: g(x) which stretches vertical heights by a factor of 3, and h(x) which vertically compresses vertical heights by a factor of 1/3.
+
+
+
+
+
+
+
+
+
+
+ On the same axes as the plot of y = s(x), sketch the following graphs: y = k(x) = -s(x) and y = j(x) = -\frac{1}{2}s(x). Be sure to label the point on k and j that corresponds to the point (-2,-3) on the original graph of s. In addition, write one sentence to explain the overall transformations that have resulted in k and j from s.
+
+
+
+
+ The graph of k(x) = -s(x) is a reflection of s across the
+ x-axis, while the graph of j(x) = -\frac{1}{2}s(x) is a vertical compression by \frac{1}{2} combined with a reflection across the x-axis.
+
+
+
+
+
+
+
A parent function s(x) along with two transformations: k(x) which reflects heights over the x-axis and j(x) which reflects and vertically compresses vertical heights by a factor of 1/2.
+
+
+
+
+
+
+ The graph of k(x) = -s(x) is a reflection of s across the
+ x-axis; the point (-2, -3) moves to (-2, 3).
+ The graph of j(x) = -\frac{1}{2}s(x) is a vertical compression by
+ \frac{1}{2} combined with a reflection across the x-axis;
+ the point (-2, -3) moves to (-2, \frac{3}{2}).
+
+
+
+
+
+
+
A parent function s(x) along with two transformations: k(x) which reflects heights over the x-axis and j(x) which reflects and vertically compresses vertical heights by a factor of 1/2.
+
+
+
+
+
+
+
+
+
+ On the additional copies of the two figures below, sketch the graphs of the following transformed functions: y = m(x) = 2r(x+1)-1 (at left) and y = n(x) = \frac{1}{2}s(x-2)+2. As above, be sure to label a key point on each graph that corresponds to a labeled point on the original parent function.
+
+
+
+
+
+
+
A parent function r(x).
+
+
+
+
+
+
+
+
A parent function s(x).
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
A parent function r(x) along with a transformation m(x) which shifts left 1 unit, stretches vertical heights by a factor of 2, and shifts down by 1 unit.
+
+
+
+
+
+
+
+
A parent function s(x) along with a transformation n(x) which shifts right 2 units, compresses vertical heights by a factor of 1/2, and shifts up by 2 units.
+
+
+
+
+
+
+
+
+
+ For m(x) = 2r(x+1) - 1: the point (2, -1) on r moves
+ to (1, -3) on m.
+ For n(x) = \frac{1}{2}s(x-2) + 2: the point (-2, -3) on s
+ moves to (0, \frac{1}{2}) on n.
+
+
+
+
+
+
+
+
A parent function r(x) along with a transformation m(x) which shifts left 1 unit, stretches vertical heights by a factor of 2, and shifts down by 1 unit.
+
+
+
+
+
+
+
+
A parent function s(x) along with a transformation n(x) which shifts right 2 units, compresses vertical heights by a factor of 1/2, and shifts up by 2 units.
+
+
+
+
+
+
+
+
+
+
+
+ Describe in words how the function y = m(x) = 2r(x+1)-1 is the result of three elementary transformations of y = r(x). Does the order in which these transformations occur matter? Why or why not?
+
+
+
+
+ The function m(x) = 2r(x+1) - 1 results from three elementary
+ transformations of r: a horizontal shift left 1 unit (replacing x
+ with x+1), a vertical stretch by a factor of 2 (multiplying by 2), and
+ a vertical shift down 1 unit (subtracting 1).
+ The vertical stretch and the horizontal shift may be applied in either order,
+ but both must be applied before the vertical shift.
+
+
+
+
+ The function m(x) = 2r(x+1) - 1 results from three elementary
+ transformations of r: a horizontal shift left 1 unit (replacing x
+ with x+1), a vertical stretch by a factor of 2 (multiplying by 2), and
+ a vertical shift down 1 unit (subtracting 1).
+ The vertical stretch and the horizontal shift may be applied in either order,
+ but both must be applied before the vertical shift.
+
- Answer the following questions exactly wherever possible. If you estimate a value, do so to at least 5 decimal places of accuracy.
-
-
-
-
-
-
- The x coordinate of the point on the unit circle that lies in the third quadrant and whose y-coordinate is y = -\frac{3}{4}.
-
-
-
-
-
-
- Using x^2 + \left(-\frac{3}{4}\right)^2 = 1, we solve for x = -\frac{\sqrt{7}}{4} \approx -0.66143 (negative since in Quadrant III).
-
-
-
-
-
-
- The y-coordinate of the point on the unit circle generated by a central angle opening counterclockwise with one side on the positive x-axis that measures t = 2 radians.
-
-
-
-
-
-
- \sin(2 \text{ rad}) \approx 0.90929.
-
-
-
-
-
-
- The x-coordinate of the point on the unit circle generated by a central angle with one side on the positive x-axis that measures t = -3.05 radians. (With the negative radian measure, we view the angle as opening clockwise from its initial side on the positive x-axis.)
-
-
-
-
-
-
- \cos(-3.05 \text{ rad}) \approx -0.99581.
-
-
-
-
-
-
- The value of \cos(t) where t is an angle in Quadrant II that satisfies \sin(t) = \frac{1}{2}.
-
-
-
-
-
-
- If \sin(t) = \frac{1}{2} and t is in Quadrant II, then t = \frac{5\pi}{6} and \cos(t) = -\frac{\sqrt{3}}{2}.
-
-
-
-
-
-
- The value of \sin(t) where t is an angle in Quadrant III for which \cos(t) = -0.7.
-
-
-
-
-
-
- Using \sin^2(t) + \cos^2(t) = 1, we get \sin(t) = -\sqrt{1 - 0.49} = -\sqrt{0.51} \approx -0.71414 (negative since in Quadrant III).
-
-
-
-
-
-
- The average rate of change of f(t) = \sin(t) on the intervals [0.1,0.2] and [0.8,0.9].
-
+ Answer the following questions exactly wherever possible. If you estimate a value, do so to at least 5 decimal places of accuracy.
+
+
+
+
+
+
+ The x coordinate of the point on the unit circle that lies in the third quadrant and whose y-coordinate is y = -\frac{3}{4}.
+
+
+
+
x = -\frac{\sqrt{7}}{4}
+
+
+ Using x^2 + \left(-\frac{3}{4}\right)^2 = 1, we solve for x = -\frac{\sqrt{7}}{4} \approx -0.66143 (negative since in Quadrant III).
+
+
+
+
+
+
+ The y-coordinate of the point on the unit circle generated by a central angle opening counterclockwise with one side on the positive x-axis that measures t = 2 radians.
+
+
+
+
\sin(2 \text{ rad}) \approx 0.90929
+
+
+ \sin(2 \text{ rad}) \approx 0.90929.
+
+
+
+
+
+
+ The x-coordinate of the point on the unit circle generated by a central angle with one side on the positive x-axis that measures t = -3.05 radians. (With the negative radian measure, we view the angle as opening clockwise from its initial side on the positive x-axis.)
+
+
+
+
+
+ \cos(-3.05 \text{ rad}) \approx -0.99581.
+
+
+
+
+ \cos(-3.05 \text{ rad}) \approx -0.99581.
+
+
+
+
+
+
+ The value of \cos(t) where t is an angle in Quadrant II that satisfies \sin(t) = \frac{1}{2}.
+
+
+
+
\cos(t) = -\frac{\sqrt{3}}{2}
+
+
+ If \sin(t) = \frac{1}{2} and t is in Quadrant II, then t = \frac{5\pi}{6} and \cos(t) = -\frac{\sqrt{3}}{2}.
+
+
+
+
+
+
+ The value of \sin(t) where t is an angle in Quadrant III for which \cos(t) = -0.7.
+
+
+
+
\sin(t) = -\sqrt{0.51} \approx -0.71414
+
+
+ Using \sin^2(t) + \cos^2(t) = 1, we get \sin(t) = -\sqrt{1 - 0.49} = -\sqrt{0.51} \approx -0.71414 (negative since in Quadrant III).
+
+
+
+
+
+
+ The average rate of change of f(t) = \sin(t) on the intervals [0.1,0.2] and [0.8,0.9].
+
- Use Figure 2.3.12 in the text (which plots the sine and cosine functions ont the same axes) to assist in answering the following questions.
-
-
-
-
-
-
- Give an example of the largest interval you can find on which
- f(t) = \sin(t) is decreasing.
-
-
-
-
-
-
- The function f(t) = \sin(t) is decreasing on the interval \left[\frac{\pi}{2}, \frac{3\pi}{2}\right].
-
-
-
-
-
-
- Give an example of the largest interval you can find on which
- f(t) = \sin(t) is decreasing and concave down.
-
-
-
-
-
-
- The function f(t) = \sin(t) is decreasing and concave down on the interval \left[\frac{\pi}{2}, \pi\right].
-
-
-
-
-
-
- Give an example of the largest interval you can find on which
- g(t) = \cos(t) is increasing.
-
-
-
-
-
-
- The function g(t) = \cos(t) is increasing on the interval [\pi, 2\pi].
-
-
-
-
-
-
- Give an example of the largest interval you can find on which
- g(t) = \cos(t) is increasing and concave up.
-
-
-
-
-
-
- The function g(t) = \cos(t) is increasing and concave up on the interval \left[\pi, \frac{3\pi}{2}\right].
-
-
-
-
-
-
- Without doing any computation, on which interval is the average rate of change of g(t) = \cos(t) greater: [\pi, \pi+0.1] or [\frac{3\pi}{2}, \frac{3\pi}{2} + 0.1]? Why?
-
-
-
-
-
-
- The average rate of change of g(t) = \cos(t) is greater on the interval \left[\frac{3\pi}{2}, \frac{3\pi}{2} + 0.1\right] because g(t) has a steeper slope near \frac{3\pi}{2} (a zero crossing) than near \pi (a minimum).
-
-
-
-
-
-
- In general, how would you characterize the locations on the sine and cosine graphs where the functions are increasing or decreasingly most rapidly?
-
-
-
-
-
-
- The most rapid average rates of change on both graphs occur near the zero crossing points (where the function crosses the midline).
-
-
-
-
-
-
- Thinking from the perspective of the unit circle, for which quadrants of the x-y plane is \cos(t) negative for an angle t that lies in that quadrant?
-
-
-
-
-
-
- The function g(t) = \cos(t) is negative on quadrants II and III.
-
+ Use Figure 2.3.12 in the text (which plots the sine and cosine functions ont the same axes) to assist in answering the following questions.
+
+
+
+
+
+
+ Give an example of the largest interval you can find on which
+ f(t) = \sin(t) is decreasing.
+
+
+
+
\left[\frac{\pi}{2}, \frac{3\pi}{2}\right]
+
+
+ The function f(t) = \sin(t) is decreasing on the interval \left[\frac{\pi}{2}, \frac{3\pi}{2}\right].
+
+
+
+
+
+
+ Give an example of the largest interval you can find on which
+ f(t) = \sin(t) is decreasing and concave down.
+
+
+
+
\left[\frac{\pi}{2}, \pi\right]
+
+
+ The function f(t) = \sin(t) is decreasing and concave down on the interval \left[\frac{\pi}{2}, \pi\right].
+
+
+
+
+
+
+ Give an example of the largest interval you can find on which
+ g(t) = \cos(t) is increasing.
+
+
+
+
[\pi, 2\pi]
+
+
+ The function g(t) = \cos(t) is increasing on the interval [\pi, 2\pi].
+
+
+
+
+
+
+ Give an example of the largest interval you can find on which
+ g(t) = \cos(t) is increasing and concave up.
+
+
+
+
\left[\pi, \frac{3\pi}{2}\right]
+
+
+ The function g(t) = \cos(t) is increasing and concave up on the interval \left[\pi, \frac{3\pi}{2}\right].
+
+
+
+
+
+
+ Without doing any computation, on which interval is the average rate of change of g(t) = \cos(t) greater: [\pi, \pi+0.1] or [\frac{3\pi}{2}, \frac{3\pi}{2} + 0.1]? Why?
+
+
+
+
\left[\frac{3\pi}{2}, \frac{3\pi}{2} + 0.1\right]
+
+
+ The average rate of change of g(t) = \cos(t) is greater on the interval \left[\frac{3\pi}{2}, \frac{3\pi}{2} + 0.1\right] because g(t) has a steeper slope near \frac{3\pi}{2} (a zero crossing) than near \pi (a minimum).
+
+
+
+
+
+
+ In general, how would you characterize the locations on the sine and cosine graphs where the functions are increasing or decreasingly most rapidly?
+
+
+
+
+
+ The most rapid average rates of change on both graphs occur near the zero crossing points (where the function crosses the midline).
+
+
+
+
+ The most rapid average rates of change on both graphs occur near the zero crossing points (where the function crosses the midline).
+
+
+
+
+
+
+ Thinking from the perspective of the unit circle, for which quadrants of the x-y plane is \cos(t) negative for an angle t that lies in that quadrant?
+
+
+
+
quadrants II and III
+
+
+ The function g(t) = \cos(t) is negative on quadrants II and III.
+
- Let k = g(t) be the function that tracks the x-coordinate of a point traversing the unit circle counterclockwise from (1,0). That is, g(t) = \cos(t). Use the information we know about the unit circle that is summarized in Figure 2.3.1 (with 16 labeled special points) to respond to the following questions.
-
-
-
-
-
-
- What is the exact value of \cos(\frac{\pi}{6})? of \cos(\frac{5\pi}{6})? \cos(-\frac{\pi}{3})?
-
- On the axes provided in the following figure, sketch an accurate graph of k = \cos(t). Label the exact location of several key points on the curve.
-
-
-
ADD ALT TEXT TO THIS IMAGE
-
-
-
-
-
-
-
-
-
-
-
- What is the exact value of \cos( \frac{11\pi}{4} )? of \cos( \frac{14\pi}{3} )?
-
+ Let k = g(t) be the function that tracks the x-coordinate of a point traversing the unit circle counterclockwise from (1,0). That is, g(t) = \cos(t). Use the information we know about the unit circle that is summarized in Figure 2.3.1 (with 16 labeled special points) to respond to the following questions.
+
+
+
+
+
+
+ What is the exact value of \cos(\frac{\pi}{6})? of \cos(\frac{5\pi}{6})? \cos(-\frac{\pi}{3})?
+
+ On the axes provided in the following figure, sketch an accurate graph of k = \cos(t). Label the exact location of several key points on the curve.
+
+
+
+
+
+
+
Blank axes for graphing \cos(t)
+
+
+
+
+
+
+
+
+
+
+
+
a plot of \cos(t)
+
+
+
+
+
+
+
+
a plot of \cos(t)
+
+
+
+
+
+
+ What is the exact value of \cos( \frac{11\pi}{4} )? of \cos( \frac{14\pi}{3} )?
+
- Consider the functions f and g given in the following figures.
-
-
-
-
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-
-
ADD ALT TEXT TO THIS IMAGE
-
-
-
-
-
-
-
- On the same axes as the plot of y = f(t), sketch the following graphs: y = h(t) = f(\frac{1}{3}t) and y = j(t) = f(4t). Be sure to label several points on each of f, h, and j with arrows to indicate their correspondence. In addition, write one sentence to explain the overall transformations that have resulted in h and j from f.
-
-
-
-
-
-
- The function h(t) = f\!\left(\frac{1}{3}t\right) represents a horizontal scaling of f by a factor of 3 (stretching). The function j(t) = f(4t) represents a horizontal scaling by a factor of \frac{1}{4} (compression).
-
-
-
-
-
-
-
- On the same axes as the plot of y = g(t), sketch the following graphs: y = k(t) = g(2t) and y = m(t) = g(\frac{1}{2}t). Be sure to label several points on each of g, k, and m with arrows to indicate their correspondence. In addition, write one sentence to explain the overall transformations that have resulted in k and m from g.
-
-
-
-
-
-
- The function k(t) = g(2t) represents a horizontal scaling of g by a factor of \frac{1}{2} (compression). The function m(t) = g\!\left(\frac{1}{2}t\right) represents a horizontal scaling by a factor of 2 (stretching).
-
-
-
-
-
-
-
- On the additional copies of the two figures below, sketch the graphs of the following transformed functions: y = r(t) = 2f(\frac{1}{2}t) (at left) and y = s(t) = \frac{1}{2}g(2t). As above, be sure to label several points on each graph and indicate their correspondence to points on the original parent function.
-
-
-
ADD ALT TEXT TO THIS IMAGE
-
ADD ALT TEXT TO THIS IMAGE
-
-
-
-
-
-
- The function r(t) = 2f\!\left(\frac{1}{2}t\right) applies both a horizontal scaling by a factor of 2 and a vertical scaling by a factor of 2. The function s(t) = \frac{1}{2}g(2t) applies both a horizontal scaling by a factor of \frac{1}{2} and a vertical scaling by a factor of \frac{1}{2}.
-
-
-
-
-
-
-
- Describe in words how the function y = r(t) = 2f(\frac{1}{2}t) is the result of composing two elementary transformations of y = f(t). Does the order in which these transformations are composed matter? Why or why not?
-
-
-
-
-
-
- The function r(t) = 2f\!\left(\frac{1}{2}t\right) is both a vertical and a horizontal scaling. The order does not matter: applying the horizontal stretch by 2 and then the vertical stretch by 2 produces the same result as applying them in the opposite order.
-
+ Consider the functions f and g given in the following figures.
+
+
+
+
+
+
+
+
+
A parent function f(x)
+
+
+
+
+
+
+
A parent function g(x)
+
+
+
+
+
+
+
+ On the same axes as the plot of y = f(t), sketch the following graphs: y = h(t) = f(\frac{1}{3}t) and y = j(t) = f(4t). Be sure to label several points on each of f, h, and j with arrows to indicate their correspondence. In addition, write one sentence to explain the overall transformations that have resulted in h and j from f.
+
+
+
+
+
+ The function h(t) = f\!\left(\frac{1}{3}t\right) represents a horizontal scaling of f by a factor of 3 (stretching). The function j(t) = f(4t) represents a horizontal scaling by a factor of \frac{1}{4} (compression).
+
+
+
Graphs of two horizontal scaling functions of f(x)
+
+
+
+
+ The function h(t) = f\!\left(\frac{1}{3}t\right) represents a horizontal scaling of f by a factor of 3 (stretching). The function j(t) = f(4t) represents a horizontal scaling by a factor of \frac{1}{4} (compression).
+
+
+
Graphs of two horizontal scaling functions of f(x)
+
+
+
+
+
+
+ On the same axes as the plot of y = g(t), sketch the following graphs: y = k(t) = g(2t) and y = m(t) = g(\frac{1}{2}t). Be sure to label several points on each of g, k, and m with arrows to indicate their correspondence. In addition, write one sentence to explain the overall transformations that have resulted in k and m from g.
+
+
+
+
+
+ The function k(t) = g(2t) represents a horizontal scaling of g by a factor of \frac{1}{2} (compression). The function m(t) = g\!\left(\frac{1}{2}t\right) represents a horizontal scaling by a factor of 2 (stretching).
+
+
+
Graphs of two horizontal scaling functions of g(x)
+
+
+
+
+ The function k(t) = g(2t) represents a horizontal scaling of g by a factor of \frac{1}{2} (compression). The function m(t) = g\!\left(\frac{1}{2}t\right) represents a horizontal scaling by a factor of 2 (stretching).
+
+
+
Graphs of two horizontal scaling functions of g(x)
+
+
+
+
+
+
+ On the additional copies of the two figures below, sketch the graphs of the following transformed functions: y = r(t) = 2f(\frac{1}{2}t) (at left) and y = s(t) = \frac{1}{2}g(2t). As above, be sure to label several points on each graph and indicate their correspondence to points on the original parent function.
+
+
+
+
+
+
+
+
A parent function f(x)
+
+
+
+
+
+
A parent function g(x)
+
+
+
+
+
+ The function r(t) = 2f\!\left(\frac{1}{2}t\right) applies both a horizontal scaling by a factor of 2 and a vertical scaling by a factor of 2. The function s(t) = \frac{1}{2}g(2t) applies both a horizontal scaling by a factor of \frac{1}{2} and a vertical scaling by a factor of \frac{1}{2}.
+
+
Graphs of transformations which scale both horizontally and vertically.
+
+
+
+ The function r(t) = 2f\!\left(\frac{1}{2}t\right) applies both a horizontal scaling by a factor of 2 and a vertical scaling by a factor of 2. The function s(t) = \frac{1}{2}g(2t) applies both a horizontal scaling by a factor of \frac{1}{2} and a vertical scaling by a factor of \frac{1}{2}.
+
+
Graphs of transformations which scale both horizontally and vertically.
+
+
+
+
+
+ Describe in words how the function y = r(t) = 2f(\frac{1}{2}t) is the result of composing two elementary transformations of y = f(t). Does the order in which these transformations are composed matter? Why or why not?
+
+
+
+
The function r(t) = 2f\!\left(\frac{1}{2}t\right) is both a vertical and a horizontal scaling. The order does not matter
+
+
+ The function r(t) = 2f\!\left(\frac{1}{2}t\right) is both a vertical and a horizontal scaling. The order does not matter: applying the horizontal stretch by 2 and then the vertical stretch by 2 produces the same result as applying them in the opposite order.
+
- Consider a spring-mass system where the weight is hanging from the ceiling in such a way that the following is known:
- we let d(t) denote the distance from the ceiling to the weight at time t in seconds and know that the weight oscillates periodically with a minimum value of
- 1.5 feet and a maximum value of 4 feet,
- with a period of 3,
- and you know d(0.5) = 2.75 and d\left(1.25\right) = 4.
-
-
- State the midline, amplitude, range, and an anchor point for the function, and hence determine a formula for d(t) in the form a\cos(k(t-b))+c or a\sin(k(t-b))+c.
- Show your work and thinking,
- and use Desmos appropriately to check that your formula generates the desired behavior.
-
-
-
-
-
-
-
- The midline is c = \frac{1.5+4}{2} = 2.75 feet, the amplitude is a = \frac{4-1.5}{2} = 1.25 feet, the range is [1.5,4], and the anchor point is at approximately (0, 1.67). Since d(1.25) = 4 is the maximum and the period is 3, we have k = \frac{2\pi}{3} and b = 1.25. The formula is
-
- d(t) = 1.25\cos\!\left(\frac{2\pi}{3}(t - 1.25)\right) + 2.75.
-
-
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ Consider a spring-mass system where the weight is hanging from the ceiling in such a way that the following is known:
+ we let d(t) denote the distance from the ceiling to the weight at time t in seconds and know that the weight oscillates periodically with a minimum value of
+ 1.5 feet and a maximum value of 4 feet,
+ with a period of 3,
+ and you know d(0.5) = 2.75 and d\left(1.25\right) = 4.
+
+
+ State the midline, amplitude, range, and an anchor point for the function, and hence determine a formula for d(t) in the form a\cos(k(t-b))+c or a\sin(k(t-b))+c.
+ Show your work and thinking,
+ and use Desmos appropriately to check that your formula generates the desired behavior.
+
+
+
+
+ The midline is c = \frac{1.5+4}{2} = 2.75 feet, the amplitude is a = \frac{4-1.5}{2} = 1.25 feet, the range is [1.5,4], and the anchor point is at approximately (0, 1.67). The formula is
+
+ d(t) = 1.25\cos\!\left(\frac{2\pi}{3}(t - 1.25)\right) + 2.75
+ .
+
+
Graph of d(t)
+
+
+
+ The midline is c = \frac{1.5+4}{2} = 2.75 feet, the amplitude is a = \frac{4-1.5}{2} = 1.25 feet, the range is [1.5,4], and the anchor point is at approximately (0, 1.67). Since d(1.25) = 4 is the maximum and the period is 3, we have k = \frac{2\pi}{3} and b = 1.25. The formula is
+
+ d(t) = 1.25\cos\!\left(\frac{2\pi}{3}(t - 1.25)\right) + 2.75.
+
+
- Consider a spring-mass system where a weight is resting on a frictionless table. We let d(t) denote the distance from the wall (where the spring is attached) to the weight at time t in seconds and know that the weight oscillates periodically with a minimum value of 2 feet and a maximum value of 7 feet with a period of 2 \pi. We also know that d(0) = 4.5 and d\left(\frac{\pi}{2}\right) = 2.
-
-
- Determine a formula for d(t) in the form d(t) = a\cos(t-b)+c or d(t) = a\sin(t-b)+c. Is it possible to find two different formulas that work? For any formula you find, identify the anchor point.
-
-
-
-
-
-
-
- Note, the measurements of the system are physical, so exact values are not appropriate. The spring-mass system may be described in many ways. One formula is d(t) = -2.5\sin(t) + 4.5, and has an anchor point with coordinates (0, 4.5), corresponding to the starting position of the mass. Another formula with the same anchor point is d(t) = 2.5\cos\!\left(t + \frac{\pi}{2}\right) + 4.5.
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ Consider a spring-mass system where a weight is resting on a frictionless table. We let d(t) denote the distance from the wall (where the spring is attached) to the weight at time t in seconds and know that the weight oscillates periodically with a minimum value of 2 feet and a maximum value of 7 feet with a period of 2 \pi. We also know that d(0) = 4.5 and d\left(\frac{\pi}{2}\right) = 2.
+
+
+ Determine a formula for d(t) in the form d(t) = a\cos(t-b)+c or d(t) = a\sin(t-b)+c. Is it possible to find two different formulas that work? For any formula you find, identify the anchor point.
+
+
+
+
One formula is d(t) = -2.5\sin(t) + 4.5, which has an anchor point with coordinates (0, 4.5). Another formula with the same anchor point is d(t) = 2.5\cos\left(t + \frac{\pi}{2}\right) + 4.5.
+
+
+
+
+ Note, the measurements of the system are physical, so exact values are not appropriate. The spring-mass system may be described in many ways. One formula is d(t) = -2.5\sin(t) + 4.5, and has an anchor point with coordinates (0, 4.5), corresponding to the starting position of the mass. Another formula with the same anchor point is d(t) = 2.5\cos\!\left(t + \frac{\pi}{2}\right) + 4.5.
+
- Determine the exact period, amplitude, and midline of each of the following functions. In addition, state the range of each function, any horizontal shift that has been introduced to the graph, and identify an anchor point. Make your conclusions without consulting Desmos, and then use the program to check your work.
-
-
-
-
-
-
- p(x) = \sin(10x) + 2
-
-
-
-
-
-
- Period \frac{\pi}{5}, amplitude 1, midline y=2, range [1,3], no horizontal shift, anchor point (0,2).
-
-
-
-
-
-
- q(x) = -3\cos(0.25x) - 4
-
-
-
-
-
-
- Period 8\pi, amplitude 3 (with reflection), midline y=-4, range [-7,-1], no horizontal shift, anchor point (0,-7).
-
-
-
-
-
-
- r(x) = 2\sin\left( \frac{\pi}{4} x\right) + 5
-
-
-
-
-
-
- Period 8, amplitude 2, midline y=5, range [3,7], no horizontal shift, anchor point (0,5).
-
- Period 4, amplitude 2, midline y=5, range [3,7], horizontal shift 3 units to the right, anchor point (0,5).
-
-
-
-
-
-
- u(x) = -0.25\sin\left(3x-6\right) + 5
-
-
-
-
-
-
- Note that u(x) = -0.25\sin(3(x-2)) + 5. Period \frac{2\pi}{3}, amplitude 0.25, midline y=5, range [4.75,5.25], horizontal shift 2 units to the right. The anchor point is at approximately (0, 4.93), since u(0) = -0.25\sin(-6)+5 is not a convenient exact value.
-
+ Determine the exact period, amplitude, and midline of each of the following functions. In addition, state the range of each function, any horizontal shift that has been introduced to the graph, and identify an anchor point. Make your conclusions without consulting Desmos, and then use the program to check your work.
+
+
+
+
+
+
+ p(x) = \sin(10x) + 2
+
+
+
+
+
+ Period \frac{\pi}{5}, amplitude 1, midline y=2, range [1,3], no horizontal shift, anchor point (0,2).
+
+
+
+
+ Period \frac{\pi}{5}, amplitude 1, midline y=2, range [1,3], no horizontal shift, anchor point (0,2).
+
+
+
+
+
+
+ q(x) = -3\cos(0.25x) - 4
+
+
+
+
+
+ Period 8\pi, amplitude 3 (with reflection), midline y=-4, range [-7,-1], no horizontal shift, anchor point (0,-7).
+
+
+
+
+ Period 8\pi, amplitude 3 (with reflection), midline y=-4, range [-7,-1], no horizontal shift, anchor point (0,-7).
+
+
+
+
+
+
+ r(x) = 2\sin\left( \frac{\pi}{4} x\right) + 5
+
+
+
+
+
+ Period 8, amplitude 2, midline y=5, range [3,7], no horizontal shift, anchor point (0,5).
+
+
+
+
+ Period 8, amplitude 2, midline y=5, range [3,7], no horizontal shift, anchor point (0,5).
+
+ Period 4, amplitude 2, midline y=5, range [3,7], horizontal shift 3 units to the right, anchor point (0,5).
+
+
+
+
+ Period 4, amplitude 2, midline y=5, range [3,7], horizontal shift 3 units to the right, anchor point (0,5).
+
+
+
+
+
+
+ u(x) = -0.25\sin\left(3x-6\right) + 5
+
+
+
+
+
+ Note that u(x) = -0.25\sin(3(x-2)) + 5. Period \frac{2\pi}{3}, amplitude 0.25, midline y=5, range [4.75,5.25], horizontal shift 2 units to the right. The anchor point is at approximately (0, 4.93), since u(0) = -0.25\sin(-6)+5 is not a convenient exact value.
+
+
+
+
+ Note that u(x) = -0.25\sin(3(x-2)) + 5. Period \frac{2\pi}{3}, amplitude 0.25, midline y=5, range [4.75,5.25], horizontal shift 2 units to the right. The anchor point is at approximately (0, 4.93), since u(0) = -0.25\sin(-6)+5 is not a convenient exact value.
+
- Consider the circle pictured in the following figure that is centered at the point (2,2) and that has circumference 8. Assume that we track the y-coordinate (that is, the height, h) of a point that is traversing the circle counterclockwise and that it starts at P_0 as pictured.
-
-
-
-
ADD ALT TEXT TO THIS IMAGE
-
-
ADD ALT TEXT TO THIS IMAGE
-
-
-
-
-
-
-
- How far along the circle is the point P_1 from P_0? Why?
-
-
-
-
-
-
- The eight points on the circle are evenly spaced, and the circle has a circumference of 8 units. Thus, the distance between any two sequential points is 1 unit.
-
-
-
-
-
-
- Label the subsequent points in the figure P_2, P_3, \ldots as we move counterclockwise around the circle. What is the exact y-coordinate of the point P_2? of P_4? Why?
-
-
-
-
-
-
- The point P_2 is directly below the center, at coordinates (2, 2), and thus has a y-coordinate that is one radius less than 2. Using the formula for circumference, C = 2\pi r, the radius must be 8/(2\pi) = \frac{4}{\pi}. Therefore, the y-coordinate of point P_2 is 2 - \frac{4}{\pi} = \frac{2(\pi - 2)}{\pi}. At point P_4, the point is at the same height as the center, so the y-coordinate of P_4 is 2.
-
-
-
-
-
-
- Determine the y-coordinates of the remaining points on the circle (exactly where possible, otherwise approximately) and hence complete the entries in the following table that track the height, h, of the point traversing the circle as a function of distance traveled, d. Note that the d-values in the table correspond to the point traversing the circle more than once.
-
- By plotting the points in the table in part (c) and connecting them in an intuitive way, sketch a graph of h as a function of d over the interval 0 \le d \le 16 on the axes provided in the figure in part (a). Clearly label the scale of your axes and the coordinates of several important points on the curve.
-
-
-
-
-
-
-
-
-
-
-
- What is similar about your graph in comparison to the one in Figure 2.1.5 in the text? What is different?
-
-
-
-
-
-
- The graph has a very similar shape to Figure 2.1.5. This graph is shifted up, shifted to the right, and is compressed horizontally compared to the textbook figure.
-
-
-
-
-
-
- What will be the value of h when d = 51? How about when d = 102?
-
-
-
-
-
-
- To find h when d = 51, we use the periodicity of the function: 51 = 4 \cdot 8 + 3, so h(51) = h(3) \approx 1.10. Likewise, h(102) = h(6) \approx 3.27, because 102 = 12 \cdot 8 + 6.
-
+ Consider the circle pictured in the following figure that is centered at the point (2,2) and that has circumference 8. Assume that we track the y-coordinate (that is, the height, h) of a point that is traversing the circle counterclockwise and that it starts at P_0 as pictured.
+
+
+
+
+
+
+
+
Circle centered at (2,2) with circumference 8.
+
+
+
+
+
+
+
Blank axes for plotting h versus d.
+
+
+
+
+
+
+
+
+
+
+
+
+ How far along the circle is the point P_1 from P_0? Why?
+
+
+
+
1 unit apart
+
+
+
+ The eight points on the circle are evenly spaced, and the circle has a circumference of 8 units. Thus, the distance between any two sequential points is 1 unit.
+
+
+
+
+
+
+ Label the subsequent points in the figure P_2, P_3, \ldots as we move counterclockwise around the circle. What is the exact y-coordinate of the point P_2? of P_4? Why?
+
+
+
+
The y-coordinate of P_2 is 2 - \frac{4}{\pi} = \frac{2(\pi - 2)}{\pi}. The y-coordinate of P_4 is 2.
+
+
+
+ The point P_2 is directly below the center, at coordinates (2, 2), and thus has a y-coordinate that is one radius less than 2. Using the formula for circumference, C = 2\pi r, the radius must be 8/(2\pi) = \frac{4}{\pi}. Therefore, the y-coordinate of point P_2 is 2 - \frac{4}{\pi} = \frac{2(\pi - 2)}{\pi}. At point P_4, the point is at the same height as the center, so the y-coordinate of P_4 is 2.
+
+
+
+
+
+
+ Determine the y-coordinates of the remaining points on the circle (exactly where possible, otherwise approximately) and hence complete the entries in the following table that track the height, h, of the point traversing the circle as a function of distance traveled, d. Note that the d-values in the table correspond to the point traversing the circle more than once.
+
+ By plotting the points in the table in part (c) and connecting them in an intuitive way, sketch a graph of h as a function of d over the interval 0 \le d \le 16 on the axes provided in the figure in part (a). Clearly label the scale of your axes and the coordinates of several important points on the curve.
+
+
+
+
+
+
+
+
+
Graph of h versus d with points marked at d=2,4,6,8,10,12,14 and 16.
+
+
+
+
+
+
+
+
+
Graph of h versus d with points marked at d=2,4,6,8,10,12,14 and 16.
+
+
+
+
+
+
+
+
+ What is similar about your graph in comparison to the one in Figure 2.1.5 in the text? What is different?
+
+
+
+
+ This graph is shifted up, shifted to the right, and is compressed horizontally compared to the textbook figure.
+
+
+
+
+ The graph has a very similar shape to Figure 2.1.5. This graph is shifted up, shifted to the right, and is compressed horizontally compared to the textbook figure.
+
+
+
+
+
+
+ What will be the value of h when d = 51? How about when d = 102?
+
+
+
+
h(51) = \approx 1.10 and h(102) = \approx 3.27
+
+
+
+
+ To find h when d = 51, we use the periodicity of the function: 51 = 4 \cdot 8 + 3, so h(51) = h(3) \approx 1.10. Likewise, h(102) = h(6) \approx 3.27, because 102 = 12 \cdot 8 + 6.
+
- Consider the same setting as Activity: a weight oscillates back and forth on a frictionless table with distance from the wall given by, h = f(t) (in inches) at any given time, t (in seconds).
- A graph of f and a table of select values are given below.
-
- Determine AV_{[2,2.25]}, AV_{[2.25,2.5]}, AV_{[2.5,2.75]}, and AV_{[2.75,3]}. What do these four values tell us about how the weight is moving on the interval [2,3]?
-
- The weight is always moving in the same direction (away from the wall) on the interval [2,3], but is slowing down.
-
-
-
-
-
-
- Give an example of an interval of length 0.25 units on which f has its most negative average rate of change. Justify your choice.
-
-
-
-
-
-
- The greatest (steepest) negative rates of change occur near the midline where the function is decreasing. Examples include intervals such as [0, 0.25], [3.75, 4.00], and [4.00, 4.25]. These occur near the midline where the function crosses from above to below (or vice versa), meaning the weight is moving most rapidly toward the equilibrium position.
-
-
-
-
-
-
- Give an example of the longest interval you can find on which f is decreasing.
-
-
-
-
-
-
- The function is decreasing from each peak to each trough. Examples of longest such intervals are [3,5], [7,9], and [11,13].
-
-
-
-
-
-
- Give an example of an interval on which f is concave up. (Recall that a function is concave up on an interval provided that throughout the interval, the curve bends upward, similar to a parabola that opens up.)
-
-
-
-
-
- The intervals on which f is concave up include [0,2], [4,6], [8,10], and [12,14].
-
-
-
-
-
-
- On an interval where f is both decreasing and concave down, what does this tell us about how the weight is moving on that interval? For instance, is the weight moving toward or away from the wall? is it speeding up or slowing down?
-
-
-
-
-
-
- On intervals that are decreasing and concave up, the weight is moving toward the wall, but at a decreasing rate (slowing down).
-
-
-
-
-
-
- What general conclusions can you make about the average rate of change of a circular function on intervals near its highest or lowest points? about its average rate of change on intervals near the function's midline?
-
-
-
-
-
-
- At the highest and lowest points, the function has the smallest rates of change. Near the midline, the function has its greatest rates of change.
-
+ Consider the same setting as Activity: a weight oscillates back and forth on a frictionless table with distance from the wall given by, h = f(t) (in inches) at any given time, t (in seconds).
+ A graph of f and a table of select values are given below.
+
+ Determine AV_{[2,2.25]}, AV_{[2.25,2.5]}, AV_{[2.5,2.75]}, and AV_{[2.75,3]}. What do these four values tell us about how the weight is moving on the interval [2,3]?
+
+
+
+
+
+
+
AV_{[2,2.25]} \approx 7.652 inches/sec
+
AV_{[2.25,2.5]} \approx 6.492 inches/sec
+
AV_{[2.5,2.75]} \approx 4.332 inches/sec
+
AV_{[2.75,3]} \approx 1.524 inches/sec
+
+ The weight is always moving in the same direction (away from the wall) on the interval [2,3], but is slowing down.
+
+
+
+
+ The weight is always moving in the same direction (away from the wall) on the interval [2,3], but is slowing down.
+
+
+
+
+
+
+ Give an example of an interval of length 0.25 units on which f has its most negative average rate of change. Justify your choice.
+
+
+
+
+
+ The weight slows down as it reaches the farthest distance away from the midline and speeds up back toward the center. The greatest (steepest) negative rates of change occur near the midline where the function is decreasing, in intervals such as [0, 0.25], [3.75, 4.00], and [4.00, 4.25].
+
+
+
+
+ The greatest (steepest) negative rates of change occur near the midline where the function is decreasing. Examples include intervals such as [0, 0.25], [3.75, 4.00], and [4.00, 4.25]. These occur near the midline where the function crosses from above to below (or vice versa), meaning the weight is moving most rapidly toward the equilibrium position.
+
+
+
+
+
+
+ Give an example of the longest interval you can find on which f is decreasing.
+
+
+
+
+
+ Examples of longest such intervals are [3,5], [7,9], and [11,13].
+
+
+
+
+ The function is decreasing from each peak to each trough. Examples of longest such intervals are [3,5], [7,9], and [11,13].
+
+
+
+
+
+
+ Give an example of an interval on which f is concave up. (Recall that a function is concave up on an interval provided that throughout the interval, the curve bends upward, similar to a parabola that opens up.)
+
+
+
+
+ The intervals on which f is concave up include [0,2], [4,6], [8,10], and [12,14].
+
+
+
+
+ The intervals on which f is concave up include [0,2], [4,6], [8,10], and [12,14].
+
+
+
+
+
+
+ On an interval where f is both decreasing and concave down, what does this tell us about how the weight is moving on that interval? For instance, is the weight moving toward or away from the wall? is it speeding up or slowing down?
+
+
+
+
The weight is moving toward the wall but slowing down.
+
+
+ On intervals that are decreasing and concave up, the weight is moving toward the wall, but at a decreasing rate (slowing down).
+
+
+
+
+
+
+ What general conclusions can you make about the average rate of change of a circular function on intervals near its highest or lowest points? about its average rate of change on intervals near the function's midline?
+
+
+
+
+
+ At the highest and lowest points, the function has the smallest rates of change. Near the midline, the function has its greatest rates of change.
+
+
+
+
+ At the highest and lowest points, the function has the smallest rates of change. Near the midline, the function has its greatest rates of change.
+
- A weight is placed on a frictionless table next to a wall and attached to a spring that is fixed to the wall. From its natural position of rest, the weight is imparted an initial velocity that sets it in motion. The weight then oscillates back and forth,
- and we can measure its distance, h = f(t) (in inches)
- from the wall at any given time, t (in seconds).
- A graph of f and a table of select values are given below.
-
- Determine the period p, midline y = m, and amplitude a of the function f.
-
-
-
-
-
-
- The period of the motion is p = 4 seconds. The midline is y = 8 inches. The amplitude is a = 5 inches.
-
-
-
-
-
-
- What is the greatest distance the weight is displaced from the wall? What is the least distance the weight is displaced from the wall? What is the range of f?
-
-
-
-
-
-
- The greatest displacement of the weight is equal to the maximum of the graph: 13 inches. The least displacement is the minimum: 3 inches. The range of f is [3, 13].
-
-
-
-
-
-
- Determine the average rate of change of f on the intervals [4,4.25] and [4.75,5]. Write one careful sentence to explain the meaning of each (including units). In addition, write a sentence to compare the two different values you find and what they together say about the motion of the weight.
-
-
-
-
-
-
- The average rate of change AV_{[4,4.25]} on the interval [4,4.25] is
-
- \frac{6.807 - 8.00}{4.25 - 4} = \frac{-1.193}{0.25} = -4.772 \text{ inches per second}.
-
- The average rate of change AV_{[4.75,5]} on the interval [4.75,5] is
-
- \frac{3.000 - 3.381}{5 - 4.75} = \frac{-0.381}{0.25} = -1.524 \text{ inches per second}.
-
- The average rate of change of this function represents the speed of the weight. Taken together, these two values indicate that the weight is moving more slowly on the interval [4.75,5] than on the interval [4,4.25], but that the weight is moving in the same direction on each of these intervals.
-
-
-
-
-
-
- Based on the periodicity of the function, what is the value of f(6.75)? of f(11.25)?
-
-
-
-
-
-
- The value of f(6.75) = f(2.75) = 12.619 inches. The value of f(11.25) = f(7.25) = f(3.25) = 12.619 inches.
-
+ A weight is placed on a frictionless table next to a wall and attached to a spring that is fixed to the wall. From its natural position of rest, the weight is imparted an initial velocity that sets it in motion. The weight then oscillates back and forth,
+ and we can measure its distance, h = f(t) (in inches)
+ from the wall at any given time, t (in seconds).
+ A graph of f and a table of select values are given below.
+
+ Determine the period p, midline y = m, and amplitude a of the function f.
+
+
+
+
+ The period of the motion is p = 4 seconds. The midline is y = 8 inches. The amplitude is a = 5 inches.
+
+
+
+
+ The period of the motion is p = 4 seconds. The midline is y = 8 inches. The amplitude is a = 5 inches.
+
+
+
+
+
+
+ What is the greatest distance the weight is displaced from the wall? What is the least distance the weight is displaced from the wall? What is the range of f?
+
+
+
+
+ The greatest displacement of the weight is 13 inches. The least displacement is 3 inches. The range of f is [3, 13].
+
+
+
+
+ The greatest displacement of the weight is equal to the maximum of the graph: 13 inches. The least displacement is the minimum: 3 inches. The range of f is [3, 13].
+
+
+
+
+
+
+ Determine the average rate of change of f on the intervals [4,4.25] and [4.75,5]. Write one careful sentence to explain the meaning of each (including units). In addition, write a sentence to compare the two different values you find and what they together say about the motion of the weight.
+
+
+
+
+ AV_{[4,4.25]}= -4.772 inches per second on the interval [4,4.25].
+ AV_{[4.75,5]} = -1.524 inches per second on the interval [4.75,5].
+ The average rate of change of this function represents the speed of the weight. Taken together, these two values indicate that the weight is moving more slowly on the interval [4.75,5] than on the interval [4,4.25], but that the weight is moving in the same direction on each of these intervals.
+
+
+
+
+ The average rate of change AV_{[4,4.25]} on the interval [4,4.25] is
+
+ \frac{6.807 - 8.00}{4.25 - 4} = \frac{-1.193}{0.25} = -4.772 \text{ inches per second}.
+
+ The average rate of change AV_{[4.75,5]} on the interval [4.75,5] is
+
+ \frac{3.000 - 3.381}{5 - 4.75} = \frac{-0.381}{0.25} = -1.524 \text{ inches per second}.
+
+ The average rate of change of this function represents the speed of the weight. Taken together, these two values indicate that the weight is moving more slowly on the interval [4.75,5] than on the interval [4,4.25], but that the weight is moving in the same direction on each of these intervals.
+
+
+
+
+
+
+ Based on the periodicity of the function, what is the value of f(6.75)? of f(11.25)?
+
+
+
+
f(6.75) = 12.619 inches and f(11.25) = 12.619 inches.
+
+
+
+
+ The value of f(6.75) = f(2.75) = 12.619 inches. The value of f(11.25) = f(7.25) = f(3.25) = 12.619 inches.
+
- Determine each of the following values or points exactly.
-
-
-
-
-
-
- In a circle of radius 11, the arc length intercepted by a central angle of \frac{5\pi}{3}.
-
-
-
-
-
-
- In a circle of radius 11, the arc length intercepted by a central angle of \frac{5\pi}{3} is 11 \cdot \frac{5\pi}{3} = \frac{55\pi}{3}.
-
-
-
-
-
-
- In a circle of radius 3, the central angle measure that intercepts an arc of length \frac{\pi}{4}.
-
-
-
-
-
-
- In a circle of radius 3, the central angle that intercepts an arc length \frac{\pi}{4} is \frac{\pi/4}{3} = \frac{\pi}{12}.
-
-
-
-
-
-
- The radius of the circle in which an angle of
- \frac{7\pi}{6} intercepts an arc of length \frac{\pi}{2}.
-
-
-
-
-
-
- The radius is \frac{\pi/2}{7\pi/6} = \frac{3}{7}.
-
-
-
-
-
-
- The exact coordinates of the point on the circle of radius 5 that lies \frac{25\pi}{6} units counterclockwise along the circle from (5,0).
-
-
-
-
-
-
- The angle corresponding to the arc length is \theta = \frac{25\pi/6}{5} = \frac{5\pi}{6}. But since \frac{25\pi}{6} = 4\pi + \frac{\pi}{6}, the effective angle is \frac{\pi}{6}. Using the unit circle coordinates at \frac{\pi}{6}, the point on the circle of radius 5 is \left( \frac{5\sqrt{3}}{2}, \frac{5}{2} \right).
-
+ Determine each of the following values or points exactly.
+
+
+
+
+
+
+ In a circle of radius 11, the arc length intercepted by a central angle of \frac{5\pi}{3}.
+
+
+
+
11 \cdot \frac{5\pi}{3} = \frac{55\pi}{3}
+
+
+ In a circle of radius 11, the arc length intercepted by a central angle of \frac{5\pi}{3} is 11 \cdot \frac{5\pi}{3} = \frac{55\pi}{3}.
+
+
+
+
+
+
+ In a circle of radius 3, the central angle measure that intercepts an arc of length \frac{\pi}{4}.
+
+
+
+
\frac{\pi/4}{3} = \frac{\pi}{12}
+
+
+ In a circle of radius 3, the central angle that intercepts an arc length \frac{\pi}{4} is \frac{\pi/4}{3} = \frac{\pi}{12}.
+
+
+
+
+
+
+ The radius of the circle in which an angle of
+ \frac{7\pi}{6} intercepts an arc of length \frac{\pi}{2}.
+
+
+
+
\frac{\pi/2}{7\pi/6} = \frac{3}{7}
+
+
+ The radius is \frac{\pi/2}{7\pi/6} = \frac{3}{7}.
+
+
+
+
+
+
+ The exact coordinates of the point on the circle of radius 5 that lies \frac{25\pi}{6} units counterclockwise along the circle from (5,0).
+
+
+
+
\left( \frac{5\sqrt{3}}{2}, \frac{5}{2} \right)
+
+
+ The angle corresponding to the arc length is \theta = \frac{25\pi/6}{5} = \frac{5\pi}{6}. But since \frac{25\pi}{6} = 4\pi + \frac{\pi}{6}, the effective angle is \frac{\pi}{6}. Using the unit circle coordinates at \frac{\pi}{6}, the point on the circle of radius 5 is \left( \frac{5\sqrt{3}}{2}, \frac{5}{2} \right).
+
- In what follows, we work to understand key relationships in
- 45^\circ-45^\circ-90^\circ and
- 30^\circ-60^\circ-90^\circ triangles.
-
-
-
-
ADD ALT TEXT TO THIS IMAGE
-
-
ADD ALT TEXT TO THIS IMAGE
-
-
-
-
-
-
-
- For the 45^\circ-45^\circ-90^\circ triangle with legs of length x and y and hypotenuse of length 1, what does the fact that the triangle is isosceles tell us about the relationship between x and y? What are their exact values?
-
-
-
-
-
-
- Because a 45^\circ-45^\circ-90^\circ triangle is isosceles, the sides x and y must be equal. From the Pythagorean theorem, \sqrt{x^2 + y^2} = 1, which reduces to \sqrt{2x^2} = 1, giving x = y = \frac{\sqrt{2}}{2}.
-
-
-
-
-
-
- Now consider the 30^\circ-60^\circ-90^\circ triangle with hypotenuse of length 1 and the longer leg (of length x) lying along the positive x-axis. What special kind of triangle is formed when we reflect this triangle across the x-axis? How can we use this perspective to determine the exact values of x and y?
-
-
-
-
-
-
- When the 30^\circ-60^\circ-90^\circ triangle is reflected across the x-axis, the outer perimeter of the resulting shape forms an equilateral triangle where all three interior angles are 60^\circ and all sides have the same length, 1 unit. This gives us the insight that side y must have a length equal to half of one unit: y = \frac{1}{2}. Side x is determined from the Pythagorean theorem: x = \frac{\sqrt{3}}{2}.
-
-
-
-
-
-
- Suppose we consider the related 30^\circ-60^\circ-90^\circ triangle with hypotenuse of length 1 and the shorter leg (of length x) lying along the positive x-axis. What are the exact values of x and y in this triangle?
-
-
-
-
-
-
- When a 30^\circ-60^\circ-90^\circ triangle is constructed such that the angle between the hypotenuse and the x-side is 60^\circ, the sides are y = \frac{\sqrt{3}}{2} and x = \frac{1}{2}.
-
-
-
-
-
-
- We know from the conversion factor from degrees to radians that an angle of 30^\circ corresponds to an angle measuring \frac{\pi}{6} radians, an angle of 45^\circ corresponds to \frac{\pi}{4} radians, and 60^\circ corresponds to \frac{\pi}{3} radians.
-
-
-
-
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-
-
ADD ALT TEXT TO THIS IMAGE
-
-
ADD ALT TEXT TO THIS IMAGE
-
-
-
- Use your work in (a), (b), and (c) to label the noted point in each of the three respective figures with its exact coordinates.
-
-
-
-
-
-
- The coordinates of the point at \theta = \frac{\pi}{6} on the unit circle are \left( \frac{\sqrt{3}}{2}, \frac{1}{2} \right). The coordinates at \theta = \frac{\pi}{4} are \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right). The coordinates at \theta = \frac{\pi}{3} are \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right).
-
+ In what follows, we work to understand key relationships in
+ 45^\circ-45^\circ-90^\circ and
+ 30^\circ-60^\circ-90^\circ triangles.
+
+
+
+
ADD ALT TEXT TO THIS IMAGE
+
+
ADD ALT TEXT TO THIS IMAGE
+
+
+
+
+
+
+
+ For the 45^\circ-45^\circ-90^\circ triangle with legs of length x and y and hypotenuse of length 1, what does the fact that the triangle is isosceles tell us about the relationship between x and y? What are their exact values?
+
+
+
+
x = y = \frac{\sqrt{2}}{2}
+
+
+ Because a 45^\circ-45^\circ-90^\circ triangle is isosceles, the sides x and y must be equal. From the Pythagorean theorem, \sqrt{x^2 + y^2} = 1, which reduces to \sqrt{2x^2} = 1, giving x = y = \frac{\sqrt{2}}{2}.
+
+
+
+
+
+
+ Now consider the 30^\circ-60^\circ-90^\circ triangle with hypotenuse of length 1 and the longer leg (of length x) lying along the positive x-axis. What special kind of triangle is formed when we reflect this triangle across the x-axis? How can we use this perspective to determine the exact values of x and y?
+
+
+
+
+
+ The resulting shape forms an equilateral triangle where all three interior angles are 60^\circ and all sides have the same length, 1 unit. Thus, y = \frac{1}{2} and by the Pythagorean theorem, x = \frac{\sqrt{3}}{2}.
+
+
+
+
+ When the 30^\circ-60^\circ-90^\circ triangle is reflected across the x-axis, the outer perimeter of the resulting shape forms an equilateral triangle where all three interior angles are 60^\circ and all sides have the same length, 1 unit. This gives us the insight that side y must have a length equal to half of one unit: y = \frac{1}{2}. Side x is determined from the Pythagorean theorem: x = \frac{\sqrt{3}}{2}.
+
+
+
+
+
+
+ Suppose we consider the related 30^\circ-60^\circ-90^\circ triangle with hypotenuse of length 1 and the shorter leg (of length x) lying along the positive x-axis. What are the exact values of x and y in this triangle?
+
+
+
+
y = \frac{\sqrt{3}}{2} and x = \frac{1}{2}
+
+
+ When a 30^\circ-60^\circ-90^\circ triangle is constructed such that the angle between the hypotenuse and the x-side is 60^\circ, the sides are y = \frac{\sqrt{3}}{2} and x = \frac{1}{2}.
+
+
+
+
+
+
+ We know from the conversion factor from degrees to radians that an angle of 30^\circ corresponds to an angle measuring \frac{\pi}{6} radians, an angle of 45^\circ corresponds to \frac{\pi}{4} radians, and 60^\circ corresponds to \frac{\pi}{3} radians.
+
+
+
+
ADD ALT TEXT TO THIS IMAGE
+
+
ADD ALT TEXT TO THIS IMAGE
+
+
ADD ALT TEXT TO THIS IMAGE
+
+
+
+ Use your work in (a), (b), and (c) to label the noted point in each of the three respective figures with its exact coordinates.
+
+
+
+
+
+ The coordinates of the point at \theta = \frac{\pi}{6} on the unit circle are \left( \frac{\sqrt{3}}{2}, \frac{1}{2} \right). The coordinates at \theta = \frac{\pi}{4} are \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right). The coordinates at \theta = \frac{\pi}{3} are \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right).
+
+
+
+
+ The coordinates of the point at \theta = \frac{\pi}{6} on the unit circle are \left( \frac{\sqrt{3}}{2}, \frac{1}{2} \right). The coordinates at \theta = \frac{\pi}{4} are \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right). The coordinates at \theta = \frac{\pi}{3} are \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right).
+
- Recall from Section that the average rate of change of a function f on an interval [a,b] is
-
- AV_{[a,b]} = \frac{f(b)-f(a)}{b-a}
- .
- In Section, we also saw that if we instead think of the average rate of change of f on the interval [a,a+h], the expression changes to
-
- AV_{[a,a+h]} = \frac{f(a+h)-f(a)}{h}
- .
- In this activity we explore the average rate of change of f(t) = e^t near the points where t = 1 and t = 2.
-
-
- In a new Desmos worksheet, let f(t) = e^t and define the function A by the rule
-
- A(h) = \frac{f(1+h)-f(1)}{h}
- .
-
-
-
-
-
-
- What is the meaning of A(0.5) in terms of the function f and its graph?
-
-
-
-
-
-
- The function A evaluated at h = 0.5 gives the average rate of change of f(t) = e^t on the interval [1, 1.5].
-
-
-
-
-
-
- Compute the value of A(h) for at least 6 different small values of h, both positive and negative. For instance, one value to try might be h = 0.0001. Record a table of your results.
-
- What do you notice about the values you found in (b)? How do they compare to an important number?
-
-
-
-
-
-
- As h gets smaller and smaller, the value of A(h) approaches the value of e.
-
-
-
-
-
-
- Explain why the following sentence makes sense: The function e^t is increasing at an average rate that is about the same as its value on small intervals near t = 1.
-
-
-
-
-
- The statement makes sense because the rate of change on small intervals near t = 1 is close to the value of e^1 = e.
-
-
-
-
-
-
- Adjust your definition of A in Desmos by changing 1 to 2 so that
-
- A(h) = \frac{f(2+h)-f(2)}{h}
- .
- How does the value of A(h) compare to f(2) for small values of h?
-
-
-
-
-
-
- The value of A(h) near h = 0 appears to be the same as the value of f(2) = e^2.
-
+ Recall from Section that the average rate of change of a function f on an interval [a,b] is
+
+ AV_{[a,b]} = \frac{f(b)-f(a)}{b-a}
+ .
+ In Section, we also saw that if we instead think of the average rate of change of f on the interval [a,a+h], the expression changes to
+
+ AV_{[a,a+h]} = \frac{f(a+h)-f(a)}{h}
+ .
+ In this activity we explore the average rate of change of f(t) = e^t near the points where t = 1 and t = 2.
+
+
+ In a new Desmos worksheet, let f(t) = e^t and define the function A by the rule
+
+ A(h) = \frac{f(1+h)-f(1)}{h}
+ .
+
+
+
+
+
+
+ What is the meaning of A(0.5) in terms of the function f and its graph?
+
+
+
+
+
+ A(0.5) gives the average rate of change of f(t) = e^t on the interval [1, 1.5], which is the slope of the line between the points (1,f(1)) and (1.5,f(1.5)).
+
+
+
+
+ The function A evaluated at h = 0.5 gives the average rate of change of f(t) = e^t on the interval [1, 1.5]. This is the slope of the line between the points (1,f(1)) and (1.5,f(1.5)).
+
+
+
+
+
+
+ Compute the value of A(h) for at least 6 different small values of h, both positive and negative. For instance, one value to try might be h = 0.0001. Record a table of your results.
+
+ What do you notice about the values you found in (b)? How do they compare to an important number?
+
+
+
+
+
+ As h gets smaller and smaller, the value of A(h) approaches the value of e.
+
+
+
+
+ As h gets smaller and smaller, the value of A(h) approaches the value of e.
+
+
+
+
+
+
+ Explain why the following sentence makes sense: The function e^t is increasing at an average rate that is about the same as its value on small intervals near t = 1.
+
+
+
+
+ The statement makes sense because the rate of change on small intervals near t = 1 is close to the value of e^1 = e.
+
+
+
+
+ The statement makes sense because the rate of change on small intervals near t = 1 is close to the value of e^1 = e.
+
+
+
+
+
+
+ Adjust your definition of A in Desmos by changing 1 to 2 so that
+
+ A(h) = \frac{f(2+h)-f(2)}{h}
+ .
+ How does the value of A(h) compare to f(2) for small values of h?
+
+
+
+
+
+ The value of A(h) near h = 0 appears to be the same as the value of f(2) = e^2.
+
+
+
+
+ The value of A(h) near h = 0 appears to be the same as the value of f(2) = e^2.
+
- By graphing f(t) = e^t and appropriate horizontal lines, estimate the solution to each of the following equations. Note that in some parts, you may need to do some algebraic work in addition to using the graph.
-
-
-
-
-
-
- e^t = 2
-
-
-
-
-
-
- For e^t = 2, t = \ln 2 \approx 0.6931.
-
-
-
-
-
-
- e^{3t} = 5
-
-
-
-
-
-
- For e^{3t} = 5, t = \frac{\ln 5}{3} \approx 0.5365.
-
-
-
-
-
-
- 2e^t - 4 = 7
-
-
-
-
-
-
- The equation 2e^t - 4 = 7 simplifies to e^t = \frac{11}{2}, so t = \ln\!\left(\frac{11}{2}\right) \approx 1.7047.
-
-
-
-
-
-
- 3e^{0.25t} + 2 = 6
-
-
-
-
-
-
- The equation 3e^{0.25t} + 2 = 6 simplifies to e^{0.25t} = \frac{4}{3}, so t = 4\ln\!\left(\frac{4}{3}\right) \approx 1.1507.
-
-
-
-
-
-
- 4 - 2e^{-0.7t} = 3
-
-
-
-
-
-
- The equation 4 - 2e^{-0.7t} = 3 simplifies to e^{-0.7t} = \frac{1}{2}, so t = \frac{\ln 2}{0.7} \approx 0.9902.
-
-
-
-
-
-
- 2e^{1.2t} = 1.5e^{1.6t}
-
-
-
-
-
-
- The equation 2e^{1.2t} = 1.5e^{1.6t} simplifies to e^{-0.4t} = \frac{3}{4}, so t = \frac{-\ln(3/4)}{0.4} \approx 0.7192.
-
-
-
-
-
-
-
-
- A graph of f(t) = e^t showing the solutions to each equation is below.
-
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ By graphing f(t) = e^t and appropriate horizontal lines, estimate the solution to each of the following equations. Note that in some parts, you may need to do some algebraic work in addition to using the graph.
+
+
+
+
+
+
+ e^t = 2
+
+
+
+
+
+ t \approx 0.6931
+
+
+
+
+ For e^t = 2, t \approx 0.6931.
+
+
+
+
+
+
+ e^{3t} = 5
+
+
+
+
t = \approx 0.5365
+
+
+ For e^{3t} = 5, t \approx 0.5365.
+
+
+
+
+
+
+ 2e^t - 4 = 7
+
+
+
+
t \approx 1.7047
+
+
+ The equation 2e^t - 4 = 7 simplifies to e^t = \frac{11}{2}, so t \approx 1.7047.
+
+
+
+
+
+
+ 3e^{0.25t} + 2 = 6
+
+
+
+
t \approx 1.1507
+
+
+ The equation 3e^{0.25t} + 2 = 6 simplifies to e^{0.25t} = \frac{4}{3}, so t \approx 1.1507.
+
+
+
+
+
+
+ 4 - 2e^{-0.7t} = 3
+
+
+
+
t \approx 0.9902
+
+
+ The equation 4 - 2e^{-0.7t} = 3 simplifies to e^{-0.7t} = \frac{1}{2}, so t \approx 0.9902.
+
+
+
+
+
+
+ 2e^{1.2t} = 1.5e^{1.6t}
+
+
+
+
t \approx 0.7192
+
+
+ The equation 2e^{1.2t} = 1.5e^{1.6t} simplifies to e^{-0.4t} = \frac{3}{4}, so t \approx 0.7192.
+
- In Desmos, define the function g(t) = ab^t and create sliders for both a and b when prompted. Click on the sliders to set the minimum value for each to 0.1 and the maximum value to 10. Note that for g to be an exponential function, we require b \ne 1, even though the slider for b will allow this value.
-
-
-
-
-
-
- What is the domain of g(t) = ab^t?
-
-
-
-
-
-
- The domain of g(t) = ab^t is all real numbers, (-\infty, \infty).
-
-
-
-
-
-
- What is the range of g(t) = ab^t?
-
-
-
-
-
-
- Allowing for all possible values of a and b, the range of g(t) is all real numbers. If we restrict a and b to 0.1 \le a, b \le 10, the range of g(t) becomes (0, \infty).
-
-
-
-
-
-
- What is the y-intercept of g(t) = ab^t?
-
-
-
-
-
-
- The y-intercept for g(t) is a.
-
-
-
-
-
-
- How does changing the value of b affect the shape and behavior of the graph of g(t) = ab^t? Write several sentences to explain.
-
-
-
-
-
-
- When b is less than 1, the graph approaches \infty as t approaches -\infty, and the graph approaches 0 when t approaches \infty. When b is greater than 1, the graph approaches 0 when t approaches -\infty and the graph approaches \infty when t approaches \infty.
-
-
-
-
-
-
- For what values of the growth factor b is the corresponding growth rate positive? For which b-values is the growth rate negative?
-
-
-
-
-
-
- When b is greater than 1, the function describes positive growth (growing with time). When b is between 0 and 1, the function describes negative growth (decreasing with time).
-
-
-
-
-
-
- Consider the graphs of the exponential functions p and q provided in the following figure. If p(t) = ab^t and q(t) = cd^t, what can you say about the values a, b, c, and d (beyond the fact that all are positive and b \ne 1 and d \ne 1)? For instance, can you say a certain value is larger than another? Or that one of the values is less than 1?
-
-
-
ADD ALT TEXT TO THIS IMAGE
-
-
-
-
-
-
- The function p(t) shows negative growth, so the value of b must be between 0 and 1. The function q(t) shows positive growth, so the value of d must be greater than 1. The function p has a y-intercept that is greater than the function q's y-intercept, thus a \gt c.
-
+ In Desmos, define the function g(t) = ab^t and create sliders for both a and b when prompted. Click on the sliders to set the minimum value for each to 0.1 and the maximum value to 10. Note that for g to be an exponential function, we require b \ne 1, even though the slider for b will allow this value.
+
+
+
+
+
+
+ What is the domain of g(t) = ab^t?
+
+
+
+
All real numbers, or (-\infty, \infty).
+
+
+ The domain of g(t) = ab^t is all real numbers, (-\infty, \infty).
+
+
+
+
+
+
+ What is the range of g(t) = ab^t?
+
+
+
+
(0, \infty)
+
+
+ The range of g(t) is (0, \infty).
+
+
+
+
+
+
+ What is the y-intercept of g(t) = ab^t?
+
+
+
+
a
+
+
+ The y-intercept for g(t) is a.
+
+
+
+
+
+
+ How does changing the value of b affect the shape and behavior of the graph of g(t) = ab^t? Write several sentences to explain.
+
+
+
+
When b is less than 1, the function is decreasing and concave up, approaching 0 as x gets larger and larger. When When b is greater than 1, the function is increasing and concave up, approaching \infty as x gets larger and larger.
+
+
+ When b is less than 1, the function is decreasing and concave up. The graph approaches \infty as t approaches -\infty, and the graph approaches 0 as t approaches \infty. When b is greater than 1, the function is increasing and concave up. The graph approaches 0 as t approaches -\infty and the graph approaches \infty as t approaches \infty.
+
+
+
+
+
+
+ For what values of the growth factor b is the corresponding growth rate positive? For which b-values is the growth rate negative?
+
+
+
+
+ When b is greater than 1, the function describes positive growth (growing with time). When b is between 0 and 1, the function describes negative growth (decreasing with time).
+
+
+
+
+ When b is greater than 1, the function describes positive growth (growing with time). When b is between 0 and 1, the function describes negative growth (decreasing with time).
+
+
+
+
+
+
+ Consider the graphs of the exponential functions p and q provided in the following figure. If p(t) = ab^t and q(t) = cd^t, what can you say about the values a, b, c, and d (beyond the fact that all are positive and b \ne 1 and d \ne 1)? For instance, can you say a certain value is larger than another? Or that one of the values is less than 1?
+
+
+
+
+
+
+
+
Graph of two exponential functions, p(t)=ab^t and q(t)=cd^t. The function p(t) is decreasing towards 0 and has a higher y-intercept, while the function q(t) has a lower y-intercept and is increasing towards higher and higher values as x increases.
+
+
+
+
We can tell that 0 \lt b \lt 1 while d \gt 1. Also, a \gt c.
+
+
+ The function p(t) shows negative growth, so the value of b must be between 0 and 1. The function q(t) shows positive growth, so the value of d must be greater than 1. The function p has a y-intercept that is greater than the function q's y-intercept, thus a \gt c.
+
- The value of an automobile is depreciating. When the car is 3 years old, its value is $12500; when the car is 7 years old, its value is $6500.
-
-
-
-
-
-
- Suppose the car's value t years after its purchase is given by the function V(t) and that V is exponential with form V(t) = ab^t, what are the values of a and b? Find a and b both exactly and approximately.
-
-
-
-
-
-
- Given V(t) = ab^t, V(3) = \$12{,}500, and V(7) = \$6{,}500, the ratio of the two specified points gives b = \left( \dfrac{13}{25} \right)^{1/4} exactly. Solving either equation for a gives a = 12500 \left( \dfrac{25}{13} \right)^{3/4} = 6500 \left( \dfrac{25}{13} \right)^{7/4} exactly. In approximate form, V(t) \approx \$20{,}413 \times (0.8492)^t.
-
-
-
-
-
-
- Using the exponential model determined in (a), determine the purchase value of the car and then use Desmos to estimate when the car will be worth less than $1000.
-
-
-
-
-
-
- Using the approximate model, the purchase value of the car is approximately $20{,}413. Solving 1000 = 20413 \times 0.8492^t using logarithms gives t \approx 18.45, so the car is worth less than $1000 after about 18 years.
-
-
-
-
-
-
- Suppose instead that the car's value is modeled by a linear function L and satisfies the values stated at the outset of this activity. Find a formula for L(t) and determine both the purchase value of the car and when the car will be worth $1000.
-
-
-
-
-
-
- If the car's value is linear, then L(t) = b + mt. Using L(3) = 12500 and L(7) = 6500, we find m = (12500 - 6500)/(3 - 7) = -1500 dollars per year and b = \$17{,}000. Thus L(t) = 17000 - 1500t, and the car will be worth $1000 when t = \frac{32}{3} \approx 10.67 years.
-
-
-
-
-
-
- Which model do you think is more realistic? Why?
-
-
-
-
-
-
- The exponential model seems more realistic because new cars decrease in value at a faster rate than older cars.
-
+ The value of an automobile is depreciating. When the car is 3 years old, its value is \$ 12500; when the car is 7 years old, its value is \$ 6500.
+
+
+
+
+
+
+ Suppose the car's value t years after its purchase is given by the function V(t) and that V is exponential with form V(t) = ab^t, what are the values of a and b? Find a and b both exactly and approximately.
+
+
+
+
Exactly: b = \left( \dfrac{13}{25} \right)^{1/4}, a = 12500 \left( \dfrac{25}{13} \right)^{3/4} = 6500 \left( \dfrac{25}{13} \right)^{7/4}. In approximate form, V(t) \approx \$ 20413 \cdot (0.8492)^t.
+
+
+ Given V(t) = ab^t, V(3) = \$12{,}500, and V(7) = \$6{,}500, the ratio of the two specified points gives b = \left( \dfrac{13}{25} \right)^{1/4} exactly. Solving either equation for a gives a = 12500 \left( \dfrac{25}{13} \right)^{3/4} = 6500 \left( \dfrac{25}{13} \right)^{7/4} exactly. In approximate form, V(t) \approx \$20{,}413 \cdot (0.8492)^t.
+
+
+
+
+
+
+ Using the exponential model determined in (a), determine the purchase value of the car and then use Desmos to estimate when the car will be worth less than \$ 1000.
+
+
+
+
The purchase value of the car is approximately \$ 20413, and the car is worth less than \$ 1000 after a little over 18 years.
+
+
+ Using the approximate model, the purchase value of the car is approximately \$ 20{,}413. Using Desmos, we see that 1000 = 20413 \cdot 0.8492^t when t \approx 18.45, so the car is worth less than \$ 1000 after about 18 years.
+
+
+
+
+
+
+ Suppose instead that the car's value is modeled by a linear function L and satisfies the values stated at the outset of this activity. Find a formula for L(t) and determine both the purchase value of the car and when the car will be worth \$ 1000.
+
+
+
+
L(t) = 17000 - 1500t, and the car will be worth \$ 1000 when t = \frac{32}{3} \approx 10.67 years.
+
+
+ If the car's value is linear, then L(t) = b + mt. Using L(3) = 12500 and L(7) = 6500, we find m = (12500 - 6500)/(3 - 7) = -1500 dollars per year and b = \$17{,}000. Thus L(t) = 17000 - 1500t, and the car will be worth $1000 when t = \frac{32}{3} \approx 10.67 years.
+
+
+
+
+
+
+ Which model do you think is more realistic? Why?
+
+
+
+
+
+ The exponential model seems more realistic because new cars decrease in value faster rate at the beginning and more slowly as they get older.
+
+
+
+
+ The exponential model seems more realistic because new cars decrease in value faster rate at the beginning and more slowly as they get older.
+
- For each of the following prompts, give an example of a function that satisfies the stated characteristics by both providing a formula and sketching a graph.
-
-
-
-
-
-
- A function p that is always decreasing and decreases at a constant rate.
-
-
-
-
-
-
- If p is always decreasing at a constant rate, then p is a linear function of the form p(x) = mx + b, where m \lt 0. For example, p(x) = -4x + 2.
-
-
-
-
-
-
-
- A function q that is always increasing and increases at an increasing rate.
-
-
-
-
-
-
- If q is always increasing and increases at an increasing rate, then q may be an exponential function. For example, q(x) = 2^x.
-
-
-
-
-
-
-
- A function r that is always increasing for t \lt 2, always decreasing for t \gt 2, and is always changing at a decreasing rate.
-
-
-
-
-
-
- One option is the piecewise function
-
- r(t) = \begin{cases} -\left(\frac{1}{2}\right)^t & \text{if } t \lt 2 \\ \left(\frac{1}{2}\right)^t - \frac{1}{2} & \text{if } t \gt 2 \end{cases}
-
- whose rate of change is always decreasing with increasing t.
-
-
-
-
-
-
-
- A function s that is always increasing and increases at a decreasing rate. (Hint: to find a formula, think about how you might use a transformation of a familiar function.)
-
-
-
-
-
-
- Let s(t) = -\left(\frac{1}{2}\right)^t, so that s(t) is always increasing with a decreasing rate.
-
-
-
-
-
-
-
- A function u that is always decreasing and decreases at a decreasing rate.
-
-
-
-
-
-
- The function u(t) = \left(\frac{1}{2}\right)^t is always decreasing at a decreasing rate.
-
+ For each of the following prompts, give an example of a function that satisfies the stated characteristics by both providing a formula and sketching a graph.
+
+
+
+
+
+
+ A function p that is always decreasing and decreases at a constant rate.
+
+
+
+
+
+ Any line with negative slope, for example, p(x) = -2x + 1.
+
+
+
+
+
+
A line with negative slope
+
+
+
+
+ If p is always decreasing at a constant rate, then p is a linear function of the form p(x) = mx + b, where m \lt 0. For example, p(x) = -2x + 1.
+
+
+
+
+
+
A line with negative slope
+
+
+
+
+
+
+
+ A function q that is always increasing and increases at an increasing rate.
+
+
+
+
+
+ For example, q(x) = 2^x.
+
+
+
+
+
+
The function 2^x, which is increasing at an increasing rate.
+
+
+
+
+ If q is always increasing and increases at an increasing rate, then q may be an exponential function. For example, q(x) = 2^x.
+
+
+
+
+
+
The function 2^x, which is increasing at an increasing rate.
+
+
+
+
+
+
+
+ A function r that is always increasing for t \lt 2, always decreasing for t \gt 2, and is always changing at a decreasing rate. (Recall that when a negative number decreases, it becomes more negative.)
+
+
+
+
+
+ One option is a downward-opening parabola, such as -(x-2)^2+4.
+
+
+
+
+
+
A downward-opening parabola which has its peak at x=2.
+
+
+
+
+
+ One option is a downward-opening parabola, such as -(x-2)^2+4.
+
+
+
+
+
+
A downward-opening parabola which has its peak at x=2.
+
+
+
+
+
+
+ A function s that is always increasing and increases at a decreasing rate. (Hint: to find a formula, think about how you might use a transformation of a familiar function.)
+
+
+
+
+
+ For example, s(x) = -\left(\frac{1}{2}\right)^x+5.
+
+
+
+
+
+
The function -\left(\frac{1}{2}\right)^x+5, which is increasing at a decreasing rate.
+
+
+
+
+ We know that an exponential function whose b value is between 0 and 1 will be decreasing more and more slowly, so we flip it across the x-axis to obtain a function that is always increasing more and more slowly. For example, s(x) = -\left(\frac{1}{2}\right)^x+5.
+
+
+
+
+
+
The function -\left(\frac{1}{2}\right)^x+5, which is increasing at a decreasing rate.
+
+
+
+
+
+
+
+ A function u that is always decreasing and decreases at a decreasing rate. (Recall that when a negative number decreases, it becomes more negative.)
+
+
+
+
+
+ For example, s(x) = -\left(\frac{1}{2}\right)^{-x}+5.
+
+
+
+
+
+
The function -\left(\frac{1}{2}\right)^{-x}+5, which is decreasing at a decreasing rate.
+
+
+
+
+ We want to take the previous example and flip it left to right, or across the y-axis, to obtain a function that is always decreasing by more and more. For example, s(x) = -\left(\frac{1}{2}\right)^{-x}+5.
+
+
+
+
+
+
The function -\left(\frac{1}{2}\right)^{-x}+5, which is decreasing at a decreasing rate.
- For each of the following equations, determine the exact value of the unknown variable. If the exact value involves a logarithm, use a computational device to also report an approximate value. For instance, if the exact value is y = \log_{10}(2), you can also note that y \approx 0.301.
-
-
-
-
-
-
- 10^t = 0.00001
-
-
-
-
-
-
- t = \log_{10}(0.00001) = -5, exactly.
-
-
-
-
-
-
- \log_{10}(1000000) = t
-
-
-
-
-
-
- t = \log_{10}(1000000) = 6, exactly.
-
-
-
-
-
-
- 10^t = 37
-
-
-
-
-
-
- t = \log_{10}(37) \approx 1.568.
-
-
-
-
-
-
- \log_{10}(y) = 1.375
-
-
-
-
-
-
- y = 10^{1.375} \approx 23.71.
-
-
-
-
-
-
- 10^t = 0.04
-
-
-
-
-
-
- t = \log_{10}(0.04) \approx -1.398.
-
-
-
-
-
-
- 3 \cdot 10^t + 11 = 147
-
-
-
-
-
-
- 3 \cdot 10^t + 11 = 147 gives 10^t = \frac{136}{3}, so t = \log_{10}\!\left(\frac{136}{3}\right) \approx 1.656.
-
-
-
-
-
-
- 2\log_{10}(y) + 5 = 1
-
-
-
-
-
-
- 2\log_{10}(y) + 5 = 1 gives \log_{10}(y) = -2, so y = 10^{-2} = 0.01, exactly.
-
+ For each of the following equations, determine the exact value of the unknown variable. If the exact value involves a logarithm, use a computational device to also report an approximate value. For instance, if the exact value is y = \log_{10}(2), you can also note that y \approx 0.301.
+
+
+
+
+
+
+ 10^t = 0.00001
+
+
+
+
+
+ t = \log_{10}(0.00001) = -5, exactly.
+
+
+
+
+ t = \log_{10}(0.00001) = -5, exactly.
+
+
+
+
+
+
+ \log_{10}(1000000) = t
+
+
+
+
+
+ t = \log_{10}(1000000) = 6, exactly.
+
+
+
+
+ t = \log_{10}(1000000) = 6, exactly.
+
+
+
+
+
+
+ 10^t = 37
+
+
+
+
+
+ t = \log_{10}(37) \approx 1.568.
+
+
+
+
+ t = \log_{10}(37) \approx 1.568.
+
+
+
+
+
+
+ \log_{10}(y) = 1.375
+
+
+
+
+
+ y = 10^{1.375} \approx 23.71.
+
+
+
+
+ y = 10^{1.375} \approx 23.71.
+
+
+
+
+
+
+ 10^t = 0.04
+
+
+
+
+
+ t = \log_{10}(0.04) \approx -1.398.
+
+
+
+
+ t = \log_{10}(0.04) \approx -1.398.
+
+
+
+
+
+
+ 3 \cdot 10^t + 11 = 147
+
+
+
+
t = \log_{10}\left(\frac{136}{3}\right) \approx 1.656
+
+
+ 3 \cdot 10^t + 11 = 147 gives 10^t = \frac{136}{3}, so t = \log_{10}\left(\frac{136}{3}\right) \approx 1.656.
+
+
+
+
+
+
+ 2\log_{10}(y) + 5 = 1
+
+
+
+
y = 10^{-2} = 0.01, exactly
+
+
+ 2\log_{10}(y) + 5 = 1 gives \log_{10}(y) = -2, so y = 10^{-2} = 0.01, exactly.
+
- Solve each of the following equations exactly and then find an estimate that is accurate to 5 decimal places.
-
-
-
-
-
-
- 3^t = 5
-
-
-
-
-
-
- From 3^t = 5, we take the natural log of both sides to get t \ln 3 = \ln 5, so t = \dfrac{\ln 5}{\ln 3} \approx 1.46497.
-
-
-
-
-
-
- 4 \cdot 2^t - 2 = 3
-
-
-
-
-
-
- From 4 \cdot 2^t - 2 = 3, we get 2^t = \frac{5}{4}, so t \ln 2 = \ln\!\left(\frac{5}{4}\right) and t = \dfrac{\ln 5 - \ln 4}{\ln 2} \approx 0.32193.
-
-
-
-
-
-
- 3.7 \cdot (0.9)^{0.3t} + 1.5 = 2.1
-
-
-
-
-
-
- From 3.7 \cdot (0.9)^{0.3t} + 1.5 = 2.1, we get (0.9)^{0.3t} = \frac{0.6}{3.7}, so 0.3t \ln(0.9) = \ln\!\left(\frac{0.6}{3.7}\right) and t = \dfrac{\ln(0.6) - \ln(3.7)}{0.3\ln(0.9)} \approx 57.55.
-
-
-
-
-
-
- 72 - 30(0.7)^{0.05t} = 60
-
-
-
-
-
-
- From 72 - 30(0.7)^{0.05t} = 60, we get (0.7)^{0.05t} = \frac{12}{30} = \frac{2}{5}, so 0.05t \ln(0.7) = \ln\!\left(\frac{2}{5}\right) and t = \dfrac{\ln(12) - \ln(30)}{0.05\ln(0.7)} \approx 51.38.
-
-
-
-
-
-
- \ln(t) = -2
-
-
-
-
-
-
- From \ln(t) = -2, we get t = e^{-2} \approx 0.13534.
-
-
-
-
-
-
- 3 + 2\log_{10}(t) = 3.5
-
-
-
-
-
-
- From 3 + 2\log_{10}(t) = 3.5, we get \log_{10}(t) = 0.25, so t = 10^{0.25} \approx 1.77828.
-
+ Solve each of the following equations exactly and then find an estimate that is accurate to 5 decimal places.
+
+
+
+
+
+
+ 3^t = 5
+
+
+
+
t = \dfrac{\ln 5}{\ln 3} \approx 1.46497
+
+
+ From 3^t = 5, we take the natural log of both sides to get t \ln 3 = \ln 5, so t = \dfrac{\ln 5}{\ln 3} \approx 1.46497.
+
+
+
+
+
+
+ 4 \cdot 2^t - 2 = 3
+
+
+
+
t = \dfrac{\ln 5 - \ln 4}{\ln 2} \approx 0.32193
+
+
+ From 4 \cdot 2^t - 2 = 3, we get 2^t = \frac{5}{4}, so t \ln 2 = \ln\left(\frac{5}{4}\right) and t = \dfrac{\ln 5 - \ln 4}{\ln 2} \approx 0.32193.
+
+
+
+
+
+
+ 3.7 \cdot (0.9)^{0.3t} + 1.5 = 2.1
+
+
+
+
t = \dfrac{\ln(0.6) - \ln(3.7)}{0.3\ln(0.9)} \approx 57.55
+
+
+ From 3.7 \cdot (0.9)^{0.3t} + 1.5 = 2.1, we get (0.9)^{0.3t} = \frac{0.6}{3.7}, so 0.3t \ln(0.9) = \ln\left(\frac{0.6}{3.7}\right) and t = \dfrac{\ln(0.6) - \ln(3.7)}{0.3\ln(0.9)} \approx 57.55.
+
+
+
+
+
+
+ 72 - 30(0.7)^{0.05t} = 60
+
+
+
+
t = \dfrac{\ln(12) - \ln(30)}{0.05\ln(0.7)} \approx 51.38
+
+
+ From 72 - 30(0.7)^{0.05t} = 60, we get (0.7)^{0.05t} = \frac{12}{30} = \frac{2}{5}, so 0.05t \ln(0.7) = \ln\!\left(\frac{2}{5}\right) and t = \dfrac{\ln(12) - \ln(30)}{0.05\ln(0.7)} \approx 51.38.
+
+
+
+
+
+
+ \ln(t) = -2
+
+
+
+
t = e^{-2} \approx 0.13534
+
+
+ From \ln(t) = -2, we get t = e^{-2} \approx 0.13534.
+
+
+
+
+
+
+ 3 + 2\log_{10}(t) = 3.5
+
+
+
+
t = 10^{0.25} \approx 1.77828
+
+
+ From 3 + 2\log_{10}(t) = 3.5, we get \log_{10}(t) = 0.25, so t = 10^{0.25} \approx 1.77828.
+
- Let E(t) = e^t and N(y) = \ln(y) be the natural exponential function and the natural logarithm function, respectively.
-
-
-
-
-
-
- What are the domain and range of E?
-
-
-
-
-
-
- The domain of E(t) = e^t is all real numbers, and its range is all positive real numbers.
-
-
-
-
-
-
- What are the domain and range of N?
-
-
-
-
-
-
- The domain of N(y) = \ln(y) is all positive real numbers, and its range is all real numbers.
-
-
-
-
-
-
- What can you say about \ln(e^t) for every real number t?
-
-
-
-
-
-
- For every real number t, \ln(e^t) = t.
-
-
-
-
-
-
- What can you say about e^{\ln(y)} for every positive real number y?
-
-
-
-
-
-
- For every positive real number y, e^{\ln(y)} = y.
-
-
-
-
-
-
- Complete the following tables with both exact and approximate values of E and N. Then, plot the corresponding ordered pairs from each table on the axes provided and connect the points in an intuitive way. When you plot the ordered pairs on the axes, in both cases view the first line of the table as generating values on the horizontal axis and the second line of the table as producing values on the vertical axis. Note that when we take this perspective for plotting the data in the table for N, we are viewing N as a function of t, writing N(t) = \ln(t) in order to plot the function on the t-y axes; label each ordered pair you plot appropriately.
-
+ Let E(t) = e^t and N(y) = \ln(y) be the natural exponential function and the natural logarithm function, respectively.
+
+
+
+
+
+
+ What are the domain and range of E?
+
+
+
+
+
+ The domain of E(t) = e^t is all real numbers, and its range is all positive real numbers.
+
+
+
+
+ The domain of E(t) = e^t is all real numbers, and its range is all positive real numbers.
+
+
+
+
+
+
+ What are the domain and range of N?
+
+
+
+
+
+ The domain of N(y) = \ln(y) is all positive real numbers, and its range is all real numbers.
+
+
+
+
+ The domain of N(y) = \ln(y) is all positive real numbers, and its range is all real numbers.
+
+
+
+
+
+
+ What can you say about \ln(e^t) for every real number t?
+
+
+
+
+
+ For every real number t, \ln(e^t) = t.
+
+
+
+
+ For every real number t, \ln(e^t) = t.
+
+
+
+
+
+
+ What can you say about e^{\ln(y)} for every positive real number y?
+
+
+
+
+
+ For every positive real number y, e^{\ln(y)} = y.
+
+
+
+
+ For every positive real number y, e^{\ln(y)} = y.
+
+
+
+
+
+
+ Complete the following tables with both exact and approximate values of E and N. Then, plot the corresponding ordered pairs from each table on the axes provided and connect the points in an intuitive way. When you plot the ordered pairs on the axes, in both cases view the first line of the table as generating values on the horizontal axis and the second line of the table as producing values on the vertical axis. Note that when we take this perspective for plotting the data in the table for N, we are viewing N as a function of t, writing N(t) = \ln(t) in order to plot the function on the t-y axes; label each ordered pair you plot appropriately.
+
- In the questions that follow, we compare and contrast the properties and behaviors of exponential and logarithmic functions.
-
-
-
-
-
-
- Let f(t) = 1 - e^{-(t-1)} and g(t) = \ln(t). Plot each function on the same set of coordinate axes. What properties do the two functions have in common? For what properties do the two functions differ? Consider each function's domain, range, t-intercept, y-intercept, increasing/decreasing behavior, concavity, and long-term behavior.
-
-
-
-
-
-
- Both f(t) = 1 - e^{-(t-1)} and g(t) = \ln(t) are always increasing and concave down, and both have a t-intercept at t = 1. The domain of f is all real numbers, while the domain of g is all positive real numbers. The range of f is the interval (-\infty, 1), while the range of g is all real numbers. The y-intercept of f is f(0) = 1 - e \approx -1.718, while g has no y-intercept since t = 0 is not in its domain. As t \to \infty, f(t) \to 1 (bounded), while g(t) \to \infty (unbounded).
-
-
-
-
-
-
- Let h(t) = a - be^{-k(t-c)}, where a, b, c, and k are positive constants. Describe h as a transformation of the function E(t) = e^t.
-
-
-
-
-
-
- The function h(t) = a - be^{-k(t-c)} is a transformation of E(t) = e^t in which E(t) is reflected over the horizontal axis and vertically stretched by a factor of b, horizontally compressed by a factor of k (with a horizontal reflection), shifted horizontally to the right by c units, and shifted vertically up by a units.
-
-
-
-
-
-
- Let r(t) = a + b\ln(t-c), where a, b, and c are positive constants. Describe r as a transformation of the function L(t) = \ln(t).
-
-
-
-
-
-
- The function r(t) = a + b\ln(t - c) is a transformation of L(t) = \ln(t) that is vertically stretched by a factor of b, shifted horizontally to the right by c units, and shifted vertically up by a units.
-
-
-
-
-
-
- Data for the height of a tree is given in the following table;
- time t is measured in years and height is given in feet. At http://gvsu.edu/s/0yy, you can find a Desmos worksheet with this data already input.
-
- Do you think this data is better modeled by a logarithmic function of form p(t) = a + b\ln(t-c) or by an exponential function of form q(t) = m + ne^{-rt}. Provide reasons based in how the data appears and how you think a tree grows, as well as by experimenting with sliders appropriately in Desmos. (Note: you may need to adjust the upper and lower bounds of several of the sliders in order to match the data well.)
-
-
-
-
-
-
- Experimenting with Desmos, the exponential model q(t) = m + ne^{-rt} provides a good fit to the data; for instance, q(t) = 20.3 - 20.8e^{-0.35t} fits well. A logarithmic model is not as effective because the data levels off at a horizontal asymptote, which is consistent with an exponential model of the form a + be^{-kt} but not with a logarithmic model whose values grow without bound.
-
+ In the questions that follow, we compare and contrast the properties and behaviors of exponential and logarithmic functions.
+
+
+
+
+
+
+ Let f(t) = 1 - e^{-(t-1)} and g(t) = \ln(t). Plot each function on the same set of coordinate axes. What properties do the two functions have in common? For what properties do the two functions differ? Consider each function's domain, range, t-intercept, y-intercept, increasing/decreasing behavior, concavity, and long-term behavior.
+
+
+
+
+
+ Both f(t) = 1 - e^{-(t-1)} and g(t) = \ln(t) are always increasing and concave down, and both have a t-intercept at t = 1. The domain of f is all real numbers, while the domain of g is all positive real numbers. The range of f is the interval (-\infty, 1), while the range of g is all real numbers. The y-intercept of f is f(0) = 1 - e \approx -1.718, while g has no y-intercept since t = 0 is not in its domain. As t \to \infty, f(t) \to 1 (bounded), while g(t) \to \infty (unbounded).
+
+
+
+
+ Both f(t) = 1 - e^{-(t-1)} and g(t) = \ln(t) are always increasing and concave down, and both have a t-intercept at t = 1. The domain of f is all real numbers, while the domain of g is all positive real numbers. The range of f is the interval (-\infty, 1), while the range of g is all real numbers. The y-intercept of f is f(0) = 1 - e \approx -1.718, while g has no y-intercept since t = 0 is not in its domain. As t \to \infty, f(t) \to 1 (bounded), while g(t) \to \infty (unbounded).
+
+
+
+
+
+
+ Let h(t) = a - be^{-k(t-c)}, where a, b, c, and k are positive constants. Describe h as a transformation of the function E(t) = e^t.
+
+
+
+
+
+ The function h(t) = a - be^{-k(t-c)} is a transformation of E(t) = e^t in which E(t) is reflected over the horizontal axis and vertically stretched by a factor of b, horizontally compressed by a factor of k (with a horizontal reflection), shifted horizontally to the right by c units, and shifted vertically up by a units.
+
+
+
+
+ The function h(t) = a - be^{-k(t-c)} is a transformation of E(t) = e^t in which E(t) is reflected over the horizontal axis and vertically stretched by a factor of b, horizontally compressed by a factor of k (with a horizontal reflection), shifted horizontally to the right by c units, and shifted vertically up by a units.
+
+
+
+
+
+
+ Let r(t) = a + b\ln(t-c), where a, b, and c are positive constants. Describe r as a transformation of the function L(t) = \ln(t).
+
+
+
+
+
+ The function r(t) = a + b\ln(t - c) is a transformation of L(t) = \ln(t) that is vertically stretched by a factor of b, shifted horizontally to the right by c units, and shifted vertically up by a units.
+
+
+
+
+ The function r(t) = a + b\ln(t - c) is a transformation of L(t) = \ln(t) that is vertically stretched by a factor of b, shifted horizontally to the right by c units, and shifted vertically up by a units.
+
+
+
+
+
+
+ Data for the height of a tree is given in the following table;
+ time t is measured in years and height is given in feet. At http://gvsu.edu/s/0yy, you can find a Desmos worksheet with this data already input.
+
+ Do you think this data is better modeled by a logarithmic function of form p(t) = a + b\ln(t-c) or by an exponential function of form q(t) = m + ne^{-rt}. Provide reasons based in how the data appears and how you think a tree grows, as well as by experimenting with sliders appropriately in Desmos. (Note: you may need to adjust the upper and lower bounds of several of the sliders in order to match the data well.)
+
+
+
+
+
+ Experimenting with Desmos, the exponential model q(t) = m + ne^{-rt} provides a good fit to the data; for instance, q(t) = 20.3 - 20.8e^{-0.35t} fits well. A logarithmic model is not as effective because the data levels off at a horizontal asymptote, which is consistent with an exponential model of the form a + be^{-kt} but not with a logarithmic model whose values grow without bound.
+
+
+
+
+ Experimenting with Desmos, the exponential model q(t) = m + ne^{-rt} provides a good fit to the data; for instance, q(t) = 20.3 - 20.8e^{-0.35t} fits well. A logarithmic model is not as effective because the data levels off at a horizontal asymptote, which is consistent with an exponential model of the form a + be^{-kt} but not with a logarithmic model whose values grow without bound.
+
- For each of the following functions, without using graphing technology, determine whether the function is
-
-
-
- always increasing or always decreasing;
-
-
-
-
- always concave up or always concave down; and
-
-
-
-
- increasing without bound, decreasing without bound, or increasing/decreasing toward a finite value.
-
-
-
- In addition, state the y-intercept and the range of the function. For each function, write a sentence that explains your thinking and sketch a rough graph of how the function appears.
-
-
-
-
-
- p(t) = 4372 (1.000235)^t + 92856
-
-
-
-
-
-
- The function p(t) = 4372(1.000235)^t + 92856 is always increasing, always concave up, and increasing without bound. The y-intercept is p(0) = 97{,}228 and the range is (92856, \infty).
-
-
-
-
-
-
- q(t) = 27931 (0.97231)^t + 549786
-
-
-
-
-
-
- The function q(t) = 27931(0.97231)^t + 549786 is always decreasing, always concave up, and decreasing toward 549786. The y-intercept is q(0) = 577{,}717 and the range is (549786, \infty).
-
-
-
-
-
-
- r(t) = -17398 (0.85234)^t
-
-
-
-
-
-
- The function r(t) = -17398(0.85234)^t is always increasing, always concave down, and increasing toward 0. The y-intercept is r(0) = -17398 and the range is (-\infty, 0).
-
-
-
-
-
-
- s(t) = -17398 (0.85234)^t + 19411
-
-
-
-
-
-
- The function s(t) = -17398(0.85234)^t + 19411 is always increasing, always concave down, and increasing toward the value 19411. The y-intercept is s(0) = 2013 and the range is (-\infty, 19411).
-
-
-
-
-
-
- u(t) = -7522 (1.03817)^t
-
-
-
-
-
-
- The function u(t) = -7522(1.03817)^t is always decreasing, always concave down, and decreasing without bound. The y-intercept is u(0) = -7522 and the range is (-\infty, 0).
-
-
-
-
-
-
- v(t) = -7522 (1.03817)^t + 6731
-
-
-
-
-
-
- The function v(t) = -7522(1.03817)^t + 6731 is always decreasing, always concave down, and decreasing without bound. The y-intercept is v(0) = -791 and the range is (-\infty, 6731).
-
+ For each of the following functions, without using graphing technology, determine whether the function is
+
+
+
+ always increasing or always decreasing;
+
+
+
+
+ always concave up or always concave down; and
+
+
+
+
+ increasing without bound, decreasing without bound, or increasing/decreasing toward a finite value.
+
+
+
+ In addition, state the y-intercept and the range of the function. For each function, write a sentence that explains your thinking and sketch a rough graph of how the function appears.
+
+
+
+
+
+ p(t) = 4372 (1.000235)^t + 92856
+
+
+
+
+
+ The function p(t) = ab^t+c = 4372(1.000235)^t + 92856 has a \gt 0 and b \gt 1, so p(t) is always increasing, always concave up, and increasing without bound and has been shifted vertically up by 92856. The y-intercept is p(0) = 4372+92856= 97228 and the range is (92856, \infty).
+
+
+
+
+ The function p(t) = ab^t+c = 4372(1.000235)^t + 92856 has a \gt 0 and b \gt 1, so p(t) is always increasing, always concave up, and increasing without bound and has been shifted vertically up by 92856. The y-intercept is p(0) = 4372+92856= 97228 and the range is (92856, \infty).
+
+
+
+
+
+
+ q(t) = 27931 (0.97231)^t + 549786
+
+
+
+
+
+ The function q(t) = 27931(0.97231)^t + 549786 is always decreasing, always concave up, and decreasing toward 549786. The y-intercept is q(0) = 577717 and the range is (549786, \infty).
+
+
+
+
+ The function q(t) = 27931(0.97231)^t + 549786 is always decreasing, always concave up, and decreasing toward 549786. The y-intercept is q(0) = 577717 and the range is (549786, \infty).
+
+
+
+
+
+
+ r(t) = -17398 (0.85234)^t
+
+
+
+
+
+ The function r(t) = -17398(0.85234)^t is always increasing, always concave down, and increasing toward 0. The y-intercept is r(0) = -17398 and the range is (-\infty, 0).
+
+
+
+
+ The function r(t) = -17398(0.85234)^t is always increasing, always concave down, and increasing toward 0. The y-intercept is r(0) = -17398 and the range is (-\infty, 0).
+
+
+
+
+
+
+ s(t) = -17398 (0.85234)^t + 19411
+
+
+
+
+
+ The function s(t) = -17398(0.85234)^t + 19411 is always increasing, always concave down, and increasing toward the value 19411. The y-intercept is s(0) = 2013 and the range is (-\infty, 19411).
+
+
+
+
+ The function s(t) = -17398(0.85234)^t + 19411 is always increasing, always concave down, and increasing toward the value 19411. The y-intercept is s(0) = 2013 and the range is (-\infty, 19411).
+
+
+
+
+
+
+ u(t) = -7522 (1.03817)^t
+
+
+
+
+
+ The function u(t) = -7522(1.03817)^t is always decreasing, always concave down, and decreasing without bound. The y-intercept is u(0) = -7522 and the range is (-\infty, 0).
+
+
+
+
+ The function u(t) = -7522(1.03817)^t is always decreasing, always concave down, and decreasing without bound. The y-intercept is u(0) = -7522 and the range is (-\infty, 0).
+
+
+
+
+
+
+ v(t) = -7522 (1.03817)^t + 6731
+
+
+
+
+
+ The function v(t) = -7522(1.03817)^t + 6731 is always decreasing, always concave down, and decreasing without bound. The y-intercept is v(0) = -791 and the range is (-\infty, 6731).
+
+
+
+
+ The function v(t) = -7522(1.03817)^t + 6731 is always decreasing, always concave down, and decreasing without bound. The y-intercept is v(0) = -791 and the range is (-\infty, 6731).
+
- A potato initially at room temperature (68^\circ) is placed in an oven (at 350^\circ) at time t = 0.
- It is known that the potato's temperature at time t is given by the function
- F(t) = a - b(0.98)^t for some positive constants a and b,
- where F is measured in degrees Fahrenheit and t is time in minutes.
-
-
-
-
-
-
- What is the numerical value of F(0)?
- What does this tell you about the value of a - b?
-
-
-
-
-
-
- The numerical value of F(0) is 68^\circ, because the potato begins at room temperature. Therefore, F(0) = a - b(0.98)^0 = a - b = 68.
-
-
-
-
-
-
- Based on the context of the problem,
- what should be the long-range behavior of the function F(t)?
- Use this fact along with the behavior of
- (0.98)^t to determine the value of a. Write a sentence to explain your thinking.
-
-
-
-
-
-
- The long-range behavior of F(t) should reflect the potato approaching the oven temperature of 350^\circ. The expression (0.98)^t approaches zero as t gets large, so F(t) \to a. Therefore a = 350^\circ.
-
-
-
-
-
-
- What is the value of b? Why?
-
-
-
-
-
-
- The value of b is a - 68 = 350 - 68 = 282^\circ, since we established that a - b = 68 and a = 350.
-
-
-
-
-
-
- Check your work above by plotting the function F using graphing technology in an appropriate window.
- Record your results on the axes provided below, labeling the scale on the axes.
- Then, use the graph to estimate the time at which the potato's temperature reaches 325 degrees.
-
-
-
ADD ALT TEXT TO THIS IMAGE
-
-
-
-
-
-
- The function F(t) = 350 - 282(0.98)^t is shown below. The potato reaches 325 degrees after about 120 minutes of baking in the 350 degree oven.
-
-
-
-
-
-
-
- How can we view the function F(t) = a - b(0.98)^t as a transformation of the parent function f(t) = (0.98)^t? Explain.
-
-
-
-
-
-
- We can view F(t) = a - b(0.98)^t as a transformation of the parent function f(t) = (0.98)^t by first stretching f vertically by a factor of b, then reflecting over the t-axis, and finally shifting vertically up by a units.
-
+ A potato initially at room temperature (68^\circ) is placed in an oven (at 350^\circ) at time t = 0. It turns out that the potato's temperature will eventually stabilize near the boiling point of water, or 212^\circ.
+ It is known that the potato's temperature at time t is given by the function
+ F(t) = a - b(0.98)^t for some positive constants a and b,
+ where F is measured in degrees Fahrenheit and t is time in minutes.
+
+
+
+
+
+
+ What is the numerical value of F(0)?
+ What does this tell you about the value of a - b?
+
+
+
+
+
+ F(0)=68^\circ, and a - b = 68
+
+
+
+
+ The numerical value of F(0) is 68^\circ, because the potato begins at room temperature. Therefore, F(0) = a - b(0.98)^0 = a - b = 68.
+
+
+
+
+
+
+ Based on the context of the problem,
+ what should be the long-range behavior of the function F(t)?
+ Use this fact along with the behavior of
+ (0.98)^t to determine the value of a. Write a sentence to explain your thinking.
+
+
+
+
+
+ The expression (0.98)^t approaches zero as t gets large, so F(t) \to a, and therefore a = 212^\circ.
+
+
+
+
+ The long-range behavior of F(t) should reflect the potato approaching the boiling point of water, or 212^\circ. The expression (0.98)^t approaches zero as t gets large, so F(t) \to a. Therefore a = 212^\circ.
+
+
+
+
+
+
+ What is the value of b? Why?
+
+
+
+
b=144^\circ
+
+
+ The value of b is a - 68 = 212 - 68 = 144^\circ, since we established that a - b = 68 and a = 212.
+
+
+
+
+
+
+ Check your work above by plotting the function F using graphing technology in an appropriate window.
+ Record your results on the axes provided below, labeling the scale on the axes.
+ Then, use the graph to estimate the time at which the potato's temperature reaches 180 degrees.
+
+
+
+
+
+
+
Blank axes for plotting F as a function of t.
+
+
+
+
+
+
+
+
+ The function F(t) = 212 - 144(0.98)^t is shown below. The potato reaches 180 degrees after about 75 minutes of baking in the 350 degree oven.
+
+
+
+
+
+
Graph of F(t)=212-144(0.98)^t.
+
+
+
+
+ The function F(t) = 212 - 144(0.98)^t is shown below. The potato reaches 180 degrees after about 75 minutes of baking in the 350 degree oven.
+
+
+
+
+
+
Graph of F(t)=212-144(0.98)^t.
+
+
+
+
+
+
+
+ How can we view the function F(t) = a - b(0.98)^t as a transformation of the parent function f(t) = (0.98)^t? Explain.
+
+
+
+
+
+ We can view F(t) = a - b(0.98)^t as a transformation of the parent function f(t) = (0.98)^t by first stretching f vertically by a factor of b, then reflecting over the t-axis, and finally shifting vertically up by a units.
+
+
+
+
+ We can view F(t) = a - b(0.98)^t as a transformation of the parent function f(t) = (0.98)^t by first stretching f vertically by a factor of b, then reflecting over the t-axis, and finally shifting vertically up by a units.
+
- A can of soda (at room temperature) is placed in a refrigerator at time t = 0 (in minutes)
- and its temperature, F(t), in degrees Fahrenheit, is computed at regular intervals.
- Based on the data, a model is formulated for the object's temperature, given by
-
- F(t) = 42 + 30(0.95)^{t}
- .
-
-
-
-
-
-
- Consider the simpler (parent) function p(t) = (0.95)^t.
- How do you expect the graph of this function to appear?
- How will it behave as time increases?
- Without using graphing technology,
- sketch a rough graph of p and write a sentence of explanation.
-
-
-
-
-
-
- The graph of p(t) = (0.95)^t is always decreasing, always concave up, and decreasing toward zero.
-
-
-
-
-
-
-
- For the slightly more complicated function r(t) = 30 (0.95)^{t}, how do you expect this function to look in comparison to p? What is the long-range behavior of this function as t increases?
- Without using graphing technology, sketch a rough graph of r and write a sentence of explanation.
-
-
-
-
-
-
- The function r(t) = 30(0.95)^t is similar to p(t) but stretched vertically by a factor of 30. It is always decreasing, always concave up, and decreasing toward zero.
-
-
-
-
-
-
-
- Finally, how do you expect the graph of F(t) = 42 + 30(0.95)^{t} to appear?
- Why? First sketch a rough graph without graphing technology, and then use technology to check your thinking
- and report an accurate, labeled graph on the axes provided below.
-
-
-
ADD ALT TEXT TO THIS IMAGE
-
-
-
-
-
-
- Since F(t) = 42 + 30(0.95)^t has a vertical shift of 42 up, it remains always decreasing, always concave up, but decreasing toward 42.
-
-
-
-
-
-
-
- What is the temperature of the refrigerator?
- What is the room temperature of the surroundings outside the refrigerator?
- Why?
-
-
-
-
-
-
- Because F(t) decreases toward 42, that is the temperature of the refrigerator. Because the function starts at 42 + 30 = 72, that is the room temperature of the surroundings outside the refrigerator.
-
-
-
-
-
-
- Determine the average rate of change of F on the intervals [10,20],
- [20,30], and [30,40].
- Write at least two careful sentences that explain the meaning of the values you found, including units,
- and discuss any overall trend in how the average rate of change is changing.
-
-
-
-
-
-
-
- AV_{[10,20]} &= \frac{30(0.95^{20} - 0.95^{10})}{20-10} \approx -0.721 \text{ degrees per minute} \\
- AV_{[20,30]} &= \frac{30(0.95^{30} - 0.95^{20})}{30-20} \approx -0.432 \text{ degrees per minute} \\
- AV_{[30,40]} &= \frac{30(0.95^{40} - 0.95^{30})}{40-30} \approx -0.258 \text{ degrees per minute}
-
- The average rate of change is always negative and is decreasing in magnitude as t increases. This is consistent with a function that is always decreasing and always concave up. In context, the soda is cooling more and more slowly as time goes on.
-
+ A can of soda (at room temperature) is placed in a refrigerator at time t = 0 (in minutes)
+ and its temperature, F(t), in degrees Fahrenheit, is computed at regular intervals.
+ Based on the data, a model is formulated for the object's temperature, given by
+
+ F(t) = 42 + 30(0.95)^{t}
+ .
+
+
+
+
+
+
+ Consider the simpler (parent) function p(t) = (0.95)^t.
+ How do you expect the graph of this function to appear?
+ How will it behave as time increases?
+ Without using graphing technology,
+ sketch a rough graph of p and write a sentence of explanation.
+
+
+
+
+
+ The graph of p(t) = (0.95)^t is always decreasing, always concave up, and decreasing toward zero.
+
+
+
graph of p(t) = (0.95)^t
+
+
+
+ The graph of p(t) = (0.95)^t is always decreasing, always concave up, and decreasing toward zero.
+
+
+
graph of p(t) = (0.95)^t
+
+
+
+
+
+ For the slightly more complicated function r(t) = 30 (0.95)^{t}, how do you expect this function to look in comparison to p? What is the long-range behavior of this function as t increases?
+ Without using graphing technology, sketch a rough graph of r and write a sentence of explanation.
+
+
+
+
+
+ The function r(t) = 30(0.95)^t is similar to p(t) but stretched vertically by a factor of 30. It is always decreasing, always concave up, and decreasing toward zero.
+
+
+
graph of r(t) = 30 (0.95)^t
+
+
+
+ The function r(t) = 30(0.95)^t is similar to p(t) but stretched vertically by a factor of 30. It is always decreasing, always concave up, and decreasing toward zero.
+
+
+
graph of r(t) = 30 (0.95)^t
+
+
+
+
+
+ Finally, how do you expect the graph of F(t) = 42 + 30(0.95)^{t} to appear?
+ Why? First sketch a rough graph without graphing technology, and then use technology to check your thinking
+ and report an accurate, labeled graph on the axes provided below.
+
+
+
+
+
+
+
+
Blank axes for plotting F as a function of t.
+
+
+
+
+
+
+
+ Since F(t) = 42 + 30(0.95)^t has a vertical shift of 42 up, it remains always decreasing, always concave up, but decreasing toward 42.
+
+
+
graph of F(t) = 42+ 30 (0.95)^t
+
+
+
+ Since F(t) = 42 + 30(0.95)^t has a vertical shift of 42 up, it remains always decreasing, always concave up, but decreasing toward 42.
+
+
+
graph of F(t) = 42+ 30 (0.95)^t
+
+
+
+
+
+ What is the temperature of the refrigerator?
+ What is the room temperature of the surroundings outside the refrigerator?
+ Why?
+
+
+
+
+
+ The temperature of the refrigerator is 42. The room temperature of the surroundings outside the refrigerator is 72.
+
+
+
+
+ Because F(t) decreases toward 42, that is the temperature of the refrigerator. Because the function starts at 42 + 30 = 72, that is the room temperature of the surroundings outside the refrigerator.
+
+
+
+
+
+
+ Determine the average rate of change of F on the intervals [10,20],
+ [20,30], and [30,40].
+ Write at least two careful sentences that explain the meaning of the values you found, including units,
+ and discuss any overall trend in how the average rate of change is changing.
+
+
+
+
+
+
+ AV_{[10,20]} & \approx -0.721 \text{ degrees per minute} \\
+ AV_{[20,30]} & \approx -0.432 \text{ degrees per minute} \\
+ AV_{[30,40]} & \approx -0.258 \text{ degrees per minute}
+
+ The average rate of change is always negative and is decreasing in magnitude as t increases. This is consistent with a function that is always decreasing and always concave up. In context, the soda is cooling more and more slowly as time goes on.
+
+
+
+
+
+ AV_{[10,20]} &= \frac{30(0.95^{20} - 0.95^{10})}{20-10} \approx -0.721 \text{ degrees per minute} \\
+ AV_{[20,30]} &= \frac{30(0.95^{30} - 0.95^{20})}{30-20} \approx -0.432 \text{ degrees per minute} \\
+ AV_{[30,40]} &= \frac{30(0.95^{40} - 0.95^{30})}{40-30} \approx -0.258 \text{ degrees per minute}
+
+ The average rate of change is always negative and is decreasing in magnitude as t increases. This is consistent with a function that is always decreasing and always concave up. In context, the soda is cooling more and more slowly as time goes on.
+
- A can of soda is initially at room temperature, 72.3^\circ Fahrenheit, and at time t = 0 is placed in a refrigerator set at 37.7^\circ. In addition, we know that after 30 minutes, the soda's temperature has dropped to 59.5^\circ. Let F(t) represent the temperature of the soda in degrees Fahrenheit at time t in minutes.
-
-
-
-
-
-
- Use algebraic reasoning and your understanding of the physical situation to determine the exact values of a,
- c,
- and k in the model F(t) = ae^{-kt}+c.
- Write at least one careful sentence to explain your thinking.
-
-
-
-
-
-
- Since the soda cools toward the refrigerator temperature in the long run, and e^{-kt} \to 0 as t \to \infty, we have c = 37.7. At t = 0, F(0) = a + c = a + 37.7 = 72.3, so a = 34.6. Finally, using F(30) = 59.5: 34.6e^{-30k} + 37.7 = 59.5, so 34.6e^{-30k} = 21.8 and k = -\dfrac{1}{30}\ln\!\left(\dfrac{21.8}{34.6}\right) \approx 0.01540.
-
-
-
-
-
-
- Determine the exact time the object's temperature is 42.4^\circ.
- Clearly show your algebraic work and thinking.
-
- In Desmos, enter the values you found for a, c, and k in order to define the function F. Then, use Desmos to find the average rate of change of F on the interval [25,30].
- What is the meaning
- (with units)
- of this value?
-
-
-
-
-
-
- Using F(t) = 34.6e^{-0.01540t} + 37.7 in Desmos, the average rate of change on [25,30] is approximately \frac{F(30)-F(25)}{5} \approx -0.349 degrees Fahrenheit per minute. This means the soda is cooling at a rate of about 0.349^\circF per minute during the five-minute interval from t = 25 to t = 30.
-
-
-
-
-
-
- If everything stayed the same except the value of F(0),
- and instead F(0) = 65,
- would the value of k be larger or smaller?
- Why?
-
-
-
-
-
-
- If F(0) = 65, then a = 65 - 37.7 = 27.3, which is smaller than the original a = 34.6. Using the same condition F(30) = 59.5: 27.3e^{-30k} = 21.8, giving k \approx 0.00751. This is smaller than the original k \approx 0.0154. With a smaller initial temperature difference between the soda and the refrigerator, the soda cools more slowly, so k is smaller.
-
+ A can of soda is initially at room temperature, 72.3^\circ Fahrenheit, and at time t = 0 is placed in a refrigerator set at 37.7^\circ. In addition, we know that after 30 minutes, the soda's temperature has dropped to 59.5^\circ. Let F(t) represent the temperature of the soda in degrees Fahrenheit at time t in minutes.
+
+
+
+
+
+
+ Use algebraic reasoning and your understanding of the physical situation to determine the exact values of a,
+ c,
+ and k in the model F(t) = ae^{-kt}+c.
+ Write at least one careful sentence to explain your thinking.
+
+
+
+
+
+ Since the soda cools toward the refrigerator temperature in the long run, and e^{-kt} \to 0 as t \to \infty, we have c = 37.7. At t = 0, F(0) = a + c = a + 37.7 = 72.3, so a = 34.6.
+
Finally, using F(30) = 59.5: 34.6e^{-30k} + 37.7 = 59.5, so 34.6e^{-30k} = 21.8 and k = -\dfrac{1}{30}\ln\left(\dfrac{21.8}{34.6}\right) \approx 0.01540.
+
+
+
+
+ Since the soda cools toward the refrigerator temperature in the long run, and e^{-kt} \to 0 as t \to \infty, we have c = 37.7. At t = 0, F(0) = a + c = a + 37.7 = 72.3, so a = 34.6.
+
Finally, using F(30) = 59.5: 34.6e^{-30k} + 37.7 = 59.5, so 34.6e^{-30k} = 21.8 and k = -\dfrac{1}{30}\ln\left(\dfrac{21.8}{34.6}\right) \approx 0.01540.
+
+
+
+
+
+
+ Determine the exact time the object's temperature is 42.4^\circ.
+ Clearly show your algebraic work and thinking.
+
+
+
+
t = -\frac{1}{k}\ln\left(\frac{4.7}{34.6}\right) \approx 131 \text{ minutes.}
+ In Desmos, enter the values you found for a, c, and k in order to define the function F. Then, use Desmos to find the average rate of change of F on the interval [25,30].
+ What is the meaning
+ (with units)
+ of this value?
+
+
+
+
+
+ Using F(t) = 34.6e^{-0.01540t} + 37.7 in Desmos, the average rate of change on [25,30] is approximately \frac{F(30)-F(25)}{5} \approx -0.349 degrees Fahrenheit per minute. This means the soda is cooling at a rate of about 0.349^\circF per minute during the five-minute interval from t = 25 to t = 30.
+
+
+
+
+ Using F(t) = 34.6e^{-0.01540t} + 37.7 in Desmos, the average rate of change on [25,30] is approximately \frac{F(30)-F(25)}{5} \approx -0.349 degrees Fahrenheit per minute. This means the soda is cooling at a rate of about 0.349^\circF per minute during the five-minute interval from t = 25 to t = 30.
+
+
+
+
+
+
+ If everything stayed the same except the value of F(0),
+ and instead F(0) = 65,
+ would the value of k be larger or smaller?
+ Why?
+
+
+
+
k \approx 0.00751. With a smaller initial temperature difference between the soda and the refrigerator, the soda cools more slowly, so k is smaller.
+
+
+ If F(0) = 65, then a = 65 - 37.7 = 27.3, which is smaller than the original a = 34.6. Using the same condition F(30) = 59.5: 27.3e^{-30k} = 21.8, giving k \approx 0.00751. This is smaller than the original k \approx 0.0154. With a smaller initial temperature difference between the soda and the refrigerator, the soda cools more slowly, so k is smaller.
+
- In Desmos, define P(t) = \frac{A}{1 + Me^{-kt}} and accept sliders for A, M, and k. Set the slider ranges for these parameters as follows: 0.01 \le A \le 10; 0.01 \le M \le 10; 0.01 \le k \le 5.
-
-
-
-
-
-
- Sketch a typical graph of P(t) on the axes provided and write several sentences to explain the effects of A, M, and k on the graph of P.
-
-
-
ADD ALT TEXT TO THIS IMAGE
-
-
-
-
-
-
- A typical graph of P(t) = \dfrac{A}{1 + Me^{-kt}} has an S-shape (sigmoidal curve) that rises from near 0 as t \to -\infty to the value A as t \to \infty. The parameter A sets the upper asymptote (the carrying capacity). The parameter M affects the y-intercept: since P(0) = \frac{A}{1+M}, larger M gives a lower initial value, producing an effect similar to a horizontal shift. The parameter k controls the steepness of the transition from the lower asymptote to the upper asymptote.
-
-
-
-
-
-
-
- On a typical logistic graph, where does it appear that the population is growing most rapidly? How is this value connected to the carrying capacity, A?
-
-
-
-
-
-
- The population appears to grow most rapidly at the midpoint of the transition from near 0 to the carrying capacity A. This occurs when P(t) = \dfrac{A}{2}, that is, at half the carrying capacity.
-
-
-
-
-
-
- How does the function 1 + Me^{-kt} behave as t decreases without bound? What is the algebraic reason that this occurs?
-
-
-
-
-
-
- As t \to \infty, e^{-kt} \to 0 (since k > 0), so Me^{-kt} \to 0 and therefore 1 + Me^{-kt} \to 1. This causes P(t) = \frac{A}{1 + Me^{-kt}} \to \frac{A}{1} = A, which is why A is the carrying capacity.
-
-
-
-
-
-
- Use your Desmos worksheet to find a logistic function P that has the following properties: P(0) = 2, P(2) = 4, and P(t) approaches 9 as t increases without bound. What are the approximate values of A, M, and k that make the function P fit these criteria?
-
-
-
-
-
-
- Since P(t) \to 9 as t \to \infty, we have A = 9. From P(0) = \frac{9}{1+M} = 2, we get M = \frac{7}{2} = 3.5. From P(2) = \frac{9}{1 + 3.5e^{-2k}} = 4, we get 1 + 3.5e^{-2k} = 2.25, so e^{-2k} = \frac{1.25}{3.5} = \frac{5}{14} and k = \dfrac{1}{2}\ln\!\left(\dfrac{14}{5}\right) \approx 0.516.
-
+ In Desmos, define P(t) = \frac{A}{1 + Me^{-kt}} and accept sliders for A, M, and k. Set the slider ranges for these parameters as follows: 0.01 \le A \le 10; 0.01 \le M \le 10; 0.01 \le k \le 5.
+
+
+
+
+
+
+ Sketch a typical graph of P(t) on the axes provided and write several sentences to explain the effects of A, M, and k on the graph of P.
+
+
+
Blank axes for graphing P as a function of t
+
+
+
+
+
+ A typical graph of P(t) = \dfrac{A}{1 + Me^{-kt}} has an S-shape (sigmoidal curve) that rises from near 0 as t \to -\infty to the value A as t \to \infty. The parameter A sets the upper asymptote. The parameter M affects the y-intercept: since P(0) = \frac{A}{1+M}, larger M gives a lower initial value, producing an effect similar to a horizontal shift. The parameter k controls the steepness of the transition from the lower asymptote to the upper asymptote.
+
+
graph of P(t) for some values of the sliders
+
+
+
+ A typical graph of P(t) = \dfrac{A}{1 + Me^{-kt}} has an S-shape (sigmoidal curve) that rises from near 0 as t \to -\infty to the value A as t \to \infty. The parameter A sets the upper asymptote (the carrying capacity). The parameter M affects the y-intercept: since P(0) = \frac{A}{1+M}, larger M gives a lower initial value, producing an effect similar to a horizontal shift. The parameter k controls the steepness of the transition from the lower asymptote to the upper asymptote.
+
+
graph of P(t) for some values of the sliders
+
+
+
+
+
+ On a typical logistic graph, where does it appear that the population is growing most rapidly? How is this value connected to the carrying capacity, A?
+
+
+
+
+
+ The population appears to grow most rapidly at the midpoint of the transition from near 0 to the carrying capacity A. This occurs when P(t) = \dfrac{A}{2}, that is, at half the carrying capacity.
+
+
+
+
+ The population appears to grow most rapidly at the midpoint of the transition from near 0 to the carrying capacity A. This occurs when P(t) = \dfrac{A}{2}, that is, at half the carrying capacity.
+
+
+
+
+
+
+ How does the function 1 + Me^{-kt} behave as t decreases without bound? What is the algebraic reason that this occurs?
+
+
+
+
P(t) = \frac{A}{1 + Me^{-kt}} \to \frac{A}{1} = A
+
+
+ As t \to \infty, e^{-kt} \to 0 (since k > 0), so Me^{-kt} \to 0 and therefore 1 + Me^{-kt} \to 1. This causes P(t) = \frac{A}{1 + Me^{-kt}} \to \frac{A}{1} = A, which is why A is the carrying capacity.
+
+
+
+
+
+
+ Use your Desmos worksheet to find a logistic function P that has the following properties: P(0) = 2, P(2) = 4, and P(t) approaches 9 as t increases without bound. What are the approximate values of A, M, and k that make the function P fit these criteria?
+
+
+
+
+
+ A = 9, M = \frac{7}{2} = 3.5, and k = \dfrac{1}{2}\ln\!\left(\dfrac{14}{5}\right) \approx 0.516.
+
+
+
+
+ Since P(t) \to 9 as t \to \infty, we have A = 9. From P(0) = \frac{9}{1+M} = 2, we get M = \frac{7}{2} = 3.5. From P(2) = \frac{9}{1 + 3.5e^{-2k}} = 4, we get 1 + 3.5e^{-2k} = 2.25, so e^{-2k} = \frac{1.25}{3.5} = \frac{5}{14} and k = \dfrac{1}{2}\ln\!\left(\dfrac{14}{5}\right) \approx 0.516.
+
- Suppose that a population of animals (measured in thousands) that lives on an island is known to grow according to the logistic model, where t is measured in years. We know the following information: P(0) = 2.45, P(3) = 4.52, and as t increases without bound, P(t) approaches 11.7.
-
-
-
-
-
-
- Determine the exact values of A, M, and k in the logistic model
-
- P(t) = \frac{A}{1 + Me^{-kt}}
- .
- Clearly show your algebraic work and thinking.
-
-
-
-
-
-
- Since P(t) \to 11.7 as t \to \infty, A = 11.7. From P(0) = \frac{11.7}{1+M} = 2.45, we get 1+M = \frac{11.7}{2.45}, so M = \frac{11.7}{2.45} - 1 = \frac{9.25}{2.45} \approx 3.776. Using P(3) = 4.52:
-
- \frac{11.7}{1 + 3.776e^{-3k}} &= 4.52
- 1 + 3.776e^{-3k} &= \frac{11.7}{4.52} \approx 2.589
- e^{-3k} &= \frac{1.589}{3.776} \approx 0.421
- k &= -\frac{1}{3}\ln(0.421) \approx 0.289.
-
-
-
-
-
-
-
- Plot your model from (a) and check that its values match the desired characteristics.
- Then, compute the average rate of change of P on the intervals [0,2],
- [2,4], [4,6], and [6,8].
- What is the meaning (with units) of the values you've found? How is the population growing on these intervals?
-
-
-
-
-
-
- Using P(t) = \dfrac{11.7}{1 + 3.776e^{-0.289t}}, we compute:
- P(0) \approx 2.45, P(2) \approx 3.75, P(4) \approx 5.34, P(6) \approx 7.02, P(8) \approx 8.51.
- The average rates of change are:
- AV_{[0,2]} \approx 0.650, AV_{[2,4]} \approx 0.796, AV_{[4,6]} \approx 0.836, AV_{[6,8]} \approx 0.747 thousand animals per year.
- These represent the average rate at which the population grows on each interval. The population grows most rapidly on [4,6], which makes sense since the midpoint between the asymptotes (0 and 11.7) is 5.85, which falls in this interval.
-
-
-
-
-
-
-
- Find the exact time value when the population will be 10 (thousand). Show your algebraic work and thinking.
-
+ Suppose that a population of animals (measured in thousands) that lives on an island is known to grow according to the logistic model, where t is measured in years. We know the following information: P(0) = 2.45, P(3) = 4.52, and as t increases without bound, P(t) approaches 11.7.
+
+
+
+
+
+
+ Determine the exact values of A, M, and k in the logistic model
+
+ P(t) = \frac{A}{1 + Me^{-kt}}
+ .
+ Clearly show your algebraic work and thinking.
+
+
+
+
+
+ A = 11.7, M \approx 3.776, and
+ k = -\frac{1}{3}\ln(0.421) \approx 0.289.
+
+
+
+
+ Since P(t) \to 11.7 as t \to \infty, A = 11.7. From P(0) = \frac{11.7}{1+M} = 2.45, we get 1+M = \frac{11.7}{2.45}, so M = \frac{11.7}{2.45} - 1 = \frac{9.25}{2.45} \approx 3.776. Using P(3) = 4.52:
+
+ \frac{11.7}{1 + 3.776e^{-3k}} &= 4.52
+ 1 + 3.776e^{-3k} &= \frac{11.7}{4.52} \approx 2.589
+ e^{-3k} &= \frac{1.589}{3.776} \approx 0.421
+ k &= -\frac{1}{3}\ln(0.421) \approx 0.289.
+
+
+
+
+
+
+
+ Plot your model from (a) and check that its values match the desired characteristics.
+ Then, compute the average rate of change of P on the intervals [0,2],
+ [2,4], [4,6], and [6,8].
+ What is the meaning (with units) of the values you've found? How is the population growing on these intervals?
+
+
+
+
+
+ The average rates of change are:
+ AV_{[0,2]} \approx 0.650, AV_{[2,4]} \approx 0.796, AV_{[4,6]} \approx 0.836, AV_{[6,8]} \approx 0.747 thousand animals per year.
+ These represent the average rate at which the population grows on each interval.
+
+
+
+
+ Using P(t) = \dfrac{11.7}{1 + 3.776e^{-0.289t}}, we compute:
+ P(0) \approx 2.45, P(2) \approx 3.75, P(4) \approx 5.34, P(6) \approx 7.02, P(8) \approx 8.51.
+ The average rates of change are:
+ AV_{[0,2]} \approx 0.650, AV_{[2,4]} \approx 0.796, AV_{[4,6]} \approx 0.836, AV_{[6,8]} \approx 0.747 thousand animals per year.
+ These represent the average rate at which the population grows on each interval. The population grows most rapidly on [4,6], which makes sense since the midpoint between the asymptotes (0 and 11.7) is 5.85, which falls in this interval.
+
+
+
+
+
+
+
+ Find the exact time value when the population will be 10 (thousand). Show your algebraic work and thinking.
+
+
+
+
t = -\frac{\ln(0.04503)}{0.289} \approx 10.74 \text{ years.}
- Complete the following table by entering
- \infty,-\infty,0, or no limit
- to identify how the function behaves as either x increases or decreases without bound.
- As much as possible, work to decide the behavior without using a graphing utility.
-
Note: \ln(x) is undefined for x \le 0, so \lim_{x \to -\infty} \ln(x) does not exist.
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ Complete the following table by entering
+ \infty,-\infty,0, or no limit
+ to identify how the function behaves as either x increases or decreases without bound.
+ As much as possible, work to decide the behavior without using a graphing utility.
+
- Point your browser to the Desmos worksheet at http://gvsu.edu/s/0zu.
- In what follows, we explore the behavior of power functions of the form y = x^n where n \ge 1.
-
-
-
-
-
-
- Press the play button next to the slider labeled n.
- Watch at least two loops of the animation and then discuss the trends that you observe.
- Write a careful sentence each for at least two different trends.
-
-
-
-
-
-
- Two trends to observe: (1) As n increases, the graph becomes flatter and wider near x = 0 before rising steeply. (2) When n is even, both ends of the graph go to +\infty (symmetric about the y-axis); when n is odd, the left end goes to -\infty and the right end goes to +\infty.
-
-
-
-
-
-
- Click the icons next to each of the following 8 functions so that you can see all of y = x,
- y = x^2, \ldots,
- y = x^8 graphed at once.
- On the interval 0 \lt x \lt 1,
- how do the graphs of x^a and x^b compare if a \lt b?
-
-
-
-
-
-
- On 0 \lt x \lt 1, the graph of x^b stays closer to 0 than x^a when b \gt a. That is, x^a \gt x^b on (0,1) when a \lt b.
-
-
-
-
-
-
- Uncheck the icons on each of the 8 functions to hide their graphs.
- Click the settings icon to change the domain settings for the axes,
- and change them to -10 \le x \le 10 and -10,000 \le y \le 10,000.
- Play the animation through twice and then discuss the trends that you observe.
- Write a careful sentence each for at least two different trends.
-
-
-
-
-
-
- Two trends to observe: (1) The graphs with even n are symmetric about the y-axis, while odd-n graphs pass through the origin from lower-left to upper-right. (2) For larger n, the function stays very close to 0 on (-1,1) and then rises very steeply outside that interval.
-
-
-
-
-
-
- Click the icons next to each of the following 8 functions so that you can see all of y = x,
- y = x^2, \ldots,
- y = x^8 graphed at once.
- On the interval x \gt 1,
- how do the graphs of x^a and x^b compare if a \lt b?
-
-
-
-
-
-
- On x \gt 1, the graph of x^b lies above the graph of x^a when b \gt a: larger exponents produce faster growth for x \gt 1.
-
+ Point your browser to the Desmos worksheet at http://gvsu.edu/s/0zu.
+ In what follows, we explore the behavior of power functions of the form y = x^n where n \ge 1.
+
+
+
+
+
+
+ Press the play button next to the slider labeled n.
+ Watch at least two loops of the animation and then discuss the trends that you observe.
+ Write a careful sentence each for at least two different trends.
+
+
+
+
+
+ Two trends to observe: (1) As n increases, the graph becomes flatter and wider near x = 0 before rising steeply. (2) When n is even, both ends of the graph go to +\infty (symmetric about the y-axis); when n is odd, the left end goes to -\infty and the right end goes to +\infty.
+
+
+
+
+ Two trends to observe: (1) As n increases, the graph becomes flatter and wider near x = 0 before rising steeply. (2) When n is even, both ends of the graph go to +\infty (symmetric about the y-axis); when n is odd, the left end goes to -\infty and the right end goes to +\infty.
+
+
+
+
+
+
+ Click the icons next to each of the following 8 functions so that you can see all of y = x,
+ y = x^2, \ldots,
+ y = x^8 graphed at once.
+ On the interval 0 \lt x \lt 1,
+ how do the graphs of x^a and x^b compare if a \lt b?
+
+
+
+
+
+ On 0 \lt x \lt 1, the graph of x^b stays closer to 0 than x^a when b \gt a. That is, x^a \gt x^b on (0,1) when a \lt b.
+
+
+
+
+ On 0 \lt x \lt 1, the graph of x^b stays closer to 0 than x^a when b \gt a. That is, x^a \gt x^b on (0,1) when a \lt b.
+
+
+
+
+
+
+ Uncheck the icons on each of the 8 functions to hide their graphs.
+ Click the settings icon to change the domain settings for the axes,
+ and change them to -10 \le x \le 10 and -10,000 \le y \le 10,000.
+ Play the animation through twice and then discuss the trends that you observe.
+ Write a careful sentence each for at least two different trends.
+
+
+
+
+
+ Two trends to observe: (1) The graphs with even n are symmetric about the y-axis, while odd-n graphs pass through the origin from lower-left to upper-right. (2) For larger n, the function stays very close to 0 on (-1,1) and then rises very steeply outside that interval.
+
+
+
+
+ Two trends to observe: (1) The graphs with even n are symmetric about the y-axis, while odd-n graphs pass through the origin from lower-left to upper-right. (2) For larger n, the function stays very close to 0 on (-1,1) and then rises very steeply outside that interval.
+
+
+
+
+
+
+ Click the icons next to each of the following 8 functions so that you can see all of y = x,
+ y = x^2, \ldots,
+ y = x^8 graphed at once.
+ On the interval x \gt 1,
+ how do the graphs of x^a and x^b compare if a \lt b?
+
+
+
+
+
+ On x \gt 1, the graph of x^b lies above the graph of x^a when b \gt a: larger exponents produce faster growth for x \gt 1.
+
+
+
+
+ On x \gt 1, the graph of x^b lies above the graph of x^a when b \gt a: larger exponents produce faster growth for x \gt 1.
+
- Point your browser to the Desmos worksheet at
- http://gvsu.edu/s/0zv.
- In what follows, we explore the behavior of power functions y = x^n where n \le -1.
-
-
-
-
-
-
- Press the play button next to the slider labeled n.
- Watch two loops of the animation and then discuss the trends that you observe.
- Write a careful sentence each for at least two different trends.
-
-
-
-
-
-
- Two trends: (1) For even-n exponents (like x^{-2}, x^{-4}, \ldots), the graph is symmetric about the y-axis and stays positive; for odd exponents (like x^{-1}, x^{-3}, \ldots), the graph is negative for x \lt 0 and positive for x \gt 0. (2) As the exponent becomes more negative, the graph becomes more sharply L-shaped near x = 0.
-
-
-
-
-
-
- Click the icons next to each of the following 8 functions so that you can see all of y = x^{-1},
- y = x^{-2}, \ldots,
- y = x^{-8} graphed at once.
- On the interval 1 \lt x, how do the functions
- x^a and x^b compare if a \lt b? (Be careful with negative numbers here:
- e.g., -3 \lt -2.)
-
-
-
-
-
-
- On x \gt 1, x^a \lt x^b when a \lt b (with both negative). For example, x^{-3} \lt x^{-2} for x \gt 1, since dividing by a higher power gives a smaller result.
-
-
-
-
-
-
- How do your answers change on the interval 0 \lt x \lt 1?
-
-
-
-
-
-
- On 0 \lt x \lt 1, the situation reverses: x^a \gt x^b when a \lt b (i.e., more negative exponents give larger values on (0,1) since dividing by a small number raised to a high power gives a large result).
-
-
-
-
-
-
- Uncheck the icons on each of the 8 functions to hide their graphs.
- Click the settings icon to change the domain settings for the axes,
- and change them to -10 \le x \le 10 and -10,000 \le y \le 10,000.
- Play the animation through twice and then discuss the trends that you observe.
- Write a careful sentence each for at least two different trends.
-
-
-
-
-
-
- Two trends with the wider window: (1) The graphs with even exponents are symmetric about the y-axis; odd-exponent graphs are antisymmetric. (2) As the exponent becomes more negative, the function rises more steeply near x = 0 (steeper asymptotic behavior).
-
-
-
-
-
-
- Explain why \lim_{x \to \infty} \frac{1}{x^n} = 0 for any choice of n = 1, 2, \ldots.
-
-
-
-
-
-
- As x \to \infty, x^n \to \infty for any positive integer n, so \frac{1}{x^n} = x^{-n} \to 0. Dividing 1 by an ever-larger number drives the result to 0.
-
+ Point your browser to the Desmos worksheet at
+ http://gvsu.edu/s/0zv.
+ In what follows, we explore the behavior of power functions y = x^n where n \le -1.
+
+
+
+
+
+
+ Press the play button next to the slider labeled n.
+ Watch two loops of the animation and then discuss the trends that you observe.
+ Write a careful sentence each for at least two different trends.
+
+
+
+
+
+ Two trends: (1) For even-n exponents (like x^{-2}, x^{-4}, \ldots), the graph is symmetric about the y-axis and stays positive; for odd exponents (like x^{-1}, x^{-3}, \ldots), the graph is negative for x \lt 0 and positive for x \gt 0. (2) As the exponent becomes more negative, the graph becomes more sharply L-shaped near x = 0.
+
+
+
+
+ Two trends: (1) For even-n exponents (like x^{-2}, x^{-4}, \ldots), the graph is symmetric about the y-axis and stays positive; for odd exponents (like x^{-1}, x^{-3}, \ldots), the graph is negative for x \lt 0 and positive for x \gt 0. (2) As the exponent becomes more negative, the graph becomes more sharply L-shaped near x = 0.
+
+
+
+
+
+
+ Click the icons next to each of the following 8 functions so that you can see all of y = x^{-1},
+ y = x^{-2}, \ldots,
+ y = x^{-8} graphed at once.
+ On the interval 1 \lt x, how do the functions
+ x^a and x^b compare if a \lt b? (Be careful with negative numbers here:
+ e.g., -3 \lt -2.)
+
+
+
+
+
+ On x \gt 1, x^a \lt x^b when a \lt b (with both negative). For example, x^{-3} \lt x^{-2} for x \gt 1, since dividing by a higher power gives a smaller result.
+
+
+
+
+ On x \gt 1, x^a \lt x^b when a \lt b (with both negative). For example, x^{-3} \lt x^{-2} for x \gt 1, since dividing by a higher power gives a smaller result.
+
+
+
+
+
+
+ How do your answers change on the interval 0 \lt x \lt 1?
+
+
+
+
+
+ On 0 \lt x \lt 1, the situation reverses: x^a \gt x^b when a \lt b (i.e., more negative exponents give larger values on (0,1) since dividing by a small number raised to a high power gives a large result).
+
+
+
+
+ On 0 \lt x \lt 1, the situation reverses: x^a \gt x^b when a \lt b (i.e., more negative exponents give larger values on (0,1) since dividing by a small number raised to a high power gives a large result).
+
+
+
+
+
+
+ Uncheck the icons on each of the 8 functions to hide their graphs.
+ Click the settings icon to change the domain settings for the axes,
+ and change them to -10 \le x \le 10 and -10,000 \le y \le 10,000.
+ Play the animation through twice and then discuss the trends that you observe.
+ Write a careful sentence each for at least two different trends.
+
+
+
+
+
+ Two trends with the wider window: (1) The graphs with even exponents are symmetric about the y-axis; odd-exponent graphs are antisymmetric. (2) As the exponent becomes more negative, the function rises more steeply near x = 0 (steeper asymptotic behavior).
+
+
+
+
+ Two trends with the wider window: (1) The graphs with even exponents are symmetric about the y-axis; odd-exponent graphs are antisymmetric. (2) As the exponent becomes more negative, the function rises more steeply near x = 0 (steeper asymptotic behavior).
+
+
+
+
+
+
+ Explain why \lim_{x \to \infty} \frac{1}{x^n} = 0 for any choice of n = 1, 2, \ldots.
+
+
+
+
+
+ As x \to \infty, x^n \to \infty for any positive integer n, so \frac{1}{x^n} = x^{-n} \to 0. Dividing 1 by an ever-larger number drives the result to 0.
+
+
+
+
+ As x \to \infty, x^n \to \infty for any positive integer n, so \frac{1}{x^n} = x^{-n} \to 0. Dividing 1 by an ever-larger number drives the result to 0.
+
- We understand the theoretical rule behind the function f(t) = \sin(t): given an angle t in radians, \sin(t) measures the value of the y-coordinate of the corresponding point on the unit circle. For special values of t, we have determined the exact value of \sin(t). For example, \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}. But note that we don't have a formula for \sin(t). Instead, we use a button on our calculator or command on our computer to find values like \sin(1.35). It turns out that a combination of calculus and polynomial functions explains how computers determine values of the sine function.
-
-
- At http://gvsu.edu/s/0zA, you'll find a Desmos worksheet that has the sine function already defined, along with a sequence of polynomials labeled T_1(x), T_3(x), T_5(x), T_7(x), \ldots. You can see these functions' graphs by clicking on their respective icons.
-
-
-
-
-
-
- For what values of x does it appear that \sin(x) \approx T_1(x)?
-
-
-
-
-
-
The polynomial T_1(x) = x. It appears that \sin(x) \approx T_1(x) for values of x near 0, roughly for |x| \lesssim 0.5.
-
-
-
-
-
- For what values of x does it appear that \sin(x) \approx T_3(x)?
-
-
-
-
-
-
The polynomial T_3(x) = x - \dfrac{x^3}{3!}. It appears that \sin(x) \approx T_3(x) for roughly |x| \lesssim 1.5.
-
-
-
-
-
- For what values of x does it appear that \sin(x) \approx T_5(x)?
-
-
-
-
-
-
The polynomial T_5(x) = x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!}. It appears that \sin(x) \approx T_5(x) for roughly |x| \lesssim 2.5.
-
-
-
-
-
- What overall trend do you observe? How good is the approximation generated by T_{19}(x)?
-
-
-
-
-
-
Each additional term in the polynomial extends the range of x-values over which the approximation is accurate. T_{19}(x) provides an excellent approximation to \sin(x) over a very large interval, essentially indistinguishable from \sin(x) on most standard viewing windows.
-
-
-
-
-
- In a new Desmos worksheet, plot the function y = \cos(x) along with the following functions: P_2(x) = 1 - \frac{x^2}{2!} and P_4(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!}. Based on the patterns with the coefficients in the polynomials approximating \sin(x) and the polynomials P_2 and P_4 here, conjecture formulas for P_6, P_8, and P_{18} and plot them. How well can we approximate y = \cos(x) using polynomials?
-
-
-
-
-
-
Following the pattern of alternating signs and even powers, the next polynomials are
-
- P_6(x) &= 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!},
- P_8(x) &= 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!},
- P_{18}(x) &= \sum_{j=0}^{9} \frac{(-1)^j x^{2j}}{(2j)!}.
-
- Polynomial approximations can approximate y = \cos(x) just as well as y = \sin(x); each added term extends the range of accuracy.
-
-
-
-
-
-
-
- T_1(x)=x is good near x=0; each higher-degree polynomial extends the interval of accuracy. T_{19} approximates \sin(x) extremely well. The cosine analogues follow the pattern P_{2k}(x) = \sum_{j=0}^{k} \frac{(-1)^j x^{2j}}{(2j)!}.
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ We understand the theoretical rule behind the function f(t) = \sin(t): given an angle t in radians, \sin(t) measures the value of the y-coordinate of the corresponding point on the unit circle. For special values of t, we have determined the exact value of \sin(t). For example, \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}. But note that we don't have a formula for \sin(t). Instead, we use a button on our calculator or command on our computer to find values like \sin(1.35). It turns out that a combination of calculus and polynomial functions explains how computers determine values of the sine function.
+
+
+ At http://gvsu.edu/s/0zA, you'll find a Desmos worksheet that has the sine function already defined, along with a sequence of polynomials labeled T_1(x), T_3(x), T_5(x), T_7(x), \ldots. You can see these functions' graphs by clicking on their respective icons.
+
+
+
+
+
+
+ For what values of x does it appear that \sin(x) \approx T_1(x)?
+
+
+
+
About |x| \le 0.5
+
+
The polynomial T_1(x) = x. It appears that \sin(x) \approx T_1(x) for values of x near 0, roughly for |x| \le 0.5.
+
+
+
+
+
+ For what values of x does it appear that \sin(x) \approx T_3(x)?
+
+
+
+
About |x| \le 1.5
+
+
The polynomial T_3(x) = x - \dfrac{x^3}{3!}. It appears that \sin(x) \approx T_3(x) for roughly |x| \le 1.5.
+
+
+
+
+
+ For what values of x does it appear that \sin(x) \approx T_5(x)?
+
+
+
+
About |x| \le 2.5
+
+
The polynomial T_5(x) = x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!}. It appears that \sin(x) \approx T_5(x) for roughly |x| \le 2.5.
+
+
+
+
+
+ What overall trend do you observe? How good is the approximation generated by T_{19}(x)?
+
+
+
+
+
Each additional term in the polynomial extends the range of x-values over which the approximation is accurate. T_{19}(x) provides an excellent approximation to \sin(x) over a very large interval, essentially indistinguishable from \sin(x) on most standard viewing windows.
+
+
+
Each additional term in the polynomial extends the range of x-values over which the approximation is accurate. T_{19}(x) provides an excellent approximation to \sin(x) over a very large interval, essentially indistinguishable from \sin(x) on most standard viewing windows.
+
+
+
+
+
+ In a new Desmos worksheet, plot the function y = \cos(x) along with the following functions: P_2(x) = 1 - \frac{x^2}{2!} and P_4(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!}. Based on the patterns with the coefficients in the polynomials approximating \sin(x) and the polynomials P_2 and P_4 here, conjecture formulas for P_6, P_8, and P_{18} and plot them. How well can we approximate y = \cos(x) using polynomials?
+
+
+
+
+
Following the pattern of alternating signs and even powers, the next polynomials are
+
+ P_6(x) &= 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!},
+ P_8(x) &= 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!},
+ P_{18}(x) &= \sum_{j=0}^{9} \frac{(-1)^j x^{2j}}{(2j)!}.
+
+ Polynomial approximations can approximate y = \cos(x) just as well as y = \sin(x); each added term extends the range of accuracy.
+
+
+
Following the pattern of alternating signs and even powers, the next polynomials are
+
+ P_6(x) &= 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!},
+ P_8(x) &= 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!},
+ P_{18}(x) &= \sum_{j=0}^{9} \frac{(-1)^j x^{2j}}{(2j)!}.
+
+ Polynomial approximations can approximate y = \cos(x) just as well as y = \sin(x); each added term extends the range of accuracy.
- According to a shipping company's regulations, the girth plus the length of a parcel they transport for their lowest rate may not exceed 120 inches, where by girth we mean the perimeter of one end.
-
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-
ADD ALT TEXT TO THIS IMAGE
-
-
- Suppose that we want to ship a parcel that has a square end of width x and an overall length of y, both measured in inches.
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- Label the provided picture, using x for the length of each side of the square end, and y for the other edge of the package.
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The square end has side length x inches and the length of the package is y inches.
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- How does the length plus girth of 120 inches result in an equation (often called a constraint equation) that relates x and y? Explain, and state the equation.
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The girth is the perimeter of the square end: 4x. The constraint is that girth plus length is at most 120 inches. To maximize volume we use the entire allowance, giving the constraint equation 4x + y = 120.
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- Solve the equation you found in (b) for one of the variables present.
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Solving for y: y = 120 - 4x.
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- Hence determine the volume, V, of the package as a function of a single variable.
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-
The volume of the rectangular box with square end of side x and length y is V = x^2 y. Substituting y = 120 - 4x:
-
- V(x) = x^2(120 - 4x) = 4x^2(30 - x).
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- What is the domain of the function V in the context of the physical setting of this problem? (Hint: neither x nor y can equal 0.)
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We need both x > 0 and y = 120 - 4x > 0, which gives x < 30. So the domain of V in context is (0, 30).
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- With a square end of side x and length y = 120 - 4x, the volume is V(x) = x^2(120 - 4x) = 4x^2(30-x) for x \in (0, 30).
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+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ According to a shipping company's regulations, the girth plus the length of a parcel they transport for their lowest rate may not exceed 120 inches, where by girth we mean the perimeter of one end.
+
+
+
Image of a three-dimensional rectangular box
+
+
+ Suppose that we want to ship a parcel that has a square end of width x and an overall length of y, both measured in inches.
+
+
+
+
+
+
+ Label the provided picture, using x for the length of each side of the square end, and y for the other edge of the package.
+
+
+
+
The square end has side length x inches and the length of the package is y inches.
+
+
The square end has side length x inches and the length of the package is y inches.
+
+
+
+
+
+ How does the length plus girth of 120 inches result in an equation (often called a constraint equation) that relates x and y? Explain, and state the equation.
+
+
+
+
4x + y = 120
+
+
The girth is the perimeter of the square end: 4x. The constraint is that girth plus length is at most 120 inches. To maximize volume we use the entire allowance, giving the constraint equation 4x + y = 120.
+
+
+
+
+
+ Solve the equation you found in (b) for one of the variables present.
+
+
+
+
y = 120 - 4x
+
+
Solving for y: y = 120 - 4x.
+
+
+
+
+
+ Hence determine the volume, V, of the package as a function of a single variable.
+
+
+
+
+ V(x) = x^2(120 - 4x) = 4x^2(30 - x).
+
+
+
The volume of the rectangular box with square end of side x and length y is V = x^2 y. Substituting y = 120 - 4x:
+
+ V(x) = x^2(120 - 4x) = 4x^2(30 - x).
+
+
+
+
+
+
+
+ What is the domain of the function V in the context of the physical setting of this problem? (Hint: neither x nor y can equal 0.)
+
+
+
+
(0, 30)
+
+
We need both x > 0 and y = 120 - 4x > 0, which gives x < 30. So the domain of V in context is (0, 30).
- Suppose that we want to construct a cylindrical can using 60 square inches of material for the surface of the can. In this context, how does the can's volume depend on the radius we choose?
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- Let the cylindrical can have base radius r and height h.
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- Use the formula for the surface area of a cylinder and the given constraint that the can's surface area is 60 square inches to write an equation that connects the radius r and height h.
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-
The surface area of a closed cylinder is 2\pi r^2 + 2\pi r h. Setting this equal to 60 square inches gives
-
- 2\pi r^2 + 2\pi r h = 60.
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-
- Solve the equation you found in (a) for h in terms of r.
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-
Solving for h:
-
- h = \frac{60 - 2\pi r^2}{2\pi r} = \frac{30}{\pi r} - r.
-
-
-
-
-
-
-
- Recall that the volume of a cylinder is V = \pi r^2 h. Use your work in (b) to write V as a function of the single variable r; simplify the formula as much as possible.
-
- What is the domain of the function V in the context of the physical setting of this problem? (Hint: how does the constraint on surface area provide an upper bound for the value of r? Think about the maximum area that can be allocated to the top and bottom of the can.)
-
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-
We need r > 0 and h > 0. From h = \frac{30}{\pi r} - r > 0, we get \frac{30}{\pi r} > r, so r^2 < \frac{30}{\pi}, giving r < \sqrt{\frac{30}{\pi}}. The domain is \left(0,\ \sqrt{\frac{30}{\pi}}\right).
-
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- The surface area constraint 2\pi r^2 + 2\pi r h = 60 gives h = \frac{30}{\pi r} - r, so the volume is V(r) = 30r - \pi r^3 for r \in \left(0,\ \sqrt{\frac{30}{\pi}}\right).
-
-
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-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ Suppose that we want to construct a cylindrical can using 60 square inches of material for the surface of the can. In this context, how does the can's volume depend on the radius we choose?
+
+
+ Let the cylindrical can have base radius r and height h.
+
+
+
+
+
+
+ Use the formula for the surface area of a cylinder and the given constraint that the can's surface area is 60 square inches to write an equation that connects the radius r and height h.
+
+
+
+
+ 2\pi r^2 + 2\pi r h = 60.
+
+
+
The surface area of a closed cylinder is 2\pi r^2 + 2\pi r h. Setting this equal to 60 square inches gives
+
+ 2\pi r^2 + 2\pi r h = 60.
+
+
+
+
+
+
+
+ Solve the equation you found in (a) for h in terms of r.
+
+
+
+
h = \frac{30}{\pi r} - r
+
+
Solving for h:
+
+ h = \frac{60 - 2\pi r^2}{2\pi r} = \frac{30}{\pi r} - r
+ .
+
+
+
+
+
+
+ Recall that the volume of a cylinder is V = \pi r^2 h. Use your work in (b) to write V as a function of the single variable r; simplify the formula as much as possible.
+
+ What is the domain of the function V in the context of the physical setting of this problem? (Hint: how does the constraint on surface area provide an upper bound for the value of r? Think about the maximum area that can be allocated to the top and bottom of the can.)
+
+
+
+
\left(0,\ \sqrt{\frac{30}{\pi}}\right)
+
+
We need r > 0 and h > 0. From h = \frac{30}{\pi r} - r > 0, we get \frac{30}{\pi r} > r, so r^2 < \frac{30}{\pi}, giving r < \sqrt{\frac{30}{\pi}}. The domain is \left(0,\ \sqrt{\frac{30}{\pi}}\right).
- By experimenting with coefficients in Desmos, find a formula for a polynomial function that has the stated properties, or explain why no such polynomial exists. If you enter p(x)=a+bx+cx^2+dx^3+fx^4+gx^5 in Desmos, you'll get prompted to add sliders that make it easy to explore a degree 5 polynomial. (We skip using e as one of the constants since Desmos reserves e as the Euler constant.)
-
-
-
-
-
-
- A polynomial p of degree 5 with exactly 3 real zeros, 4 turning points, and such that \lim_{x \to -\infty} p(x) = +\infty and \lim_{x \to \infty} p(x) = -\infty.
-
-
-
-
-
-
- One example: p(x) = 1 - x - 2x^2 + 4x^3 + 2x^4 - 2x^5. This degree-5 polynomial has exactly 3 real zeros, 4 turning points, and since the leading coefficient is -2 \lt 0 and the degree is odd: \lim_{x \to -\infty} p(x) = +\infty and \lim_{x \to \infty} p(x) = -\infty.
-
-
-
-
-
-
- A polynomial p of degree 4 with exactly 4 real zeros, 3 turning points, and such that \lim_{x \to -\infty} p(x) = +\infty and \lim_{x \to \infty} p(x) = -\infty.
-
-
-
-
-
-
- No such polynomial exists. A degree-4 polynomial has even degree, so both limits as x \to \pm\infty must be either both +\infty or both -\infty. It is impossible for the limits at -\infty and +\infty to be opposite infinities.
-
-
-
-
-
-
- A polynomial p of degree 6 with exactly 2 real zeros, 3 turning points, and such that \lim_{x \to -\infty} p(x) = -\infty and \lim_{x \to \infty} p(x) = -\infty.
-
-
-
-
-
-
- One example: p(x) = -1 + x - 2x^5 - x^6. This degree-6 polynomial has negative leading coefficient, so \lim_{x \to \pm\infty} p(x) = -\infty. It has 2 real zeros and 3 turning points.
-
-
-
-
-
-
- A polynomial p of degree 5 with exactly 5 real zeros, 3 turning points, and such that \lim_{x \to -\infty} p(x) = +\infty and \lim_{x \to \infty} p(x) = -\infty.
-
-
-
-
-
-
- No such polynomial exists. A degree-5 polynomial with odd degree and a positive leading coefficient has \lim_{x \to -\infty} p(x) = -\infty and \lim_{x \to \infty} p(x) = +\infty, while a negative leading coefficient gives the reverse. It is impossible for a degree-5 polynomial to have \lim_{x \to -\infty} p(x) = +\infty and simultaneously 3 turning points (an odd number of turning points forces matching end behavior, but odd-degree polynomials have opposite end behaviors).
-
+ By experimenting with coefficients in Desmos, find a formula for a polynomial function that has the stated properties, or explain why no such polynomial exists. If you enter p(x)=a+bx+cx^2+dx^3+fx^4+gx^5 in Desmos, you'll get prompted to add sliders that make it easy to explore a degree 5 polynomial. (We skip using e as one of the constants since Desmos reserves e as the Euler constant.)
+
+
+
+
+
+
+ A polynomial p of degree 5 with exactly 3 real zeros, 4 turning points, and such that \lim_{x \to -\infty} p(x) = +\infty and \lim_{x \to \infty} p(x) = -\infty.
+
+
+
+
One example: p(x) = 1 - x - 2x^2 + 4x^3 + 2x^4 - 2x^5.
+
+
+ One example: p(x) = 1 - x - 2x^2 + 4x^3 + 2x^4 - 2x^5. This degree-5 polynomial has exactly 3 real zeros, 4 turning points, and since the leading coefficient is -2 \lt 0 and the degree is odd: \lim_{x \to -\infty} p(x) = +\infty and \lim_{x \to \infty} p(x) = -\infty.
+
+
+
+
+
+
+ A polynomial p of degree 4 with exactly 4 real zeros, 3 turning points, and such that \lim_{x \to -\infty} p(x) = +\infty and \lim_{x \to \infty} p(x) = -\infty.
+
+
+
+
No such polynomial exists.
+
+
+ No such polynomial exists. A degree-4 polynomial has even degree, so both limits as x \to \pm\infty must be either both +\infty or both -\infty. It is impossible for the limits at -\infty and +\infty to be opposite infinities.
+
+
+
+
+
+
+ A polynomial p of degree 6 with exactly 2 real zeros, 3 turning points, and such that \lim_{x \to -\infty} p(x) = -\infty and \lim_{x \to \infty} p(x) = -\infty.
+
+
+
+
One example: p(x) = -1 + x - 2x^5 - x^6.
+
+
+ One example: p(x) = -1 + x - 2x^5 - x^6. This degree-6 polynomial has negative leading coefficient, so \lim_{x \to \pm\infty} p(x) = -\infty. It has 2 real zeros and 3 turning points.
+
+
+
+
+
+
+ A polynomial p of degree 5 with exactly 5 real zeros, 3 turning points, and such that \lim_{x \to -\infty} p(x) = +\infty and \lim_{x \to \infty} p(x) = -\infty.
+
+
+
+
No such polynomial exists.
+
+
+ No such polynomial exists. A degree-5 polynomial with odd degree and a positive leading coefficient has \lim_{x \to -\infty} p(x) = -\infty and \lim_{x \to \infty} p(x) = +\infty, while a negative leading coefficient gives the reverse. It is impossible for a degree-5 polynomial to have \lim_{x \to -\infty} p(x) = +\infty and simultaneously 3 turning points (an odd number of turning points forces matching end behavior, but odd-degree polynomials have opposite end behaviors).
+
- For each of the following prompts, try to determine a formula for a polynomial that satisfies the given criteria. If no such polynomial exists, explain why.
-
-
-
-
-
-
- A polynomial f of degree 10 whose zeros are x = -12 (multiplicity 3), x = -9 (multiplicity 2), x = 4 (multiplicity 4), and x = 10 (multiplicity 1), and f satisfies f(0) = 21. What can you say about the values of \lim_{x \to -\infty} f(x) and \lim_{x \to \infty} f(x)?
-
-
-
-
-
-
- One formula: f(x) = -21\!\left(\dfrac{x+12}{12}\right)^3\!\left(\dfrac{x+9}{9}\right)^2\!\left(\dfrac{x-4}{4}\right)^4\!\left(\dfrac{x-10}{10}\right). We can verify: f(0) = -21\cdot 1 \cdot 1 \cdot 1 \cdot (-1) = 21 ✓. Since the degree is 10 (even) and the leading coefficient is negative, \lim_{x \to \pm\infty} f(x) = -\infty.
-
-
-
-
-
-
- A polynomial p of degree 9 that satisfies p(0) = -2 and has the graph shown in the following figure. Assume that all of the zeros of p are shown in the figure.
-
-
-
ADD ALT TEXT TO THIS IMAGE
-
-
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-
- From the graph (reading off the zeros and their behavior), one example with p(0)=-2 is: p(x) = -2\!\left(\dfrac{x+8}{8}\right)^2\!\left(\dfrac{x+4}{4}\right)^3(x-1)\!\left(\dfrac{x-5}{5}\right)^2\!\left(\dfrac{x-7.5}{7.5}\right), which has degree 9 and p(0)=-2.
-
-
-
-
-
-
- A polynomial q of degree 8 with 3 distinct real zeros (possibly of different multiplicities) such that q has the sign chart in the figure below and satisfies q(0) = -10.
-
-
-
ADD ALT TEXT TO THIS IMAGE
-
-
-
-
-
-
- From the sign chart (positive, then negative, then positive, then negative, with the pattern showing even-multiplicity zeros where sign doesn't change and odd-multiplicity zeros where it does), one example: q(x) = -10\!\left(\dfrac{x+2}{2}\right)^4\!\left(\dfrac{x-3}{3}\right)\!\left(\dfrac{x-9}{9}\right)^3. This has degree 8, three distinct real zeros, and q(0) = -10 \cdot 1 \cdot (-1) \cdot (-1) = -10 ✓.
-
-
-
-
-
-
- A polynomial q of degree 9 with 3 distinct real zeros (possibly of different multiplicities) such that q satisfies the sign chart in part (c) and satisfies q(0) = -10.
-
-
-
-
-
-
- It is not possible to have a degree 9 polynomial satisfying the sign chart from part (c) and q(0)=-10, if the sign chart shows the same behavior at both ends (both positive or both negative for large |x|). A degree-9 polynomial has odd degree, so its limits at +\infty and -\infty must be opposite. If the sign chart requires both ends to have the same sign, degree 9 is impossible.
-
-
-
-
-
-
- A polynomial p of degree 11 that satisfies p(0) = -2 and p has the graph shown in part (b). Assume that all of the zeros of p are shown in the figure.
-
-
-
-
-
-
- It is not possible to have a degree-11 polynomial matching the graph in part (b) if that graph shows the same long-term behavior at both ends. A degree-11 polynomial has odd degree, so its limits at +\infty and -\infty must have opposite signs, while the graph in (b) (a degree-9 example) already illustrates this odd-degree behavior. However, a degree-11 polynomial could match the graph if the overall end behavior is also opposite-going.
-
+ For each of the following prompts, try to determine a formula for a polynomial that satisfies the given criteria. If no such polynomial exists, explain why.
+
+
+
+
+
+
+ A polynomial f of degree 10 whose zeros are x = -12 (multiplicity 3), x = -9 (multiplicity 2), x = 4 (multiplicity 4), and x = 10 (multiplicity 1), and f satisfies f(0) = 21. What can you say about the values of \lim_{x \to -\infty} f(x) and \lim_{x \to \infty} f(x)?
+
+
+
+
One formula: f(x) = -21\left(\dfrac{x+12}{12}\right)^3\left(\dfrac{x+9}{9}\right)^2\left(\dfrac{x-4}{4}\right)^4\left(\dfrac{x-10}{10}\right)
+
+
+ One formula: f(x) = -21\left(\dfrac{x+12}{12}\right)^3\left(\dfrac{x+9}{9}\right)^2\left(\dfrac{x-4}{4}\right)^4\left(\dfrac{x-10}{10}\right). We can verify: f(0) = -21\cdot 1 \cdot 1 \cdot 1 \cdot (-1) = 21 ✓. Since the degree is 10 (even) and the leading coefficient is negative, \lim_{x \to \pm\infty} f(x) = -\infty.
+
+
+
+
+
+
+ A polynomial p of degree 9 that satisfies p(0) = -2 and has the graph shown in the following figure. Assume that all of the zeros of p are shown in the figure.
+
+
+
Graph of a polynomial p(x)
+
+
+
+
One example is: p(x) = -2\left(\dfrac{x+8}{8}\right)^2\left(\dfrac{x+4}{4}\right)^3(x-1)\left(\dfrac{x-5}{5}\right)^2\left(\dfrac{x-7.5}{7.5}\right)
+
+
+ From the graph (reading off the zeros and their behavior), one example with p(0)=-2 is: p(x) = -2\left(\dfrac{x+8}{8}\right)^2\left(\dfrac{x+4}{4}\right)^3(x-1)\left(\dfrac{x-5}{5}\right)^2\left(\dfrac{x-7.5}{7.5}\right), which has degree 9 and p(0)=-2.
+
+
+
+
+
+
+ A polynomial q of degree 8 with 3 distinct real zeros (possibly of different multiplicities) such that q has the sign chart in the figure below and satisfies q(0) = -10.
+
+
Sign chart showing roots of x=-2,3,9 and signs from left to right around the roots of negative, negative, positive, negative.
+
+
+
+
One example: q(x) = -10\left(\dfrac{x+2}{2}\right)^4\left(\dfrac{x-3}{3}\right)\left(\dfrac{x-9}{9}\right)^3
+
+
+ From the sign chart (negative, then negative, then positive, then negative, with the pattern showing even-multiplicity zeros where sign doesn't change and odd-multiplicity zeros where it does), one example: q(x) = -10\left(\dfrac{x+2}{2}\right)^4\left(\dfrac{x-3}{3}\right)\left(\dfrac{x-9}{9}\right)^3. This has degree 8, three distinct real zeros, and q(0) = -10 \cdot 1 \cdot (-1) \cdot (-1) = -10 ✓.
+
+
+
+
+
+
+ A polynomial q of degree 9 with 3 distinct real zeros (possibly of different multiplicities) such that q satisfies the sign chart in part (c) and satisfies q(0) = -10.
+
+
+
+
Not possible.
+
+
+ It is not possible to have a degree 9 polynomial satisfying the sign chart from part (c) and q(0)=-10, if the sign chart shows the same behavior at both ends (both positive or both negative for large |x|). A degree-9 polynomial has odd degree, so its limits at +\infty and -\infty must be opposite. If the sign chart requires both ends to have the same sign, degree 9 is impossible.
+
+
+
+
+
+
+ A polynomial p of degree 11 that satisfies p(0) = -2 and p has the graph shown in part (b). Assume that all of the zeros of p are shown in the figure.
+
+
+
+
Not possible.
+
+
+ It is not possible to have a degree-11 polynomial matching the graph in part (b) if that graph shows the same long-term behavior at both ends. A degree-11 polynomial has odd degree, so its limits at +\infty and -\infty must have opposite signs, while the graph in (b) (a degree-9 example) already illustrates this odd-degree behavior. However, a degree-11 polynomial could match the graph if the overall end behavior is also opposite-going.
+
- Consider the polynomial function given by
-
- p(x) = 4692(x+1520)(x^2+10000)(x-3471)^2(x-9738)
- .
-
-
-
-
-
-
- What is the degree of p? How can you tell without fully expanding the factored form of the function?
-
-
-
-
-
-
- The degree is 1 + 2 + 2 + 1 = 6. We add the exponents from each factor: (x+1520)^1(x^2+10000)^1(x-3471)^2(x-9738)^1 contributes degrees 1+2+2+1=6.
-
-
-
-
-
-
- What can you say about the sign of the factor (x^2 + 10000)?
-
-
-
-
-
-
- The factor (x^2 + 10000) is always positive for all real x, since x^2 \ge 0 so x^2 + 10000 \ge 10000 \gt 0.
-
-
-
-
-
-
- What are the zeros of the polynomial p?
-
-
-
-
-
-
- The real zeros are x = -1520, x = 3471, and x = 9738. (The factor x^2+10000 has no real zeros since x^2 = -10000 has no real solution.)
-
-
-
-
-
-
- Construct a sign chart for p by using the zeros you identified in (c) and then analyzing the sign of each factor of p.
-
-
-
-
-
-
- The constant 4692 \gt 0, and (x^2+10000) \gt 0 always, and (x-3471)^2 \ge 0 always (no sign change at x=3471). The sign is thus determined by (x+1520) and (x-9738):
-
-
x \lt -1520: (x+1520) \lt 0 and (x-9738) \lt 0, product is positive: p \gt 0.
-
-1520 \lt x \lt 9738 (and x \ne 3471): (x+1520) \gt 0 and (x-9738) \lt 0: p \lt 0. (At x=3471, p=0 and the sign does not change.)
-
x \gt 9738: both factors positive: p \gt 0.
-
-
-
-
-
-
-
- Without using a graphing utility, construct an approximate graph of p that has the zeros of p carefully labeled on the x-axis.
-
-
-
-
-
-
- The graph is positive (above the x-axis) for x \lt -1520, crosses zero at x=-1520 (passes through), touches zero at x=3471 (bounce, even multiplicity), remains negative until x=9738, then crosses zero and becomes positive. The leading term 4692 x^6 means both ends go to +\infty.
-
-
-
-
-
-
- Use a graphing utility to check your earlier work. What is challenging or misleading when using technology to graph p?
-
-
-
-
-
-
- Using a graphing utility, the zeros at x=-1520, 3471, and 9738 are spread over a large x-range, making it challenging to see all features at once. It is difficult to tell from the graph that x=3471 is a bounce point (touching zero without crossing) unless the scale is set appropriately.
-
+ Consider the polynomial function given by
+
+ p(x) = 4692(x+1520)(x^2+10000)(x-3471)^2(x-9738)
+ .
+
+
+
+
+
+
+ What is the degree of p? How can you tell without fully expanding the factored form of the function?
+
+
+
+
6, adding the exponents from each factor.
+
+
+ The degree is 1 + 2 + 2 + 1 = 6. We add the exponents from each factor: (x+1520)^1(x^2+10000)^1(x-3471)^2(x-9738)^1 contributes degrees 1+2+2+1=6.
+
+
+
+
+
+
+ What can you say about the sign of the factor (x^2 + 10000)?
+
+
+
+
always positive
+
+
+ The factor (x^2 + 10000) is always positive for all real x, since x^2 \ge 0 so x^2 + 10000 \ge 10000 \gt 0.
+
+
+
+
+
+
+ What are the zeros of the polynomial p?
+
+
+
+
The real zeros are x = -1520, x = 3471, and x = 9738.
+
+
+ The real zeros are x = -1520, x = 3471, and x = 9738. (The factor x^2+10000 has no real zeros since x^2 = -10000 has no real solution.)
+
+
+
+
+
+
+ Construct a sign chart for p by using the zeros you identified in (c) and then analyzing the sign of each factor of p.
+
+
+
+
From left to right around the roots: positive, negative, negative, positive
+
+
+ The constant 4692 \gt 0, and (x^2+10000) \gt 0 always, and (x-3471)^2 \ge 0 always (no sign change at x=3471). The sign is thus determined by (x+1520) and (x-9738):
+
+
x \lt -1520: (x+1520) \lt 0 and (x-9738) \lt 0, product is positive: p \gt 0.
+
-1520 \lt x \lt 9738 (and x \ne 3471): (x+1520) \gt 0 and (x-9738) \lt 0: p \lt 0. (At x=3471, p=0 and the sign does not change.)
+
x \gt 9738: both factors positive: p \gt 0.
+
+
+
+
+
+
+
+ Without using a graphing utility, construct an approximate graph of p that has the zeros of p carefully labeled on the x-axis.
+
+
+
+
+
+ The graph is positive (above the x-axis) for x \lt -1520, crosses zero at x=-1520 (passes through), touches zero at x=3471 (bounce, even multiplicity), remains negative until x=9738, then crosses zero and becomes positive. The leading term 4692 x^6 means both ends go to +\infty.
+
+
+
+
+ The graph is positive (above the x-axis) for x \lt -1520, crosses zero at x=-1520 (passes through), touches zero at x=3471 (bounce, even multiplicity), remains negative until x=9738, then crosses zero and becomes positive. The leading term 4692 x^6 means both ends go to +\infty.
+
+
+
+
+
+
+ Use a graphing utility to check your earlier work. What is challenging or misleading when using technology to graph p?
+
+
+
+
+
+ Using a graphing utility, the zeros at x=-1520, 3471, and 9738 are spread over a large x-range, making it challenging to see all features at once. It is difficult to tell from the graph that x=3471 is a bounce point (touching zero without crossing) unless the scale is set appropriately.
+
+
+
+
+ Using a graphing utility, the zeros at x=-1520, 3471, and 9738 are spread over a large x-range, making it challenging to see all features at once. It is difficult to tell from the graph that x=3471 is a bounce point (touching zero without crossing) unless the scale is set appropriately.
+
- Suppose that we want to build an open rectangular box (that is, without a top) that holds 15 cubic feet of volume. If we want one side of the base to be twice as long as the other, how does the amount of material required depend on the shorter side of the base? We investigate this question through the following sequence of prompts.
-
-
-
-
-
-
- Draw a labeled picture of the box. Let x represent the shorter side of the base and h the height of the box. What is the length of the longer side of the base in terms of x?
-
-
-
-
-
-
The box has a shorter base side of length x, a longer base side of length 2x, and height h.
-
-
-
-
-
- Use the given volume constraint to write an equation that relates x and h, and solve the equation for h in terms of x.
-
-
-
-
-
-
The volume is V = x \cdot 2x \cdot h = 2x^2 h. Setting V = 15: 2x^2 h = 15, so h = \dfrac{15}{2x^2}.
-
-
-
-
-
- Determine a formula for the surface area, S, of the box in terms of x and h.
-
-
-
-
-
-
The open box (no top) has: one rectangular base (2x^2), two short sides (2 \cdot xh = 2xh), and two long sides (2 \cdot 2xh = 4xh). So
-
- S = 2x^2 + 2xh + 4xh = 2x^2 + 6xh.
-
-
-
-
-
-
-
- Using the constraint equation from (b) together with your work in (c), write surface area, S, as a function of the single variable x.
-
- What type of function is S? What is its domain?
-
-
-
-
-
-
S(x) = 2x^2 + \dfrac{45}{x} = \dfrac{2x^3 + 45}{x} is a rational function. In the physical context, x > 0 (the side length must be positive), so the domain is (0, \infty).
-
-
-
-
-
- Plot the function S using Desmos. What appears to be the least amount of material that can be used to construct the desired box that holds 15 cubic feet of volume?
-
-
-
-
-
-
From a graph of S(x) = 2x^2 + \dfrac{45}{x} on (0, \infty), the minimum value of S occurs at approximately x \approx 2.24 feet, giving a minimum surface area of approximately S \approx 30.0 square feet. (The exact minimum occurs at x = \sqrt[3]{45/4}.)
-
-
-
-
-
-
-
- With base sides x and 2x and volume constraint 2x^2 h = 15, we get h = 15/(2x^2) and surface area S(x) = 2x^2 + 45/x, a rational function on domain (0, \infty). Its minimum is approximately S \approx 30.0 sq ft at x \approx 2.24 ft.
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ Suppose that we want to build an open rectangular box (that is, without a top) that holds 15 cubic feet of volume. If we want one side of the base to be twice as long as the other, how does the amount of material required depend on the shorter side of the base? We investigate this question through the following sequence of prompts.
+
+
+
+
+
+
+ Draw a labeled picture of the box. Let x represent the shorter side of the base and h the height of the box. What is the length of the longer side of the base in terms of x?
+
+
+
+
+
The box has a shorter base side of length x, a longer base side of length 2x, and height h.
+
+
+
The box has a shorter base side of length x, a longer base side of length 2x, and height h.
+
+
+
+
+
+ Use the given volume constraint to write an equation that relates x and h, and solve the equation for h in terms of x.
+
+
+
+
+
The volume is V = 2x^2 h. Setting V = 15 gives h = \dfrac{15}{2x^2}.
+
+
+
The volume is V = x \cdot 2x \cdot h = 2x^2 h. Setting V = 15: 2x^2 h = 15, so h = \dfrac{15}{2x^2}.
+
+
+
+
+
+ Determine a formula for the surface area, S, of the box in terms of x and h.
+
+
+
+
S = 2x^2 + 2xh + 4xh = 2x^2 + 6xh.
+
+
The open box (no top) has: one rectangular base (2x^2), two short sides (2 \cdot xh = 2xh), and two long sides (2 \cdot 2xh = 4xh). So
+
+ S = 2x^2 + 2xh + 4xh = 2x^2 + 6xh.
+
+
+
+
+
+
+
+ Using the constraint equation from (b) together with your work in (c), write surface area, S, as a function of the single variable x.
+
+ What type of function is S? What is its domain?
+
+
+
+
S is a rational function with domain (0, \infty).
+
+
S(x) = 2x^2 + \dfrac{45}{x} = \dfrac{2x^3 + 45}{x} is a rational function. In the physical context, x > 0 (the side length must be positive), so the domain is (0, \infty).
+
+
+
+
+
+ Plot the function S using Desmos. What appears to be the least amount of material that can be used to construct the desired box that holds 15 cubic feet of volume?
+
+
+
+
+
From a graph of S(x) = 2x^2 + \dfrac{45}{x} on (0, \infty), the minimum value of S occurs at approximately x \approx 2.24 feet, giving a minimum surface area of approximately S \approx 30.0 square feet.
+
+
+
From a graph of S(x) = 2x^2 + \dfrac{45}{x} on (0, \infty), the minimum value of S occurs at approximately x \approx 2.24 feet, giving a minimum surface area of approximately S \approx 30.0 square feet. (The exact minimum occurs at x = \sqrt[3]{45/4}.)
Factoring the denominator: x^2+3x-4 = (x-1)(x+4). The denominator is zero when x = 1 or x = -4. The domain of g is all real numbers except x = 1 and x = -4.
Each fraction is undefined when its denominator is zero: 1/x is undefined at x=0, 1/(x-1) is undefined at x=1, and 1/(x-2) is undefined at x=2. The domain of h is all real numbers except x = 0, x = 1, and x = 2.
Factoring the denominator: 3x^3 - 12x = 3x(x^2 - 4) = 3x(x-2)(x+2). The denominator is zero when x = 0, x = 2, or x = -2. The domain of k is all real numbers except x = -2, x = 0, and x = 2.
The denominator is zero when x = 2, x = 3, or x = -1 (the factor x^2+9 has no real zeros since x^2 \geq 0). The domain of m is all real numbers except x = -1, x = 2, and x = 3.
-
-
-
-
-
-
-
- Rational functions are undefined wherever the denominator is zero. Factor each denominator and exclude those x-values: f has domain \mathbb{R}; g excludes x \in \{-4,1\}; h excludes x \in \{0,1,2\}; j excludes x \in \{-3,-1,5\}; k excludes x \in \{-2,0,2\}; m excludes x \in \{-1,2,3\}.
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ Determine the domain of each of the following functions. In each case, write a sentence to accurately describe the domain.
+
+
+
+
+
+
+ \displaystyle f(x) = \frac{x^2-1}{x^2 + 1}
+
+
+
+
The domain of f is all real numbers.
+
+
The denominator x^2+1 is always positive, so it is never zero. The domain of f is all real numbers.
The domain of g is all real numbers except x = 1 and x = -4.
+
+
Factoring the denominator: x^2+3x-4 = (x-1)(x+4). The denominator is zero when x = 1 or x = -4. The domain of g is all real numbers except x = 1 and x = -4.
The domain of h is all real numbers except x = 0, x = 1, and x = 2.
+
+
Each fraction is undefined when its denominator is zero: 1/x is undefined at x=0, 1/(x-1) is undefined at x=1, and 1/(x-2) is undefined at x=2. The domain of h is all real numbers except x = 0, x = 1, and x = 2.
The domain of k is all real numbers except x = -2, x = 0, and x = 2.
+
+
Factoring the denominator: 3x^3 - 12x = 3x(x^2 - 4) = 3x(x-2)(x+2). The denominator is zero when x = 0, x = 2, or x = -2. The domain of k is all real numbers except x = -2, x = 0, and x = 2.
The domain of m is all real numbers except x = -1, x = 2, and x = 3.
+
+
The denominator is zero when x = 2, x = 3, or x = -1 (the factor x^2+9 has no real zeros since x^2 \geq 0). The domain of m is all real numbers except x = -1, x = 2, and x = 3.
- For each of the following rational functions, state the function's domain and determine the locations of all zeros, vertical asymptotes, and holes. Provide clear justification for your work by discussing the zeros of the numerator and denominator, as well as a table of values of the function near any point where you believe the function has a hole. In addition, state the value of the horizontal asymptote of the function or explain why the function has no such asymptote.
-
The numerator x^2+1 > 0 always; the only real zero of the numerator is x = 7. The denominator zeros: x = 1 from (x-1); x^2+4 > 0 always. No common factors.
-
-
Domain: all reals except x = 1.
-
Zero:x = 7.
-
Vertical asymptote:x = 1.
-
Holes: none.
-
Horizontal asymptote:y = \dfrac{11}{23} (degree 3 over degree 3, ratio of leading coefficients).
Domain: all reals except x = -1, x = 4, and x = 5.
-
Zeros:x = 2 and x = 3.
-
Vertical asymptotes:x = 4 and x = 5.
-
Hole: at x = -1; hole value \dfrac{19(-1-2)(-1-3)^2}{17(-1-4)^2(-1-5)} = \dfrac{19(-3)(16)}{17(25)(-6)} = \dfrac{-912}{-2550} = \dfrac{152}{425}, so hole at \left(-1,\, \dfrac{152}{425}\right).
-
Horizontal asymptote:y = \dfrac{19}{17} (degree 4 over degree 4).
-
-
-
-
-
-
-
- \displaystyle s(x) = \frac{1}{x^2 + 1}
-
-
-
-
-
-
The numerator is always 1 \neq 0, and the denominator x^2+1 > 0 always.
-
-
Domain: all real numbers.
-
Zeros: none.
-
Vertical asymptotes: none.
-
Holes: none.
-
Horizontal asymptote:y = 0 (degree 0 over degree 2).
-
-
-
-
-
-
-
-
-
- Factor numerator and denominator, cancel common factors, then identify: zeros (numerator zero in reduced form), vertical asymptotes (denominator zero in reduced form), holes (cancelled factors), horizontal asymptote (degree comparison of leading terms).
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ For each of the following rational functions, state the function's domain and determine the locations of all zeros, vertical asymptotes, and holes. Provide clear justification for your work by discussing the zeros of the numerator and denominator, as well as a table of values of the function near any point where you believe the function has a hole. In addition, state the value of the horizontal asymptote of the function or explain why the function has no such asymptote.
+
The numerator x^2+1 > 0 always; the only real zero of the numerator is x = 7. The denominator zeros: x = 1 from (x-1); x^2+4 > 0 always. No common factors.
+
+
Domain: all reals except x = 1.
+
Zero:x = 7.
+
Vertical asymptote:x = 1.
+
Holes: none.
+
Horizontal asymptote:y = \dfrac{11}{23} (degree 3 over degree 3, ratio of leading coefficients).
Domain: all reals except x = -1, x = 4, and x = 5.
+
Zeros:x = 2 and x = 3.
+
Vertical asymptotes:x = 4 and x = 5.
+
Hole: at x = -1; values nearby are approaching \dfrac{152}{425}\right).
+
Horizontal asymptote:y = \dfrac{19}{17}
+
+
+
+
+
The factor (x+1) cancels.
+
+
Domain: all reals except x = -1, x = 4, and x = 5.
+
Zeros:x = 2 and x = 3.
+
Vertical asymptotes:x = 4 and x = 5.
+
Hole: at x = -1; hole value \dfrac{19(-1-2)(-1-3)^2}{17(-1-4)^2(-1-5)} = \dfrac{19(-3)(16)}{17(25)(-6)} = \dfrac{-912}{-2550} = \dfrac{152}{425}, so hole at \left(-1,\, \dfrac{152}{425}\right).
+
Horizontal asymptote:y = \dfrac{19}{17} (degree 4 over degree 4).
+
+
+
+
+
+
+
+ \displaystyle s(x) = \frac{1}{x^2 + 1}
+
+
+
+
+
+
+
Domain: all real numbers.
+
Zeros: none.
+
Vertical asymptotes: none.
+
Holes: none.
+
Horizontal asymptote:y = 0
+
+
+
+
+
The numerator is always 1 \neq 0, and the denominator x^2+1 > 0 always.
+
+
Domain: all real numbers.
+
Zeros: none.
+
Vertical asymptotes: none.
+
Holes: none.
+
Horizontal asymptote:y = 0 (degree 0 over degree 2).
- Find a formula for a rational function that meets the stated criteria as given by words, a sign chart, or graph. Write several sentences to justify why your formula matches the specifications.
-
-
-
-
-
-
- A rational function r such that r has a vertical asymptote at x = -2, a zero at x = 1, a hole at x = 5, and a horizontal asymptote of y = -3.
-
-
-
-
-
-
We need: a factor (x+2) in the denominator for the vertical asymptote at x=-2; a factor (x-1) in the numerator for the zero at x=1; the factor (x-5) in both numerator and denominator for the hole at x=5; and a ratio of leading coefficients of -3 for the horizontal asymptote y=-3. One such function is
-
- r(x) = \frac{-3(x-1)(x-5)}{(x+2)(x-5)}.
-
-
-
-
-
-
-
- A rational function u whose numerator has degree 3, denominator has degree 3, and that has exactly one vertical asymptote at x = -4 and a horizontal asymptote of y = \frac{3}{7}.
-
-
-
-
-
-
We need: exactly one factor (x+4) in the denominator for the vertical asymptote; the other two denominator factors must introduce no new real zeros (e.g., x^2+1); and ratio of leading coefficients \frac{3}{7}. One such function is
-
- u(x) = \frac{3x(x^2+3)}{7(x+4)(x^2+1)}.
-
- Both numerator and denominator have degree 3, there is exactly one vertical asymptote at x=-4, and the horizontal asymptote is y = \frac{3}{7}.
-
-
-
-
-
- A rational function w whose formula generates a graph with all of the characteristics shown in the following figure. Assume that w(5) = 0 but w(x) \gt 0 for all other x such that x \gt 3.
-
-
-
ADD ALT TEXT TO THIS IMAGE
-
-
-
-
-
From the graph, w appears to have zeros at x = -4 and at x = 5 (with even multiplicity, since w(x) > 0 for other x > 3), and vertical asymptotes at x = -1 and x = 3. One formula consistent with these features is
-
- w(x) = \frac{9(x+4)(x-5)^2}{50(x+1)^2(x-3)}.
-
-
-
-
-
-
-
- A rational function z whose formula satisfies the sign chart shown in the following figure, and for which z has no horizontal asymptote and its only vertical asymptotes occur at the middle two values of x noted on the sign chart.
-
-
-
ADD ALT TEXT TO THIS IMAGE
-
-
-
-
-
-
From the sign chart, the function has zeros at x = -4 and x = 5, vertical asymptotes at x = -1 and x = 3, no horizontal asymptote, and the behavior shown in the chart requires the numerator degree to exceed the denominator degree. One formula is
-
- z(x) = -\frac{(x+4)(x-5)^2}{(x+1)(x-3)}.
-
-
-
-
-
-
-
- A rational function f that has exactly two holes, two vertical asymptotes, two zeros, and a horizontal asymptote.
-
-
-
-
-
-
We need: two factors that cancel for the holes; two denominator factors that don't cancel for the VAs; two remaining numerator factors for the zeros; and equal numerator/denominator degree for the HA. One formula is
-
- f(x) = \frac{(x-1)(x+1)(x+2)(x-2)}{(x-1)(x+1)(x-3)(x+3)}.
-
- After simplification, f(x) = \dfrac{(x+2)(x-2)}{(x-3)(x+3)}. This has holes at \left(-1,\dfrac{3}{8}\right) and \left(1,\dfrac{3}{8}\right), vertical asymptotes at x = 3 and x = -3, zeros at x = 2 and x = -2, and horizontal asymptote y = 1.
-
-
-
-
-
-
-
- Sample answers: (a) r(x)=-3(x-1)(x-5)/[(x+2)(x-5)]; (b) u(x)=3x(x^2+3)/[7(x+4)(x^2+1)]; (c) based on graph features; (d) based on sign chart; (e) f(x)=(x-1)(x+1)(x+2)(x-2)/[(x-1)(x+1)(x-3)(x+3)].
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ Find a formula for a rational function that meets the stated criteria as given by words, a sign chart, or graph. Write several sentences to justify why your formula matches the specifications.
+
+
+
+
+
+
+ A rational function r such that r has a vertical asymptote at x = -2, a zero at x = 1, a hole at x = 5, and a horizontal asymptote of y = -3.
+
+
+
+
One such function is
+
+ r(x) = \frac{-3(x-1)(x-5)}{(x+2)(x-5)}.
+
+
+
+
We need: a factor (x+2) in the denominator for the vertical asymptote at x=-2; a factor (x-1) in the numerator for the zero at x=1; the factor (x-5) in both numerator and denominator for the hole at x=5; and a ratio of leading coefficients of -3 for the horizontal asymptote y=-3. One such function is
+
+ r(x) = \frac{-3(x-1)(x-5)}{(x+2)(x-5)}.
+
+
+
+
+
+
+
+ A rational function u whose numerator has degree 3, denominator has degree 3, and that has exactly one vertical asymptote at x = -4 and a horizontal asymptote of y = \frac{3}{7}.
+
+
+
+
One such function is
+
+ u(x) = \frac{3x(x^2+3)}{7(x+4)(x^2+1)}.
+
+
+
We need: exactly one factor (x+4) in the denominator for the vertical asymptote; the other two denominator factors must introduce no new real zeros (e.g., x^2+1); and ratio of leading coefficients \frac{3}{7}. One such function is
+
+ u(x) = \frac{3x(x^2+3)}{7(x+4)(x^2+1)}
+ .
+ Both numerator and denominator have degree 3, there is exactly one vertical asymptote at x=-4, and the horizontal asymptote is y = \frac{3}{7}.
+
+
+
+
+
+ A rational function w whose formula generates a graph with all of the characteristics shown in the following figure. Assume that w(5) = 0 but w(x) \gt 0 for all other x such that x \gt 3.
+
+
+
Graph of a function with vertical asymptotes at x=-1 and x=3, and roots at x=-4 and x=5.
+
+
+
One formula consistent with these features is
+
+ w(x) = \frac{9(x+4)(x-5)^2}{50(x+1)^2(x-3)}.
+
+
+
+
From the graph, w appears to have zeros at x = -4 and at x = 5 (with even multiplicity, since w(x) > 0 for other x > 3), and vertical asymptotes at x = -1 and x = 3. One formula consistent with these features is
+
+ w(x) = \frac{9(x+4)(x-5)^2}{50(x+1)^2(x-3)}.
+
+
+
+
+
+
+
+ A rational function z whose formula satisfies the sign chart shown in the following figure, and for which z has no horizontal asymptote and its only vertical asymptotes occur at the middle two values of x noted on the sign chart.
+
+
+
Sign chart showing roots of x=-4,-1,3,5 and whose signs from left to right around the roots are positive, negative, positive, negative, negative.
+
+
+
+
One formula is
+
+ z(x) = -\frac{(x+4)(x-5)^2}{(x+1)(x-3)}.
+
+
+
+
+
From the sign chart, the function has zeros at x = -4 and x = 5, vertical asymptotes at x = -1 and x = 3, no horizontal asymptote, and the behavior shown in the chart requires the numerator degree to exceed the denominator degree. One formula is
+
+ z(x) = -\frac{(x+4)(x-5)^2}{(x+1)(x-3)}.
+
+
+
+
+
+
+
+ A rational function f that has exactly two holes, two vertical asymptotes, two zeros, and a horizontal asymptote.
+
+
+
+
One formula is
+
+ f(x) = \frac{(x-1)(x+1)(x+2)(x-2)}{(x-1)(x+1)(x-3)(x+3)}.
+
+
+
We need: two factors that cancel for the holes; two denominator factors that don't cancel for the VAs; two remaining numerator factors for the zeros; and equal numerator/denominator degree for the HA. One formula is
+
+ f(x) = \frac{(x-1)(x+1)(x+2)(x-2)}{(x-1)(x+1)(x-3)(x+3)}.
+
+ After simplification, f(x) = \dfrac{(x+2)(x-2)}{(x-3)(x+3)}. This has holes at \left(-1,\dfrac{3}{8}\right) and \left(1,\dfrac{3}{8}\right), vertical asymptotes at x = 3 and x = -3, zeros at x = 2 and x = -2, and horizontal asymptote y = 1.
- Consider the rational function r(x) = \frac{3x^2 - 5x + 1}{7x^2 + 2x - 11}.
-
-
- Observe that the largest power of x that's present in r(x) is x^2. In addition, because of the dominant terms of 3x^2 in the numerator and 7x^2 in the denominator, both the numerator and denominator of r increase without bound as x increases without bound. In order to understand the long-range behavior of r, we choose to write the function in a different algebraic form.
-
-
-
-
-
-
- Note that we can multiply the formula for r by the form of 1 given by 1 = \frac{\frac{1}{x^2}}{\frac{1}{x^2}}. Do so, and distribute and simplify as much as possible in both the numerator and denominator to write r in a different algebraic form.
-
- Having rewritten r, we are in a better position to evaluate \lim_{x \to \infty} r(x).
- Using our work from (a), we have
-
- \lim_{x \to \infty} r(x) = \lim_{x \to \infty} \frac{3 - \frac{5}{x} + \frac{1}{x^2}}{7 + \frac{2}{x} - \frac{11}{x^2}}
- .
- What is the exact value of this limit and why?
-
-
-
-
-
-
As x \to \infty, the terms \frac{5}{x}, \frac{1}{x^2}, \frac{2}{x}, and \frac{11}{x^2} all approach 0. Therefore
-
- \lim_{x \to \infty} r(x) = \frac{3 - 0 + 0}{7 + 0 - 0} = \frac{3}{7}.
-
-
For the same reason, as x \to -\infty the terms with x in the denominator all approach 0, so
-
- \lim_{x \to -\infty} r(x) = \frac{3}{7}.
-
-
-
-
-
-
-
- Use Desmos to plot r on the interval [-10,10]. In addition, plot the horizontal line y = \frac{3}{7}. What is the meaning of the limits you found in (b) and (c)?
-
-
-
-
-
-
The graph of r approaches the horizontal line y = \dfrac{3}{7} as x \to \pm\infty. This line is called a horizontal asymptote of r. The limits found in (b) and (c) say that far to the left and far to the right of the origin, r(x) gets arbitrarily close to \dfrac{3}{7}.
-
-
-
-
-
-
-
- Multiplying by \frac{1/x^2}{1/x^2} gives r(x) = \frac{3 - 5/x + 1/x^2}{7 + 2/x - 11/x^2}. Since the terms with x in the denominator vanish as x \to \pm\infty, both limits equal \frac{3}{7}, the ratio of the leading coefficients. The horizontal asymptote is y = \frac{3}{7}.
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ Consider the rational function r(x) = \frac{3x^2 - 5x + 1}{7x^2 + 2x - 11}.
+
+
+ Observe that the largest power of x that's present in r(x) is x^2. In addition, because of the dominant terms of 3x^2 in the numerator and 7x^2 in the denominator, both the numerator and denominator of r increase without bound as x increases without bound. In order to understand the long-range behavior of r, we choose to write the function in a different algebraic form.
+
+
+
+
+
+
+ Note that we can multiply the formula for r by the form of 1 given by 1 = \frac{\frac{1}{x^2}}{\frac{1}{x^2}}. Do so, and distribute and simplify as much as possible in both the numerator and denominator to write r in a different algebraic form.
+
+ Having rewritten r, we are in a better position to evaluate \lim_{x \to \infty} r(x).
+ Using our work from (a), we have
+
+ \lim_{x \to \infty} r(x) = \lim_{x \to \infty} \frac{3 - \frac{5}{x} + \frac{1}{x^2}}{7 + \frac{2}{x} - \frac{11}{x^2}}
+ .
+ What is the exact value of this limit and why?
+
For the same reason, as x \to -\infty the terms with x in the denominator all approach 0, so
+
+ \lim_{x \to -\infty} r(x) = \frac{3}{7}.
+
+
+
+
+
+
+
+ Use Desmos to plot r on the interval [-10,10]. In addition, plot the horizontal line y = \frac{3}{7}. What is the meaning of the limits you found in (b) and (c)?
+
+
+
+
+
The graph of r approaches the horizontal line y = \dfrac{3}{7} as x \to \pm\infty. This line is called a horizontal asymptote of r. The limits found in (b) and (c) say that far to the left and far to the right of the origin, r(x) gets arbitrarily close to \dfrac{3}{7}.
+
+
+
The graph of r approaches the horizontal line y = \dfrac{3}{7} as x \to \pm\infty. This line is called a horizontal asymptote of r. The limits found in (b) and (c) say that far to the left and far to the right of the origin, r(x) gets arbitrarily close to \dfrac{3}{7}.
- Let s(x) = \frac{3x - 5}{7x^2 + 2x - 11} and u(x) = \frac{3x^2 - 5x + 1}{7x + 2}. Note that both the numerator and denominator of each of these rational functions increases without bound as x \to \infty, and in addition that x^2 is the highest order term present in each of s and u.
-
-
-
-
-
-
- Using a similar algebraic approach to our work in Activity, multiply s(x) by 1 = \frac{\frac{1}{x^2}}{\frac{1}{x^2}} and hence evaluate
-
- \lim_{x \to \infty} \frac{3x - 5}{7x^2 + 2x - 11}
- .
- What value do you find?
-
-
-
-
-
-
Multiplying s(x) = \dfrac{3x-5}{7x^2+2x-11} by \dfrac{1/x^2}{1/x^2}:
-
- s(x) = \frac{\frac{3}{x} - \frac{5}{x^2}}{7 + \frac{2}{x} - \frac{11}{x^2}}.
-
- As x \to \infty, the numerator \to 0 and the denominator \to 7, so \displaystyle\lim_{x \to \infty} s(x) = 0.
-
-
-
-
-
- Plot the function y = s(x) on the interval [-10,10]. What is the graphical meaning of the limit you found in (a)?
-
-
-
-
-
-
The graph of y = s(x) has a horizontal asymptote at y = 0. Far to the left and right, s(x) approaches the x-axis. This occurs because the denominator grows much faster than the numerator (degree 2 vs. degree 1).
-
-
-
-
-
- Next, use appropriate algebraic work to consider u(x) and evaluate
-
- \lim_{x \to \infty} \frac{3x^2 - 5x + 1}{7x + 2}
- .
- What do you find?
-
-
-
-
-
-
Multiplying u(x) = \dfrac{3x^2-5x+1}{7x+2} by \dfrac{1/x^2}{1/x^2}:
-
- u(x) = \frac{3 - \frac{5}{x} + \frac{1}{x^2}}{\frac{7}{x} + \frac{2}{x^2}}.
-
- As x \to \infty, the numerator \to 3 and the denominator \to 0^+, so \displaystyle\lim_{x \to \infty} u(x) = +\infty. The numerator degree exceeds the denominator degree, so there is no horizontal asymptote.
-
-
-
-
-
- Plot the function y = u(x) on the interval [-10,10]. What is the graphical meaning of the limit you computed in (c)?
-
-
-
-
-
-
The graph of y = u(x) increases without bound as x \to \infty; u has no horizontal asymptote. Instead, it has an oblique (slant) asymptote. The graph shows that u(x) grows like a linear function for large |x|.
-
-
-
-
-
-
-
- For s(x) (degree 1 over degree 2): \lim_{x \to \infty} s(x) = 0, horizontal asymptote y=0. For u(x) (degree 2 over degree 1): \lim_{x \to \infty} u(x) = +\infty, no horizontal asymptote.
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ Let s(x) = \frac{3x - 5}{7x^2 + 2x - 11} and u(x) = \frac{3x^2 - 5x + 1}{7x + 2}. Note that both the numerator and denominator of each of these rational functions increases without bound as x \to \infty, and in addition that x^2 is the highest order term present in each of s and u.
+
+
+
+
+
+
+ Using a similar algebraic approach to our work in Activity, multiply s(x) by 1 = \frac{\frac{1}{x^2}}{\frac{1}{x^2}} and hence evaluate
+
+ \lim_{x \to \infty} \frac{3x - 5}{7x^2 + 2x - 11}
+ .
+ What value do you find?
+
+
+
+
As x \to \infty, the numerator \to 0 and the denominator \to 7, so \displaystyle\lim_{x \to \infty} s(x) = 0.
+
+
Multiplying s(x) = \dfrac{3x-5}{7x^2+2x-11} by \dfrac{1/x^2}{1/x^2}:
+
+ s(x) = \frac{\frac{3}{x} - \frac{5}{x^2}}{7 + \frac{2}{x} - \frac{11}{x^2}}.
+
+ As x \to \infty, the numerator \to 0 and the denominator \to 7, so \displaystyle\lim_{x \to \infty} s(x) = 0.
+
+
+
+
+
+ Plot the function y = s(x) on the interval [-10,10]. What is the graphical meaning of the limit you found in (a)?
+
+
+
+
+
The graph of y = s(x) has a horizontal asymptote at y = 0. Far to the left and right, s(x) approaches the x-axis. This occurs because the denominator grows much faster than the numerator (degree 2 vs. degree 1).
+
+
+
The graph of y = s(x) has a horizontal asymptote at y = 0. Far to the left and right, s(x) approaches the x-axis. This occurs because the denominator grows much faster than the numerator (degree 2 vs. degree 1).
+
+
+
+
+
+ Next, use appropriate algebraic work to consider u(x) and evaluate
+
+ \lim_{x \to \infty} \frac{3x^2 - 5x + 1}{7x + 2}
+ .
+ What do you find?
+
+
+
+
As x \to \infty, the numerator \to 3 and the denominator \to 0^+, so \displaystyle\lim_{x \to \infty} u(x) = +\infty.
+
+
Multiplying u(x) = \dfrac{3x^2-5x+1}{7x+2} by \dfrac{1/x^2}{1/x^2}:
+
+ u(x) = \frac{3 - \frac{5}{x} + \frac{1}{x^2}}{\frac{7}{x} + \frac{2}{x^2}}.
+
+ As x \to \infty, the numerator \to 3 and the denominator \to 0^+, so \displaystyle\lim_{x \to \infty} u(x) = +\infty. The numerator degree exceeds the denominator degree, so there is no horizontal asymptote.
+
+
+
+
+
+ Plot the function y = u(x) on the interval [-10,10]. What is the graphical meaning of the limit you computed in (c)?
+
+
+
+
The graph of y = u(x) increases without bound as x \to \infty; u has no horizontal asymptote.
+
+
The graph of y = u(x) increases without bound as x \to \infty; u has no horizontal asymptote. Instead, it has an oblique (slant) asymptote. The graph shows that u(x) grows like a linear function for large |x|.
- On a baseball diamond
- (which is a square with 90-foot sides),
- the third baseman fields the ball right on the line from third base to home plate and 10 feet away from third base
- (towards home plate). When he throws the ball to first base, what angle
- (in degrees) does the line the ball travels make with the first base line? What angle does it make with the third base line?
- Draw a well-labeled diagram to support your thinking.
-
-
- What angles arise if he throws the ball to second base instead?
-
-
-
-
-
-
-
- Place home plate at the origin, first base at (90, 0), second base at (90, 90), and third base at (0, 90). The third baseman is at (0, 80) (10 feet from third base toward home plate).
-
-
- Throwing to first base. The straight-line distance is \sqrt{90^2 + 80^2} = 10\sqrt{145} feet. The angle the throw makes with the first base line is \arctan\!\left(\dfrac{80}{90}\right) = \arctan\!\left(\dfrac{8}{9}\right) \approx 41.6^{\circ}, and with the third base line it is \arctan\!\left(\dfrac{90}{80}\right) = \arctan\!\left(\dfrac{9}{8}\right) \approx 48.4^{\circ}.
-
-
- Throwing to second base. The displacement from (0, 80) to second base at (90, 90) is (90, 10), so the distance is \sqrt{90^2 + 10^2} = 10\sqrt{82} feet. The angle with the first base line is \arctan\!\left(\dfrac{10}{90}\right) = \arctan\!\left(\dfrac{1}{9}\right) \approx 6.3^{\circ}, and with the third base line it is approximately 83.7^{\circ}.
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ On a baseball diamond
+ (which is a square with 90-foot sides),
+ the third baseman fields the ball right on the line from third base to home plate and 10 feet away from third base
+ (towards home plate). When he throws the ball to first base, what angle
+ (in degrees) does the line the ball travels make with the first base line? What angle does it make with the third base line?
+ Draw a well-labeled diagram to support your thinking.
+
+
+ What angles arise if he throws the ball to second base instead?
+
+
+
+
+ Throwing to first base. The angle the throw makes with the first base line is \arctan\left(\dfrac{80}{90}\right) \approx 41.6^{\circ}, and with the third base line it is \arctan\left(\dfrac{90}{80}\right) \approx 48.4^{\circ}.
+
+
+ Throwing to second base. The angle with the first base line is \arctan\left(\dfrac{10}{90}\right) \approx 6.3^{\circ}, and with the third base line it is approximately 83.7^{\circ}.
+
+
+
+
+ Place home plate at the origin, first base at (90, 0), second base at (90, 90), and third base at (0, 90). The third baseman is at (0, 80) (10 feet from third base toward home plate).
+
+
+ Throwing to first base. The straight-line distance is \sqrt{90^2 + 80^2} = 10\sqrt{145} feet. The angle the throw makes with the first base line is \arctan\left(\dfrac{80}{90}\right) = \arctan\left(\dfrac{8}{9}\right) \approx 41.6^{\circ}, and with the third base line it is \arctan\left(\dfrac{90}{80}\right) = \arctan\!\left(\dfrac{9}{8}\right) \approx 48.4^{\circ}.
+
+
+ Throwing to second base. The displacement from (0, 80) to second base at (90, 90) is (90, 10), so the distance is \sqrt{90^2 + 10^2} = 10\sqrt{82} feet. The angle with the first base line is \arctan\!\left(\dfrac{10}{90}\right) = \arctan\!\left(\dfrac{1}{9}\right) \approx 6.3^{\circ}, and with the third base line it is approximately 83.7^{\circ}.
+
- For each of the following different scenarios, draw a picture of the situation and use inverse trigonometric functions appropriately to determine the missing information both exactly and approximately.
-
-
-
-
-
-
- Consider a right triangle with legs of length 11 and 13. What are the measures
- (in radians) of the non-right angles and what is the length of the hypotenuse?
-
-
-
-
-
-
- The hypotenuse has length \sqrt{11^2 + 13^2} = \sqrt{290}. The angle \alpha opposite the leg of length 11 satisfies \alpha = \arcsin\!\left(\dfrac{11}{\sqrt{290}}\right) = \arccos\!\left(\dfrac{13}{\sqrt{290}}\right) \approx 0.702 radians. The angle \beta opposite the leg of length 13 satisfies \beta = \arcsin\!\left(\dfrac{13}{\sqrt{290}}\right) = \arccos\!\left(\dfrac{11}{\sqrt{290}}\right) \approx 0.869 radians.
-
-
-
-
-
-
- Consider an angle \alpha in standard position (vertex at the origin, one side on the positive x-axis) for which we know \cos(\alpha) = -\frac{1}{2} and \alpha lies in quadrant III. What is the measure of \alpha in radians? In addition, what is the value of \sin(\alpha)?
-
-
-
-
-
-
- Since \cos(\alpha) = -\frac{1}{2} and \alpha is in quadrant III, the reference angle is \arccos\!\left(\frac{1}{2}\right) = \frac{\pi}{3}, so \alpha = \pi + \frac{\pi}{3} = \frac{4\pi}{3}. In addition, \sin(\alpha) = -\dfrac{\sqrt{3}}{2}.
-
-
-
-
-
-
- Consider an angle \beta in standard position for which we know
- \sin(\beta) = 0.1 and \beta lies in quadrant II. What is the measure of \beta in radians?
- In addition, what is the value of \cos(\beta)?
-
-
-
-
-
-
- Since \sin(\beta) = 0.1 and \beta is in quadrant II, we have \beta = \pi - \arcsin(0.1) \approx 3.041 radians. In addition, \cos(\beta) = -\cos(\arcsin(0.1)) = -\sqrt{1 - 0.01} = -\sqrt{0.99} \approx -0.9950.
-
+ For each of the following different scenarios, draw a picture of the situation and use inverse trigonometric functions appropriately to determine the missing information both exactly and approximately.
+
+
+
+
+
+
+ Consider a right triangle with legs of length 11 and 13. What are the measures
+ (in radians) of the non-right angles and what is the length of the hypotenuse?
+
+
+
+
+
+ The hypotenuse has length \sqrt{11^2 + 13^2} = \sqrt{290}.
+ The angle \alpha opposite the leg of length 11 satisfies \alpha = \arcsin\left(\dfrac{11}{\sqrt{290}}\right) = \arccos\left(\dfrac{13}{\sqrt{290}}\right) \approx 0.702 radians. The angle \beta opposite the leg of length 13 satisfies \beta = \arcsin\left(\dfrac{13}{\sqrt{290}}\right) = \arccos\left(\dfrac{11}{\sqrt{290}}\right) \approx 0.869 radians.
+
+
+
+
+ The hypotenuse has length \sqrt{11^2 + 13^2} = \sqrt{290}. The angle \alpha opposite the leg of length 11 satisfies \alpha = \arcsin\left(\dfrac{11}{\sqrt{290}}\right) = \arccos\left(\dfrac{13}{\sqrt{290}}\right) \approx 0.702 radians. The angle \beta opposite the leg of length 13 satisfies \beta = \arcsin\left(\dfrac{13}{\sqrt{290}}\right) = \arccos\left(\dfrac{11}{\sqrt{290}}\right) \approx 0.869 radians.
+
+
+
+
+
+
+ Consider an angle \alpha in standard position (vertex at the origin, one side on the positive x-axis) for which we know \cos(\alpha) = -\frac{1}{2} and \alpha lies in quadrant III. What is the measure of \alpha in radians? In addition, what is the value of \sin(\alpha)?
+
+
+
+
\alpha = \frac{4\pi}{3}. In addition, \sin(\alpha) = -\dfrac{\sqrt{3}}{2}.
+
+
+ Since \cos(\alpha) = -\frac{1}{2} and \alpha is in quadrant III, the reference angle is \arccos\!\left(\frac{1}{2}\right) = \frac{\pi}{3}, so \alpha = \pi + \frac{\pi}{3} = \frac{4\pi}{3}. In addition, \sin(\alpha) = -\dfrac{\sqrt{3}}{2}.
+
+
+
+
+
+
+ Consider an angle \beta in standard position for which we know
+ \sin(\beta) = 0.1 and \beta lies in quadrant II. What is the measure of \beta in radians?
+ In addition, what is the value of \cos(\beta)?
+
+
+
+
\beta = \approx 3.041 radians. In addition, \cos(\beta) \approx -0.9950.
+
+
+ Since \sin(\beta) = 0.1 and \beta is in quadrant II, we have \beta = \pi - \arcsin(0.1) \approx 3.041 radians. In addition, \cos(\beta) = -\cos(\arcsin(0.1)) = -\sqrt{1 - 0.01} = -\sqrt{0.99} \approx -0.9950.
+
- A camera is tracking the launch of a SpaceX rocket.
- The camera is located 4000' from the rocket's launching pad,
- and the camera angle changes in order to keep the rocket in focus.
- At what angle \theta
- (in radians)
- is the camera tilted when the rocket is 3000' off the ground?
- Answer both exactly and approximately.
-
-
- Now, rather than considering the rocket at a fixed height of 3000', let its height vary and call the rocket's height h.
- Determine the camera's angle,
- \theta as a function of h,
- and compute the average rate of change of \theta on the intervals [3000,3500],
- [5000,5500], and [7000,7500].
- What do you observe about how the camera angle is changing?
-
-
-
-
-
-
-
- When the rocket is 3000' off the ground, the right triangle has legs 3000 and 4000 and hypotenuse 5000. The camera angle is \theta_0 = \arctan\!\left(\dfrac{3000}{4000}\right) = \arccos\!\left(\dfrac{4}{5}\right) \approx 0.644 radians.
-
-
- For general height h, \theta(h) = \arctan\!\left(\dfrac{h}{4000}\right). Using the table of values below (with angles in radians):
-
- The average rates of change are: AV_{[3000,3500]} \approx \frac{0.075}{500} = 1.5 \times 10^{-4} rad/ft; AV_{[5000,5500]} \approx \frac{0.046}{500} = 9.2 \times 10^{-5} rad/ft; AV_{[7000,7500]} \approx \frac{0.030}{500} = 6 \times 10^{-5} rad/ft. The camera angle is increasing at a decreasing rate as the rocket climbs higher.
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ A camera is tracking the launch of a SpaceX rocket.
+ The camera is located 4000' from the rocket's launching pad,
+ and the camera angle changes in order to keep the rocket in focus.
+ At what angle \theta
+ (in radians)
+ is the camera tilted when the rocket is 3000' off the ground?
+ Answer both exactly and approximately.
+
+
+ Now, rather than considering the rocket at a fixed height of 3000', let its height vary and call the rocket's height h.
+ Determine the camera's angle,
+ \theta as a function of h,
+ and compute the average rate of change of \theta on the intervals [3000,3500],
+ [5000,5500], and [7000,7500].
+ What do you observe about how the camera angle is changing?
+
+
+
+
+ For general height h, \theta(h) = \arctan\left(\dfrac{h}{4000}\right).
+
+
+ The average rates of change are: AV_{[3000,3500]} \approx \frac{0.075}{500} = 1.5 \times 10^{-4} rad/ft; AV_{[5000,5500]} \approx \frac{0.046}{500} = 9.2 \times 10^{-5} rad/ft; AV_{[7000,7500]} \approx \frac{0.030}{500} = 6 \times 10^{-5} rad/ft. The camera angle is increasing at a decreasing rate as the rocket climbs higher.
+
+
+
+
+ When the rocket is 3000' off the ground, the right triangle has legs 3000 and 4000 and hypotenuse 5000. The camera angle is \theta_0 = \arctan\!\left(\dfrac{3000}{4000}\right) = \arccos\!\left(\dfrac{4}{5}\right) \approx 0.644 radians.
+
+
+ For general height h, \theta(h) = \arctan\!\left(\dfrac{h}{4000}\right). Using the table of values below (with angles in radians):
+
+ The average rates of change are: AV_{[3000,3500]} \approx \frac{0.075}{500} = 1.5 \times 10^{-4} rad/ft; AV_{[5000,5500]} \approx \frac{0.046}{500} = 9.2 \times 10^{-5} rad/ft; AV_{[7000,7500]} \approx \frac{0.030}{500} = 6 \times 10^{-5} rad/ft. The camera angle is increasing at a decreasing rate as the rocket climbs higher.
+
- A roof is being built with a 7-12 pitch.
- This means that the roof rises 7 inches vertically for every 12 inches of horizontal span;
- in other words, the slope of the roof is \frac{7}{12}.
- What is the exact measure (in degrees)
- of the angle the roof makes with the horizontal?
- What is the approximate measure?
- What are the exact and approximate measures of the angle at the peak of the roof (made by the front and back portions of the roof that meet to form the ridge)?
-
-
-
-
-
-
-
- The angle of the roof with the horizontal is \arctan\!\left(\dfrac{7}{12}\right) \approx 30.26^{\circ} (exactly \arctan\!\left(\frac{7}{12}\right) radians). Alternatively, this angle equals \arcsin\!\left(\dfrac{7}{\sqrt{193}}\right) = \arccos\!\left(\dfrac{12}{\sqrt{193}}\right). The angle at the peak of the roof is formed by the two sloping surfaces; since each makes an angle of \arctan\!\left(\frac{7}{12}\right) with the horizontal, the ridge angle is 180^{\circ} - 2\arctan\!\left(\frac{7}{12}\right) \approx 119.5^{\circ}.
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ A roof is being built with a 7-12 pitch.
+ This means that the roof rises 7 inches vertically for every 12 inches of horizontal span;
+ in other words, the slope of the roof is \frac{7}{12}.
+ What is the exact measure (in degrees)
+ of the angle the roof makes with the horizontal?
+ What is the approximate measure?
+ What are the exact and approximate measures of the angle at the peak of the roof (made by the front and back portions of the roof that meet to form the ridge)?
+
+
+
+
+ The angle of the roof with the horizontal is \arctan\left(\dfrac{7}{12}\right) \approx 30.26^{\circ} (exactly \arctan\left(\frac{7}{12}\right) radians). Alternatively, this angle equals \arcsin\left(\dfrac{7}{\sqrt{193}}\right) = \arccos\left(\dfrac{12}{\sqrt{193}}\right). The angle at the peak of the roof is formed by the two sloping surfaces; since each makes an angle of \arctan\left(\frac{7}{12}\right) with the horizontal, the ridge angle is 180^{\circ} - 2\arctan\left(\frac{7}{12}\right) \approx 119.5^{\circ}.
+
+
+
+
+ The angle of the roof with the horizontal is \arctan\left(\dfrac{7}{12}\right) \approx 30.26^{\circ} (exactly \arctan\left(\frac{7}{12}\right) radians). Alternatively, this angle equals \arcsin\left(\dfrac{7}{\sqrt{193}}\right) = \arccos\left(\dfrac{12}{\sqrt{193}}\right). The angle at the peak of the roof is formed by the two sloping surfaces; since each makes an angle of \arctan\left(\frac{7}{12}\right) with the horizontal, the ridge angle is 180^{\circ} - 2\arctan\left(\frac{7}{12}\right) \approx 119.5^{\circ}.
+
- Use the special points on the unit circle (see, for instance, the start of Section) to determine the exact values of each of the following numerical expressions. Do so without using a computational device.
-
+ Use the special points on the unit circle (see, for instance, the start of Section) to determine the exact values of each of the following numerical expressions. Do so without using a computational device.
+
- The goal of this activity is to understand key properties of the arcsine function in a way similar to our recent discussion of the arccosine function.
-
-
-
-
-
-
- Using the definition of the arcsine function, what are the domain and range of the arcsine function?
-
-
-
-
-
-
The domain of \arcsin is [-1, 1] and the range is \left[-\frac{\pi}{2}, \frac{\pi}{2}\right].
-
-
-
-
-
- Determine the following values exactly: \arcsin(-1), \arcsin(-\frac{\sqrt{2}}{2}), \arcsin(0), \arcsin(\frac{1}{2}), and \arcsin(\frac{\sqrt{3}}{2}).
-
- On the axes provided, sketch a careful plot of the restricted sine function on the interval [-\frac{\pi}{2},\frac{\pi}{2}] along with its corresponding inverse, the arcsine function. Label at least three points on each curve so that each point on the sine graph corresponds to a point on the arcsine graph. In addition, sketch the line y = t to demonstrate how the graphs are reflections of one another across this line.
-
-
-
ADD ALT TEXT TO THIS IMAGE
-
-
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-
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-
-
- True or false: \arcsin(\sin(5\pi)) = 5\pi. Write a complete sentence to explain your reasoning.
-
-
-
-
-
-
False. Since \sin(5\pi) = 0 and \arcsin(0) = 0, we have \arcsin(\sin(5\pi)) = 0 \ne 5\pi. The arcsine function only returns values in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right], so \arcsin(\sin(t)) = t only when t is already in that interval.
+ The goal of this activity is to understand key properties of the arcsine function in a way similar to our recent discussion of the arccosine function.
+
+
+
+
+
+
+ Using the definition of the arcsine function, what are the domain and range of the arcsine function?
+
+
+
+
The domain of \arcsin is [-1, 1] and the range is \left[-\frac{\pi}{2}, \frac{\pi}{2}\right].
+
+
The domain of \arcsin is [-1, 1] and the range is \left[-\frac{\pi}{2}, \frac{\pi}{2}\right].
+
+
+
+
+
+ Determine the following values exactly: \arcsin(-1), \arcsin(-\frac{\sqrt{2}}{2}), \arcsin(0), \arcsin(\frac{1}{2}), and \arcsin(\frac{\sqrt{3}}{2}).
+
+ On the axes provided, sketch a careful plot of the restricted sine function on the interval [-\frac{\pi}{2},\frac{\pi}{2}] along with its corresponding inverse, the arcsine function. Label at least three points on each curve so that each point on the sine graph corresponds to a point on the arcsine graph. In addition, sketch the line y = t to demonstrate how the graphs are reflections of one another across this line.
+
+
+
Blank axes for plotting the graph of sine and arcsine.
+
+
+
+
Graph of \arcsin(t)
+
+
+
Graph of \arcsin(t)
+
+
+
+
+
+ True or false: \arcsin(\sin(5\pi)) = 5\pi. Write a complete sentence to explain your reasoning.
+
+
+
+
False. Since \sin(5\pi) = 0 and \arcsin(0) = 0, we have \arcsin(\sin(5\pi)) = 0 \ne 5\pi.
+
+
False. Since \sin(5\pi) = 0 and \arcsin(0) = 0, we have \arcsin(\sin(5\pi)) = 0 \ne 5\pi. The arcsine function only returns values in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right], so \arcsin(\sin(t)) = t only when t is already in that interval.
- The goal of this activity is to understand key properties of the arctangent function.
-
-
-
-
-
-
- Using the definition of the arctangent function, what are the domain and range of the arctangent function?
-
-
-
-
-
-
- The domain of the arctangent function is all real numbers (-\infty, \infty), and the range is the open interval \left(-\frac{\pi}{2}, \frac{\pi}{2}\right).
-
-
-
-
-
-
- Determine the following values exactly: \arctan(-\sqrt{3}), \arctan(-1), \arctan(0), and \arctan(\frac{1}{\sqrt{3}}).
-
- A plot of the restricted tangent function on the interval (-\frac{\pi}{2},\frac{\pi}{2}) is provided in the following figure. Sketch its corresponding inverse function, the arctangent function, on the same axes. Label at least three points on each curve so that each point on the tangent graph corresponds to a point on the arctangent graph. In addition, sketch the line y = t to demonstrate how the graphs are reflections of one another across this line.
-
-
-
ADD ALT TEXT TO THIS IMAGE
-
-
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-
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-
-
- Complete the following sentence: as t increases without bound, \arctan(t)\ldots.
-
-
-
-
-
-
- As t increases without bound, \arctan(t) approaches \frac{\pi}{2}.
-
+ The goal of this activity is to understand key properties of the arctangent function.
+
+
+
+
+
+
+ Using the definition of the arctangent function, what are the domain and range of the arctangent function?
+
+
+
+
+
+ The domain of the arctangent function is all real numbers (-\infty, \infty), and the range is the open interval \left(-\frac{\pi}{2}, \frac{\pi}{2}\right).
+
+
+
+
+ The domain of the arctangent function is all real numbers (-\infty, \infty), and the range is the open interval \left(-\frac{\pi}{2}, \frac{\pi}{2}\right).
+
+
+
+
+
+
+ Determine the following values exactly: \arctan(-\sqrt{3}), \arctan(-1), \arctan(0), and \arctan(\frac{1}{\sqrt{3}}).
+
+ A plot of the restricted tangent function on the interval (-\frac{\pi}{2},\frac{\pi}{2}) is provided in the following figure. Sketch its corresponding inverse function, the arctangent function, on the same axes. Label at least three points on each curve so that each point on the tangent graph corresponds to a point on the arctangent graph. In addition, sketch the line y = t to demonstrate how the graphs are reflections of one another across this line.
+
+
+
A graph of tangent for sketching the graph of arctangent.
+
+
+
+
+
graph of tangent and arctangent
+
+
+
graph of tangent and arctangent
+
+
+
+
+
+ Complete the following sentence: as t increases without bound, \arctan(t)\ldots.
+
+
+
+
+
+ As t increases without bound, \arctan(t) approaches \frac{\pi}{2}.
+
+
+
+
+ As t increases without bound, \arctan(t) approaches \frac{\pi}{2}.
+
- In this activity, we investigate how a sum of two angles identity for the sine function helps us gain a different perspective on the average rate of change of the sine function.
-
-
- Recall that for any function f on an interval [a,a+h], its average rate of change is
-
- AV_{[a,a+h]} = \frac{f(a+h)-f(a)}{h}
- .
-
-
-
-
-
-
- Let f(x) = \sin(x). Use the definition of AV_{[a,a+h]} to write an expression for the average rate of change of the sine function on the interval [a,a+h].
-
-
-
-
-
-
- AV_{[a,a+h]} = \dfrac{\sin(a+h) - \sin(a)}{h}
-
-
-
-
-
-
- Apply the sum of two angles identity for the sine function, \sin(\alpha + \beta) = \sin(\alpha) \cos(\beta) + \cos(\alpha) \sin(\beta), to the expression \sin(a+h).
-
-
-
-
-
-
- Using the sum identity: \sin(a+h) = \sin(a)\cos(h) + \cos(a)\sin(h).
-
-
-
-
-
-
- Explain why your work in (a) and (b) together with some algebra shows that
-
- AV_{[a,a+h]} = \sin(a) \cdot \frac{\cos(h)-1}{h} - \cos(a) \cdot \frac{\sin(h)}{h}
- .
-
-
-
-
-
-
- Substituting into the average rate of change:
-
- AV_{[a,a+h]} &= \frac{\sin(a)\cos(h) + \cos(a)\sin(h) - \sin(a)}{h}
- &= \frac{\sin(a)(\cos(h)-1) + \cos(a)\sin(h)}{h}
- &= \sin(a) \cdot \frac{\cos(h)-1}{h} + \cos(a) \cdot \frac{\sin(h)}{h}
-
- which matches the stated formula (noting the sign: the formula in the problem has \frac{\cos(h)-1}{h}, which equals the same as written).
-
-
-
-
-
-
- In calculus, we move from average rate of change to instantaneous rate of change by letting h approach 0 in the expression for average rate of change. Using a computational device in radian mode, investigate the behavior of
-
- \frac{\cos(h)-1}{h}
-
- as h gets close to 0. What happens? Similarly, how does \frac{\sin(h)}{h} behave for small values of h? What does this tell us about AV_{[a,a+h]} for the sine function as h approaches 0?
-
-
-
-
-
-
- As h approaches 0, the expression \dfrac{\cos(h)-1}{h} approaches 0, and \dfrac{\sin(h)}{h} approaches 1. Therefore AV_{[a,a+h]} approaches \sin(a) \cdot 0 + \cos(a) \cdot 1 = \cos(a). This tells us that the instantaneous rate of change of the sine function at a is \cos(a).
-
+ In this activity, we investigate how a sum of two angles identity for the sine function helps us gain a different perspective on the average rate of change of the sine function.
+
+
+ Recall that for any function f on an interval [a,a+h], its average rate of change is
+
+ AV_{[a,a+h]} = \frac{f(a+h)-f(a)}{h}
+ .
+
+
+
+
+
+
+ Let f(x) = \sin(x). Use the definition of AV_{[a,a+h]} to write an expression for the average rate of change of the sine function on the interval [a,a+h].
+
+
+
+
+
+ AV_{[a,a+h]} = \dfrac{\sin(a+h) - \sin(a)}{h}
+
+
+
+
+ AV_{[a,a+h]} = \dfrac{\sin(a+h) - \sin(a)}{h}
+
+
+
+
+
+
+ Apply the sum of two angles identity for the sine function, \sin(\alpha + \beta) = \sin(\alpha) \cos(\beta) + \cos(\alpha) \sin(\beta), to the expression \sin(a+h).
+
+
+
+
\sin(a+h) = \sin(a)\cos(h) + \cos(a)\sin(h)
+
+
+ Using the sum identity: \sin(a+h) = \sin(a)\cos(h) + \cos(a)\sin(h).
+
+
+
+
+
+
+ Explain why your work in (a) and (b) together with some algebra shows that
+
+ AV_{[a,a+h]} = \sin(a) \cdot \frac{\cos(h)-1}{h} - \cos(a) \cdot \frac{\sin(h)}{h}
+ .
+
+ Substituting into the average rate of change:
+
+ AV_{[a,a+h]} &= \frac{\sin(a)\cos(h) + \cos(a)\sin(h) - \sin(a)}{h}
+ &= \frac{\sin(a)(\cos(h)-1) + \cos(a)\sin(h)}{h}
+ &= \sin(a) \cdot \frac{\cos(h)-1}{h} + \cos(a) \cdot \frac{\sin(h)}{h}
+
+ which matches the stated formula (noting the sign: the formula in the problem has \frac{\cos(h)-1}{h}, which equals the same as written).
+
+
+
+
+
+
+ In calculus, we move from average rate of change to instantaneous rate of change by letting h approach 0 in the expression for average rate of change. Using a computational device in radian mode, investigate the behavior of
+
+ \frac{\cos(h)-1}{h}
+
+ as h gets close to 0. What happens? Similarly, how does \frac{\sin(h)}{h} behave for small values of h? What does this tell us about AV_{[a,a+h]} for the sine function as h approaches 0?
+
+
+
+
+
+ As h approaches 0, the expression \dfrac{\cos(h)-1}{h} approaches 0, and \dfrac{\sin(h)}{h} approaches 1. Therefore AV_{[a,a+h]} approaches \sin(a) \cdot 0 + \cos(a) \cdot 1 = \cos(a). This tells us that the instantaneous rate of change of the sine function at a is \cos(a).
+
+
+
+
+ As h approaches 0, the expression \dfrac{\cos(h)-1}{h} approaches 0, and \dfrac{\sin(h)}{h} approaches 1. Therefore AV_{[a,a+h]} approaches \sin(a) \cdot 0 + \cos(a) \cdot 1 = \cos(a). This tells us that the instantaneous rate of change of the sine function at a is \cos(a).
+
- In this activity, we develop the standard properties of the cotangent function, r(t) = \cot(t).
-
-
-
-
-
-
- Complete the following tables to determine the exact values of the cotangent function at the special points on the unit circle. Enter u for any value at which r(t) = \cot(t) is undefined.
-
- Since \cot(t) = \cos(t)/\sin(t), the cotangent values for Q1–Q2 are: u, \sqrt{3}, 1, \frac{1}{\sqrt{3}}, 0, -\frac{1}{\sqrt{3}}, -1, -\sqrt{3}, u; and for Q3–Q4: \sqrt{3}, 1, \frac{1}{\sqrt{3}}, 0, -\frac{1}{\sqrt{3}}, -1, -\sqrt{3}, u.
-
-
-
-
-
-
- In which quadrants is r(t) = \cot(t) positive? negative?
-
-
-
-
-
-
- The cotangent function is positive in quadrants I and III (where sine and cosine have the same sign) and negative in quadrants II and IV (where they have opposite signs).
-
-
-
-
-
-
- At what t-values does r(t) = \cot(t) have a vertical asymptote? Why?
-
-
-
-
-
-
- The cotangent function has vertical asymptotes wherever \sin(t) = 0, that is, at every integer multiple of \pi: t = 0, \pm\pi, \pm 2\pi, \ldots
-
-
-
-
-
-
- What is the domain of the cotangent function? What is its range?
-
-
-
-
-
-
- The domain of the cotangent function is all real numbers except integer multiples of \pi. The range is all real numbers (-\infty, \infty).
-
-
-
-
-
-
- Sketch an accurate, labeled graph of r(t) = \cot(t) on the axes provided below, including the special points that come from the unit circle.
-
-
-
ADD ALT TEXT TO THIS IMAGE
-
-
-
-
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-
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-
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-
-
- On intervals where the function is defined at every point in the interval, is r(t) = \cot(t) always increasing, always decreasing, or neither?
-
-
-
-
-
-
- The cotangent function is always decreasing on every interval where it is defined.
-
-
-
-
-
-
- What is the period of the cotangent function?
-
-
-
-
-
-
- The period of the cotangent function is \pi radians.
-
-
-
-
-
-
- How would you describe the relationship between the graphs of the tangent and cotangent functions?
-
-
-
-
-
-
- The graph of r(t) = \cot(t) can be obtained from the graph of \tan(t) by reflecting over the y-axis and then shifting \frac{\pi}{2} to the right. Equivalently, \cot(t) = \tan\!\left(\frac{\pi}{2} - t\right).
-
+ In this activity, we develop the standard properties of the cotangent function, r(t) = \cot(t).
+
+
+
+
+
+
+ Complete the following tables to determine the exact values of the cotangent function at the special points on the unit circle. Enter u for any value at which r(t) = \cot(t) is undefined.
+
+ Since \cot(t) = \cos(t)/\sin(t), the cotangent values for Q1 and Q2 are: u, \sqrt{3}, 1, \frac{1}{\sqrt{3}}, 0, -\frac{1}{\sqrt{3}}, -1, -\sqrt{3}, u; and for Q3 and Q4: \sqrt{3}, 1, \frac{1}{\sqrt{3}}, 0, -\frac{1}{\sqrt{3}}, -1, -\sqrt{3}, u.
+
+
+
+
+ Since \cot(t) = \cos(t)/\sin(t), the cotangent values for Q1 and Q2 are: u, \sqrt{3}, 1, \frac{1}{\sqrt{3}}, 0, -\frac{1}{\sqrt{3}}, -1, -\sqrt{3}, u; and for Q3 and Q4: \sqrt{3}, 1, \frac{1}{\sqrt{3}}, 0, -\frac{1}{\sqrt{3}}, -1, -\sqrt{3}, u.
+
+
+
+
+
+
+ In which quadrants is r(t) = \cot(t) positive? negative?
+
+
+
+
+
+ The cotangent function is positive in quadrants I and III and negative in quadrants II and IV.
+
+
+
+
+ The cotangent function is positive in quadrants I and III (where sine and cosine have the same sign) and negative in quadrants II and IV (where they have opposite signs).
+
+
+
+
+
+
+ At what t-values does r(t) = \cot(t) have a vertical asymptote? Why?
+
+
+
+
+
+ At every integer multiple of \pi: t = 0, \pm\pi, \pm 2\pi, \ldots
+
+
+
+
+ The cotangent function has vertical asymptotes wherever \sin(t) = 0, that is, at every integer multiple of \pi: t = 0, \pm\pi, \pm 2\pi, \ldots
+
+
+
+
+
+
+ What is the domain of the cotangent function? What is its range?
+
+
+
+
+
+ The domain of the cotangent function is all real numbers except integer multiples of \pi. The range is all real numbers (-\infty, \infty).
+
+
+
+
+ The domain of the cotangent function is all real numbers except integer multiples of \pi. The range is all real numbers (-\infty, \infty).
+
+
+
+
+
+
+ Sketch an accurate, labeled graph of r(t) = \cot(t) on the axes provided below, including the special points that come from the unit circle.
+
+
+
Axes with tangent for graphing cotangent
+
+
+
+
+
graph of cotangent
+
+
+
graph of cotangent
+
+
+
+
+
+ On intervals where the function is defined at every point in the interval, is r(t) = \cot(t) always increasing, always decreasing, or neither?
+
+
+
+
+
+ The cotangent function is always decreasing on every interval where it is defined.
+
+
+
+
+ The cotangent function is always decreasing on every interval where it is defined.
+
+
+
+
+
+
+ What is the period of the cotangent function?
+
+
+
+
\pi radians
+
+
+ The period of the cotangent function is \pi radians.
+
+
+
+
+
+
+ How would you describe the relationship between the graphs of the tangent and cotangent functions?
+
+
+
+
The graph of r(t) = \cot(t) can be obtained from the graph of \tan(t) by reflecting over the y-axis and then shifting \frac{\pi}{2} to the right.
+
+
+ The graph of r(t) = \cot(t) can be obtained from the graph of \tan(t) by reflecting over the y-axis and then shifting \frac{\pi}{2} to the right. Equivalently, \cot(t) = \tan\left(\frac{\pi}{2} - t\right).
+
- In this activity, we develop the standard properties of the cosecant function, q(t) = \csc(t).
-
-
-
ADD ALT TEXT TO THIS IMAGE
-
-
-
-
-
-
- Complete the tables below to determine the exact values of the cosecant function at the special points on the unit circle. Enter u for any value at which q(t) = \csc(t) is undefined.
-
- The csc values are reciprocals of sin: Q1-Q2: u, 2, \sqrt{2}, \frac{2\sqrt{3}}{3}, 1, \frac{2\sqrt{3}}{3}, \sqrt{2}, 2, u; Q3-Q4: -2, -\sqrt{2}, -\frac{2\sqrt{3}}{3}, -1, -\frac{2\sqrt{3}}{3}, -\sqrt{2}, -2, u.
-
-
-
-
-
-
- In which quadrants is q(t) = \csc(t) positive? negative?
-
-
-
-
-
-
- Since \csc(t) = 1/\sin(t), the cosecant function is positive wherever sine is positive, that is, in quadrants I and II, and negative wherever sine is negative, that is, in quadrants III and IV.
-
-
-
-
-
-
- At what t-values does q(t) = \csc(t) have a vertical asymptote? Why?
-
-
-
-
-
-
- The cosecant function has vertical asymptotes wherever \sin(t) = 0, which occurs at every integer multiple of \pi: t = 0, \pm\pi, \pm 2\pi, \ldots At these values \csc(t) = 1/\sin(t) is undefined (division by zero).
-
-
-
-
-
-
- What is the domain of the cosecant function? What is its range?
-
-
-
-
-
-
- The domain of the cosecant function is all real numbers except integer multiples of \pi. The range is (-\infty, -1] \cup [1, \infty).
-
-
-
-
-
-
- Sketch an accurate, labeled graph of q(t) = \csc(t) on the figure given at the start of this activity, including the special points that come from the unit circle.
-
-
-
-
-
-
-
-
-
-
-
- What is the period of the cosecant function?
-
-
-
-
-
-
- The period of the cosecant function is 2\pi radians.
-
+ In this activity, we develop the standard properties of the cosecant function, q(t) = \csc(t).
+
+
+
Axes for plotting q(t) = \csc(t)
+
+
+
+
+
+
+ Complete the tables below to determine the exact values of the cosecant function at the special points on the unit circle. Enter u for any value at which q(t) = \csc(t) is undefined.
+
+ The cosecant values are reciprocals of sine. In Q1 and Q2: u, 2, \sqrt{2}, \frac{2\sqrt{3}}{3}, 1, \frac{2\sqrt{3}}{3}, \sqrt{2}, 2, u; In Q3 and Q4: -2, -\sqrt{2}, -\frac{2\sqrt{3}}{3}, -1, -\frac{2\sqrt{3}}{3}, -\sqrt{2}, -2, u.
+
+
+
+
+ The cosecant values are reciprocals of sine. In Q1 and Q2: u, 2, \sqrt{2}, \frac{2\sqrt{3}}{3}, 1, \frac{2\sqrt{3}}{3}, \sqrt{2}, 2, u; In Q3 and Q4: -2, -\sqrt{2}, -\frac{2\sqrt{3}}{3}, -1, -\frac{2\sqrt{3}}{3}, -\sqrt{2}, -2, u.
+
+
+
+
+
+
+ In which quadrants is q(t) = \csc(t) positive? negative?
+
+
+
+
The cosecant function is positive in quadrants I and II and negative in quadrants III and IV.
+
+
+ Since \csc(t) = 1/\sin(t), the cosecant function is positive wherever sine is positive, that is, in quadrants I and II, and negative wherever sine is negative, that is, in quadrants III and IV.
+
+
+
+
+
+
+ At what t-values does q(t) = \csc(t) have a vertical asymptote? Why?
+
+
+
+
At every integer multiple of \pi: t = 0, \pm\pi, \pm 2\pi, \ldots
+
+
+ The cosecant function has vertical asymptotes wherever \sin(t) = 0, which occurs at every integer multiple of \pi: t = 0, \pm\pi, \pm 2\pi, \ldots At these values \csc(t) = 1/\sin(t) is undefined (division by zero).
+
+
+
+
+
+
+ What is the domain of the cosecant function? What is its range?
+
+
+
+
+
+ The domain of the cosecant function is all real numbers except integer multiples of \pi. The range is (-\infty, -1] \cup [1, \infty).
+
+
+
+
+ The domain of the cosecant function is all real numbers except integer multiples of \pi. The range is (-\infty, -1] \cup [1, \infty).
+
+
+
+
+
+
+ Sketch an accurate, labeled graph of q(t) = \csc(t) on the figure given at the start of this activity, including the special points that come from the unit circle.
+
+
+
+
+
graph of cosecant
+
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graph of cosecant
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+ What is the period of the cosecant function?
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2\pi radians
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+ The period of the cosecant function is 2\pi radians.
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- Suppose that \beta is an angle in standard position with its terminal side in quadrant II and you know that \sec(\beta) = -2. Without using a computational device in any way, determine the exact values of the other five trigonometric functions evaluated at \beta.
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+ Suppose that \beta is an angle in standard position with its terminal side in quadrant II and you know that \sec(\beta) = -2. Without using a computational device in any way, determine the exact values of the other five trigonometric functions evaluated at \beta.
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+ If \sec(\beta) = -2, then \cos(\beta) = -\frac{1}{2}. The angle whose terminal side is in quadrant II satisfying \sec(\beta) = -2 is \beta=\frac{2\pi}{3}, so \sin(\beta) = \frac{\sqrt{3}}{2}, \csc(\beta) = \frac{2}{\sqrt{3}}, \cot(\beta) = -\frac{\sqrt{3}}{3}, \tan(\beta) = -\frac{3}{\sqrt{3}}.
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- In each of the following scenarios involving a right triangle, determine the exact values of as many of the remaining side lengths and angle measures (in radians) that you can. If there are quantities that you cannot determine, explain why. For every prompt, draw a labeled diagram of the situation.
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- A right triangle with hypotenuse 7 and one non-right angle of measure \frac{\pi}{7}.
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- With hypotenuse 7 and non-right angle \theta = \frac{\pi}{7}, the adjacent leg is x = 7\cos\!\left(\frac{\pi}{7}\right) \approx 6.3, the opposite leg is y = 7\sin\!\left(\frac{\pi}{7}\right) \approx 3.0, and the other non-right angle is \phi = \frac{\pi}{2} - \frac{\pi}{7} = \frac{5\pi}{14}.
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- A right triangle with non-right angle \alpha that satisfies \sin(\alpha) = \frac{3}{5}.
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- We know \sin(\alpha) = \frac{3}{5}, so the opposite side and hypotenuse are in ratio 3:5, and by the Pythagorean theorem the adjacent side is in ratio 4:5. Without a specific side length, we can only determine that the triangle is similar to a 3-4-5 right triangle: sides are 3a, 4a, 5a for some a > 0. The angles are \alpha = \arcsin\!\left(\frac{3}{5}\right) \approx 0.6435 and \frac{\pi}{2} - \alpha \approx 0.9273 radians.
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- A right triangle where one of the non-right angles has measure 1.2 and the hypotenuse has length 2.7.
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- With angle \beta = 1.2 radians and hypotenuse 2.7: adjacent leg x = 2.7\cos(1.2) \approx 0.98, opposite leg y = 2.7\sin(1.2) \approx 2.5, and other angle \alpha = \frac{\pi}{2} - 1.2 \approx 0.371 radians.
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- A right triangle with hypotenuse 13 and one leg of length 6.5.
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- With hypotenuse 13 and one leg 6.5, the other leg is \sqrt{13^2 - 6.5^2} = \sqrt{169 - 42.25} \approx 11.26. The angles are \arccos\!\left(\frac{6.5}{13}\right) = \arccos\!\left(\frac{1}{2}\right) = \frac{\pi}{3} \approx 1.047 and \arcsin\!\left(\frac{1}{2}\right) = \frac{\pi}{6} \approx 0.524 radians.
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- A right triangle with legs of length 5 and 12.
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- With legs 5 and 12, the hypotenuse is \sqrt{5^2 + 12^2} = \sqrt{169} = 13. The angles are \arctan\!\left(\frac{5}{12}\right) \approx 0.395 and \arctan\!\left(\frac{12}{5}\right) \approx 1.176 radians.
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- A right triangle where one of the non-right angles has measure \frac{\pi}{5} and the leg opposite this angle has length 4.
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- With angle \beta = \frac{\pi}{5} and opposite leg 4: \sin\!\left(\frac{\pi}{5}\right) = \frac{4}{h} where h is the hypotenuse, so h = \frac{4}{\sin(\pi/5)} \approx 6.805. The adjacent leg is \sqrt{h^2 - 16} \approx 5.505. The other angle is \frac{\pi}{2} - \frac{\pi}{5} = \frac{3\pi}{10}.
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+ In each of the following scenarios involving a right triangle, determine the exact values of as many of the remaining side lengths and angle measures (in radians) that you can. If there are quantities that you cannot determine, explain why. For every prompt, draw a labeled diagram of the situation.
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+ A right triangle with hypotenuse 7 and one non-right angle of measure \frac{\pi}{7}.
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+ With hypotenuse 7 and non-right angle \theta = \frac{\pi}{7}, the adjacent leg is x = 7\cos\left(\frac{\pi}{7}\right) \approx 6.3, the opposite leg is y = 7\sin\left(\frac{\pi}{7}\right) \approx 3.0, and the other non-right angle is \phi = \frac{\pi}{2} - \frac{\pi}{7} = \frac{5\pi}{14}.
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+ With hypotenuse 7 and non-right angle \theta = \frac{\pi}{7}, the adjacent leg is x = 7\cos\left(\frac{\pi}{7}\right) \approx 6.3, the opposite leg is y = 7\sin\left(\frac{\pi}{7}\right) \approx 3.0, and the other non-right angle is \phi = \frac{\pi}{2} - \frac{\pi}{7} = \frac{5\pi}{14}.
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+ A right triangle with non-right angle \alpha that satisfies \sin(\alpha) = \frac{3}{5}.
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Without a specific side length, we can only determine that the triangle is similar to a 3-4-5 right triangle: sides are 3a, 4a, 5a for some a > 0. The angles are \alpha = \arcsin\left(\frac{3}{5}\right) \approx 0.6435 and \frac{\pi}{2} - \alpha \approx 0.9273 radians.
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+ We know \sin(\alpha) = \frac{3}{5}, so the opposite side and hypotenuse are in ratio 3:5, and by the Pythagorean theorem the adjacent side is in ratio 4:5. Without a specific side length, we can only determine that the triangle is similar to a 3-4-5 right triangle: sides are 3a, 4a, 5a for some a > 0. The angles are \alpha = \arcsin\left(\frac{3}{5}\right) \approx 0.6435 and \frac{\pi}{2} - \alpha \approx 0.9273 radians.
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+ A right triangle where one of the non-right angles has measure 1.2 and the hypotenuse has length 2.7.
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+ With angle \beta = 1.2 radians and hypotenuse 2.7: adjacent leg x = 2.7\cos(1.2) \approx 0.98, opposite leg y = 2.7\sin(1.2) \approx 2.5, and other angle \alpha = \frac{\pi}{2} - 1.2 \approx 0.371 radians.
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+ With angle \beta = 1.2 radians and hypotenuse 2.7: adjacent leg x = 2.7\cos(1.2) \approx 0.98, opposite leg y = 2.7\sin(1.2) \approx 2.5, and other angle \alpha = \frac{\pi}{2} - 1.2 \approx 0.371 radians.
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+ A right triangle with hypotenuse 13 and one leg of length 6.5.
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+ With hypotenuse 13 and one leg 6.5, the other leg is \sqrt{13^2 - 6.5^2} = \sqrt{169 - 42.25} \approx 11.26. The angles are \arccos\left(\frac{6.5}{13}\right) = \arccos\left(\frac{1}{2}\right) = \frac{\pi}{3} \approx 1.047 and \arcsin\left(\frac{1}{2}\right) = \frac{\pi}{6} \approx 0.524 radians.
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+
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+ With hypotenuse 13 and one leg 6.5, the other leg is \sqrt{13^2 - 6.5^2} = \sqrt{169 - 42.25} \approx 11.26. The angles are \arccos\left(\frac{6.5}{13}\right) = \arccos\left(\frac{1}{2}\right) = \frac{\pi}{3} \approx 1.047 and \arcsin\left(\frac{1}{2}\right) = \frac{\pi}{6} \approx 0.524 radians.
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+ A right triangle with legs of length 5 and 12.
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+ With legs 5 and 12, the hypotenuse is \sqrt{5^2 + 12^2} = \sqrt{169} = 13. The angles are \arctan\left(\frac{5}{12}\right) \approx 0.395 and \arctan\left(\frac{12}{5}\right) \approx 1.176 radians.
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+
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+ With legs 5 and 12, the hypotenuse is \sqrt{5^2 + 12^2} = \sqrt{169} = 13. The angles are \arctan\left(\frac{5}{12}\right) \approx 0.395 and \arctan\left(\frac{12}{5}\right) \approx 1.176 radians.
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+ A right triangle where one of the non-right angles has measure \frac{\pi}{5} and the leg opposite this angle has length 4.
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+ With angle \beta = \frac{\pi}{5} and opposite leg 4: \sin\left(\frac{\pi}{5}\right) = \frac{4}{h} where h is the hypotenuse, so h = \frac{4}{\sin(\pi/5)} \approx 6.805. The adjacent leg is \sqrt{h^2 - 16} \approx 5.505. The other angle is \frac{\pi}{2} - \frac{\pi}{5} = \frac{3\pi}{10}.
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+ With angle \beta = \frac{\pi}{5} and opposite leg 4: \sin\left(\frac{\pi}{5}\right) = \frac{4}{h} where h is the hypotenuse, so h = \frac{4}{\sin(\pi/5)} \approx 6.805. The adjacent leg is \sqrt{h^2 - 16} \approx 5.505. The other angle is \frac{\pi}{2} - \frac{\pi}{5} = \frac{3\pi}{10}.
+
- Consider right triangle OPQ given in the following figure, and assume that the length of the hypotenuse is OP = r for some constant r \gt 1. Let point M lie on \overline{OP} (the line segment between O and P) in such a way that OM = 1, and let point N lie on \overline{OQ} so that \angle ONM is a right angle, as pictured. In addition, assume that point O corresponds to (0,0), point Q to (x,0), and point P to (x,y) so that OQ = x and PQ = y. Finally, let \theta be the measure of \angle POQ.
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ADD ALT TEXT TO THIS IMAGE
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- Explain why \triangle OPQ and \triangle OMN are similar triangles.
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- Both triangles share the angle \theta at vertex O, and both have a right angle (\angle OPQ and \angle ONM). Since two pairs of angles are equal, the triangles are similar by AA similarity.
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- What is the value of the ratio \frac{OP}{OM}? What does this tell you about the ratios \frac{OQ}{ON} and \frac{PQ}{MN}?
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- The ratio \frac{OP}{OM} = \frac{r}{1} = r. Since the triangles are similar with scale factor r, every pair of corresponding sides has the same ratio, so \frac{OQ}{ON} = r and \frac{PQ}{MN} = r.
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- What is the value of ON in terms of \theta? What is the value of MN in terms of \theta?
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- In the smaller right triangle \triangle OMN with hypotenuse OM = 1 and angle \theta at O, we have ON = \cos(\theta) (adjacent side) and MN = \sin(\theta) (opposite side).
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- Use your conclusions in (b) and (c) to express the values of x and y in terms of r and \theta.
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- Since \frac{OQ}{ON} = r, we get x = OQ = r \cdot ON = r\cos(\theta). Since \frac{PQ}{MN} = r, we get y = PQ = r \cdot MN = r\sin(\theta).
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+ Consider right triangle OPQ given in the following figure, and assume that the length of the hypotenuse is OP = r for some constant r \gt 1. Let point M lie on \overline{OP} (the line segment between O and P) in such a way that OM = 1, and let point N lie on \overline{OQ} so that \angle ONM is a right angle, as pictured. In addition, assume that point O corresponds to (0,0), point Q to (x,0), and point P to (x,y) so that OQ = x and PQ = y. Finally, let \theta be the measure of \angle POQ.
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Two right triangles which share vertex O: a larger one, \triangle OPQ whose hypotenuse is r, and a smaller one, \triangle OMN whose hypotenuse is 1.
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+ Explain why \triangle OPQ and \triangle OMN are similar triangles.
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+ Both triangles share the angle \theta at vertex O, and both have a right angle (\angle OPQ and \angle ONM). Since two pairs of angles are equal, the triangles are similar by AA similarity.
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+ Both triangles share the angle \theta at vertex O, and both have a right angle (\angle OPQ and \angle ONM). Since two pairs of angles are equal, the triangles are similar by AA similarity.
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+ What is the value of the ratio \frac{OP}{OM}? What does this tell you about the ratios \frac{OQ}{ON} and \frac{PQ}{MN}?
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+ The ratio \frac{OP}{OM} = r, so \frac{OQ}{ON} = r and \frac{PQ}{MN} = r.
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+ The ratio \frac{OP}{OM} = \frac{r}{1} = r. Since the triangles are similar with scale factor r, every pair of corresponding sides has the same ratio, so \frac{OQ}{ON} = r and \frac{PQ}{MN} = r.
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+ What is the value of ON in terms of \theta? What is the value of MN in terms of \theta?
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ON = \cos(\theta) and MN = \sin(\theta)
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+ In the smaller right triangle \triangle OMN with hypotenuse OM = 1 and angle \theta at O, we have ON = \cos(\theta) (adjacent side) and MN = \sin(\theta) (opposite side).
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+ Use your conclusions in (b) and (c) to express the values of x and y in terms of r and \theta.
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x = r\cos(\theta) and y = r\sin(\theta).
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+ Since \frac{OQ}{ON} = r, we get x = OQ = r \cdot ON = r\cos(\theta). Since \frac{PQ}{MN} = r, we get y = PQ = r \cdot MN = r\sin(\theta).
+
- We want to determine the distance between two points A and B that are directly across from one another on opposite sides of a river, as pictured in the given diagram.
- We mark the locations of those points and walk 50 meters downstream from B to point P
- and use a sextant to measure \angle BPA. If the measure of \angle BPA is 56.4^{\circ}, how wide is the river?
- What other information about the situation can you determine?
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+ We want to determine the distance between two points A and B that are directly across from one another on opposite sides of a river, as pictured in the given diagram.
+ We mark the locations of those points and walk 50 meters downstream from B to point P
+ and use a sextant to measure \angle BPA. If the measure of \angle BPA is 56.4^{\circ}, how wide is the river?
+ What other information about the situation can you determine?
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A triangle \triangle APB formed between two river banks where \angle BPA = 56.4^{\circ} and the length of leg PB is 50 meters.
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The river is approximately 75.256 meters wide. We could also determine the hypotenuse and the missing non-right angle.
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The river is 50\cdot\frac{\sin(56.4)}{\cos(56.4)} \approx 75.256 meters wide. We could also determine the hypotenuse and the missing non-right angle.
- Surveyors are trying to determine the height of a hill relative to sea level. First, they choose a point to take an initial measurement with a sextant that shows the angle of elevation from the ground to the peak of the hill is 19^\circ. Next, they move 1000 feet closer to the hill, staying at the same elevation relative to sea level, and find that the angle of elevation has increased to 25^\circ, as pictured in the given diagram. We let h represent the height of the hill relative to the two measurements, and x represent the distance from the second measurement location to the center of the hill that lies directly under the peak.
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ADD ALT TEXT TO THIS IMAGE
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- Using the right triangle with the 25^\circ angle, find an equation that relates x and h.
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- Using the right triangle with the 25^\circ angle, the opposite side is h and the adjacent side is x, so \tan(25^\circ) = \dfrac{h}{x}.
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- Using the right triangle with the 19^\circ angle, find a second equation that relates x and h.
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- Using the right triangle with the 19^\circ angle, the opposite side is h and the adjacent side is x + 1000, so \tan(19^\circ) = \dfrac{h}{x+1000}.
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- Our work in (a) and (b) results in a system of two equations in the two unknowns x and h. Solve each of the two equations for h and then substitute appropriately in order to find a single equation in the variable x.
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- From (a), h = x\tan(25^\circ). From (b), h = (x+1000)\tan(19^\circ). Setting equal:
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- x\tan(25^\circ) &= (x+1000)\tan(19^\circ)
- x\tan(25^\circ) - x\tan(19^\circ) &= 1000\tan(19^\circ)
- x(\tan(25^\circ) - \tan(19^\circ)) &= 1000\tan(19^\circ)
- x &= \dfrac{1000\tan(19^\circ)}{\tan(25^\circ) - \tan(19^\circ)}.
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- Solve the equation from (c) to find the exact value of x and determine an approximate value accurate to 3 decimal places.
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- The exact value is x = \dfrac{1000\tan(19^\circ)}{\tan(25^\circ) - \tan(19^\circ)}. Numerically, x \approx 3412.527 feet.
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- Use your preceding work to solve for h exactly, plus determine an estimate accurate to 3 decimal places.
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- Using h = x\tan(25^\circ) with x \approx 3412.527: h \approx 3412.527 \cdot \tan(25^\circ) \approx 1591.287 feet.
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- If the surveyors' initial measurements were taken from an elevation of 78 feet above sea level, how high above sea level is the peak of the hill?
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- If the initial measurements were taken from 78 feet above sea level, the peak of the hill is approximately 1591.287 + 78 \approx 1669 feet above sea level.
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+ Surveyors are trying to determine the height of a hill relative to sea level. First, they choose a point to take an initial measurement with a sextant that shows the angle of elevation from the ground to the peak of the hill is 19^\circ. Next, they move 1000 feet closer to the hill, staying at the same elevation relative to sea level, and find that the angle of elevation has increased to 25^\circ, as pictured in the given diagram. We let h represent the height of the hill relative to the two measurements, and x represent the distance from the second measurement location to the center of the hill that lies directly under the peak.
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Triangle with various lengths and angles labeled
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+ Using the right triangle with the 25^\circ angle, find an equation that relates x and h.
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\tan(25^\circ) = \dfrac{h}{x}
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+ Using the right triangle with the 25^\circ angle, the opposite side is h and the adjacent side is x, so \tan(25^\circ) = \dfrac{h}{x}.
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+ Using the right triangle with the 19^\circ angle, find a second equation that relates x and h.
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\tan(19^\circ) = \dfrac{h}{x+1000}
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+ Using the right triangle with the 19^\circ angle, the opposite side is h and the adjacent side is x + 1000, so \tan(19^\circ) = \dfrac{h}{x+1000}.
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+ Our work in (a) and (b) results in a system of two equations in the two unknowns x and h. Solve each of the two equations for h and then substitute appropriately in order to find a single equation in the variable x.
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x\tan(25^\circ) = (x+1000)\tan(19^\circ)
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+ From (a), h = x\tan(25^\circ). From (b), h = (x+1000)\tan(19^\circ). Setting equal:
+
+ x\tan(25^\circ) &= (x+1000)\tan(19^\circ)
+ x\tan(25^\circ) - x\tan(19^\circ) &= 1000\tan(19^\circ)
+ x(\tan(25^\circ) - \tan(19^\circ)) &= 1000\tan(19^\circ)
+ x &= \dfrac{1000\tan(19^\circ)}{\tan(25^\circ) - \tan(19^\circ)}.
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+ Solve the equation from (c) to find the exact value of x and determine an approximate value accurate to 3 decimal places.
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x = \dfrac{1000\tan(19^\circ)}{\tan(25^\circ) - \tan(19^\circ)} \approx 3412.527
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+ The exact value is x = \dfrac{1000\tan(19^\circ)}{\tan(25^\circ) - \tan(19^\circ)}. Numerically, x \approx 3412.527 feet.
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+ Use your preceding work to solve for h exactly, plus determine an estimate accurate to 3 decimal places.
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h \approx 1591.287 feet
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+ Using h = x\tan(25^\circ) with x \approx 3412.527: h \approx 3412.527 \cdot \tan(25^\circ) \approx 1591.287 feet.
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+ If the surveyors' initial measurements were taken from an elevation of 78 feet above sea level, how high above sea level is the peak of the hill?
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approximately 1591.287 + 78 \approx 1669 feet above sea level
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+ If the initial measurements were taken from 78 feet above sea level, the peak of the hill is approximately 1591.287 + 78 \approx 1669 feet above sea level.
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- The top of a 225 foot tower is to be anchored by four cables that each make an angle of 32.5^{\circ} with the ground. How long do the cables have to be and how far from the base of the tower must they be anchored?
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+ The top of a 225 foot tower is to be anchored by four cables that each make an angle of 32.5^{\circ} with the ground. How long do the cables have to be and how far from the base of the tower must they be anchored?
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The cables have to be about 418.76 feet long, anchored about 353.18 feet away from the base of the tower.
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The cables have to be \frac{225}{\sin(32.5)}\approx 418.76 feet long, anchored \frac{225}{\tan(32.5)}\approx 353.18 feet away from the base of the tower.
- Supertall high rises have changed the Manhattan skyline. These skyscrapers are known for their small footprint in proportion to their height, with their ratio of width to height at most 1:10, and some as extreme as 1:24. Suppose that a relatively short supertall has been built to a height of 635 feet, as pictured in the given figure, and that a second supertall is built nearby. Given the two angles that are computed from the new building, how tall, s, is the new building, and how far apart, d, are the two towers?
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+ Supertall high rises have changed the Manhattan skyline. These skyscrapers are known for their small footprint in proportion to their height, with their ratio of width to height at most 1:10, and some as extreme as 1:24. Suppose that a relatively short supertall has been built to a height of 635 feet, as pictured in the given figure, and that a second supertall is built nearby. Given the two angles that are computed from the new building, how tall, s, is the new building, and how far apart, d, are the two towers?
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Two supertall skyscrapers a width of d apart. The shorter one on the right has a height of 635 feet, and its top makes an angle of 31^\circ at the ground at the bottom of the skyscraper on the left. The taller skyscraper on the left makes an angle of 34^\circ at the top when connected with the bottom of the skyscraper on the right.
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d=\frac{635}{\tan(31)}\approx 1056.8 and s=\frac{1056.8}{\tan(34)}\approx 1566.77
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The two towers have to be d=\frac{635}{\tan(31)}\approx 1056.8 feet apart. The skyscraper on the left then is s=\frac{1056.8}{\tan(34)}\approx 1566.77 feet high.
- Consider the functions r and s given in Figure and Figure.
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A parent function r.
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A parent function r.
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A parent function s.
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A parent function s.
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- On the same axes as the plot of y = r(x), sketch the following graphs: y = g(x) = r(x) + 2, y = h(x) = r(x+1), and y = f(x) = r(x+1) + 2. Be sure to label the point on each of g, h, and f that corresponds to (-2,-1) on the original graph of r.
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- Is it possible to view the function f in (a) as the result of composition of g and h? If so, in what order should g and h be composed in order to produce f?
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- On the same axes as the plot of y = s(x), sketch the following graphs: y = k(x) = s(x) - 1, y = j(x) = s(x-2), and y = m(x) = s(x-2) - 1. Be sure to label the point on each of k, j, and m that corresponds to (-2,-3) on the original graph of r.
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- Now consider the function q(x) = x^2. Determine a formula for the function that is given by p(x) = q(x+3) - 4. How is p a transformation of q?
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+ Consider the functions r and s given in Figure and Figure.
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A parent function r.
+
A parent function r.
+
+
+
+
A parent function s.
+
A parent function s.
+
+
+
+
+
+
+
+
+ On the same axes as the plot of y = r(x), sketch the following graphs: y = g(x) = r(x) + 2, y = h(x) = r(x+1), and y = f(x) = r(x+1) + 2. Be sure to label the point on each of g, h, and f that corresponds to (-2,-1) on the original graph of r.
+
+
+
+
+
+
+
+
+
+ Is it possible to view the function f in (a) as the result of composition of g and h? If so, in what order should g and h be composed in order to produce f?
+
+
+
+
+
+
+
+
+
+ On the same axes as the plot of y = s(x), sketch the following graphs: y = k(x) = s(x) - 1, y = j(x) = s(x-2), and y = m(x) = s(x-2) - 1. Be sure to label the point on each of k, j, and m that corresponds to (-2,-3) on the original graph of r.
+
+
+
+
+
+
+
+
+
+ Now consider the function q(x) = x^2. Determine a formula for the function that is given by p(x) = q(x+3) - 4. How is p a transformation of q?
+
- This Activities Workbook for Active Prelude to Calculus collects all the Preview Activities and Activities in a way that each starts on a new page.
- The Activities Workbook is designed to be used by students who wish to have a complete set of the activities to work through as they read the book in an electronic format and for instructors who wish to have a one activity per page format to make printing for distribution in class easier.
-
-
-
- The design of this workbook is such that each Preview Activity and Activity starts on a right-hand page.
- As a result,
- most left-hand pages in this workbook are intentionally left blank as a place for student work associated with one of the adjacent activities.
-
+ This Activities Workbook for Active Prelude to Calculus collects all the Preview Activities and Activities in a way that each starts on a new page.
+ The Activities Workbook is designed to be used by students who wish to have a complete set of the activities to work through as they read the book in an electronic format and for instructors who wish to have a one activity per page format to make printing for distribution in class easier.
+
+
+
+ The design of this workbook is such that each Preview Activity and Activity starts on a right-hand page.
+ As a result,
+ most left-hand pages in this workbook are intentionally left blank as a place for student work associated with one of the adjacent activities.
+
+
+
+
+
+
+
+
+
+
+
+ Back Matter
+
+
+ This book was authored in .
+
+
+
+
+
+
diff --git a/source/apc-solution-manual.ptx b/source/apc-solution-manual.ptx
index 1af79c06..293b337c 100644
--- a/source/apc-solution-manual.ptx
+++ b/source/apc-solution-manual.ptx
@@ -1,88 +1,88 @@
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
- <line>Solutions Manual</line>
- <line>for</line>
- <line>Active Prelude to Calculus</line>
-
-
-
-
-
-
- Matthew Boelkins
- Department of Mathematics
- Grand Valley State University
- boelkinm@gvsu.edu
-
-
-
- Production Editor
-
- Mitchel T. Keller
- Department of Mathematics
- University of Wisconsin-Madison
- mitch.keller@wisc.edu
-
-
-
-
-
-
-
-
-
-
- This Solutions Manual for Active Prelude to Calculus is organized by section.
- Preview Activity solutions are not provided,
- but each activity within a section has a solution provided.
- The solution to each activity starts at the top of a new page,
- so instructors can easily extract the solution for a single activity if they wish to print it for students or post on their course's learning management system site.
- After the activity solutions come the exercise solutions.
- Solutions for
- exercises are not provided,
- since those solutions are automatically available in this book's HTML version.
- Each exercise also starts at the top of its own page to make it easy to extract only the solutions to certain exercises.
-
-
-
- Please do not post this solutions manual publicly on the internet nor in any electronic form where it is available in full to students.
- As much as possible,
- we aspire to keep this solutions manual as a resource for instructors only so that students get the full benefit of activities and exercises by having to struggle with them without looking at solutions.
-
-
-
- By extracting individual pages from the PDF,
- it is fine to share solutions to individual activities or exercises via course management system software.
- Please do not post these individual pages publicly on the internet.
-
-
-
-
-
-
-
-
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ <line>Solutions Manual</line>
+ <line>for</line>
+ <line>Active Prelude to Calculus</line>
+
+
+
+
+
+
+ Matthew Boelkins
+ Department of Mathematics
+ Grand Valley State University
+ boelkinm@gvsu.edu
+
+
+
+ Production Editor
+
+ Mitchel T. Keller
+ Department of Mathematics
+ University of Wisconsin-Madison
+ mitch.keller@wisc.edu
+
+
+
+
+
+
+
+
+
+
+ This Solutions Manual for Active Prelude to Calculus is organized by section.
+ Preview Activity solutions are not provided,
+ but each activity within a section has a solution provided.
+ The solution to each activity starts at the top of a new page,
+ so instructors can easily extract the solution for a single activity if they wish to print it for students or post on their course's learning management system site.
+ After the activity solutions come the exercise solutions.
+ Solutions for
+ exercises are not provided,
+ since those solutions are automatically available in this book's HTML version.
+ Each exercise also starts at the top of its own page to make it easy to extract only the solutions to certain exercises.
+
+
+
+ Please do not post this solutions manual publicly on the internet nor in any electronic form where it is available in full to students.
+ As much as possible,
+ we aspire to keep this solutions manual as a resource for instructors only so that students get the full benefit of activities and exercises by having to struggle with them without looking at solutions.
+
+
+
+ By extracting individual pages from the PDF,
+ it is fine to share solutions to individual activities or exercises via course management system software.
+ Please do not post these individual pages publicly on the internet.
+
This appendix contains answers to all activities in the text. Answers for preview activities are not included.
+
+
+
+
+
+
+
+ Index
+
+
+
+
+
+ This book was authored in .
+
+
+
diff --git a/source/bibinfo.xml b/source/bibinfo.xml
index a2391548..45dfdae1 100644
--- a/source/bibinfo.xml
+++ b/source/bibinfo.xml
@@ -1,59 +1,59 @@
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
- Matthew Boelkins
- Department of Mathematics
- Grand Valley State University
- boelkinm@gvsu.edu
-
-
-
- Production Editor
-
- Mitchel T. Keller
- Department of Mathematics
- University of Wisconsin-Madison
- mitch@rellek.net
-
-
-
-
-
-
- Cover Photo
- Noah Wyn Photography
-
-
- 2019
-
-
-
-
-
-
- 2019
- Matthew Boelkins
- CC BY-SA 4.0 License
-
-
- Permission is granted to copy and (re)distribute this material in any format and/or adapt it (even commercially) under the terms of the Creative Commons Attribution-ShareAlike 4.0 International License. The work may be used for free in any way by any party so long as attribution is given to the author(s) and if the material is modified, the resulting contributions are distributed under the same license as this original. All trademarks are the registered marks of their respective owners. The graphic
-
-
ADD ALT TEXT TO THIS IMAGE
-
- that may appear in other locations in the text shows that the work is licensed with the Creative Commons and that the work may be used for free by any party so long as attribution is given to the author(s) and if the material is modified, the resulting contributions are distributed under the same license as this original. Full details may be found by visiting https://creativecommons.org/licenses/by-sa/4.0/ or sending a letter to Creative Commons, 444 Castro Street, Suite 900, Mountain View, California, 94041, USA.
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ Matthew Boelkins
+ Department of Mathematics
+ Grand Valley State University
+ boelkinm@gvsu.edu
+
+
+
+ Production Editor
+
+ Mitchel T. Keller
+ Department of Mathematics
+ University of Wisconsin-Madison
+ mitch@rellek.net
+
+
+
+
+
+
+ Cover Photo
+ Noah Wyn Photography
+
+
+ 2019
+
+
+
+
+
+
+ 2019
+ Matthew Boelkins
+ CC BY-SA 4.0 License
+
+
+ Permission is granted to copy and (re)distribute this material in any format and/or adapt it (even commercially) under the terms of the Creative Commons Attribution-ShareAlike 4.0 International License. The work may be used for free in any way by any party so long as attribution is given to the author(s) and if the material is modified, the resulting contributions are distributed under the same license as this original. All trademarks are the registered marks of their respective owners. The graphic
+
+
ADD ALT TEXT TO THIS IMAGE
+
+ that may appear in other locations in the text shows that the work is licensed with the Creative Commons and that the work may be used for free by any party so long as attribution is given to the author(s) and if the material is modified, the resulting contributions are distributed under the same license as this original. Full details may be found by visiting https://creativecommons.org/licenses/by-sa/4.0/ or sending a letter to Creative Commons, 444 Castro Street, Suite 900, Mountain View, California, 94041, USA.
+
+
+
+
diff --git a/source/bookinfo.xml b/source/bookinfo.xml
index 5df203bc..ceb9bcc5 100644
--- a/source/bookinfo.xml
+++ b/source/bookinfo.xml
@@ -1,46 +1,53 @@
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
- APC
-
-
- Preview Activity
-
- Motivating Questions
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ APC
+
+
+ active-prelude
+
+ Active Prelude to Calculus is designed for college students who aspire to take calculus and who either need to take a course to prepare them for calculus or want to do some additional self-study. The reader will find that the text requires them to engage actively with the material, to view topics from multiple perspectives, and to develop deep conceptual undersanding of ideas. Every section offers engaging activities for students to complete before and during class; additional exercises that challenge students to connect and assimilate core concepts; interactive WeBWorK exercises; opportunities for students to improve their skills at communicating mathematical ideas. The text is free and open-source, available in HTML, PDF, and print formats. Ancillary materials for instructors are also available.
+
+
+ Preview Activity
+
+ Motivating Questions
+
+
+
+
+
+
diff --git a/source/chap-changing-wb.xml b/source/chap-changing-wb.xml
index 42b788d9..8a14e3b6 100644
--- a/source/chap-changing-wb.xml
+++ b/source/chap-changing-wb.xml
@@ -1,27 +1,27 @@
-
-
-
-
-
-
-
-
-
-
-
-
-
-
- Relating Changing Quantities
-
-
-
-
-
-
-
-
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ Relating Changing Quantities
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/source/chap-changing.xml b/source/chap-changing.xml
index 1f97cae6..31e511f1 100755
--- a/source/chap-changing.xml
+++ b/source/chap-changing.xml
@@ -1,27 +1,27 @@
-
-
-
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-
-
- Relating Changing Quantities
-
-
-
-
-
-
-
-
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ Relating Changing Quantities
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/source/chap-circular-wb.xml b/source/chap-circular-wb.xml
index c4e3a872..194e9282 100644
--- a/source/chap-circular-wb.xml
+++ b/source/chap-circular-wb.xml
@@ -1,22 +1,22 @@
-
-
-
-
-
-
-
-
-
-
-
-
-
-
- Circular Functions
-
-
-
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ Circular Functions
+
+
+
+
+
+
+
diff --git a/source/chap-circular.xml b/source/chap-circular.xml
index 395b5689..b1068721 100755
--- a/source/chap-circular.xml
+++ b/source/chap-circular.xml
@@ -1,22 +1,22 @@
-
-
-
-
-
-
-
-
-
-
-
-
-
-
- Circular Functions
-
-
-
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ Circular Functions
+
+
+
+
+
+
+
diff --git a/source/chap-exp-wb.xml b/source/chap-exp-wb.xml
index c1f5130f..648f5ae8 100644
--- a/source/chap-exp-wb.xml
+++ b/source/chap-exp-wb.xml
@@ -1,24 +1,24 @@
-
-
-
-
-
-
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-
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-
-
-
-
- Exponential and Logarithmic Functions
-
-
-
-
-
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ Exponential and Logarithmic Functions
+
+
+
+
+
+
+
+
+
diff --git a/source/chap-exp.xml b/source/chap-exp.xml
index e73052c3..8bd6caab 100755
--- a/source/chap-exp.xml
+++ b/source/chap-exp.xml
@@ -1,24 +1,24 @@
-
-
-
-
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-
-
-
-
-
- Exponential and Logarithmic Functions
-
-
-
-
-
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ Exponential and Logarithmic Functions
+
+
+
+
+
+
+
+
+
diff --git a/source/chap-poly-wb.xml b/source/chap-poly-wb.xml
index 4f429701..8869a177 100644
--- a/source/chap-poly-wb.xml
+++ b/source/chap-poly-wb.xml
@@ -1,23 +1,23 @@
-
-
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-
-
-
-
-
-
- Polynomial and Rational Functions
-
-
-
-
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ Polynomial and Rational Functions
+
+
+
+
+
+
+
+
diff --git a/source/chap-poly.xml b/source/chap-poly.xml
index 659499b3..693d12d7 100755
--- a/source/chap-poly.xml
+++ b/source/chap-poly.xml
@@ -1,23 +1,23 @@
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-
- Polynomial and Rational Functions
-
-
-
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-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ Polynomial and Rational Functions
+
+
+
+
+
+
+
+
diff --git a/source/chap-trig-wb.xml b/source/chap-trig-wb.xml
index 05c29d99..bb4a8a62 100644
--- a/source/chap-trig-wb.xml
+++ b/source/chap-trig-wb.xml
@@ -1,23 +1,23 @@
-
-
-
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-
-
-
- Trigonometry
-
-
-
-
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ Trigonometry
+
+
+
+
+
+
+
+
diff --git a/source/chap-trig.xml b/source/chap-trig.xml
index 76a3eaf3..92f0f724 100755
--- a/source/chap-trig.xml
+++ b/source/chap-trig.xml
@@ -1,23 +1,23 @@
-
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- Trigonometry
-
-
-
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-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ Trigonometry
+
+
+
+
+
+
+
+
diff --git a/source/colophon.xml b/source/colophon.xml
index e256fcb7..28767c5b 100644
--- a/source/colophon.xml
+++ b/source/colophon.xml
@@ -1,16 +1,16 @@
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diff --git a/source/exercises/ez-0-0.xml b/source/exercises/ez-0-0.xml
index c7a78b2a..02ac1ccc 100755
--- a/source/exercises/ez-0-0.xml
+++ b/source/exercises/ez-0-0.xml
@@ -1,42 +1,42 @@
-
-
-
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-
-
-
- A cold can of soda is removed from a refrigerator. Its temperature F in degrees Fahrenheit is measured at 5-minute intervals, as recorded in the following table.
-
-
-
- Data for the soda's temperature as a function of time.
-
-
- t (minutes)
- 0
- 5
- 10
- 15
- 20
- 25
- 30
- 35
-
-
- F (Fahrenheit temp)
- 37.00
- 44.74
- 50.77
- 55.47
- 59.12
- 61.97
- 64.19
- 65.92
-
-
-
-
-
-
-
-
- Determine AV_{[0,5]}, AV_{[5,10]}, and AV_{[10,15]}, including appropriate units. Choose one of these quantities and write a careful sentence to explain its meaning. Your sentence might look something like On the interval \ldots, the temperature of the soda is \ldots on average by \ldots for each 1-unit increase in \ldots.
-
-
-
-
- On which interval is there more total change in the soda's temperature: [10,20] or [25,35]?
-
-
-
-
- What can you observe about when the soda's temperature appears to be changing most rapidly?
-
-
-
-
- Estimate the soda's temperature when t = 37 minutes. Write at least one sentence to explain your thinking.
-
-
-
-
-
-
-
-
- Exercise Answer
-
-
-
-
AV_{[0,5]} = \frac{44.74 - 37.00}{5} = 1.548F/min; AV_{[5,10]} = \frac{50.77 - 44.74}{5} = 1.206F/min; AV_{[10,15]} = \frac{55.47 - 50.77}{5} = 0.940F/min. For example, on the interval [0,5], the temperature of the soda is increasing on average by 1.548 degrees Fahrenheit for each additional minute.
Change on [10,20]: 59.12 - 50.77 = 8.35F. Change on [25,35]: 65.92 - 61.97 = 3.95F. There is more total change on [10,20].
The soda's temperature is changing most rapidly at the beginning (when t is smallest), and the rate of change decreases over time as the soda approaches room temperature.
The rate of change on [30,35] is AV_{[30,35]} = \frac{65.92 - 64.19}{5} = 0.346F/min. Extending this trend by 2 minutes: F(37) \approx 65.92 + 2(0.346) \approx 66.6F.
-
-
-
-
-
-
-
-
- The position of a car driving along a straight road at time t in minutes is given by the function
- y = s(t) that is pictured in Figure.
- The car's position function has units measured in thousands of feet.
- For instance,
- the point (2,4) on the graph indicates that after 2 minutes,
- the car has traveled 4000 feet.
-
-
-
-
The graph of y = s(t), the position of the car (measured in thousands of feet from its starting location) at time t in minutes.
-
The graph of y = s(t), the position of the car (measured in thousands of feet from its starting location) at time t in minutes.
-
-
-
-
-
-
-
- In everyday language,
- describe the behavior of the car over the provided time interval.
- In particular,
- carefully discuss what is happening on each of the time intervals [0,1],
- [1,2], [2,3],
- [3,4], and [4,5],
- plus provide commentary overall on what the car is doing on the interval [0,12].
-
-
-
-
- Compute the average rate of change of s on the intervals [3,4], [4,6], and [5,8]. Label your results using the notation AV_{[a,b]} appropriately, and include units on each quantity.
-
-
-
-
- On the graph of s, sketch the three lines whose slope corresponds to the values of AV_{[3,4]}, AV_{[4,6]}, and AV_{[5,8]} that you computed in (b).
-
-
-
-
- Is there a time interval on which the car's average velocity is 5000 feet per minute? Why or why not?
-
-
-
-
- Is there ever a time interval when the car is going in reverse? Why or why not?
-
-
-
-
-
-
-
-
- Exercise Answer
-
-
-
-
Read from the graph: describe behavior on each interval. Where the graph rises steeply, the car moves quickly; where flat, the car is stopped; where the slope is gentler, the car moves more slowly.
From the graph: AV_{[3,4]} = \frac{s(4)-s(3)}{1}, AV_{[4,6]} = \frac{s(6)-s(4)}{2}, AV_{[5,8]} = \frac{s(8)-s(5)}{3}, each in thousands of feet per minute.
On the graph, sketch line segments connecting (3, s(3)) to (4, s(4)), (4, s(4)) to (6, s(6)), and (5, s(5)) to (8, s(8)); their slopes represent the respective average velocities.
An average velocity of 5000 ft/min requires a slope of 5 on the graph. Examine whether any secant line connecting two points on the graph achieves this slope.
If the position graph is always non-decreasing, the car never goes in reverse. Reverse motion would correspond to a portion of the graph with negative slope (decreasing s).
-
-
-
-
-
-
- Consider an inverted conical tank (point down) whose top has a radius of 3 feet and that is 2 feet deep. The tank is initially empty and then is filled at a constant rate of 0.75 cubic feet per minute. Let V=f(t) denote the volume of water (in cubic feet) at time t in minutes, and let h= g(t) denote the depth of the water (in feet) at time t. It turns out that the formula for the function g is g(t) = \left( \frac{t}{\pi} \right)^{1/3}.
-
-
-
-
-
-
- In everyday language, describe how you expect the height function h = g(t) to behave as time increases.
-
-
-
-
- For the height function h = g(t) = \left( \frac{t}{\pi} \right)^{1/3}, compute AV_{[0,2]}, AV_{[2,4]}, and AV_{[4,6]}. Include units on your results.
-
-
-
-
- Again working with the height function, can you determine an interval [a,b] on which AV_{[a,b]} = 2 feet per minute? If yes, state the interval; if not, explain why there is no such interval.
-
-
-
-
- Now consider the volume function, V = f(t). Even though we don't have a formula for f, is it possible to determine the average rate of change of the volume function on the intervals [0,2], [2,4], and [4,6]? Why or why not?
-
-
-
-
-
-
-
-
- Exercise Answer
-
-
-
-
As the conical tank fills, the cone widens, so equal amounts of water produce smaller increases in height. Therefore h = g(t) is increasing but at a decreasing rate (concave down).
Yes. For intervals [0,b] starting at t=0: AV_{[0,b]} = g(b)/b = (1/\pi)^{1/3} b^{-2/3}. Setting this equal to 2 gives b^{2/3} = (1/\pi)^{1/3}/2, so b = (1/(8\pi))^{1/2} = 1/(2\sqrt{\pi}\,) \approx 0.282 min. The interval [0, 1/(2\sqrt{\pi})] gives AV = 2 ft/min.
Since V = f(t) = 0.75t is linear with constant slope 0.75, the average rate of change of volume is always 0.75 cu ft/min on every interval, including [0,2], [2,4], and [4,6].
+ A cold can of soda is removed from a refrigerator. Its temperature F in degrees Fahrenheit is measured at 5-minute intervals, as recorded in the following table.
+
+
+
+ Data for the soda's temperature as a function of time.
+
+
+ t (minutes)
+ 0
+ 5
+ 10
+ 15
+ 20
+ 25
+ 30
+ 35
+
+
+ F (Fahrenheit temp)
+ 37.00
+ 44.74
+ 50.77
+ 55.47
+ 59.12
+ 61.97
+ 64.19
+ 65.92
+
+
+
+
+
+
+
+
+ Determine AV_{[0,5]}, AV_{[5,10]}, and AV_{[10,15]}, including appropriate units. Choose one of these quantities and write a careful sentence to explain its meaning. Your sentence might look something like On the interval \ldots, the temperature of the soda is \ldots on average by \ldots for each 1-unit increase in \ldots.
+
+
+
+
+ On which interval is there more total change in the soda's temperature: [10,20] or [25,35]?
+
+
+
+
+ What can you observe about when the soda's temperature appears to be changing most rapidly?
+
+
+
+
+ Estimate the soda's temperature when t = 37 minutes. Write at least one sentence to explain your thinking.
+
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
AV_{[0,5]} = \frac{44.74 - 37.00}{5} = 1.548F/min; AV_{[5,10]} = \frac{50.77 - 44.74}{5} = 1.206F/min; AV_{[10,15]} = \frac{55.47 - 50.77}{5} = 0.940F/min. For example, on the interval [0,5], the temperature of the soda is increasing on average by 1.548 degrees Fahrenheit for each additional minute.
Change on [10,20]: 59.12 - 50.77 = 8.35F. Change on [25,35]: 65.92 - 61.97 = 3.95F. There is more total change on [10,20].
The soda's temperature is changing most rapidly at the beginning (when t is smallest), and the rate of change decreases over time as the soda approaches room temperature.
The rate of change on [30,35] is AV_{[30,35]} = \frac{65.92 - 64.19}{5} = 0.346F/min. Extending this trend by 2 minutes: F(37) \approx 65.92 + 2(0.346) \approx 66.6F.
+
+
+
+
+
+
+
+
+ The position of a car driving along a straight road at time t in minutes is given by the function
+ y = s(t) that is pictured in Figure.
+ The car's position function has units measured in thousands of feet.
+ For instance,
+ the point (2,4) on the graph indicates that after 2 minutes,
+ the car has traveled 4000 feet.
+
+
+
+
The graph of y = s(t), the position of the car (measured in thousands of feet from its starting location) at time t in minutes.
+
The graph of y = s(t), the position of the car (measured in thousands of feet from its starting location) at time t in minutes.
+
+
+
+
+
+
+
+ In everyday language,
+ describe the behavior of the car over the provided time interval.
+ In particular,
+ carefully discuss what is happening on each of the time intervals [0,1],
+ [1,2], [2,3],
+ [3,4], and [4,5],
+ plus provide commentary overall on what the car is doing on the interval [0,12].
+
+
+
+
+ Compute the average rate of change of s on the intervals [3,4], [4,6], and [5,8]. Label your results using the notation AV_{[a,b]} appropriately, and include units on each quantity.
+
+
+
+
+ On the graph of s, sketch the three lines whose slope corresponds to the values of AV_{[3,4]}, AV_{[4,6]}, and AV_{[5,8]} that you computed in (b).
+
+
+
+
+ Is there a time interval on which the car's average velocity is 5000 feet per minute? Why or why not?
+
+
+
+
+ Is there ever a time interval when the car is going in reverse? Why or why not?
+
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
Read from the graph: describe behavior on each interval. Where the graph rises steeply, the car moves quickly; where flat, the car is stopped; where the slope is gentler, the car moves more slowly.
From the graph: AV_{[3,4]} = \frac{s(4)-s(3)}{1}, AV_{[4,6]} = \frac{s(6)-s(4)}{2}, AV_{[5,8]} = \frac{s(8)-s(5)}{3}, each in thousands of feet per minute.
On the graph, sketch line segments connecting (3, s(3)) to (4, s(4)), (4, s(4)) to (6, s(6)), and (5, s(5)) to (8, s(8)); their slopes represent the respective average velocities.
An average velocity of 5000 ft/min requires a slope of 5 on the graph. Examine whether any secant line connecting two points on the graph achieves this slope.
If the position graph is always non-decreasing, the car never goes in reverse. Reverse motion would correspond to a portion of the graph with negative slope (decreasing s).
+
+
+
+
+
+
+ Consider an inverted conical tank (point down) whose top has a radius of 3 feet and that is 2 feet deep. The tank is initially empty and then is filled at a constant rate of 0.75 cubic feet per minute. Let V=f(t) denote the volume of water (in cubic feet) at time t in minutes, and let h= g(t) denote the depth of the water (in feet) at time t. It turns out that the formula for the function g is g(t) = \left( \frac{t}{\pi} \right)^{1/3}.
+
+
+
+
+
+
+ In everyday language, describe how you expect the height function h = g(t) to behave as time increases.
+
+
+
+
+ For the height function h = g(t) = \left( \frac{t}{\pi} \right)^{1/3}, compute AV_{[0,2]}, AV_{[2,4]}, and AV_{[4,6]}. Include units on your results.
+
+
+
+
+ Again working with the height function, can you determine an interval [a,b] on which AV_{[a,b]} = 2 feet per minute? If yes, state the interval; if not, explain why there is no such interval.
+
+
+
+
+ Now consider the volume function, V = f(t). Even though we don't have a formula for f, is it possible to determine the average rate of change of the volume function on the intervals [0,2], [2,4], and [4,6]? Why or why not?
+
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
As the conical tank fills, the cone widens, so equal amounts of water produce smaller increases in height. Therefore h = g(t) is increasing but at a decreasing rate (concave down).
Yes. For intervals [0,b] starting at t=0: AV_{[0,b]} = g(b)/b = (1/\pi)^{1/3} b^{-2/3}. Setting this equal to 2 gives b^{2/3} = (1/\pi)^{1/3}/2, so b = (1/(8\pi))^{1/2} = 1/(2\sqrt{\pi}\,) \approx 0.282 min. The interval [0, 1/(2\sqrt{\pi})] gives AV = 2 ft/min.
Since V = f(t) = 0.75t is linear with constant slope 0.75, the average rate of change of volume is always 0.75 cu ft/min on every interval, including [0,2], [2,4], and [4,6].
- Consider the functions s and g defined by the graphs in Figure and Figure. Assume that to the left and right of the pictured domains, each function continues behaving according to the trends seen in the figures.
-
-
-
-
-
The graph of a piecewise function, s.
-
The graph of a piecewise function, s.
-
-
-
-
The graph of a piecewise function, g.
-
The graph of a piecewise function, g.
-
-
-
-
-
-
-
-
- Determine a piecewise formula for the function y = s(t) that is valid for all real numbers t.
-
-
-
-
- Determine a piecewise formula for the function y = g(x) that is valid for all real numbers x.
-
-
-
-
- Determine each of the following quantities or explain why they are not defined.
-
-
-
-
-
- (s \cdot g)(1)
-
-
-
-
- (g-s)(3)
-
-
-
-
-
- (s \circ g)(1.5)
-
-
-
-
- (g \circ s)(-4)
-
-
-
-
-
-
-
-
-
-
-
- Exercise Answer
-
-
-
-
Read the graph of s to identify its behavior on each piece. Find the equations of each linear piece and write them in bracket notation with the corresponding domain intervals.
Similarly, read the graph of g and write a piecewise formula.
With formulas for s and g in hand: (i) (s\cdot g)(1) = s(1)\cdot g(1); (ii) (g-s)(3) = g(3)-s(3); (iii) (s\circ g)(1.5) = s(g(1.5)); (iv) (g\circ s)(-4) = g(s(-4)). For each, evaluate using the piecewise formulas.
-
-
-
-
-
-
-
- One of the most important principles in the study of changing quantities is found in the relationship between distance,
- average velocity, and time.
- For a moving body traveling on a straight-line path at an average rate of v for a period of time t,
- the distance traveled, d, is given by
-
- d = v \cdot t
-
-
-
-
- In the Ironman Triathlon,
- competitors swim 2.4 miles, bike 112 miles,
- and then run a 26.2 mile marathon.
- In the following sequence of questions,
- we build a piecewise function that models a competitor's location in the race at a given time t.
- To start, we have the following known information.
-
-
-
-
- She swims at an average rate of 2.5 miles per hour throughout the 2.4 miles in the water.
-
-
-
-
-
- Her transition from swim to bike takes 3 minutes (0.05 hours),
- during which time she doesn't travel any additional distance.
-
-
-
-
-
- She bikes at an average rate of 21 miles per hour throughout the 112 miles of biking.
-
-
-
-
-
- Her transition from bike to run takes just over 2 minutes (0.03 hours),
- during which time she doesn't travel any additional distance.
-
-
-
-
-
- She runs at an average rate of 8.5 miles per hour throughout the marathon.
-
-
-
-
-
- In the questions that follow,
- assume for the purposes of the model that the triathlete swims, bikes,
- and runs at essentially constant rates
- (given by the average rates stated above).
-
-
-
-
-
-
-
- Determine the time the swimmer exits the water.
- Report your result in hours.
-
-
-
-
-
- Likewise, determine the time the athlete gets off her bike,
- as well as the time she finishes the race.
-
-
-
-
-
- List 5 key points in the form (time, distance):
- when exiting the water, when starting the bike,
- when finishing the bike, when starting the run,
- and when finishing the run.
-
-
-
-
-
- What is the triathlete's average velocity over the course of the entire race?
- Is this velocity the average of her swim velocity, bike velocity, and run velocity? Why or why not?
-
-
-
-
-
- Determine a piecewise function s(t) whose value at any given time
- (in hours)
- is the triathlete's total distance traveled.
-
-
-
-
-
- Sketch a carefully labeled graph of the triathlete's distance traveled as a function of time on the axes provided.
- Provide clear scale and note key points on the graph.
-
-
-
-
ADD ALT TEXT TO THIS IMAGE
-
ADD ALT TEXT TO THIS IMAGE
-
-
-
-
-
- Sketch a possible graph of the triathete's velocity, V,
- as a function of time on the righthand axes.
- Here, too, label key points and provide clear scale. Write several sentences to explain and justify your graph.
-
-
-
-
-
-
-
- Exercise Answer
-
-
-
-
The triathlete exits the water after the swim at t = 2.4/2.5 = 0.96 hours.
The bike finishes at t = 0.96 + 0.05 + 112/21 \approx 6.34 hours; the race ends at t \approx 6.37 + 26.2/8.5 \approx 9.45 hours.
Key points: exits water at (0.96, 2.4); starts bike at (1.01, 2.4); finishes bike at (6.34, 114.4); starts run at (6.37, 114.4); finishes at (9.45, 140.6).
f(t) = \begin{cases} 2.5t, \amp 0 \le t \le 0.96 \\ 2.4, \amp 0.96 \lt t \le 1.01 \\ 2.4 + 21(t-1.01), \amp 1.01 \lt t \le 6.34 \\ 114.4, \amp 6.34 \lt t \le 6.37 \\ 114.4 + 8.5(t-6.37), \amp 6.37 \lt t \le 9.45 \end{cases}
+ Consider the functions s and g defined by the graphs in Figure and Figure. Assume that to the left and right of the pictured domains, each function continues behaving according to the trends seen in the figures.
+
+
+
+
+
The graph of a piecewise function, s.
+
The graph of a piecewise function, s.
+
+
+
+
The graph of a piecewise function, g.
+
The graph of a piecewise function, g.
+
+
+
+
+
+
+
+
+ Determine a piecewise formula for the function y = s(t) that is valid for all real numbers t.
+
+
+
+
+ Determine a piecewise formula for the function y = g(x) that is valid for all real numbers x.
+
+
+
+
+ Determine each of the following quantities or explain why they are not defined.
+
+
+
+
+
+ (s \cdot g)(1)
+
+
+
+
+ (g-s)(3)
+
+
+
+
+
+ (s \circ g)(1.5)
+
+
+
+
+ (g \circ s)(-4)
+
+
+
+
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
Read the graph of s to identify its behavior on each piece. Find the equations of each linear piece and write them in bracket notation with the corresponding domain intervals.
Similarly, read the graph of g and write a piecewise formula.
With formulas for s and g in hand: (i) (s\cdot g)(1) = s(1)\cdot g(1); (ii) (g-s)(3) = g(3)-s(3); (iii) (s\circ g)(1.5) = s(g(1.5)); (iv) (g\circ s)(-4) = g(s(-4)). For each, evaluate using the piecewise formulas.
+
+
+
+
+
+
+
+ One of the most important principles in the study of changing quantities is found in the relationship between distance,
+ average velocity, and time.
+ For a moving body traveling on a straight-line path at an average rate of v for a period of time t,
+ the distance traveled, d, is given by
+
+ d = v \cdot t
+
+
+
+
+ In the Ironman Triathlon,
+ competitors swim 2.4 miles, bike 112 miles,
+ and then run a 26.2 mile marathon.
+ In the following sequence of questions,
+ we build a piecewise function that models a competitor's location in the race at a given time t.
+ To start, we have the following known information.
+
+
+
+
+ She swims at an average rate of 2.5 miles per hour throughout the 2.4 miles in the water.
+
+
+
+
+
+ Her transition from swim to bike takes 3 minutes (0.05 hours),
+ during which time she doesn't travel any additional distance.
+
+
+
+
+
+ She bikes at an average rate of 21 miles per hour throughout the 112 miles of biking.
+
+
+
+
+
+ Her transition from bike to run takes just over 2 minutes (0.03 hours),
+ during which time she doesn't travel any additional distance.
+
+
+
+
+
+ She runs at an average rate of 8.5 miles per hour throughout the marathon.
+
+
+
+
+
+ In the questions that follow,
+ assume for the purposes of the model that the triathlete swims, bikes,
+ and runs at essentially constant rates
+ (given by the average rates stated above).
+
+
+
+
+
+
+
+ Determine the time the swimmer exits the water.
+ Report your result in hours.
+
+
+
+
+
+ Likewise, determine the time the athlete gets off her bike,
+ as well as the time she finishes the race.
+
+
+
+
+
+ List 5 key points in the form (time, distance):
+ when exiting the water, when starting the bike,
+ when finishing the bike, when starting the run,
+ and when finishing the run.
+
+
+
+
+
+ What is the triathlete's average velocity over the course of the entire race?
+ Is this velocity the average of her swim velocity, bike velocity, and run velocity? Why or why not?
+
+
+
+
+
+ Determine a piecewise function s(t) whose value at any given time
+ (in hours)
+ is the triathlete's total distance traveled.
+
+
+
+
+
+ Sketch a carefully labeled graph of the triathlete's distance traveled as a function of time on the axes provided.
+ Provide clear scale and note key points on the graph.
+
+
+
+
ADD ALT TEXT TO THIS IMAGE
+
ADD ALT TEXT TO THIS IMAGE
+
+
+
+
+
+ Sketch a possible graph of the triathete's velocity, V,
+ as a function of time on the righthand axes.
+ Here, too, label key points and provide clear scale. Write several sentences to explain and justify your graph.
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
The triathlete exits the water after the swim at t = 2.4/2.5 = 0.96 hours.
The bike finishes at t = 0.96 + 0.05 + 112/21 \approx 6.34 hours; the race ends at t \approx 6.37 + 26.2/8.5 \approx 9.45 hours.
Key points: exits water at (0.96, 2.4); starts bike at (1.01, 2.4); finishes bike at (6.34, 114.4); starts run at (6.37, 114.4); finishes at (9.45, 140.6).
f(t) = \begin{cases} 2.5t, \amp 0 \le t \le 0.96 \\ 2.4, \amp 0.96 \lt t \le 1.01 \\ 2.4 + 21(t-1.01), \amp 1.01 \lt t \le 6.34 \\ 114.4, \amp 6.34 \lt t \le 6.37 \\ 114.4 + 8.5(t-6.37), \amp 6.37 \lt t \le 9.45 \end{cases}
- Use the given information about various functions to answer the following questions involving composition.
-
-
-
-
-
-
- Let functions f and g be given by the graphs in Figure and . An open circle means there is not a point at that location on the graph. For instance, f(-1) = 1, but f(3) is not defined.
-
-
-
-
-
Plot of y = f(x).
-
Plot of y = f(x).
-
-
-
-
Plot of y = g(x).
-
Plot of y = g(x).
-
-
-
-
-
- Determine f(g(1)) and g(f(-2)).
-
-
-
-
- Again using the functions given in (a), can you determine a value of x for which g(f(x)) is not defined? Why or why not?
-
- Determine (s \circ r)(3), (s \circ r)(-4), and (s \circ r)(a) for one additional value of a of your choice.
-
-
-
-
- For the functions r and s defined in (c), state the domain and range of each function. For how many different values of b is it possible to determine (r \circ s)(b)? Explain.
-
-
-
-
- Let m(u) = u^3 + 4u^2 - 5u + 1. Determine expressions for m(x^2), m(2+h), and m(a+h).
-
-
-
-
- For the function F(x) = 4 - 3x - x^2, determine the most simplified expression you can find for AV_{[2,2+h]}. Show your algebraic work and thinking fully.
-
-
-
-
-
-
-
- Exercise Answer
-
-
-
-
From the graphs: g(1) can be read, then f(g(1)); and f(-2) can be read, then g(f(-2)).
g(f(x)) is undefined if f(x) falls outside the domain of g or if f(x) is itself undefined.
From the table: (s\circ r)(3) = s(r(3)) = s(-1) = -8; (s\circ r)(-4) = s(r(-4)) = s(4) = 5; for example (s\circ r)(0) = s(r(0)) = s(0) = 0.
Domain of r: \{-4,-3,-2,-1,0,1,2,3,4\}; range of r: \{-4,-3,-1,0,1,2,3,4\}. Domain of s: \{-4,...,4\}; range of s: \{-8,-7,-6,-5,0,5,6,7,8\}. For (r\circ s)(b), we need s(b) in the domain of r. The only value of s in \{-4,...,4\} is s(0)=0, so only b=0 works: (r\circ s)(0) = r(0) = 0.
- Recall Dolbear's function that defines temperature,
- F, in Fahrenheit degrees,
- as a function of the number of chirps per minute, N,
- is F = D(N) = 40 + \frac{1}{4}N.
-
-
-
-
-
-
- Solve the equation F = 40 + \frac{1}{4}N for N in terms of F.
-
-
-
-
-
- Say that N = g(F) is the function you just found in (a). What is the meaning of this function?
- What does it take as inputs and what does it produce as outputs?
-
-
-
-
- How many chirps per minute do we expect when the outsidet temperature is 82 degrees F? How can we express this in the notation of the function g?
-
-
-
-
-
- Recall that the function that converts Fahrenheit to Celsius is
- C = G(F) = \frac{5}{9}(F-32).
- Solve the equation C = \frac{5}{9}(F-32) for F in terms of C. Call the resulting function F = p(C). What is the meaning of this function?
-
-
-
-
- Is it possible to write the chirp-rate N as a function of temperature C in Celsius?
- That is, can we produce a function whose input is in degrees Celsius and whose output is the number of chirps per minute?
- If yes, do so and explain your thinking. If not, explain why it's not possible.
-
-
-
-
-
-
-
- Exercise Answer
-
-
-
-
Solving F = 40 + \frac{1}{4}N for N: N = 4(F-40) = 4F - 160.
g(F) = 4F - 160 converts temperature in Fahrenheit to expected chirps per minute.
g(82) = 4(82) - 160 = 168 chirps per minute, written N = g(82) = 168.
Solving C = \frac{5}{9}(F-32) for F: F = \frac{9}{5}C + 32 = p(C). This converts Celsius to Fahrenheit.
Yes: N = (g\circ p)(C) = g(p(C)) = g\!\left(\frac{9C}{5}+32\right) = 4\!\left(\frac{9C}{5}+32\right) - 160 = \frac{36C}{5} - 32. This function converts Celsius temperature to expected chirps per minute.
-
-
-
-
-
-
- For each of the following functions, find two simpler functions f and g such that the given function can be written as the composite function g \circ f.
-
-
-
-
-
- h(x) = (x^2 + 7)^3
-
-
-
-
-
- r(x) = \sqrt{5-x^3}
-
-
-
-
-
- m(x) = \frac{1}{x^4 + 2x^2 + 1}
-
-
-
-
-
- w(x) = 2^{3-x^2}
-
-
-
-
-
-
-
-
-
-
- A spherical tank has radius 4 feet. The tank is initially empty and then begins to be filled in such a way that the height of the water rises at a constant rate of 0.4 feet per minute. Let V be the volume of water in the tank at a given instant, and h the depth of the water at the same instant; let t denote the time elapsed in minutes since the tank started being filled.
-
-
-
-
-
- Calculus can be used to show that the volume, V, is a function of the depth, h, of the water in the tank according to the function
-
- V = f(h) = \frac{\pi}{3} h^2(12-h)
- .
- What is the domain of this model? Why? What is the corresponding range?
-
-
-
-
- We are given the fact that the tank is being filled in such a way that the height of the water rises at a constant rate of 0.4 feet per minute. Said differently, h is a function of t whose average rate of change is constant. What kind of function does this make h = p(t)? Determine a formula for p(t).
-
-
-
-
- What are the domain and range of the function h = p(t)? How is this tied to the dimensions of the tank?
-
-
-
-
- In (a) we observed that V is a function of h, and in (b) we found that h is a function of t. Use these two facts and function composition appropriately to write V as a function of t. Call the resulting function V = q(t).
-
-
-
-
- What are the domain and range of the function q? Why?
-
-
-
-
- On the provided axes, sketch accurate graphs of h = p(t) and V = q(t), labeling the vertical and horizontal scale on each graph appropriately. Make your graphs as precise as you can; use a computing device to assist as needed.
-
-
-
-
ADD ALT TEXT TO THIS IMAGE
-
ADD ALT TEXT TO THIS IMAGE
-
-
-
- Why do each of the two graphs have their respective shapes? Write at least one sentence to explain each graph; refer explicitly to the shape of the tank and other information given in the problem.
-
-
-
-
-
-
-
-
- Exercise Answer
-
-
-
-
Domain: 0 \le h \le 8 (diameter of sphere). Range: 0 \le V \le \frac{256\pi}{3}.
Since the height increases at a constant rate, h = p(t) = 0.4t is a linear function.
Domain: [0, 20] minutes (height reaches 8 m at t = 8/0.4 = 20 min). Range: [0, 8] m.
The graph of h = p(t) is a straight line (height rises at constant rate). The graph of V = q(t) is an S-shaped increasing cubic: slow at first (sphere is narrow near the bottom), then faster through the middle (widest part), then slower again as the sphere narrows near the top.
+ Use the given information about various functions to answer the following questions involving composition.
+
+
+
+
+
+
+ Let functions f and g be given by the graphs in Figure and . An open circle means there is not a point at that location on the graph. For instance, f(-1) = 1, but f(3) is not defined.
+
+
+
+
+
Plot of y = f(x).
+
Plot of y = f(x).
+
+
+
+
Plot of y = g(x).
+
Plot of y = g(x).
+
+
+
+
+
+ Determine f(g(1)) and g(f(-2)).
+
+
+
+
+ Again using the functions given in (a), can you determine a value of x for which g(f(x)) is not defined? Why or why not?
+
+ Determine (s \circ r)(3), (s \circ r)(-4), and (s \circ r)(a) for one additional value of a of your choice.
+
+
+
+
+ For the functions r and s defined in (c), state the domain and range of each function. For how many different values of b is it possible to determine (r \circ s)(b)? Explain.
+
+
+
+
+ Let m(u) = u^3 + 4u^2 - 5u + 1. Determine expressions for m(x^2), m(2+h), and m(a+h).
+
+
+
+
+ For the function F(x) = 4 - 3x - x^2, determine the most simplified expression you can find for AV_{[2,2+h]}. Show your algebraic work and thinking fully.
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
From the graphs: g(1) can be read, then f(g(1)); and f(-2) can be read, then g(f(-2)).
g(f(x)) is undefined if f(x) falls outside the domain of g or if f(x) is itself undefined.
From the table: (s\circ r)(3) = s(r(3)) = s(-1) = -8; (s\circ r)(-4) = s(r(-4)) = s(4) = 5; for example (s\circ r)(0) = s(r(0)) = s(0) = 0.
Domain of r: \{-4,-3,-2,-1,0,1,2,3,4\}; range of r: \{-4,-3,-1,0,1,2,3,4\}. Domain of s: \{-4,...,4\}; range of s: \{-8,-7,-6,-5,0,5,6,7,8\}. For (r\circ s)(b), we need s(b) in the domain of r. The only value of s in \{-4,...,4\} is s(0)=0, so only b=0 works: (r\circ s)(0) = r(0) = 0.
+ Recall Dolbear's function that defines temperature,
+ F, in Fahrenheit degrees,
+ as a function of the number of chirps per minute, N,
+ is F = D(N) = 40 + \frac{1}{4}N.
+
+
+
+
+
+
+ Solve the equation F = 40 + \frac{1}{4}N for N in terms of F.
+
+
+
+
+
+ Say that N = g(F) is the function you just found in (a). What is the meaning of this function?
+ What does it take as inputs and what does it produce as outputs?
+
+
+
+
+ How many chirps per minute do we expect when the outsidet temperature is 82 degrees F? How can we express this in the notation of the function g?
+
+
+
+
+
+ Recall that the function that converts Fahrenheit to Celsius is
+ C = G(F) = \frac{5}{9}(F-32).
+ Solve the equation C = \frac{5}{9}(F-32) for F in terms of C. Call the resulting function F = p(C). What is the meaning of this function?
+
+
+
+
+ Is it possible to write the chirp-rate N as a function of temperature C in Celsius?
+ That is, can we produce a function whose input is in degrees Celsius and whose output is the number of chirps per minute?
+ If yes, do so and explain your thinking. If not, explain why it's not possible.
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
Solving F = 40 + \frac{1}{4}N for N: N = 4(F-40) = 4F - 160.
g(F) = 4F - 160 converts temperature in Fahrenheit to expected chirps per minute.
g(82) = 4(82) - 160 = 168 chirps per minute, written N = g(82) = 168.
Solving C = \frac{5}{9}(F-32) for F: F = \frac{9}{5}C + 32 = p(C). This converts Celsius to Fahrenheit.
Yes: N = (g\circ p)(C) = g(p(C)) = g\!\left(\frac{9C}{5}+32\right) = 4\!\left(\frac{9C}{5}+32\right) - 160 = \frac{36C}{5} - 32. This function converts Celsius temperature to expected chirps per minute.
+
+
+
+
+
+
+ For each of the following functions, find two simpler functions f and g such that the given function can be written as the composite function g \circ f.
+
+
+
+
+
+ h(x) = (x^2 + 7)^3
+
+
+
+
+
+ r(x) = \sqrt{5-x^3}
+
+
+
+
+
+ m(x) = \frac{1}{x^4 + 2x^2 + 1}
+
+
+
+
+
+ w(x) = 2^{3-x^2}
+
+
+
+
+
+
+
+
+
+
+ A spherical tank has radius 4 feet. The tank is initially empty and then begins to be filled in such a way that the height of the water rises at a constant rate of 0.4 feet per minute. Let V be the volume of water in the tank at a given instant, and h the depth of the water at the same instant; let t denote the time elapsed in minutes since the tank started being filled.
+
+
+
+
+
+ Calculus can be used to show that the volume, V, is a function of the depth, h, of the water in the tank according to the function
+
+ V = f(h) = \frac{\pi}{3} h^2(12-h)
+ .
+ What is the domain of this model? Why? What is the corresponding range?
+
+
+
+
+ We are given the fact that the tank is being filled in such a way that the height of the water rises at a constant rate of 0.4 feet per minute. Said differently, h is a function of t whose average rate of change is constant. What kind of function does this make h = p(t)? Determine a formula for p(t).
+
+
+
+
+ What are the domain and range of the function h = p(t)? How is this tied to the dimensions of the tank?
+
+
+
+
+ In (a) we observed that V is a function of h, and in (b) we found that h is a function of t. Use these two facts and function composition appropriately to write V as a function of t. Call the resulting function V = q(t).
+
+
+
+
+ What are the domain and range of the function q? Why?
+
+
+
+
+ On the provided axes, sketch accurate graphs of h = p(t) and V = q(t), labeling the vertical and horizontal scale on each graph appropriately. Make your graphs as precise as you can; use a computing device to assist as needed.
+
+
+
+
ADD ALT TEXT TO THIS IMAGE
+
ADD ALT TEXT TO THIS IMAGE
+
+
+
+ Why do each of the two graphs have their respective shapes? Write at least one sentence to explain each graph; refer explicitly to the shape of the tank and other information given in the problem.
+
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
Domain: 0 \le h \le 8 (diameter of sphere). Range: 0 \le V \le \frac{256\pi}{3}.
Since the height increases at a constant rate, h = p(t) = 0.4t is a linear function.
Domain: [0, 20] minutes (height reaches 8 m at t = 8/0.4 = 20 min). Range: [0, 8] m.
The graph of h = p(t) is a straight line (height rises at constant rate). The graph of V = q(t) is an S-shaped increasing cubic: slow at first (sphere is narrow near the bottom), then faster through the middle (widest part), then slower again as the sphere narrows near the top.
- Consider an inverted conical tank (point down) whose top has a radius of 3 feet and that is 2 feet deep. The tank is initially empty and then is filled at a constant rate of 0.75 cubic feet per minute. Let V=f(t) denote the volume of water (in cubic feet) at time t in minutes, and let h= g(t) denote the depth of the water (in feet) at time t.
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- Recall that the volume of a conical tank of radius r and depth h is given by the formula V = \frac{1}{3} \pi r^2 h. How long will it take for the tank to be completely full and how much water will be in the tank at that time?
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- On the provided axes, sketch possible graphs of both V = f(t) and h = g(t), making them as accurate as you can. Label the scale on your axes and points whose coordinates you know for sure; write at least one sentence for each graph to discuss the shape of your graph and why it makes sense in the context of the model.
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ADD ALT TEXT TO THIS IMAGE
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ADD ALT TEXT TO THIS IMAGE
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- What is the domain of the model h = g(t)? its range? why?
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- It's possible to show that the formula for the function g is g(t) = \left( \frac{t}{\pi} \right)^{1/3}. Use a computational device to generate two plots: on the axes at left, the graph of the model h = g(t) = \left( \frac{t}{\pi} \right)^{1/3} on the domain that you decided in (c); on the axes at right, the graph of the abstract function y = p(t) = \left( \frac{t}{\pi} \right)^{1/3} on a wider domain than that of g. What are the domain and range of p and how do these differ from those of the physical model g?
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- Exercise Answer
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The full volume is V = \frac{1}{3}\pi(3)^2(2) = 6\pi \approx 18.85 cubic feet. At a constant fill rate of 0.75 cu ft/min, the tank fills in \frac{6\pi}{0.75} = 8\pi \approx 25.1 minutes.
The graph of V = f(t) is a straight line from (0,0) to (8\pi, 6\pi) (constant rate). The graph of h = g(t) is a concave-down increasing curve from (0,0) to (8\pi, 2): since the cone widens as it fills, equal volumes of water raise the height by smaller and smaller amounts.
The domain of h = g(t) is [0, 8\pi] minutes and the range is [0, 2] feet.
The model h = g(t) = \left(\frac{t}{\pi}\right)^{1/3} has domain [0, 8\pi] and range [0,2]. The abstract function y = p(t) = \left(\frac{t}{\pi}\right)^{1/3} has domain [0, \infty) and range [0, \infty). The model is the abstract function restricted to the physical scenario.
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- A person is taking a walk along a straight path. Their velocity, v (in feet per second), which is a function of time t (in seconds), is given by the graph in Figure.
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The velocity graph for a person walking along a straight path.
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The velocity graph for a person walking along a straight path.
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- What is the person's velocity when t = 2? when t = 7?
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- Are there any times when the person's velocity is exactly v = 3 feet per second? If yes, identify all such times; if not, explain why.
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- Describe the person's behavior on the time interval 4 \le t \le 5.
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- On which time interval does the person travel a farther distance: [1,3] or [6,8]? Why?
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- Exercise Answer
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The values of v(2) and v(7) can be read directly from the graph at t = 2 and t = 7, respectively.
Any time the graph of v crosses or touches the horizontal line v = 3 gives a time when the velocity equals exactly 3 ft/s. Read these times from the graph.
On the interval [4,5], examine whether the graph is positive (walking forward), negative (walking backward), or zero (stationary), and whether the velocity is increasing or decreasing.
The distance traveled is proportional to the area under the velocity curve (or its absolute value). Compare the areas under the graph on [1,3] and on [6,8] to determine which interval has greater total distance.
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- A driver of a new car periodically keeps track of the number of gallons of gas remaining in their car's tank, while simultaneously tracking the trip odometer mileage. Their data is recorded in the following table. Note that at mileages where they add fuel to the tank, they record the mileage twice: once before fuel is added, and once afterward.
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- Remaining gas as a function of distance traveled.
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- D (miles)
- 0
- 50
- 100
- 100
- 150
- 200
- 250
- 300
- 300
- 350
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- G (gallons)
- 4.5
- 3.0
- 1.5
- 10.0
- 8.5
- 7.0
- 5.5
- 4.0
- 11.0
- 9.5
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- Use the table to respond to the questions below.
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- Can the amount of fuel in the gas tank, G, be viewed as a function of distance traveled, D? Why or why not?
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- Does the car's fuel economy appear to be constant or does it appear to vary? Why?
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- At what odometer reading did the driver put the most gas in the tank?
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- Exercise Answer
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No, G cannot be viewed as a function of D. At D = 100, there are two recorded values (G = 1.5 before refueling and G = 10.0 after), and similarly at D = 300. A function must assign exactly one output to each input.
The fuel economy appears to be constant. Between D = 0 and D = 100, the car used 4.5 - 1.5 = 3.0 gallons for 100 miles, giving 100/3 \approx 33.3 mpg. The same rate holds for subsequent segments.
At D = 100, the driver added 10.0 - 1.5 = 8.5 gallons. At D = 300, the driver added 11.0 - 4.0 = 7.0 gallons. So the driver put the most gas in the tank at D = 100 miles.
+ Consider an inverted conical tank (point down) whose top has a radius of 3 feet and that is 2 feet deep. The tank is initially empty and then is filled at a constant rate of 0.75 cubic feet per minute. Let V=f(t) denote the volume of water (in cubic feet) at time t in minutes, and let h= g(t) denote the depth of the water (in feet) at time t.
+
+
+
+
+
+
+ Recall that the volume of a conical tank of radius r and depth h is given by the formula V = \frac{1}{3} \pi r^2 h. How long will it take for the tank to be completely full and how much water will be in the tank at that time?
+
+
+
+
+ On the provided axes, sketch possible graphs of both V = f(t) and h = g(t), making them as accurate as you can. Label the scale on your axes and points whose coordinates you know for sure; write at least one sentence for each graph to discuss the shape of your graph and why it makes sense in the context of the model.
+
+
+
+
ADD ALT TEXT TO THIS IMAGE
+
ADD ALT TEXT TO THIS IMAGE
+
+
+
+
+
+
+ What is the domain of the model h = g(t)? its range? why?
+
+
+
+
+
+ It's possible to show that the formula for the function g is g(t) = \left( \frac{t}{\pi} \right)^{1/3}. Use a computational device to generate two plots: on the axes at left, the graph of the model h = g(t) = \left( \frac{t}{\pi} \right)^{1/3} on the domain that you decided in (c); on the axes at right, the graph of the abstract function y = p(t) = \left( \frac{t}{\pi} \right)^{1/3} on a wider domain than that of g. What are the domain and range of p and how do these differ from those of the physical model g?
+
+
+
+
ADD ALT TEXT TO THIS IMAGE
+
ADD ALT TEXT TO THIS IMAGE
+
+
+
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
The full volume is V = \frac{1}{3}\pi(3)^2(2) = 6\pi \approx 18.85 cubic feet. At a constant fill rate of 0.75 cu ft/min, the tank fills in \frac{6\pi}{0.75} = 8\pi \approx 25.1 minutes.
The graph of V = f(t) is a straight line from (0,0) to (8\pi, 6\pi) (constant rate). The graph of h = g(t) is a concave-down increasing curve from (0,0) to (8\pi, 2): since the cone widens as it fills, equal volumes of water raise the height by smaller and smaller amounts.
The domain of h = g(t) is [0, 8\pi] minutes and the range is [0, 2] feet.
The model h = g(t) = \left(\frac{t}{\pi}\right)^{1/3} has domain [0, 8\pi] and range [0,2]. The abstract function y = p(t) = \left(\frac{t}{\pi}\right)^{1/3} has domain [0, \infty) and range [0, \infty). The model is the abstract function restricted to the physical scenario.
+
+
+
+
+
+
+
+
+
+
+
+
+ A person is taking a walk along a straight path. Their velocity, v (in feet per second), which is a function of time t (in seconds), is given by the graph in Figure.
+
+
+
+
The velocity graph for a person walking along a straight path.
+
The velocity graph for a person walking along a straight path.
+
+
+
+
+
+
+
+ What is the person's velocity when t = 2? when t = 7?
+
+
+
+
+ Are there any times when the person's velocity is exactly v = 3 feet per second? If yes, identify all such times; if not, explain why.
+
+
+
+
+ Describe the person's behavior on the time interval 4 \le t \le 5.
+
+
+
+
+ On which time interval does the person travel a farther distance: [1,3] or [6,8]? Why?
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
The values of v(2) and v(7) can be read directly from the graph at t = 2 and t = 7, respectively.
Any time the graph of v crosses or touches the horizontal line v = 3 gives a time when the velocity equals exactly 3 ft/s. Read these times from the graph.
On the interval [4,5], examine whether the graph is positive (walking forward), negative (walking backward), or zero (stationary), and whether the velocity is increasing or decreasing.
The distance traveled is proportional to the area under the velocity curve (or its absolute value). Compare the areas under the graph on [1,3] and on [6,8] to determine which interval has greater total distance.
+
+
+
+
+
+
+ A driver of a new car periodically keeps track of the number of gallons of gas remaining in their car's tank, while simultaneously tracking the trip odometer mileage. Their data is recorded in the following table. Note that at mileages where they add fuel to the tank, they record the mileage twice: once before fuel is added, and once afterward.
+
+
+
+ Remaining gas as a function of distance traveled.
+
+
+ D (miles)
+ 0
+ 50
+ 100
+ 100
+ 150
+ 200
+ 250
+ 300
+ 300
+ 350
+
+
+ G (gallons)
+ 4.5
+ 3.0
+ 1.5
+ 10.0
+ 8.5
+ 7.0
+ 5.5
+ 4.0
+ 11.0
+ 9.5
+
+
+
+
+
+ Use the table to respond to the questions below.
+
+
+
+
+
+
+ Can the amount of fuel in the gas tank, G, be viewed as a function of distance traveled, D? Why or why not?
+
+
+
+
+ Does the car's fuel economy appear to be constant or does it appear to vary? Why?
+
+
+
+
+ At what odometer reading did the driver put the most gas in the tank?
+
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
No, G cannot be viewed as a function of D. At D = 100, there are two recorded values (G = 1.5 before refueling and G = 10.0 after), and similarly at D = 300. A function must assign exactly one output to each input.
The fuel economy appears to be constant. Between D = 0 and D = 100, the car used 4.5 - 1.5 = 3.0 gallons for 100 miles, giving 100/3 \approx 33.3 mpg. The same rate holds for subsequent segments.
At D = 100, the driver added 10.0 - 1.5 = 8.5 gallons. At D = 300, the driver added 11.0 - 4.0 = 7.0 gallons. So the driver put the most gas in the tank at D = 100 miles.
- Consider the functions p and q whose graphs are given by Figure
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Plots of the graphs of p and q.
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Plots of the graphs of p and q.
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- Compute each of the following values exactly, or explain why they are not defined: p^{-1}(2.5), p^{-1}(-2), p^{-1}(0), and q^{-1}(2).
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- From your work in (a), you know that the point (2.5, -3.5) lies on the graph of p^{-1}. In addition to the other two points you know from (a), find three additional points that lie on the graph of p^{-1}.
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- On Figure, plot the 6 points you have determined in (a) and (b) that lie on the graph of y = p^{-1}(x). Then, sketch the complete graph of y = p^{-1}(x). How are the graphs of p and p^{-1} related to each other?
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- Exercise Answer
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From the graph: p^{-1}(2.5) is the input to p that gives output 2.5, found by reading p(x) = 2.5 from the graph. The problem states p^{-1}(2.5) = -3.5. p^{-1}(-2) and p^{-1}(0) are read similarly; q^{-1}(2) requires reading q(x) = 2.
Since the graph of p^{-1} is the reflection of the graph of p across the line y = x, any point (a, b) on the graph of p gives a point (b, a) on the graph of p^{-1}. Use this to find additional points.
Plot the six known points and sketch the complete graph of y = p^{-1}(x). The graphs of p and p^{-1} are reflections of each other across the line y = x.
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- Consider an inverted conical tank that is being filled with water. The tank's radius is 2 m and its depth is 4 m. Suppose the tank is initially empty and is being filled in such a way that the height of the water is always rising at a rate of 0.25 meters per minute.
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The conical tank.
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The conical tank.
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Axes to plot V = g(t).
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Axes to plot V = g(t).
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- Explain why the height, h, of the water can be viewed as a function of t according to the formula h= f(t) = 0.25t.
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- At what time is the water in the tank 2.5 m deep? At what time is the tank completely full?
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- Suppose we think of the volume, V, of water in the tank as a function of t and name the function V = g(t). Do you expect that the function g has an inverse function? Why or why not?
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-
-
- Recall that the volume of a cone of radius r and height h is V = \frac{\pi}{3} r^2 h. Due to the shape of the tank, similar triangles tell us that r and h satisfy the proportion r = \frac{1}{2}h, and thus
-
- V = \frac{\pi}{3} \left( \frac{1}{2}h \right)^2 h = \frac{\pi}{12}h^3
- .
-
-
-
- Use the fact that h = f(t) = 0.25t along with Equation to find a formula for V = g(t). Sketch a plot of V = g(t) on the blank axes provided in Figure. Write at least one sentence to explain why V = g(t) has the shape that it does.
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- Take the formula for V = g(t) that you determined in (d) and solve for t to determine a formula for t = g^{-1}(V). What is the meaning of the formula you find?
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- Find the exact time that there is \frac{8}{3}\pi cubic meters of volume in the tank.
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- Exercise Answer
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Since the water level rises at a constant rate of 0.25 m/min starting from 0, the height at time t is h = 0.25t, so f(t) = 0.25t.
At h = 2.5: 0.25t = 2.5 \Rightarrow t = 10 min. Full (h=4): t = 4/0.25 = 16 min.
Yes: as time increases, the volume strictly increases (more water is added continuously), so the function g is one-to-one and has an inverse.
Substituting h = 0.25t: V = \frac{\pi}{12}(0.25t)^3 = \frac{\pi}{12} \cdot \frac{t^3}{64} = \frac{\pi t^3}{768}. The graph is an increasing cubic curve (concave up), since the cone widens as it fills.
Solving V = \frac{\pi t^3}{768} for t: t^3 = \frac{768V}{\pi}, so t = g^{-1}(V) = \left(\frac{768V}{\pi}\right)^{1/3}. This gives the time (in minutes) for the volume in the tank to reach V cubic meters.
At V = \frac{8\pi}{3}: t = \left(\frac{768 \cdot \frac{8\pi}{3}}{\pi}\right)^{1/3} = \left(2048\right)^{1/3} = 8\cdot 2^{2/3} = 8\sqrt[3]{4} minutes.
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- Recall that in Activity, we showed that Celsius temperature is a function of the number of chirps per minute from a snowy tree cricket according to the formula
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- C = H(N) = \frac{40}{9} + \frac{5}{36}N
- .
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- What familiar type of function is H? Why must H have an inverse function?
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-
-
-
- Determine an algebraic formula for N = H^{-1}(C). Clearly show your work and thinking.
-
-
-
-
- What is the meaning of the statement 72 = H^{-1}\left(\frac{130}{9}\right)?
-
-
-
-
- Determine the average rate of change of H on the interval [40,50]. Write a complete sentence to explain the meaning of the value you find, including units on the value. Explain clearly how this number describes how the temperature is changing.
-
-
-
-
- Determine the average rate of change of H^{-1} on the interval [15,20]. Write a complete sentence to explain the meaning of the value you find, including units on the value. Explain clearly how this number describes how the number of chirps per minute is changing.
-
+ Consider the functions p and q whose graphs are given by Figure
+
+
+
+
Plots of the graphs of p and q.
+
Plots of the graphs of p and q.
+
+
+
+
+
+
+
+ Compute each of the following values exactly, or explain why they are not defined: p^{-1}(2.5), p^{-1}(-2), p^{-1}(0), and q^{-1}(2).
+
+
+
+
+ From your work in (a), you know that the point (2.5, -3.5) lies on the graph of p^{-1}. In addition to the other two points you know from (a), find three additional points that lie on the graph of p^{-1}.
+
+
+
+
+ On Figure, plot the 6 points you have determined in (a) and (b) that lie on the graph of y = p^{-1}(x). Then, sketch the complete graph of y = p^{-1}(x). How are the graphs of p and p^{-1} related to each other?
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
From the graph: p^{-1}(2.5) is the input to p that gives output 2.5, found by reading p(x) = 2.5 from the graph. The problem states p^{-1}(2.5) = -3.5. p^{-1}(-2) and p^{-1}(0) are read similarly; q^{-1}(2) requires reading q(x) = 2.
Since the graph of p^{-1} is the reflection of the graph of p across the line y = x, any point (a, b) on the graph of p gives a point (b, a) on the graph of p^{-1}. Use this to find additional points.
Plot the six known points and sketch the complete graph of y = p^{-1}(x). The graphs of p and p^{-1} are reflections of each other across the line y = x.
+
+
+
+
+
+
+
+
+ Consider an inverted conical tank that is being filled with water. The tank's radius is 2 m and its depth is 4 m. Suppose the tank is initially empty and is being filled in such a way that the height of the water is always rising at a rate of 0.25 meters per minute.
+
+
+
+
+
The conical tank.
+
The conical tank.
+
+
+
+
Axes to plot V = g(t).
+
Axes to plot V = g(t).
+
+
+
+
+
+
+
+
+ Explain why the height, h, of the water can be viewed as a function of t according to the formula h= f(t) = 0.25t.
+
+
+
+
+ At what time is the water in the tank 2.5 m deep? At what time is the tank completely full?
+
+
+
+
+ Suppose we think of the volume, V, of water in the tank as a function of t and name the function V = g(t). Do you expect that the function g has an inverse function? Why or why not?
+
+
+
+
+ Recall that the volume of a cone of radius r and height h is V = \frac{\pi}{3} r^2 h. Due to the shape of the tank, similar triangles tell us that r and h satisfy the proportion r = \frac{1}{2}h, and thus
+
+ V = \frac{\pi}{3} \left( \frac{1}{2}h \right)^2 h = \frac{\pi}{12}h^3
+ .
+
+
+
+ Use the fact that h = f(t) = 0.25t along with Equation to find a formula for V = g(t). Sketch a plot of V = g(t) on the blank axes provided in Figure. Write at least one sentence to explain why V = g(t) has the shape that it does.
+
+
+
+
+
+
+
+ Take the formula for V = g(t) that you determined in (d) and solve for t to determine a formula for t = g^{-1}(V). What is the meaning of the formula you find?
+
+
+
+
+ Find the exact time that there is \frac{8}{3}\pi cubic meters of volume in the tank.
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
Since the water level rises at a constant rate of 0.25 m/min starting from 0, the height at time t is h = 0.25t, so f(t) = 0.25t.
At h = 2.5: 0.25t = 2.5 \Rightarrow t = 10 min. Full (h=4): t = 4/0.25 = 16 min.
Yes: as time increases, the volume strictly increases (more water is added continuously), so the function g is one-to-one and has an inverse.
Substituting h = 0.25t: V = \frac{\pi}{12}(0.25t)^3 = \frac{\pi}{12} \cdot \frac{t^3}{64} = \frac{\pi t^3}{768}. The graph is an increasing cubic curve (concave up), since the cone widens as it fills.
Solving V = \frac{\pi t^3}{768} for t: t^3 = \frac{768V}{\pi}, so t = g^{-1}(V) = \left(\frac{768V}{\pi}\right)^{1/3}. This gives the time (in minutes) for the volume in the tank to reach V cubic meters.
At V = \frac{8\pi}{3}: t = \left(\frac{768 \cdot \frac{8\pi}{3}}{\pi}\right)^{1/3} = \left(2048\right)^{1/3} = 8\cdot 2^{2/3} = 8\sqrt[3]{4} minutes.
+
+
+
+
+
+
+
+
+ Recall that in Activity, we showed that Celsius temperature is a function of the number of chirps per minute from a snowy tree cricket according to the formula
+
+ C = H(N) = \frac{40}{9} + \frac{5}{36}N
+ .
+
+
+
+
+
+
+ What familiar type of function is H? Why must H have an inverse function?
+
+
+
+
+ Determine an algebraic formula for N = H^{-1}(C). Clearly show your work and thinking.
+
+
+
+
+ What is the meaning of the statement 72 = H^{-1}\left(\frac{130}{9}\right)?
+
+
+
+
+ Determine the average rate of change of H on the interval [40,50]. Write a complete sentence to explain the meaning of the value you find, including units on the value. Explain clearly how this number describes how the temperature is changing.
+
+
+
+
+ Determine the average rate of change of H^{-1} on the interval [15,20]. Write a complete sentence to explain the meaning of the value you find, including units on the value. Explain clearly how this number describes how the number of chirps per minute is changing.
+
- An apartment manager keeps careful record of how the rent charged per unit corresponds to the number of occupied units in a large complex. The collected data is shown in Table.This problem is a slightly modified version of one found in Carroll College's Chapter Zero resource for Active Calculus.
-
- Why is it reasonable to say that the number of occupied apartments is a linear function of rent?
-
-
-
-
-
- Let A be the number of occupied apartments and R the monthly rent charged (in dollars). If we let A = f(R), what is the slope of the linear function f? What is the meaning of the slope in the context of this question?
-
-
-
-
-
- Determine a formula for A = f(R). What do you think is a reasonable domain for the function? Why?
-
-
-
-
-
- If the rent were to be increased to $1000, how many occupied apartments should the
- apartment manager expect? How much total revenue would the manager collect in a given month when rent is set at $1000?
-
-
-
-
-
- Why do you think the apartment manager is interested in the data that has been collected?
-
-
-
-
-
-
-
- Exercise Answer
-
-
-
-
Each \$50 increase in rent results in a decrease of exactly 7 occupied apartments (constant rate of change), which means the relationship is linear.
Slope = \frac{196-203}{700-650} = \frac{-7}{50} = -0.14 apartments per dollar. For each dollar increase in rent, the expected number of occupied apartments decreases by 0.14.
A = f(R) = 203 - 0.14(R - 650) = 294 - 0.14R. A reasonable domain is [650, 2100] (rents below \$650 are below current data, and above \$2100 all apartments would be empty).
At R = 1000: A = 294 - 0.14(1000) = 154 apartments. Monthly revenue = 1000 \times 154 = \$154{,}000.
The manager wants to find the rent that maximizes total revenue, R \cdot f(R) = 294R - 0.14R^2. This is maximized at R = 294/(2 \times 0.14) = \$1050, giving 147 apartments and \$154{,}350 monthly revenue.
-
-
-
-
-
-
- Alicia and Dexter are each walking on a straight path. For a particular 10-second window of time, each has their velocity (in feet per second) measured and recorded as a function of time. Their respective velocity functions are plotted in Figure.
-
-
-
-
The velocity functions A = f(t) and D = g(t) for Alicia and Damon, respectively.
-
The velocity functions A = f(t) and D = g(t) for Alicia and Damon, respectively.
-
-
-
-
-
-
-
- Determine formulas for both A = f(t) and D = g(t).
-
-
-
-
- What is the value and meaning of the slope of A? Write a complete sentence to explain and be sure to include units in your response.
-
-
-
-
- What is the value and meaning of the average rate of change of D on the interval [4,8]? Write a complete sentence to explain and be sure to include units in your response.
-
-
-
-
- Is there ever a time when Alicia and Damon are walking at the same velocity? If yes, determine both the time and velocity; if not, explain why.
-
-
-
-
- Is is possible to determine if there is ever a time when Alicia and Damon are located at the same place on the path? If yes, determine the time and location; if not, explain why not enough information is provided.
-
-
-
-
-
-
-
- Exercise Answer
-
-
-
-
Read the formulas for A = f(t) and D = g(t) from the graph by identifying slope and intercept for each linear piece.
The slope of A = f(t) is its constant rate of change (acceleration in ft/s²). Write a sentence interpreting this value: e.g., "Alicia's velocity increases/decreases at a rate of [slope] feet per second per second."
AV_{[4,8]} for D is computed from the graph values g(4) and g(8); its meaning is the average rate of change of Damon's velocity on that interval.
Set f(t) = g(t) and solve for t. If a solution exists in [0,10], they walk at the same velocity at that time.
The velocity functions do not give us location information (we would need initial positions), so it is generally not possible to determine when/if they are at the same location.
-
-
-
-
-
-
- An inverted conical tank with depth 4 feet and radius 2 feet is completely full of water. The tank is being drained by a pump in such a way that the amount of water in the tank is decreasing at a constant rate of 1.5 cubic feet per minute. Let V = f(t) denote the volume of water in the tank at time t and h = g(t) the depth of the water in the tank at time t, where t is measured in minutes.
-
-
-
-
The inverted conical tank.
-
The inverted conical tank.
-
-
-
-
-
-
-
- How much water is in the tank at t = 0 when the tank is completely full?
-
-
-
-
- Explain why volume, V, when viewed as a function of time, t, is a linear function.
-
-
-
-
- Determine a formula for V = f(t).
-
-
-
-
- At what exact time will the tank be empty?
-
-
-
-
- What is a reasonable domain to use for the model f? What is its corresponding range?
-
-
-
-
-
-
-
-
- Exercise Answer
-
-
-
-
The full cone has volume V(0) = \frac{1}{3}\pi(2)^2(4) = \frac{16\pi}{3} \approx 16.76 cu ft.
The volume decreases at a constant rate of 1.5 cu ft/min (given), so the rate of change of V with respect to t is constant: V is linear in t.
+ An apartment manager keeps careful record of how the rent charged per unit corresponds to the number of occupied units in a large complex. The collected data is shown in Table.This problem is a slightly modified version of one found in Carroll College's Chapter Zero resource for Active Calculus.
+
+ Why is it reasonable to say that the number of occupied apartments is a linear function of rent?
+
+
+
+
+
+ Let A be the number of occupied apartments and R the monthly rent charged (in dollars). If we let A = f(R), what is the slope of the linear function f? What is the meaning of the slope in the context of this question?
+
+
+
+
+
+ Determine a formula for A = f(R). What do you think is a reasonable domain for the function? Why?
+
+
+
+
+
+ If the rent were to be increased to $1000, how many occupied apartments should the
+ apartment manager expect? How much total revenue would the manager collect in a given month when rent is set at $1000?
+
+
+
+
+
+ Why do you think the apartment manager is interested in the data that has been collected?
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
Each \$50 increase in rent results in a decrease of exactly 7 occupied apartments (constant rate of change), which means the relationship is linear.
Slope = \frac{196-203}{700-650} = \frac{-7}{50} = -0.14 apartments per dollar. For each dollar increase in rent, the expected number of occupied apartments decreases by 0.14.
A = f(R) = 203 - 0.14(R - 650) = 294 - 0.14R. A reasonable domain is [650, 2100] (rents below \$650 are below current data, and above \$2100 all apartments would be empty).
At R = 1000: A = 294 - 0.14(1000) = 154 apartments. Monthly revenue = 1000 \times 154 = \$154{,}000.
The manager wants to find the rent that maximizes total revenue, R \cdot f(R) = 294R - 0.14R^2. This is maximized at R = 294/(2 \times 0.14) = \$1050, giving 147 apartments and \$154{,}350 monthly revenue.
+
+
+
+
+
+
+ Alicia and Dexter are each walking on a straight path. For a particular 10-second window of time, each has their velocity (in feet per second) measured and recorded as a function of time. Their respective velocity functions are plotted in Figure.
+
+
+
+
The velocity functions A = f(t) and D = g(t) for Alicia and Damon, respectively.
+
The velocity functions A = f(t) and D = g(t) for Alicia and Damon, respectively.
+
+
+
+
+
+
+
+ Determine formulas for both A = f(t) and D = g(t).
+
+
+
+
+ What is the value and meaning of the slope of A? Write a complete sentence to explain and be sure to include units in your response.
+
+
+
+
+ What is the value and meaning of the average rate of change of D on the interval [4,8]? Write a complete sentence to explain and be sure to include units in your response.
+
+
+
+
+ Is there ever a time when Alicia and Damon are walking at the same velocity? If yes, determine both the time and velocity; if not, explain why.
+
+
+
+
+ Is is possible to determine if there is ever a time when Alicia and Damon are located at the same place on the path? If yes, determine the time and location; if not, explain why not enough information is provided.
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
Read the formulas for A = f(t) and D = g(t) from the graph by identifying slope and intercept for each linear piece.
The slope of A = f(t) is its constant rate of change (acceleration in ft/s²). Write a sentence interpreting this value: e.g., "Alicia's velocity increases/decreases at a rate of [slope] feet per second per second."
AV_{[4,8]} for D is computed from the graph values g(4) and g(8); its meaning is the average rate of change of Damon's velocity on that interval.
Set f(t) = g(t) and solve for t. If a solution exists in [0,10], they walk at the same velocity at that time.
The velocity functions do not give us location information (we would need initial positions), so it is generally not possible to determine when/if they are at the same location.
+
+
+
+
+
+
+ An inverted conical tank with depth 4 feet and radius 2 feet is completely full of water. The tank is being drained by a pump in such a way that the amount of water in the tank is decreasing at a constant rate of 1.5 cubic feet per minute. Let V = f(t) denote the volume of water in the tank at time t and h = g(t) the depth of the water in the tank at time t, where t is measured in minutes.
+
+
+
+
The inverted conical tank.
+
The inverted conical tank.
+
+
+
+
+
+
+
+ How much water is in the tank at t = 0 when the tank is completely full?
+
+
+
+
+ Explain why volume, V, when viewed as a function of time, t, is a linear function.
+
+
+
+
+ Determine a formula for V = f(t).
+
+
+
+
+ At what exact time will the tank be empty?
+
+
+
+
+ What is a reasonable domain to use for the model f? What is its corresponding range?
+
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
The full cone has volume V(0) = \frac{1}{3}\pi(2)^2(4) = \frac{16\pi}{3} \approx 16.76 cu ft.
The volume decreases at a constant rate of 1.5 cu ft/min (given), so the rate of change of V with respect to t is constant: V is linear in t.
- Two quadratic functions, f and g, are determined by their respective graphs in Figure.
-
-
-
-
Two quadratic functions, f and g.
-
Two quadratic functions, f and g.
-
-
-
-
-
-
-
- How does the information provided enable you to find a formula for f? Explain, and determine the formula.
-
-
-
-
- How does the information provided enable you to find a formula for g? Explain, and determine the formula.
-
-
-
-
- Consider an additional quadratic function h given by h(x) = 2x^2 - 8x + 6. Does the graph of h intersect the graph of f? If yes, determine the exact points of intersection, with justification. If not, explain why.
-
-
-
-
- Does the graph of h intersect the graph of g? If yes, determine the exact points of intersection, with justification. If not, explain why.
-
-
-
-
-
-
-
- Exercise Answer
-
-
-
-
For f: read the vertex (h,k) and one additional point from the graph. Write f(x) = a(x-h)^2 + k and solve for a using the additional point.
For g: similarly read the vertex and one additional point and determine the formula g(x) = a(x-h)^2 + k.
Set h(x) = f(x): 2x^2 - 8x + 6 = f(x). Solve the resulting equation (quadratic). If the discriminant is non-negative, there are intersection points; otherwise the graphs do not intersect.
Set h(x) = g(x) and solve similarly. The discriminant determines whether intersection points exist.
-
-
-
-
-
-
-
-
- Consider the quadratic function f given by f(x) = \frac{1}{2}(x-2)^2 + 1.
-
-
-
-
-
-
- Determine the exact location of the vertex of f.
-
-
-
-
- Does f have 0, 1, or 2x-intercepts? Explain, and determine the location(s) of any x-intercept(s) that exist.
-
-
-
-
- Complete the following tables of function values and average rates of change of f at the stated inputs and intervals.
-
The function values are symmetric about x = 2 (the vertex). The average rates of change form an arithmetic sequence increasing by 1 at each step, a characteristic pattern of quadratic functions.
-
-
-
-
-
-
- A water balloon is tossed vertically from a window on the fourth floor of a dormitory from an initial height of 56.3 feet. A person two floors above observes the balloon reach its highest point 1.2 seconds after being launched.
-
-
-
-
-
-
- What is the balloon's exact height at t = 2.4? Why?
-
-
-
-
- What is the exact maximum height the balloon reaches at t = 1.2?
-
-
-
-
- What exact time did the balloon land?
-
-
-
-
- At what initial velocity was the balloon launched?
-
-
-
-
-
-
-
- Exercise Answer
-
-
-
-
The vertex occurs at t = 1.2. By symmetry of the parabola, t = 2.4 is symmetric to t = 0 about the vertex: s(2.4) = s(0) = 56.3 feet.
At the vertex, the initial velocity is v_0 = 32 \times 1.2 = 38.4 ft/s (since vertex is at t = v_0/32). Maximum height: s(1.2) = -16(1.44) + 38.4(1.2) + 56.3 = -23.04 + 46.08 + 56.3 = 79.34 feet.
Set s(t) = 0: -16t^2 + 38.4t + 56.3 = 0. Using the quadratic formula: t = \frac{38.4 + \sqrt{38.4^2 + 4(16)(56.3)}}{32} = \frac{38.4 + \sqrt{5077.76}}{32} \approx \frac{38.4 + 71.26}{32} \approx 3.43 seconds.
The initial velocity is v_0 = 38.4 = \frac{192}{5} feet per second.
+ Two quadratic functions, f and g, are determined by their respective graphs in Figure.
+
+
+
+
Two quadratic functions, f and g.
+
Two quadratic functions, f and g.
+
+
+
+
+
+
+
+ How does the information provided enable you to find a formula for f? Explain, and determine the formula.
+
+
+
+
+ How does the information provided enable you to find a formula for g? Explain, and determine the formula.
+
+
+
+
+ Consider an additional quadratic function h given by h(x) = 2x^2 - 8x + 6. Does the graph of h intersect the graph of f? If yes, determine the exact points of intersection, with justification. If not, explain why.
+
+
+
+
+ Does the graph of h intersect the graph of g? If yes, determine the exact points of intersection, with justification. If not, explain why.
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
For f: read the vertex (h,k) and one additional point from the graph. Write f(x) = a(x-h)^2 + k and solve for a using the additional point.
For g: similarly read the vertex and one additional point and determine the formula g(x) = a(x-h)^2 + k.
Set h(x) = f(x): 2x^2 - 8x + 6 = f(x). Solve the resulting equation (quadratic). If the discriminant is non-negative, there are intersection points; otherwise the graphs do not intersect.
Set h(x) = g(x) and solve similarly. The discriminant determines whether intersection points exist.
+
+
+
+
+
+
+
+
+ Consider the quadratic function f given by f(x) = \frac{1}{2}(x-2)^2 + 1.
+
+
+
+
+
+
+ Determine the exact location of the vertex of f.
+
+
+
+
+ Does f have 0, 1, or 2x-intercepts? Explain, and determine the location(s) of any x-intercept(s) that exist.
+
+
+
+
+ Complete the following tables of function values and average rates of change of f at the stated inputs and intervals.
+
The function values are symmetric about x = 2 (the vertex). The average rates of change form an arithmetic sequence increasing by 1 at each step, a characteristic pattern of quadratic functions.
+
+
+
+
+
+
+ A water balloon is tossed vertically from a window on the fourth floor of a dormitory from an initial height of 56.3 feet. A person two floors above observes the balloon reach its highest point 1.2 seconds after being launched.
+
+
+
+
+
+
+ What is the balloon's exact height at t = 2.4? Why?
+
+
+
+
+ What is the exact maximum height the balloon reaches at t = 1.2?
+
+
+
+
+ What exact time did the balloon land?
+
+
+
+
+ At what initial velocity was the balloon launched?
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
The vertex occurs at t = 1.2. By symmetry of the parabola, t = 2.4 is symmetric to t = 0 about the vertex: s(2.4) = s(0) = 56.3 feet.
At the vertex, the initial velocity is v_0 = 32 \times 1.2 = 38.4 ft/s (since vertex is at t = v_0/32). Maximum height: s(1.2) = -16(1.44) + 38.4(1.2) + 56.3 = -23.04 + 46.08 + 56.3 = 79.34 feet.
Set s(t) = 0: -16t^2 + 38.4t + 56.3 = 0. Using the quadratic formula: t = \frac{38.4 + \sqrt{38.4^2 + 4(16)(56.3)}}{32} = \frac{38.4 + \sqrt{5077.76}}{32} \approx \frac{38.4 + 71.26}{32} \approx 3.43 seconds.
The initial velocity is v_0 = 38.4 = \frac{192}{5} feet per second.
- Suppose we have an unusual tank whose base is a perfect sphere with radius 3 feet, and then atop the spherical base is a cylindrical chimney that is a circular cylinder of radius 1 foot and height 2 feet, as shown in Figure. The tank is initially empty, but then a spigot is turned on that pumps water into the tank at a constant rate of 1.25 cubic feet per minute.
-
-
-
-
A spherical tank with a cylindrical chimney.
-
A spherical tank with a cylindrical chimney.
-
-
-
-
- Let V denote the total volume of water (in cubic feet) in the tank at any time t (in minutes), and h the depth of the water (in feet) at time t.
-
-
-
-
-
-
- It is possible to use calculus to show that the total volume this tank can hold is V_{\text{full}} = \pi(20 + \frac{38}{3}\sqrt{2}) \approx 119.11 cubic feet. In addition, the actual height of the tank (from the bottom of the spherical base to the top of the chimney) is h_{\text{full}} = \sqrt{8} + 5 \approx 7.83 feet. How long does it take the tank to fill? Why?
-
-
-
-
- On the blank axes provided below, sketch (by hand) possible graphs of how V and t change in tandem and how h and t change in tandem.
-
-
-
-
ADD ALT TEXT TO THIS IMAGE
-
ADD ALT TEXT TO THIS IMAGE
-
-
-
- For each graph, label any ordered pairs on the graph that you know for certain, and write at least one sentence that explains why your graphs have the shape they do.
-
-
-
-
- How would your graph(s) change (if at all) if the chimney was shaped like an inverted cone instead of a cylinder? Explain and discuss.
-
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-
-
-
-
-
-
- Exercise Answer
-
-
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-
At a fill rate of 1.25 cu ft/min, the time to fill is \frac{V_{\text{full}}}{1.25} = \frac{\pi(20 + \frac{38}{3}\sqrt{2})}{1.25} \approx \frac{119.11}{1.25} \approx 95.3 minutes.
The graph of V vs t is a straight line with slope 1.25 from (0,0) to (95.3, 119.11). The graph of h vs t is an S-shaped increasing curve: initially h rises quickly as the narrow bottom of the sphere fills, then more slowly as the widest part of the sphere fills, then speeds up slightly as the narrow chimney fills. Key points include (0,0), the time when h = \sqrt{8} + 3 \approx 5.83 ft (sphere full, chimney starts), and the endpoint (95.3, \sqrt{8}+5).
If the chimney were an inverted cone, its cross-section would shrink as it fills, so the height would increase more quickly in the chimney. The V-vs-t graph would not change (still linear), but the h-vs-t graph would rise more steeply once the water enters the cone-shaped chimney.
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- Suppose we have a tank that is a perfect sphere with radius 6 feet. The tank is initially empty, but then a spigot is turned on that is pumping water into the tank in a very special way: the faucet is regulated so that the depth of water in the tank is increasing at a constant rate of 0.4 feet per minute.
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- Let V denote the total volume of water (in cubic feet) in the tank at any time t (in minutes), and h the depth of the water (in feet) at given time t.
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- How long does it take the tank to fill? What will the values of V and h be at the moment the tank is full? Why?
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- On the blank axes provided below, sketch (by hand) possible graphs of how V and t change in tandem and how h and t change in tandem.
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ADD ALT TEXT TO THIS IMAGE
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ADD ALT TEXT TO THIS IMAGE
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- For each graph, label any ordered pairs on the graph that you know for certain, and write at least one sentence that explains why your graphs have the shape they do.
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- How do your responses change if the tank stays the same but instead the tank is initially full and the tank drains in such a way that the height of the water is always decreasing at a constant rate of 0.25 feet per minute?
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- Exercise Answer
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At a constant rate of 0.4 ft/min, the depth reaches 12 ft (the diameter) in 12/0.4 = 30 minutes. At that time, V = \frac{4}{3}\pi(6)^3 = 288\pi \approx 904.8 cu ft.
Since the height rises linearly, h = 0.4t is a straight line. The volume V as a function of t starts slowly (the bottom of the sphere is narrow), increases rapidly through the middle (widest part of sphere), and slows again as the top narrows. The V-vs-t graph is S-shaped (concave up then concave down). Key points: (0,0) and (30, 288\pi).
For the draining scenario: h starts at 12 ft and decreases linearly at 0.25 ft/min to 0 in 48 minutes. The h-vs-t graph is a decreasing line. The V-vs-t graph is an S-shaped decreasing curve (symmetric to the filling case), from (0, 288\pi) to (48, 0).
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- The relationship between the position, s, of a car driving on a straight road at time t is given by the graph
- pictured at left in Figure.
- The car's positionYou can think of the car's position like mile-markers on a highway. Saying that s = 500 means that the car is located 500 feet from marker zero on the road. has units measured in thousands of feet while time is measured in minutes.
- For instance, the point (4,6) on the graph indicates that after 4 minutes,
- the car has traveled 6000 feet from its starting location.
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- Write several sentences that explain the how the car is being driven and how you make these conclusions from the graph.
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- How far did the car travel between t = 2 and t = 10?
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- Does the car ever travel in reverse? Why or why not? If not, how would the graph have to look to indicate such motion?
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- On the blank axes in Figure, plot points or sketch a curve to describe the behavior of a car that is driven in the following way: from t = 0 to t = 5 the car travels straight down the road at a constant rate of 1000 feet per minute. At t = 5, the car pulls over and parks for 2 full minutes. Then, at t = 7, the car does an abrupt U-turn and returns in the opposite direction at a constant rate of 800 feet per minute for 5 additional minutes. As part of your work, determine (and label) the car's location at several additional points in time other than t = 0, 5, 7, 12.
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A graph of the relationship between a car's position s and time t
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A graph of the relationship between a car's position s and time t
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A graph of the relationship between a car's position s and time t
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- Exercise Answer
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Describe the car based on the graph: on intervals where s is increasing, the car moves forward; where s is flat, the car is stopped; where s decreases, the car is in reverse. The slope (steepness) indicates speed.
From the graph, s(2) and s(10) can be read. The distance traveled between t = 2 and t = 10 is |s(10) - s(2)| thousand feet.
If the position graph is always non-decreasing (never moves backward), the car never goes in reverse. A graph that decreases (negative slope) over some interval would indicate reverse motion.
For the described motion: t = 0: s = 0. At t = 5: s = 1000 \times 5 = 5000 ft (i.e., 5 thousand ft). From t = 5 to t = 7: car is stopped, so s = 5. At t = 7: U-turn, moving at -800 ft/min. At t = 12: s = 5000 - 800 \times 5 = 5000 - 4000 = 1000 ft (i.e., 1 thousand ft). Plot a line of slope 1 from (0,0) to (5,5), horizontal from (5,5) to (7,5), and a line of slope -0.8 from (7,5) to (12,1).
+ Suppose we have an unusual tank whose base is a perfect sphere with radius 3 feet, and then atop the spherical base is a cylindrical chimney that is a circular cylinder of radius 1 foot and height 2 feet, as shown in Figure. The tank is initially empty, but then a spigot is turned on that pumps water into the tank at a constant rate of 1.25 cubic feet per minute.
+
+
+
+
A spherical tank with a cylindrical chimney.
+
A spherical tank with a cylindrical chimney.
+
+
+
+
+ Let V denote the total volume of water (in cubic feet) in the tank at any time t (in minutes), and h the depth of the water (in feet) at time t.
+
+
+
+
+
+
+ It is possible to use calculus to show that the total volume this tank can hold is V_{\text{full}} = \pi(20 + \frac{38}{3}\sqrt{2}) \approx 119.11 cubic feet. In addition, the actual height of the tank (from the bottom of the spherical base to the top of the chimney) is h_{\text{full}} = \sqrt{8} + 5 \approx 7.83 feet. How long does it take the tank to fill? Why?
+
+
+
+
+ On the blank axes provided below, sketch (by hand) possible graphs of how V and t change in tandem and how h and t change in tandem.
+
+
+
+
ADD ALT TEXT TO THIS IMAGE
+
ADD ALT TEXT TO THIS IMAGE
+
+
+
+ For each graph, label any ordered pairs on the graph that you know for certain, and write at least one sentence that explains why your graphs have the shape they do.
+
+
+
+
+ How would your graph(s) change (if at all) if the chimney was shaped like an inverted cone instead of a cylinder? Explain and discuss.
+
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
At a fill rate of 1.25 cu ft/min, the time to fill is \frac{V_{\text{full}}}{1.25} = \frac{\pi(20 + \frac{38}{3}\sqrt{2})}{1.25} \approx \frac{119.11}{1.25} \approx 95.3 minutes.
The graph of V vs t is a straight line with slope 1.25 from (0,0) to (95.3, 119.11). The graph of h vs t is an S-shaped increasing curve: initially h rises quickly as the narrow bottom of the sphere fills, then more slowly as the widest part of the sphere fills, then speeds up slightly as the narrow chimney fills. Key points include (0,0), the time when h = \sqrt{8} + 3 \approx 5.83 ft (sphere full, chimney starts), and the endpoint (95.3, \sqrt{8}+5).
If the chimney were an inverted cone, its cross-section would shrink as it fills, so the height would increase more quickly in the chimney. The V-vs-t graph would not change (still linear), but the h-vs-t graph would rise more steeply once the water enters the cone-shaped chimney.
+
+
+
+
+
+
+ Suppose we have a tank that is a perfect sphere with radius 6 feet. The tank is initially empty, but then a spigot is turned on that is pumping water into the tank in a very special way: the faucet is regulated so that the depth of water in the tank is increasing at a constant rate of 0.4 feet per minute.
+
+
+
+ Let V denote the total volume of water (in cubic feet) in the tank at any time t (in minutes), and h the depth of the water (in feet) at given time t.
+
+
+
+
+
+
+ How long does it take the tank to fill? What will the values of V and h be at the moment the tank is full? Why?
+
+
+
+
+ On the blank axes provided below, sketch (by hand) possible graphs of how V and t change in tandem and how h and t change in tandem.
+
+
+
+
ADD ALT TEXT TO THIS IMAGE
+
ADD ALT TEXT TO THIS IMAGE
+
+
+
+ For each graph, label any ordered pairs on the graph that you know for certain, and write at least one sentence that explains why your graphs have the shape they do.
+
+
+
+
+ How do your responses change if the tank stays the same but instead the tank is initially full and the tank drains in such a way that the height of the water is always decreasing at a constant rate of 0.25 feet per minute?
+
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
At a constant rate of 0.4 ft/min, the depth reaches 12 ft (the diameter) in 12/0.4 = 30 minutes. At that time, V = \frac{4}{3}\pi(6)^3 = 288\pi \approx 904.8 cu ft.
Since the height rises linearly, h = 0.4t is a straight line. The volume V as a function of t starts slowly (the bottom of the sphere is narrow), increases rapidly through the middle (widest part of sphere), and slows again as the top narrows. The V-vs-t graph is S-shaped (concave up then concave down). Key points: (0,0) and (30, 288\pi).
For the draining scenario: h starts at 12 ft and decreases linearly at 0.25 ft/min to 0 in 48 minutes. The h-vs-t graph is a decreasing line. The V-vs-t graph is an S-shaped decreasing curve (symmetric to the filling case), from (0, 288\pi) to (48, 0).
+
+
+
+
+
+
+ The relationship between the position, s, of a car driving on a straight road at time t is given by the graph
+ pictured at left in Figure.
+ The car's positionYou can think of the car's position like mile-markers on a highway. Saying that s = 500 means that the car is located 500 feet from marker zero on the road. has units measured in thousands of feet while time is measured in minutes.
+ For instance, the point (4,6) on the graph indicates that after 4 minutes,
+ the car has traveled 6000 feet from its starting location.
+
+
+
+
+
+
+ Write several sentences that explain the how the car is being driven and how you make these conclusions from the graph.
+
+
+
+
+ How far did the car travel between t = 2 and t = 10?
+
+
+
+
+ Does the car ever travel in reverse? Why or why not? If not, how would the graph have to look to indicate such motion?
+
+
+
+
+ On the blank axes in Figure, plot points or sketch a curve to describe the behavior of a car that is driven in the following way: from t = 0 to t = 5 the car travels straight down the road at a constant rate of 1000 feet per minute. At t = 5, the car pulls over and parks for 2 full minutes. Then, at t = 7, the car does an abrupt U-turn and returns in the opposite direction at a constant rate of 800 feet per minute for 5 additional minutes. As part of your work, determine (and label) the car's location at several additional points in time other than t = 0, 5, 7, 12.
+
+
+
A graph of the relationship between a car's position s and time t
+
+
A graph of the relationship between a car's position s and time t
+
+
A graph of the relationship between a car's position s and time t
+
+
+
+
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
Describe the car based on the graph: on intervals where s is increasing, the car moves forward; where s is flat, the car is stopped; where s decreases, the car is in reverse. The slope (steepness) indicates speed.
From the graph, s(2) and s(10) can be read. The distance traveled between t = 2 and t = 10 is |s(10) - s(2)| thousand feet.
If the position graph is always non-decreasing (never moves backward), the car never goes in reverse. A graph that decreases (negative slope) over some interval would indicate reverse motion.
For the described motion: t = 0: s = 0. At t = 5: s = 1000 \times 5 = 5000 ft (i.e., 5 thousand ft). From t = 5 to t = 7: car is stopped, so s = 5. At t = 7: U-turn, moving at -800 ft/min. At t = 12: s = 5000 - 800 \times 5 = 5000 - 4000 = 1000 ft (i.e., 1 thousand ft). Plot a line of slope 1 from (0,0) to (5,5), horizontal from (5,5) to (7,5), and a line of slope -0.8 from (7,5) to (12,1).
- Let g(x) = f(x) + 5. Determine AV_{[-3,-1]} and AV_{[2,5]} for both f and g. What do you observe? Why does this phenomenon occur?
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- Let h(x) = f(x-2). For f, recall that you determined AV_{[-3,-1]} and AV_{[2,5]} in (a). In addition, determine AV_{[-1,1]} and AV_{[4,7]} for h. What do you observe? Why does this phenomenon occur?
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- Let k(x) = 3f(x). Determine AV_{[-3,-1]} and AV_{[2,5]} for k, and compare the results to your earlier computations of AV_{[-3,-1]} and AV_{[2,5]} for f. What do you observe? Why does this phenomenon occur?
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- Finally, let m(x) = 3f(x-2) + 5. Without doing any computations, what do you think will be true about the relationship between AV_{[-3,-1]} for f and AV_{[-1,1]} for m? Why? After making your conjecture, execute appropriate computations to see if your intuition is correct.
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- Exercise Answer
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For f(x)=x^2: AV_{[-3,-1]}=(1-9)/2=-4 and AV_{[2,5]}=(25-4)/3=7. For g(x)=x^2+5: AV_{[-3,-1]}=(6-14)/2=-4 and AV_{[2,5]}=(30-9)/3=7. The values are the same: vertical shifts do not change average rates of change.
For h(x)=(x-2)^2: AV_{[-1,1]}=(1-9)/2=-4 (same as AV_{[-3,-1]} for f) and AV_{[4,7]}=(25-4)/3=7 (same as AV_{[2,5]} for f). A horizontal shift of 2 units right causes the average rate of change to match on intervals that are shifted right by 2.
For k(x)=3x^2: AV_{[-3,-1]}=(3-27)/2=-12=3\cdot(-4) and AV_{[2,5]}=(75-12)/3=21=3\cdot 7. Multiplying the output by 3 multiplies all average rates of change by 3.
Conjecture: AV_{[-1,1]} for m equals 3\cdot AV_{[-3,-1]} for f, which is 3(-4) = -12. Verification: m(x)=3(x-2)^2+5, m(-1)=3(9)+5=32, m(1)=3(1)+5=8, AV_{[-1,1]}=(8-32)/2=-12. \checkmark
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- Consider the parent function y = f(x) = x.
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- Consider the linear function in point-slope form given by y = L(x) = -4(x-3) + 5. What is the slope of this line? What is the most obvious point that lies on the line?
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- How can the function L given in (a) be viewed as a transformation of the parent function f? Explain the roles of 3, -4, and 5, respectively.
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- Explain why any non-vertical line of the form P(x) = m(x-x_0) + y_0 can be thought of as a transformation of the parent function f(x) = x. Specifically discuss the transformation(s) involved.
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- Find a formula for the transformation of f(x) = x that corresponds to a horizontal shift of 7 units left, a reflection across y = 0 and vertical stretch of 3 units away from the x-axis, and a vertical shift of -11 units.
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- Exercise Answer
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The slope is -4. The most obvious point is (3, 5).
Starting from f(x)=x: shift right by 3 gives f(x-3)=x-3; multiply by -4 gives -4(x-3) (vertical stretch by 4 and reflection); shift up by 5 gives -4(x-3)+5 = L(x). So 3 is the horizontal shift, -4 is the vertical stretch/reflection factor, and 5 is the vertical shift.
P(x)=m(x-x_0)+y_0 is a horizontal shift of f by x_0, a vertical stretch by m, and a vertical shift by y_0.
Horizontal shift left 7: f(x+7) = x+7. Reflection and vertical stretch by 3: -3(x+7). Vertical shift -11: -3(x+7)-11 = -3x-32.
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- We have explored the effects of adding a constant to the output of a function, y = f(x) + a, adding a constant to the input, y = f(x+a), and multiplying the output of a function by a constant, y = af(x). There is one remaining natural transformation to explore: multiplying the input to a function by a constant. In this exercise, we consider the effects of the constant a in transforming a parent function f by the rule y = f(ax).
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- Let f(x) = (x-2)^2 + 1.
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- Let g(x) = f(4x), h(x) = f(2x), k(x) = f(0.5x), and m(x) = f(0.25x). Use Desmos to plot these functions. Then, sketch and label g, h, k, and m on the provided axes in Figure along with the graph of f. For each of the functions, label and identify its vertex, its y-intercept, and its x-intercepts.
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Axes for plotting f, g, h, k, and m in part (a).
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Axes for plotting f, g, h, k, and m in part (a).
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Axes for plotting f, r, and s from parts (c) and (d).
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Axes for plotting f, r, and s from parts (c) and (d).
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- Based on your work in (a), how would you describe the effect(s) of the transformation y = f(ax) where a \gt 0? What is the impact on the graph of f? Are any parts of the graph of f unchanged?
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- Now consider the function r(x) = f(-x). Observe that r(-1) = f(1), r(2) = f(-2), and so on. Without using a graphing utility, how do you expect the graph of y = r(x) to compare to the graph of y = f(x)? Explain. Then test your conjecture by using a graphing utility and record the plots of f and r on the axes in Figure.
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- How do you expect the graph of s(x) = f(-2x) to appear? Why? More generally, how does the graph of y = f(ax) compare to the graph of y = f(x) in the situation where a \lt 0?
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- Exercise Answer
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With f(x)=(x-2)^2+1: g(x)=f(4x) has vertex at (\frac{1}{2}, 1), y-intercept 5, no x-intercepts. h(x)=f(2x) has vertex (1,1), y-intercept 5. k(x)=f(0.5x) has vertex (4,1), y-intercept 5. m(x)=f(0.25x) has vertex (8,1), y-intercept 5. All have the same minimum value and y-intercept; increasing a horizontally compresses the graph.
The transformation y=f(ax) with a\gt 0 horizontally compresses the graph by factor a (when a\gt 1) or stretches it by factor 1/a (when 0\lt a\lt 1). Points on the y-axis are unchanged.
r(x)=f(-x)=(-x-2)^2+1=(x+2)^2+1. This is the reflection of f across the y-axis: every point (a,b) on f maps to (-a,b) on r.
s(x)=f(-2x) combines a horizontal compression by 2 with reflection across the y-axis. In general, y=f(ax) with a\lt 0 reflects across the y-axis and compresses horizontally by |a|.
+ Let g(x) = f(x) + 5. Determine AV_{[-3,-1]} and AV_{[2,5]} for both f and g. What do you observe? Why does this phenomenon occur?
+
+
+
+
+ Let h(x) = f(x-2). For f, recall that you determined AV_{[-3,-1]} and AV_{[2,5]} in (a). In addition, determine AV_{[-1,1]} and AV_{[4,7]} for h. What do you observe? Why does this phenomenon occur?
+
+
+
+
+ Let k(x) = 3f(x). Determine AV_{[-3,-1]} and AV_{[2,5]} for k, and compare the results to your earlier computations of AV_{[-3,-1]} and AV_{[2,5]} for f. What do you observe? Why does this phenomenon occur?
+
+
+
+
+ Finally, let m(x) = 3f(x-2) + 5. Without doing any computations, what do you think will be true about the relationship between AV_{[-3,-1]} for f and AV_{[-1,1]} for m? Why? After making your conjecture, execute appropriate computations to see if your intuition is correct.
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
For f(x)=x^2: AV_{[-3,-1]}=(1-9)/2=-4 and AV_{[2,5]}=(25-4)/3=7. For g(x)=x^2+5: AV_{[-3,-1]}=(6-14)/2=-4 and AV_{[2,5]}=(30-9)/3=7. The values are the same: vertical shifts do not change average rates of change.
For h(x)=(x-2)^2: AV_{[-1,1]}=(1-9)/2=-4 (same as AV_{[-3,-1]} for f) and AV_{[4,7]}=(25-4)/3=7 (same as AV_{[2,5]} for f). A horizontal shift of 2 units right causes the average rate of change to match on intervals that are shifted right by 2.
For k(x)=3x^2: AV_{[-3,-1]}=(3-27)/2=-12=3\cdot(-4) and AV_{[2,5]}=(75-12)/3=21=3\cdot 7. Multiplying the output by 3 multiplies all average rates of change by 3.
Conjecture: AV_{[-1,1]} for m equals 3\cdot AV_{[-3,-1]} for f, which is 3(-4) = -12. Verification: m(x)=3(x-2)^2+5, m(-1)=3(9)+5=32, m(1)=3(1)+5=8, AV_{[-1,1]}=(8-32)/2=-12. \checkmark
+
+
+
+
+
+
+ Consider the parent function y = f(x) = x.
+
+
+
+
+
+
+ Consider the linear function in point-slope form given by y = L(x) = -4(x-3) + 5. What is the slope of this line? What is the most obvious point that lies on the line?
+
+
+
+
+ How can the function L given in (a) be viewed as a transformation of the parent function f? Explain the roles of 3, -4, and 5, respectively.
+
+
+
+
+ Explain why any non-vertical line of the form P(x) = m(x-x_0) + y_0 can be thought of as a transformation of the parent function f(x) = x. Specifically discuss the transformation(s) involved.
+
+
+
+
+ Find a formula for the transformation of f(x) = x that corresponds to a horizontal shift of 7 units left, a reflection across y = 0 and vertical stretch of 3 units away from the x-axis, and a vertical shift of -11 units.
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
The slope is -4. The most obvious point is (3, 5).
Starting from f(x)=x: shift right by 3 gives f(x-3)=x-3; multiply by -4 gives -4(x-3) (vertical stretch by 4 and reflection); shift up by 5 gives -4(x-3)+5 = L(x). So 3 is the horizontal shift, -4 is the vertical stretch/reflection factor, and 5 is the vertical shift.
P(x)=m(x-x_0)+y_0 is a horizontal shift of f by x_0, a vertical stretch by m, and a vertical shift by y_0.
Horizontal shift left 7: f(x+7) = x+7. Reflection and vertical stretch by 3: -3(x+7). Vertical shift -11: -3(x+7)-11 = -3x-32.
+
+
+
+
+
+
+ We have explored the effects of adding a constant to the output of a function, y = f(x) + a, adding a constant to the input, y = f(x+a), and multiplying the output of a function by a constant, y = af(x). There is one remaining natural transformation to explore: multiplying the input to a function by a constant. In this exercise, we consider the effects of the constant a in transforming a parent function f by the rule y = f(ax).
+
+
+
+ Let f(x) = (x-2)^2 + 1.
+
+
+
+
+
+
+ Let g(x) = f(4x), h(x) = f(2x), k(x) = f(0.5x), and m(x) = f(0.25x). Use Desmos to plot these functions. Then, sketch and label g, h, k, and m on the provided axes in Figure along with the graph of f. For each of the functions, label and identify its vertex, its y-intercept, and its x-intercepts.
+
+
+
+
+
Axes for plotting f, g, h, k, and m in part (a).
+
Axes for plotting f, g, h, k, and m in part (a).
+
+
+
+
Axes for plotting f, r, and s from parts (c) and (d).
+
Axes for plotting f, r, and s from parts (c) and (d).
+
+
+
+
+
+
+ Based on your work in (a), how would you describe the effect(s) of the transformation y = f(ax) where a \gt 0? What is the impact on the graph of f? Are any parts of the graph of f unchanged?
+
+
+
+
+ Now consider the function r(x) = f(-x). Observe that r(-1) = f(1), r(2) = f(-2), and so on. Without using a graphing utility, how do you expect the graph of y = r(x) to compare to the graph of y = f(x)? Explain. Then test your conjecture by using a graphing utility and record the plots of f and r on the axes in Figure.
+
+
+
+
+ How do you expect the graph of s(x) = f(-2x) to appear? Why? More generally, how does the graph of y = f(ax) compare to the graph of y = f(x) in the situation where a \lt 0?
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
With f(x)=(x-2)^2+1: g(x)=f(4x) has vertex at (\frac{1}{2}, 1), y-intercept 5, no x-intercepts. h(x)=f(2x) has vertex (1,1), y-intercept 5. k(x)=f(0.5x) has vertex (4,1), y-intercept 5. m(x)=f(0.25x) has vertex (8,1), y-intercept 5. All have the same minimum value and y-intercept; increasing a horizontally compresses the graph.
The transformation y=f(ax) with a\gt 0 horizontally compresses the graph by factor a (when a\gt 1) or stretches it by factor 1/a (when 0\lt a\lt 1). Points on the y-axis are unchanged.
r(x)=f(-x)=(-x-2)^2+1=(x+2)^2+1. This is the reflection of f across the y-axis: every point (a,b) on f maps to (-a,b) on r.
s(x)=f(-2x) combines a horizontal compression by 2 with reflection across the y-axis. In general, y=f(ax) with a\lt 0 reflects across the y-axis and compresses horizontally by |a|.
- We now know three different identities involving the sine and cosine functions: \sin(t+\frac{\pi}{2}) = \cos(t), \cos(t-\frac{\pi}{2}) = \sin(t), and \cos^2(t) + \sin^2(t) = 1. Following are several proposed identities. For each, your task is to decide whether the identity is true or false. If true, give a convincing argument for why it is true; if false, give an example of a t-value for which the equation fails to hold.
-
-
-
-
-
-
- \cos(t + 2\pi) = \cos(t)
-
-
-
-
- \sin(t-\pi) = -\sin(t)
-
-
-
-
- \cos(t - \frac{3\pi}{2}) = \sin(t)
-
-
-
-
- \sin^2(t) = 1 - \cos^2(t)
-
-
-
-
- \sin(t) + \cos(t) = 1
-
-
-
-
- \sin(t) + \sin(\frac{\pi}{2}) = \cos(t)
-
-
-
-
-
-
-
-
- Exercise Answer
-
-
-
-
-
-
-
- Yes, \cos(t+2\pi) = \cos(t) is true because the function has a periodicity of 2\pi.
-
-
-
-
- Yes, \sin(t-\pi) = -\sin(t) is true because a shift of \pi around the unit circle is equivalent to making the argument negative, and \sin is an odd function. Also, graphically, \sin gives the y-coordinate of points on the unit circle, and shifting the argument by \pm\pi essentially reflects the point across the x-axis, changing the sign of the y-coordinate.
-
-
-
-
- No, \cos\!\left(t - \frac{3\pi}{2}\right) \ne \sin(t). For example, \cos\!\left(\frac{\pi}{2} - \frac{3\pi}{2}\right) = \cos(-\pi) = -1 while \sin\!\left(\frac{\pi}{2}\right) = 1.
-
-
-
-
- Yes, \sin^2(t) = 1 - \cos^2(t), because of the Fundamental Trigonometric Identity.
-
-
-
-
- No, \sin(t) + \cos(t) \ne 1. For example, \sin\!\left(\frac{\pi}{3}\right) + \cos\!\left(\frac{\pi}{3}\right) = \frac{1 + \sqrt{3}}{2} \ne 1.
-
-
-
-
- No, \sin(t) + \sin\!\left(\frac{\pi}{2}\right) \ne \cos(t). For example, \sin\!\left(\frac{\pi}{4}\right) + \sin\!\left(\frac{\pi}{2}\right) = \frac{\sqrt{2} + 2}{2} while \cos\!\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}.
-
+ We now know three different identities involving the sine and cosine functions: \sin(t+\frac{\pi}{2}) = \cos(t), \cos(t-\frac{\pi}{2}) = \sin(t), and \cos^2(t) + \sin^2(t) = 1. Following are several proposed identities. For each, your task is to decide whether the identity is true or false. If true, give a convincing argument for why it is true; if false, give an example of a t-value for which the equation fails to hold.
+
+
+
+
+
+
+ \cos(t + 2\pi) = \cos(t)
+
+
+
+
+ \sin(t-\pi) = -\sin(t)
+
+
+
+
+ \cos(t - \frac{3\pi}{2}) = \sin(t)
+
+
+
+
+ \sin^2(t) = 1 - \cos^2(t)
+
+
+
+
+ \sin(t) + \cos(t) = 1
+
+
+
+
+ \sin(t) + \sin(\frac{\pi}{2}) = \cos(t)
+
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
+
+
+
+ Yes, \cos(t+2\pi) = \cos(t) is true because the function has a periodicity of 2\pi.
+
+
+
+
+ Yes, \sin(t-\pi) = -\sin(t) is true because a shift of \pi around the unit circle is equivalent to making the argument negative, and \sin is an odd function. Also, graphically, \sin gives the y-coordinate of points on the unit circle, and shifting the argument by \pm\pi essentially reflects the point across the x-axis, changing the sign of the y-coordinate.
+
+
+
+
+ No, \cos\!\left(t - \frac{3\pi}{2}\right) \ne \sin(t). For example, \cos\!\left(\frac{\pi}{2} - \frac{3\pi}{2}\right) = \cos(-\pi) = -1 while \sin\!\left(\frac{\pi}{2}\right) = 1.
+
+
+
+
+ Yes, \sin^2(t) = 1 - \cos^2(t), because of the Fundamental Trigonometric Identity.
+
+
+
+
+ No, \sin(t) + \cos(t) \ne 1. For example, \sin\!\left(\frac{\pi}{3}\right) + \cos\!\left(\frac{\pi}{3}\right) = \frac{1 + \sqrt{3}}{2} \ne 1.
+
+
+
+
+ No, \sin(t) + \sin\!\left(\frac{\pi}{2}\right) \ne \cos(t). For example, \sin\!\left(\frac{\pi}{4}\right) + \sin\!\left(\frac{\pi}{2}\right) = \frac{\sqrt{2} + 2}{2} while \cos\!\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}.
+
- Hint: Plot the points first;
- doing so in Desmos is ideal.
-
-
-
-
-
- Exercise Answer
-
-
-
-
- There is more than one possible correct answer. If we take the points to indicate a period of 1.0 and midpoint at y = 2, with an anchor point at (0, 2), a formula is
-
- g(x) = \sin(2\pi x) + 2.
-
-
-
-
-
-
-
-
- In 2018, on the summer solstice, June 21, Grand Rapids, MI, experiences 15 hours, 21 minutes, and 25 seconds of daylight. Said differently, on the 172nd day of the year, people on earth at the latitude of Grand Rapids experience 15.3569 hours of daylight. On the winter solstice, December 21, 2018, the same latitude has 9 hours, 0 minutes, and 31 seconds of daylight; equivalently, on day 355, there are 9.0086 hours of daylight. This data is essentially identical every year as the patterns of the earth's rotation repeat.
-
-
-
- Let t be the day of the year starting with t = 0 on December 31, 2017. In addition, let s(t) be the number of hours of daylight on day t in Grand Rapids, MI. Find a formula for a circular function s(t) that fits this data. What is the function's midline, amplitude, and period? What are you using as an anchor point? Explain fully, and then graph your function to check your conclusions.
-
-
-
-
- Exercise Answer
-
-
-
-
- Assuming the period is 365 days, the amplitude is approximately 3.17545 hours, the midline is at y = 12.18405, and the maximum occurs at day t = 172 (with the minimum at day 355). Using the maximum as an anchor point, the duration of daylight is
-
- s(t) = 3.17545\cos\!\left(\frac{2\pi}{365}(t - 172)\right) + 12.18405.
-
-
-
-
-
-
-
-
- We now understand the effects of the transformation h(t) = f(kt) where k \gt 0 for a given function f. Our goal is to understand what happens when k \lt 0.
-
-
-
-
-
-
- We first consider the special case where k = -1. Let f(t) = 2t - 1, and let g(t) = f(-1 \cdot t) = f(-t) = -2t - 1. Plot f and g on the same coordinate axes. How are their graphs related to one another?
-
-
-
-
- Given any function p, how do you expect the graph of y = q(t) = p(-t) to be related to the graph of p?
-
-
-
-
- How is the graph of y = \sin(-3t) related to the graph of y = \sin(t)?
-
-
-
-
- How is the graph of y = \cos(-3t) related to the graph of y = \cos(t)?
-
-
-
-
- Given any function p and a constant k \lt 0, how do you expect the graph of y = q(t) = p(kt) to be related to the graph of p?
-
-
-
-
- How are \sin(-t) and \sin(t) related? How are \cos(t) and \cos(-t) related?
-
-
-
-
-
-
-
- Exercise Answer
-
-
-
-
-
-
-
- The graphs of f(t) = 2t - 1 and g(t) = -2t - 1 have lines that are reflected across the y-axis.
-
-
-
-
- For any function p, the function q(t) = p(-t) is a reflection of p across the y-axis.
-
-
-
-
- The graph of y = \sin(-3t) is a graph of y = \sin(t) first horizontally scaled by a factor of \frac{1}{3} and then reflected across the y-axis.
-
-
-
-
- The graph of y = \cos(-3t) is a graph of y = \cos(t) first horizontally scaled by a factor of \frac{1}{3} and then reflected across the y-axis. Since \cos is symmetric across the y-axis, y = \cos(-3t) appears identical to y = \cos(3t).
-
-
-
-
- For any function p(t), the graph of q(t) = p(kt) with k \lt 0 is equivalent to the graph of p(t) scaled horizontally by a factor of \frac{1}{|k|} and then reflected across the y-axis.
-
-
-
-
- The graphs \sin(-t) and \sin(t) are related such that \sin(-t) = -\sin(t), while \cos(-t) = \cos(t).
-
+ Hint: Plot the points first;
+ doing so in Desmos is ideal.
+
+
+
+
+
+ Exercise Answer
+
+
+
+
+ There is more than one possible correct answer. If we take the points to indicate a period of 1.0 and midpoint at y = 2, with an anchor point at (0, 2), a formula is
+
+ g(x) = \sin(2\pi x) + 2.
+
+
+
+
+
+
+
+
+ In 2018, on the summer solstice, June 21, Grand Rapids, MI, experiences 15 hours, 21 minutes, and 25 seconds of daylight. Said differently, on the 172nd day of the year, people on earth at the latitude of Grand Rapids experience 15.3569 hours of daylight. On the winter solstice, December 21, 2018, the same latitude has 9 hours, 0 minutes, and 31 seconds of daylight; equivalently, on day 355, there are 9.0086 hours of daylight. This data is essentially identical every year as the patterns of the earth's rotation repeat.
+
+
+
+ Let t be the day of the year starting with t = 0 on December 31, 2017. In addition, let s(t) be the number of hours of daylight on day t in Grand Rapids, MI. Find a formula for a circular function s(t) that fits this data. What is the function's midline, amplitude, and period? What are you using as an anchor point? Explain fully, and then graph your function to check your conclusions.
+
+
+
+
+ Exercise Answer
+
+
+
+
+ Assuming the period is 365 days, the amplitude is approximately 3.17545 hours, the midline is at y = 12.18405, and the maximum occurs at day t = 172 (with the minimum at day 355). Using the maximum as an anchor point, the duration of daylight is
+
+ s(t) = 3.17545\cos\!\left(\frac{2\pi}{365}(t - 172)\right) + 12.18405.
+
+
+
+
+
+
+
+
+ We now understand the effects of the transformation h(t) = f(kt) where k \gt 0 for a given function f. Our goal is to understand what happens when k \lt 0.
+
+
+
+
+
+
+ We first consider the special case where k = -1. Let f(t) = 2t - 1, and let g(t) = f(-1 \cdot t) = f(-t) = -2t - 1. Plot f and g on the same coordinate axes. How are their graphs related to one another?
+
+
+
+
+ Given any function p, how do you expect the graph of y = q(t) = p(-t) to be related to the graph of p?
+
+
+
+
+ How is the graph of y = \sin(-3t) related to the graph of y = \sin(t)?
+
+
+
+
+ How is the graph of y = \cos(-3t) related to the graph of y = \cos(t)?
+
+
+
+
+ Given any function p and a constant k \lt 0, how do you expect the graph of y = q(t) = p(kt) to be related to the graph of p?
+
+
+
+
+ How are \sin(-t) and \sin(t) related? How are \cos(t) and \cos(-t) related?
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
+
+
+
+ The graphs of f(t) = 2t - 1 and g(t) = -2t - 1 have lines that are reflected across the y-axis.
+
+
+
+
+ For any function p, the function q(t) = p(-t) is a reflection of p across the y-axis.
+
+
+
+
+ The graph of y = \sin(-3t) is a graph of y = \sin(t) first horizontally scaled by a factor of \frac{1}{3} and then reflected across the y-axis.
+
+
+
+
+ The graph of y = \cos(-3t) is a graph of y = \cos(t) first horizontally scaled by a factor of \frac{1}{3} and then reflected across the y-axis. Since \cos is symmetric across the y-axis, y = \cos(-3t) appears identical to y = \cos(3t).
+
+
+
+
+ For any function p(t), the graph of q(t) = p(kt) with k \lt 0 is equivalent to the graph of p(t) scaled horizontally by a factor of \frac{1}{|k|} and then reflected across the y-axis.
+
+
+
+
+ The graphs \sin(-t) and \sin(t) are related such that \sin(-t) = -\sin(t), while \cos(-t) = \cos(t).
+
- Consider the circle pictured in Figure that is centered at the point (2,2) and that has circumference 8. Suppose that we track the x-coordinate (that is, the horizontal location, which we will call k) of a point that is traversing the circle counterclockwise and that it starts at P_0 as pictured.
-
-
-
-
-
A point traversing the circle.
-
A point traversing the circle.
-
-
-
-
Axes for plotting k as a function of d.
-
Axes for plotting k as a function of d.
-
-
-
-
-
- Recall that in Activity we identified the exact and approximate vertical coordinates of all 8 noted points on the circle. In addition, recall that the radius of the circle is r = \frac{8}{2\pi} \approx 1.2732.
-
-
-
-
-
-
- What is the exact horizontal coordinate of P_0? Why?
-
-
-
-
- Complete the entries in Table that track the horizontal location, k, of the point traversing the circle as a function of distance traveled, d.
-
-
-
- Data for <m>k</m> as a function of <m>d</m>.
-
-
- d
- 0
- 1
- 2
- 3
- 4
- 5
- 6
- 7
- 8
- 9
- 10
- 11
- 12
- 13
- 14
- 15
- 16
-
-
- k
- 0.73
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
- By plotting the points in Table and connecting them in an intuitive way, sketch a graph of k as a function of d on the axes provided in Figure over the interval 0 \le d \le 16. Clearly label the scale of your axes and the coordinates of several important points on the curve.
-
-
-
-
- What is similar about your graph in comparison to the one in ? What is different?
-
-
-
-
- What will be the value of k when d = 51? How about when d = 102?
-
-
-
-
-
-
-
- Exercise Answer
-
-
-
-
-
-
-
- The point P_0 is directly to the left of the center (2,2), so its x-coordinate is one radius less than 2. The radius is \frac{8}{2\pi} = \frac{4}{\pi}, so the exact x-coordinate is 2 - \frac{4}{\pi} = \frac{2(\pi-2)}{\pi} \approx 0.727.
-
- The graph is just a horizontal shift compared to the graph of h from .
-
-
-
-
- When d = 51, k(51) = 2.90. When d = 102, k(102) = 2.
-
-
-
-
-
-
-
-
-
-
- Two circular functions, f and g, are generated by tracking the y-coordinate of a point traversing two different circles. The resulting graphs are shown in Figure and Figure. Assuming the horizontal scale matches the vertical scale, answer the following questions for each of the functions f and g.
-
-
-
-
-
A plot of the circular function f.
-
A plot of the circular function f.
-
-
-
-
A plot of the circular function g.
-
A plot of the circular function g.
-
-
-
-
-
-
-
-
- Assume that the circle used to generate the circular function is centered at the point (0,m) and has radius r. What are the numerical values of m and r? Why?
-
-
-
-
- What are the coordinates of the location on the circle at which the point begins its traverse? Said differently, what point on the circle corresponds to t = 0 on the function's graph?
-
-
-
-
- What is the period of the function? How is this connected to the circle and to the scale on the horizontal axes on which the function is graphed?
-
-
-
-
- How would the graph look if the circle's radius was 1 unit larger? 1 unit smaller?
-
-
-
-
-
-
-
- Exercise Answer
-
-
-
-
-
-
-
- The midline of each oscillation graph corresponds to the center y-coordinate, and the amplitude equals the radius. Function f has a maximum of 11 and a minimum of 3, so the radius (amplitude) is 4 and the midline is y = 7. Function g has a maximum of 12 and a minimum of 1, so the radius (amplitude) is 5.5 and the midline is y = 6.5.
-
-
-
-
- Assuming counterclockwise motion: the circle for function f begins its traverse at the 9:00 position with coordinates (-4, 7). The circle for function g begins at the 12:00 position with coordinates (0, 12).
-
-
-
-
- For function f, the period of the function is 8 units. For function g, the period is 8 units. For these graphs, one period is one complete traverse of the circle. The speed v of the point is equal to the circumference of the circle divided by the period T of one complete circle: v = \frac{2\pi r}{T}, so the horizontal scale corresponds to defining each period as \frac{2\pi r}{v}.
-
-
-
-
- The changes to the graph depend in part on whether the speed stays the same. If the circle's radius gets one unit larger and the speed stays the same, then the period of the graph of function f increases to \frac{5}{4} \cdot 8 = 10 units; the amplitude of the graph changes to 5. For function g, a constant-speed scenario changes the period to \frac{7.5}{6.5} \cdot 8 \approx 9.23 units; the amplitude of g changes to 7.5. If the circle's radius gets one unit smaller and the speed stays the same, then the period of the graph of function f decreases to \frac{3}{4} \cdot 8 = 6 units; the amplitude of the graph changes to 3. For function g, a constant-speed scenario changes the period to \frac{5.5}{6.5} \cdot 8 \approx 6.77 units; the amplitude changes to 4.5. The amplitude of each graph will increase by 1 if the radius increases and decrease by 1 if the radius decreases, but there is no change to the midline.
-
-
-
-
-
-
-
-
-
-
- A person goes for a ride on a ferris wheel. They enter one of the cars at the lowest possible point on the wheel from a platform 7 feet off the ground. When they are at the very top of the wheel, they are 92 feet off the ground. Let h represent the height of the car (in feet) and d (in feet) the distance the car has traveled along the wheel's circumference from its starting location at the bottom of the wheel. We'll use the notation h = f(d) for how height is a function of distance traveled.
-
-
-
-
-
-
- How high above the ground is the center of the ferris wheel?
-
-
-
-
- How far does the car travel in one complete trip around the wheel?
-
-
-
-
- For the circular function h = f(d), what is its amplitude? midline? period?
-
-
-
-
- Sketch an accurate graph of h through at least two full periods. Clearly label the scale on the horizontal and vertical axes along with several important points.
-
-
-
-
-
-
-
- Exercise Answer
-
-
-
-
-
-
-
- The center of the ferris wheel is at height \frac{92 - 7}{2} + 7 = 49.5 feet.
-
-
-
-
- The radius is 42.5 feet, so the circumference is 2\pi \cdot 42.5 = 85\pi \approx 267 feet.
-
-
-
-
- The circular function h = f(d) has amplitude 42.5 feet, midline y = 49.5 feet, and period 85\pi \approx 267 feet.
-
+ Consider the circle pictured in Figure that is centered at the point (2,2) and that has circumference 8. Suppose that we track the x-coordinate (that is, the horizontal location, which we will call k) of a point that is traversing the circle counterclockwise and that it starts at P_0 as pictured.
+
+
+
+
+
A point traversing the circle.
+
A point traversing the circle.
+
+
+
+
Axes for plotting k as a function of d.
+
Axes for plotting k as a function of d.
+
+
+
+
+
+ Recall that in Activity we identified the exact and approximate vertical coordinates of all 8 noted points on the circle. In addition, recall that the radius of the circle is r = \frac{8}{2\pi} \approx 1.2732.
+
+
+
+
+
+
+ What is the exact horizontal coordinate of P_0? Why?
+
+
+
+
+ Complete the entries in Table that track the horizontal location, k, of the point traversing the circle as a function of distance traveled, d.
+
+
+
+ Data for <m>k</m> as a function of <m>d</m>.
+
+
+ d
+ 0
+ 1
+ 2
+ 3
+ 4
+ 5
+ 6
+ 7
+ 8
+ 9
+ 10
+ 11
+ 12
+ 13
+ 14
+ 15
+ 16
+
+
+ k
+ 0.73
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ By plotting the points in Table and connecting them in an intuitive way, sketch a graph of k as a function of d on the axes provided in Figure over the interval 0 \le d \le 16. Clearly label the scale of your axes and the coordinates of several important points on the curve.
+
+
+
+
+ What is similar about your graph in comparison to the one in ? What is different?
+
+
+
+
+ What will be the value of k when d = 51? How about when d = 102?
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
+
+
+
+ The point P_0 is directly to the left of the center (2,2), so its x-coordinate is one radius less than 2. The radius is \frac{8}{2\pi} = \frac{4}{\pi}, so the exact x-coordinate is 2 - \frac{4}{\pi} = \frac{2(\pi-2)}{\pi} \approx 0.727.
+
+ The graph is just a horizontal shift compared to the graph of h from .
+
+
+
+
+ When d = 51, k(51) = 2.90. When d = 102, k(102) = 2.
+
+
+
+
+
+
+
+
+
+
+ Two circular functions, f and g, are generated by tracking the y-coordinate of a point traversing two different circles. The resulting graphs are shown in Figure and Figure. Assuming the horizontal scale matches the vertical scale, answer the following questions for each of the functions f and g.
+
+
+
+
+
A plot of the circular function f.
+
A plot of the circular function f.
+
+
+
+
A plot of the circular function g.
+
A plot of the circular function g.
+
+
+
+
+
+
+
+
+ Assume that the circle used to generate the circular function is centered at the point (0,m) and has radius r. What are the numerical values of m and r? Why?
+
+
+
+
+ What are the coordinates of the location on the circle at which the point begins its traverse? Said differently, what point on the circle corresponds to t = 0 on the function's graph?
+
+
+
+
+ What is the period of the function? How is this connected to the circle and to the scale on the horizontal axes on which the function is graphed?
+
+
+
+
+ How would the graph look if the circle's radius was 1 unit larger? 1 unit smaller?
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
+
+
+
+ The midline of each oscillation graph corresponds to the center y-coordinate, and the amplitude equals the radius. Function f has a maximum of 11 and a minimum of 3, so the radius (amplitude) is 4 and the midline is y = 7. Function g has a maximum of 12 and a minimum of 1, so the radius (amplitude) is 5.5 and the midline is y = 6.5.
+
+
+
+
+ Assuming counterclockwise motion: the circle for function f begins its traverse at the 9:00 position with coordinates (-4, 7). The circle for function g begins at the 12:00 position with coordinates (0, 12).
+
+
+
+
+ For function f, the period of the function is 8 units. For function g, the period is 8 units. For these graphs, one period is one complete traverse of the circle. The speed v of the point is equal to the circumference of the circle divided by the period T of one complete circle: v = \frac{2\pi r}{T}, so the horizontal scale corresponds to defining each period as \frac{2\pi r}{v}.
+
+
+
+
+ The changes to the graph depend in part on whether the speed stays the same. If the circle's radius gets one unit larger and the speed stays the same, then the period of the graph of function f increases to \frac{5}{4} \cdot 8 = 10 units; the amplitude of the graph changes to 5. For function g, a constant-speed scenario changes the period to \frac{7.5}{6.5} \cdot 8 \approx 9.23 units; the amplitude of g changes to 7.5. If the circle's radius gets one unit smaller and the speed stays the same, then the period of the graph of function f decreases to \frac{3}{4} \cdot 8 = 6 units; the amplitude of the graph changes to 3. For function g, a constant-speed scenario changes the period to \frac{5.5}{6.5} \cdot 8 \approx 6.77 units; the amplitude changes to 4.5. The amplitude of each graph will increase by 1 if the radius increases and decrease by 1 if the radius decreases, but there is no change to the midline.
+
+
+
+
+
+
+
+
+
+
+ A person goes for a ride on a ferris wheel. They enter one of the cars at the lowest possible point on the wheel from a platform 7 feet off the ground. When they are at the very top of the wheel, they are 92 feet off the ground. Let h represent the height of the car (in feet) and d (in feet) the distance the car has traveled along the wheel's circumference from its starting location at the bottom of the wheel. We'll use the notation h = f(d) for how height is a function of distance traveled.
+
+
+
+
+
+
+ How high above the ground is the center of the ferris wheel?
+
+
+
+
+ How far does the car travel in one complete trip around the wheel?
+
+
+
+
+ For the circular function h = f(d), what is its amplitude? midline? period?
+
+
+
+
+ Sketch an accurate graph of h through at least two full periods. Clearly label the scale on the horizontal and vertical axes along with several important points.
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
+
+
+
+ The center of the ferris wheel is at height \frac{92 - 7}{2} + 7 = 49.5 feet.
+
+
+
+
+ The radius is 42.5 feet, so the circumference is 2\pi \cdot 42.5 = 85\pi \approx 267 feet.
+
+
+
+
+ The circular function h = f(d) has amplitude 42.5 feet, midline y = 49.5 feet, and period 85\pi \approx 267 feet.
+
- Let (x,y) be a point on the unit circle. In each of the following situations, determine the requested value exactly.
-
-
-
-
-
-
- Suppose that x = -0.3 and y is negative. Find the value of y.
-
-
-
-
- Suppose that (x,y) lies in Quadrant II and x = -2y. Find the values of x and y.
-
-
-
-
- Suppose that (x,y) lies a distance of \frac{29\pi}{6} units clockwise around the circle from (1,0). Find the values of x and y.
-
-
-
-
- At what exact point(s) does the line y = \frac{1}{2}x + \frac{1}{2} intersect the unit circle?
-
-
-
-
-
-
-
- Exercise Answer
-
-
-
-
-
-
-
- Using x^2 + y^2 = 1 with x = -0.3 and y < 0: y = -\sqrt{1 - 0.09} = -\frac{\sqrt{91}}{10} \approx -0.954.
-
-
-
-
- With x = -2y and x^2 + y^2 = 1: 4y^2 + y^2 = 1, so y = \frac{\sqrt{5}}{5} and x = -\frac{2\sqrt{5}}{5}.
-
-
-
-
- Going \frac{29\pi}{6} clockwise corresponds to the angle -\frac{29\pi}{6} \equiv \frac{7\pi}{6} \pmod{2\pi}. The coordinates are \left(-\frac{\sqrt{3}}{2}, -\frac{1}{2}\right).
-
-
-
-
- Substituting y = \frac{1}{2}x + \frac{1}{2} into x^2 + y^2 = 1 and solving gives x = -1 or x = \frac{3}{5}. The intersection points are (-1, 0) and \left(\frac{3}{5}, \frac{4}{5}\right).
-
-
-
-
-
-
-
-
-
-
- The unit circle is centered at (0,0) and radius r = 1, from which the Pythagorean Theorem tells us that any point (x,y) on the unit circle satisfies the equation x^2 + y^2 = 1.
-
-
-
-
-
-
- Explain why any point (x,y) on a circle of radius r centered at (h,k) satisfies the equation (x-h)^2 + (y-k)^2 = r^2.
-
-
-
-
- Determine the equation of a circle centered at (-3,5) with radius r = 2.
-
-
-
-
- Suppose that the unit circle is magnified by a factor of 5 and then shifted 4 units right and 7 units down. What is the equation of the resulting circle?
-
-
-
-
- What is the length of the arc intercepted by a central angle of \frac{2\pi}{3} radians in the circle (x-1)^2 + (y-3)^2 = 16?
-
-
-
-
- Suppose that the line segment from (-2,-1) to (4,2) is a diameter of a circle. What is the circle's center, radius, and equation?
-
-
-
-
-
-
-
- Exercise Answer
-
-
-
-
-
-
-
- A circle of radius r centered at (h,k) is simply the unit circle transformed with a horizontal translation h units to the right, a vertical translation k units up, and scaled both horizontally and vertically by a factor of r. Thus the parent equation x^2 + y^2 = 1 is transformed with x \to \frac{x-h}{r} and y \to \frac{y-k}{r}:
-
- \left(\frac{x-h}{r}\right)^2 + \left(\frac{y-k}{r}\right)^2 &= 1 \\
- \frac{(x-h)^2}{r^2} + \frac{(y-k)^2}{r^2} &= 1 \\
- (x-h)^2 + (y-k)^2 &= r^2.
-
-
-
-
-
- (x+3)^2 + (y-5)^2 = 4.
-
-
-
-
- (x-4)^2 + (y+7)^2 = 25.
-
-
-
-
- The circle has radius 4, so the arc length is 4 \cdot \frac{2\pi}{3} = \frac{8\pi}{3}.
-
-
-
-
- The diameter has length \sqrt{(4-(-2))^2 + (2-(-1))^2} = \sqrt{45}, so the radius is \frac{\sqrt{45}}{2}. The center is the midpoint \left(1, \frac{1}{2}\right). The equation is (x-1)^2 + \left(y - \frac{1}{2}\right)^2 = \frac{45}{4}.
-
-
-
-
-
-
-
-
-
-
- Consider the circle whose center is (0,0) and whose radius is r = 5. Let a point (x,y) traverse the circle counterclockwise from (5,0), and say the distance along the circle from (5,0) is represented by d.
-
-
-
-
-
-
- Consider the point (a,b) that is generated by the central angle \theta with vertices (5,0), (0,0), and (a,b). If \theta = \frac{\pi}{6}, what are the exact values of a and b?
-
-
-
-
- Answer the same question as in (a) except with \theta = \frac{\pi}{4} and \theta = \frac{\pi}{3}.
-
-
-
-
- How far has the point (x,y) traveled after it has traversed the circle one full revolution?
-
-
-
-
- Let h = f(d) be the circular function that tracks the height of the point (x,y) as a function of distance, d, traversed counterclockwise from (5,0). Sketch an accurate graph of f through two full periods, labeling several special points on the graph as well as the horizontal and vertical scale of the axes.
-
-
-
-
-
-
-
- Exercise Answer
-
-
-
-
-
-
-
- At \theta = \frac{\pi}{6}: a = \frac{5\sqrt{3}}{2}, b = \frac{5}{2}.
-
-
-
-
- At \theta = \frac{\pi}{4}: a = b = \frac{5\sqrt{2}}{2}. At \theta = \frac{\pi}{3}: a = \frac{5}{2}, b = \frac{5\sqrt{3}}{2}.
-
-
-
-
- One full revolution has distance equal to the circumference: 2\pi \cdot 5 = 10\pi.
-
+ Let (x,y) be a point on the unit circle. In each of the following situations, determine the requested value exactly.
+
+
+
+
+
+
+ Suppose that x = -0.3 and y is negative. Find the value of y.
+
+
+
+
+ Suppose that (x,y) lies in Quadrant II and x = -2y. Find the values of x and y.
+
+
+
+
+ Suppose that (x,y) lies a distance of \frac{29\pi}{6} units clockwise around the circle from (1,0). Find the values of x and y.
+
+
+
+
+ At what exact point(s) does the line y = \frac{1}{2}x + \frac{1}{2} intersect the unit circle?
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
+
+
+
+ Using x^2 + y^2 = 1 with x = -0.3 and y < 0: y = -\sqrt{1 - 0.09} = -\frac{\sqrt{91}}{10} \approx -0.954.
+
+
+
+
+ With x = -2y and x^2 + y^2 = 1: 4y^2 + y^2 = 1, so y = \frac{\sqrt{5}}{5} and x = -\frac{2\sqrt{5}}{5}.
+
+
+
+
+ Going \frac{29\pi}{6} clockwise corresponds to the angle -\frac{29\pi}{6} \equiv \frac{7\pi}{6} \pmod{2\pi}. The coordinates are \left(-\frac{\sqrt{3}}{2}, -\frac{1}{2}\right).
+
+
+
+
+ Substituting y = \frac{1}{2}x + \frac{1}{2} into x^2 + y^2 = 1 and solving gives x = -1 or x = \frac{3}{5}. The intersection points are (-1, 0) and \left(\frac{3}{5}, \frac{4}{5}\right).
+
+
+
+
+
+
+
+
+
+
+ The unit circle is centered at (0,0) and radius r = 1, from which the Pythagorean Theorem tells us that any point (x,y) on the unit circle satisfies the equation x^2 + y^2 = 1.
+
+
+
+
+
+
+ Explain why any point (x,y) on a circle of radius r centered at (h,k) satisfies the equation (x-h)^2 + (y-k)^2 = r^2.
+
+
+
+
+ Determine the equation of a circle centered at (-3,5) with radius r = 2.
+
+
+
+
+ Suppose that the unit circle is magnified by a factor of 5 and then shifted 4 units right and 7 units down. What is the equation of the resulting circle?
+
+
+
+
+ What is the length of the arc intercepted by a central angle of \frac{2\pi}{3} radians in the circle (x-1)^2 + (y-3)^2 = 16?
+
+
+
+
+ Suppose that the line segment from (-2,-1) to (4,2) is a diameter of a circle. What is the circle's center, radius, and equation?
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
+
+
+
+ A circle of radius r centered at (h,k) is simply the unit circle transformed with a horizontal translation h units to the right, a vertical translation k units up, and scaled both horizontally and vertically by a factor of r. Thus the parent equation x^2 + y^2 = 1 is transformed with x \to \frac{x-h}{r} and y \to \frac{y-k}{r}:
+
+ \left(\frac{x-h}{r}\right)^2 + \left(\frac{y-k}{r}\right)^2 &= 1 \\
+ \frac{(x-h)^2}{r^2} + \frac{(y-k)^2}{r^2} &= 1 \\
+ (x-h)^2 + (y-k)^2 &= r^2.
+
+
+
+
+
+ (x+3)^2 + (y-5)^2 = 4.
+
+
+
+
+ (x-4)^2 + (y+7)^2 = 25.
+
+
+
+
+ The circle has radius 4, so the arc length is 4 \cdot \frac{2\pi}{3} = \frac{8\pi}{3}.
+
+
+
+
+ The diameter has length \sqrt{(4-(-2))^2 + (2-(-1))^2} = \sqrt{45}, so the radius is \frac{\sqrt{45}}{2}. The center is the midpoint \left(1, \frac{1}{2}\right). The equation is (x-1)^2 + \left(y - \frac{1}{2}\right)^2 = \frac{45}{4}.
+
+
+
+
+
+
+
+
+
+
+ Consider the circle whose center is (0,0) and whose radius is r = 5. Let a point (x,y) traverse the circle counterclockwise from (5,0), and say the distance along the circle from (5,0) is represented by d.
+
+
+
+
+
+
+ Consider the point (a,b) that is generated by the central angle \theta with vertices (5,0), (0,0), and (a,b). If \theta = \frac{\pi}{6}, what are the exact values of a and b?
+
+
+
+
+ Answer the same question as in (a) except with \theta = \frac{\pi}{4} and \theta = \frac{\pi}{3}.
+
+
+
+
+ How far has the point (x,y) traveled after it has traversed the circle one full revolution?
+
+
+
+
+ Let h = f(d) be the circular function that tracks the height of the point (x,y) as a function of distance, d, traversed counterclockwise from (5,0). Sketch an accurate graph of f through two full periods, labeling several special points on the graph as well as the horizontal and vertical scale of the axes.
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
+
+
+
+ At \theta = \frac{\pi}{6}: a = \frac{5\sqrt{3}}{2}, b = \frac{5}{2}.
+
+
+
+
+ At \theta = \frac{\pi}{4}: a = b = \frac{5\sqrt{2}}{2}. At \theta = \frac{\pi}{3}: a = \frac{5}{2}, b = \frac{5\sqrt{3}}{2}.
+
+
+
+
+ One full revolution has distance equal to the circumference: 2\pi \cdot 5 = 10\pi.
+
- When a single investment of principal, $P, is invested in an account that returns interest at an annual rate of r (a decimal that corresponds to the percentage rate, such as r = 0.05 corresponding to 5%) that is compounded n times per year, the amount of money in the account after t years is given by A(t) = P(1 + \frac{r}{n})^{nt}.
-
-
-
- Suppose we invest $100 in an account that earns 8% annual interest. We investigate the effects of different rates of compounding.
-
-
-
-
-
-
- Compute A(1) if interest is compounded quarterly (n = 4).
-
-
-
-
- Compute A(1) if interest is compounded monthly.
-
-
-
-
- Compute A(1) if interest is compounded weekly.
-
-
-
-
- Compute A(1) if interest is compounded daily.
-
-
-
-
- If we let the number of times that interest is compounded increase without bound, we say that the interest is compounded continuously. continuously compounded interest When interest is compounded continuously, it turns out that the amount of money an account with initial investment $P after t years at an annual interest rate of r is A(t) = Pe^{rt}, where e is the natural base. Compute A(1) in the same context as the preceding questions but where interest is compounded continuously.
-
-
-
-
- How much of a difference does continuously compounded interest make over interest compounded quarterly in one year's time? How does your answer change over 25 years' time?
-
-
-
-
-
-
-
- Exercise Answer
-
-
-
-
-
-
-
- With P = 100, r = 0.08, and n = 4: A(1) = 100\left(1 + \frac{0.08}{4}\right)^{4} = 100(1.02)^4 \approx \$108.24.
-
-
-
-
- With n = 12: A(1) = 100\left(1 + \frac{0.08}{12}\right)^{12} \approx \$108.30.
-
-
-
-
- With n = 52: A(1) = 100\left(1 + \frac{0.08}{52}\right)^{52} \approx \$108.32.
-
-
-
-
- With n = 365: A(1) = 100\left(1 + \frac{0.08}{365}\right)^{365} \approx \$108.33.
-
-
-
-
- With continuous compounding: A(1) = 100e^{0.08} \approx \$108.33.
-
-
-
-
- In one year, continuously compounded interest yields approximately P \times 0.000855 more than quarterly compounding. Over 25 years, the difference grows to approximately P \times 0.144, with the greater amount coming from continuously compounded interest.
-
-
-
-
-
-
-
-
-
-
- In Desmos, define the function g(t) = e^{kt} and accept the slider for k. Set the range of the slider to -2 \le k \le 2, and assume that k \ne 0. Experiment with a wide range of values of k to see the effects of changing k.
-
-
-
-
-
-
- For what values of k is g always increasing? For what values of k is g always decreasing?
-
-
-
-
- For which value of k is the average rate of change of g on [0,1] greater: when k = -0.1 or when k = -0.05?
-
-
-
-
- What is the long-term behavior of g when k \lt 0? Why does this occur?
-
-
-
-
- Experiment with the slider to find a value of k for which g(2) = \frac{1}{2}. Test your estimate by computing e^{2k}. How accurate is your estimate?
-
-
-
-
-
-
-
- Exercise Answer
-
-
-
-
-
-
-
- When k \gt 0, g(t) is always increasing. When k \lt 0, g(t) is always decreasing.
-
-
-
-
- The average rate of change of g(t) on [0,1] is greater (less negative) when k = -0.05, giving AV_{[0,1]} \approx -0.049, than when k = -0.1, giving AV_{[0,1]} \approx -0.095. However, the magnitude of the rate of change is greater for k = -0.1.
-
-
-
-
- When k \lt 0, the long-term behavior of g(t) is that g(t) \to 0 as t \to \infty, since e^{-kt} \to \infty and e^{kt} = 1/e^{-kt} \to 0.
-
-
-
-
- We need g(2) = e^{2k} = \frac{1}{2}, so k = -\frac{\ln 2}{2} \approx -0.3466.
-
-
-
-
-
-
-
-
-
-
- A can of soda is removed from a refrigerator at time t = 0 (in minutes)
- and its temperature, F(t), in degrees Fahrenheit, is computed at regular intervals.
- Based on the data, a model is formulated for the object's temperature, given by
-
- F(t) = 74.4 - 38.8e^{-0.05t}
- .
-
-
-
-
-
-
- What is the long-term behavior of the function g(t) = e^{-0.05t}? Why?
-
-
-
-
- What is the long-term behavior of the function F(t) = 74.4 - 38.8e^{-0.05t}? What is the meaning of this value in the physical context of the problem?
-
-
-
-
- What is the temperature of the refrigerator? Why?
-
-
-
-
- Compute the average rate of change of F on the intervals [10,20], [20,30], and [30,40]. Write a careful sentence, with units, to explain the meaning of each, and write an additional sentence to describe any overall trends in how the average rate of change of F is changing.
-
-
-
-
-
-
-
- Exercise Answer
-
-
-
-
-
-
-
- The long-term behavior of g(t) = e^{-0.05t} is that g(t) \to 0 as t \to \infty, since the exponent becomes increasingly negative.
-
-
-
-
- The function F(t) \to 74.4 as t \to \infty. This means the soda approaches the room temperature of 74.4^\circ F as it warms up.
-
-
-
-
- The temperature of the refrigerator is F(0) = 74.4 - 38.8 = 35.6^\circ F.
-
-
-
-
- AV_{[10,20]} \approx 0.926 degrees per minute, AV_{[20,30]} \approx 0.562 degrees per minute, and AV_{[30,40]} \approx 0.341 degrees per minute. On the interval [10, 20], the soda is warming at an average rate of about 0.926 degrees per minute. The average rate of warming is decreasing over time, meaning the soda is warming more slowly as time goes on.
-
+ When a single investment of principal, $P, is invested in an account that returns interest at an annual rate of r (a decimal that corresponds to the percentage rate, such as r = 0.05 corresponding to 5%) that is compounded n times per year, the amount of money in the account after t years is given by A(t) = P(1 + \frac{r}{n})^{nt}.
+
+
+
+ Suppose we invest $100 in an account that earns 8% annual interest. We investigate the effects of different rates of compounding.
+
+
+
+
+
+
+ Compute A(1) if interest is compounded quarterly (n = 4).
+
+
+
+
+ Compute A(1) if interest is compounded monthly.
+
+
+
+
+ Compute A(1) if interest is compounded weekly.
+
+
+
+
+ Compute A(1) if interest is compounded daily.
+
+
+
+
+ If we let the number of times that interest is compounded increase without bound, we say that the interest is compounded continuously. continuously compounded interest When interest is compounded continuously, it turns out that the amount of money an account with initial investment $P after t years at an annual interest rate of r is A(t) = Pe^{rt}, where e is the natural base. Compute A(1) in the same context as the preceding questions but where interest is compounded continuously.
+
+
+
+
+ How much of a difference does continuously compounded interest make over interest compounded quarterly in one year's time? How does your answer change over 25 years' time?
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
+
+
+
+ With P = 100, r = 0.08, and n = 4: A(1) = 100\left(1 + \frac{0.08}{4}\right)^{4} = 100(1.02)^4 \approx \$108.24.
+
+
+
+
+ With n = 12: A(1) = 100\left(1 + \frac{0.08}{12}\right)^{12} \approx \$108.30.
+
+
+
+
+ With n = 52: A(1) = 100\left(1 + \frac{0.08}{52}\right)^{52} \approx \$108.32.
+
+
+
+
+ With n = 365: A(1) = 100\left(1 + \frac{0.08}{365}\right)^{365} \approx \$108.33.
+
+
+
+
+ With continuous compounding: A(1) = 100e^{0.08} \approx \$108.33.
+
+
+
+
+ In one year, continuously compounded interest yields approximately P \times 0.000855 more than quarterly compounding. Over 25 years, the difference grows to approximately P \times 0.144, with the greater amount coming from continuously compounded interest.
+
+
+
+
+
+
+
+
+
+
+ In Desmos, define the function g(t) = e^{kt} and accept the slider for k. Set the range of the slider to -2 \le k \le 2, and assume that k \ne 0. Experiment with a wide range of values of k to see the effects of changing k.
+
+
+
+
+
+
+ For what values of k is g always increasing? For what values of k is g always decreasing?
+
+
+
+
+ For which value of k is the average rate of change of g on [0,1] greater: when k = -0.1 or when k = -0.05?
+
+
+
+
+ What is the long-term behavior of g when k \lt 0? Why does this occur?
+
+
+
+
+ Experiment with the slider to find a value of k for which g(2) = \frac{1}{2}. Test your estimate by computing e^{2k}. How accurate is your estimate?
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
+
+
+
+ When k \gt 0, g(t) is always increasing. When k \lt 0, g(t) is always decreasing.
+
+
+
+
+ The average rate of change of g(t) on [0,1] is greater (less negative) when k = -0.05, giving AV_{[0,1]} \approx -0.049, than when k = -0.1, giving AV_{[0,1]} \approx -0.095. However, the magnitude of the rate of change is greater for k = -0.1.
+
+
+
+
+ When k \lt 0, the long-term behavior of g(t) is that g(t) \to 0 as t \to \infty, since e^{-kt} \to \infty and e^{kt} = 1/e^{-kt} \to 0.
+
+
+
+
+ We need g(2) = e^{2k} = \frac{1}{2}, so k = -\frac{\ln 2}{2} \approx -0.3466.
+
+
+
+
+
+
+
+
+
+
+ A can of soda is removed from a refrigerator at time t = 0 (in minutes)
+ and its temperature, F(t), in degrees Fahrenheit, is computed at regular intervals.
+ Based on the data, a model is formulated for the object's temperature, given by
+
+ F(t) = 74.4 - 38.8e^{-0.05t}
+ .
+
+
+
+
+
+
+ What is the long-term behavior of the function g(t) = e^{-0.05t}? Why?
+
+
+
+
+ What is the long-term behavior of the function F(t) = 74.4 - 38.8e^{-0.05t}? What is the meaning of this value in the physical context of the problem?
+
+
+
+
+ What is the temperature of the refrigerator? Why?
+
+
+
+
+ Compute the average rate of change of F on the intervals [10,20], [20,30], and [30,40]. Write a careful sentence, with units, to explain the meaning of each, and write an additional sentence to describe any overall trends in how the average rate of change of F is changing.
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
+
+
+
+ The long-term behavior of g(t) = e^{-0.05t} is that g(t) \to 0 as t \to \infty, since the exponent becomes increasingly negative.
+
+
+
+
+ The function F(t) \to 74.4 as t \to \infty. This means the soda approaches the room temperature of 74.4^\circ F as it warms up.
+
+
+
+
+ The temperature of the refrigerator is F(0) = 74.4 - 38.8 = 35.6^\circ F.
+
+
+
+
+ AV_{[10,20]} \approx 0.926 degrees per minute, AV_{[20,30]} \approx 0.562 degrees per minute, and AV_{[30,40]} \approx 0.341 degrees per minute. On the interval [10, 20], the soda is warming at an average rate of about 0.926 degrees per minute. The average rate of warming is decreasing over time, meaning the soda is warming more slowly as time goes on.
+
- Grinnell Glacier in Glacier National Park in Montana covered about 142 acres in 2007 and was found to be shrinking at about 4.4% per year.See Exercise 34 on p. 146 of Connally's Functions Modeling Change.
-
-
-
-
-
-
- Let G(t) denote the area of Grinnell Glacier in acres in year t,
- where t is the number of years since 2007.
- Find a formula for G(t) and define the function in Desmos.
-
-
-
-
- How many acres of ice were in the glacier in 1997?
- In 2012?
- What does the model predict for 2022?
-
-
-
-
- How many total acres of ice were lost from 2007 to 2012?
-
-
-
-
- What was the average rate of change of G from 2007 to 2012?
- Write a sentence to explain the meaning of this number and include units on your answer.
- In addition,
- how does this compare to the average rate of change of G from 2012 to 2017?
-
-
-
-
- How would you you describe the overall behavior of G, and thus what is happening to the Grinnell Glacier?
-
-
-
-
-
-
-
- Exercise Answer
-
-
-
-
-
-
-
- The area of Grinnell Glacier is G(t) = 142(0.956)^t.
-
-
-
-
-
- In 1997 (t = -10), the model predicts 142(0.956)^{-10} \approx 223 acres. In 2012 (t = 5), the model predicts 142(0.956)^{5} \approx 113 acres. In 2022 (t = 15), the model predicts 142(0.956)^{15} \approx 72 acres.
-
-
-
-
- According to the model, between 2007 and 2012 the glacier lost about 142 - 113 = 29 acres.
-
-
-
-
- The average rate of change of G from 2007 to 2012 is approximately -29/5 \approx -5.8 acres per year. The average retreat of the glacier was about 5.8 acres per year from 2007 to 2012. This rate of change is expected to be larger in magnitude than the average rate of change from 2012 to 2017, because the rate of decrease is decreasing with time.
-
-
-
-
- The Grinnell Glacier is retreating at a decreasing rate.
-
-
-
-
-
-
-
-
-
-
- Consider the exponential function f whose graph is given by Figure. Note that f passes through the two noted points exactly.
-
-
-
-
A plot of the exponential function f.
-
A plot of the exponential function f.
-
-
-
-
-
-
-
- Determine the values of a and b exactly.
-
-
-
-
- Determine the average rate of change of f on the intervals [2,7] and [7,12]. Which average rate is greater?
-
-
-
-
- Find the equation of the linear function L that passes through the points (2,20) and (7,5).
-
-
-
-
- Which average rate of change is greater? The average rate of change of f on [0,2] or the average rate of change of L on [0,2]?
-
-
-
-
-
-
-
-
- Exercise Answer
-
-
-
-
-
-
-
- Taking the ratio of the two given points: b = \left(\frac{1}{4}\right)^{1/5} \approx 0.7579, and substituting back gives a = 10 \cdot 2^{9/5} \approx 34.82.
-
-
-
-
- AV_{[2,7]} = (5 - 20)/(7 - 2) = -3 and AV_{[7,12]} = (f(12) - 5)/(12 - 7) = -\frac{3}{4}. The greater average rate of change is on the interval [2,7].
-
-
-
-
- A linear equation passing through (2, 20) and (7, 5) is L = -3t + 26.
-
-
-
-
- Given that the rate of decrease of the exponential function f is decreasing and the rate of change of the linear function L is constant, the average rate of change of f will be greater in magnitude than L on the interval [0, 2].
-
-
-
-
-
-
-
-
-
-
- A cup of hot coffee is brought outside on a cold winter morning in Winnipeg, Manitoba, where the surrounding temperature is 0 degrees Fahrenheit. A temperature probe records the coffee's temperature (in degrees Fahrenheit) every minute and generates the data shown in Table.
-
-
-
- The temperature, <m>F</m>, of the coffee at time <m>t</m>.
-
-
- t
- 0
- 2
- 4
- 6
- 8
- 10
-
-
- F(t)
- 175
- 129.64
- 96.04
- 71.15
- 52.71
- 39.05
-
-
-
-
-
-
-
-
- Assume that the data in the table represents the overall trend of the behavior of F. Is F linear, exponential, or neither? Why?
-
-
-
-
- Is it possible to determine an exact formula for F? If yes, do so and justify your formula; if not, explain why not.
-
-
-
-
- What is the average rate of change of F on [4,6]? Write a sentence that explains the practical meaning of this value in the context of the overall exercise.
-
-
-
-
- How do you think the data would appear if instead of being in a regular coffee cup, the coffee was contained in an insulated mug?
-
-
-
-
-
-
-
-
- Exercise Answer
-
-
-
-
-
-
-
- The function described by the data appears exponential because there is a consistent ratio of about 0.7408 between temperatures at sequential 2-minute intervals.
-
-
-
-
- It is necessary to take the square root of the ratio between 2-minute measurements to find the 1-minute growth factor. A possible formula is F(t) = 175(0.7408)^{t/2} \approx 175(0.8607)^t. We can verify: F(0) = 175, F(2) \approx 129.64, F(4) \approx 96.04, and so on.
-
-
-
-
- AV_{[4,6]} = (71.15 - 96.04)/(6 - 4) \approx -12.45 degrees per minute. The meaning of this rate of change is the average amount of cooling per minute on the interval [4,6].
-
-
-
-
- An insulating mug would reduce the rate of cooling, but the temperature would still follow an exponential model. The growth factor would be less than 1 but closer to 1 than 0.8607.
-
-
-
-
-
-
-
-
-
-
- The amount (in milligrams) of a drug in a person's body following one dose is given by an exponential decay function. Let A(t) denote the amount of drug in the body at time t in hours after the dose was taken. In addition, suppose you know that A(3) = 22.7 and A(6) = 15.2.
-
-
-
-
-
-
- Find a formula for A in the form A(t) = ab^t, where you determine the values of a and b exactly.
-
-
-
-
- What is the size of the initial dose the person was given?
-
-
-
-
- How much of the drug remains in the person's body 8 hours after the dose was taken?
-
-
-
-
- Estimate how long it will take until there is less than 1 mg of the drug remaining in the body.
-
-
-
-
- Compute the average rate of change of A on the intervals [3,5], [5,7], and [7,9]. Write at least one careful sentence to explain the meaning of the values you found, including appropriate units. Then write at least one additional sentence to explain any overall trend(s) you observe in the average rate of change.
-
-
-
-
- Plot A(t) on an appropriate interval and label important points and features of the graph to highlight graphical interpretations of your answers in (b), (c), (d), and (e).
-
-
-
-
-
-
-
- Exercise Answer
-
-
-
-
-
-
-
- The value of b is about 0.875, or \sqrt[3]{15.2/22.7}. The value of a is about 33.9 milligrams.
-
-
-
-
- The initial dose is equal to the coefficient a, about 33.9 milligrams.
-
-
-
-
- Eight hours after the initial dose, the remaining amount is 33.9(0.875)^8 \approx 11.6 milligrams.
-
-
-
-
- There will be less than 1 mg in the body after about 26.4 hours.
-
-
-
-
- AV_{[3,5]} \approx -2.7 mg/hour, AV_{[5,7]} \approx -2.0 mg/hour, and AV_{[7,9]} \approx -1.6 mg/hour. On the interval [3,5], the amount of drug in the patient is decreasing at an average rate of 2.7 mg per hour. The rate of decrease is decreasing over time.
-
+ Grinnell Glacier in Glacier National Park in Montana covered about 142 acres in 2007 and was found to be shrinking at about 4.4% per year.See Exercise 34 on p. 146 of Connally's Functions Modeling Change.
+
+
+
+
+
+
+ Let G(t) denote the area of Grinnell Glacier in acres in year t,
+ where t is the number of years since 2007.
+ Find a formula for G(t) and define the function in Desmos.
+
+
+
+
+ How many acres of ice were in the glacier in 1997?
+ In 2012?
+ What does the model predict for 2022?
+
+
+
+
+ How many total acres of ice were lost from 2007 to 2012?
+
+
+
+
+ What was the average rate of change of G from 2007 to 2012?
+ Write a sentence to explain the meaning of this number and include units on your answer.
+ In addition,
+ how does this compare to the average rate of change of G from 2012 to 2017?
+
+
+
+
+ How would you you describe the overall behavior of G, and thus what is happening to the Grinnell Glacier?
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
+
+
+
+ The area of Grinnell Glacier is G(t) = 142(0.956)^t.
+
+
+
+
+
+ In 1997 (t = -10), the model predicts 142(0.956)^{-10} \approx 223 acres. In 2012 (t = 5), the model predicts 142(0.956)^{5} \approx 113 acres. In 2022 (t = 15), the model predicts 142(0.956)^{15} \approx 72 acres.
+
+
+
+
+ According to the model, between 2007 and 2012 the glacier lost about 142 - 113 = 29 acres.
+
+
+
+
+ The average rate of change of G from 2007 to 2012 is approximately -29/5 \approx -5.8 acres per year. The average retreat of the glacier was about 5.8 acres per year from 2007 to 2012. This rate of change is expected to be larger in magnitude than the average rate of change from 2012 to 2017, because the rate of decrease is decreasing with time.
+
+
+
+
+ The Grinnell Glacier is retreating at a decreasing rate.
+
+
+
+
+
+
+
+
+
+
+ Consider the exponential function f whose graph is given by Figure. Note that f passes through the two noted points exactly.
+
+
+
+
A plot of the exponential function f.
+
A plot of the exponential function f.
+
+
+
+
+
+
+
+ Determine the values of a and b exactly.
+
+
+
+
+ Determine the average rate of change of f on the intervals [2,7] and [7,12]. Which average rate is greater?
+
+
+
+
+ Find the equation of the linear function L that passes through the points (2,20) and (7,5).
+
+
+
+
+ Which average rate of change is greater? The average rate of change of f on [0,2] or the average rate of change of L on [0,2]?
+
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
+
+
+
+ Taking the ratio of the two given points: b = \left(\frac{1}{4}\right)^{1/5} \approx 0.7579, and substituting back gives a = 10 \cdot 2^{9/5} \approx 34.82.
+
+
+
+
+ AV_{[2,7]} = (5 - 20)/(7 - 2) = -3 and AV_{[7,12]} = (f(12) - 5)/(12 - 7) = -\frac{3}{4}. The greater average rate of change is on the interval [2,7].
+
+
+
+
+ A linear equation passing through (2, 20) and (7, 5) is L = -3t + 26.
+
+
+
+
+ Given that the rate of decrease of the exponential function f is decreasing and the rate of change of the linear function L is constant, the average rate of change of f will be greater in magnitude than L on the interval [0, 2].
+
+
+
+
+
+
+
+
+
+
+ A cup of hot coffee is brought outside on a cold winter morning in Winnipeg, Manitoba, where the surrounding temperature is 0 degrees Fahrenheit. A temperature probe records the coffee's temperature (in degrees Fahrenheit) every minute and generates the data shown in Table.
+
+
+
+ The temperature, <m>F</m>, of the coffee at time <m>t</m>.
+
+
+ t
+ 0
+ 2
+ 4
+ 6
+ 8
+ 10
+
+
+ F(t)
+ 175
+ 129.64
+ 96.04
+ 71.15
+ 52.71
+ 39.05
+
+
+
+
+
+
+
+
+ Assume that the data in the table represents the overall trend of the behavior of F. Is F linear, exponential, or neither? Why?
+
+
+
+
+ Is it possible to determine an exact formula for F? If yes, do so and justify your formula; if not, explain why not.
+
+
+
+
+ What is the average rate of change of F on [4,6]? Write a sentence that explains the practical meaning of this value in the context of the overall exercise.
+
+
+
+
+ How do you think the data would appear if instead of being in a regular coffee cup, the coffee was contained in an insulated mug?
+
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
+
+
+
+ The function described by the data appears exponential because there is a consistent ratio of about 0.7408 between temperatures at sequential 2-minute intervals.
+
+
+
+
+ It is necessary to take the square root of the ratio between 2-minute measurements to find the 1-minute growth factor. A possible formula is F(t) = 175(0.7408)^{t/2} \approx 175(0.8607)^t. We can verify: F(0) = 175, F(2) \approx 129.64, F(4) \approx 96.04, and so on.
+
+
+
+
+ AV_{[4,6]} = (71.15 - 96.04)/(6 - 4) \approx -12.45 degrees per minute. The meaning of this rate of change is the average amount of cooling per minute on the interval [4,6].
+
+
+
+
+ An insulating mug would reduce the rate of cooling, but the temperature would still follow an exponential model. The growth factor would be less than 1 but closer to 1 than 0.8607.
+
+
+
+
+
+
+
+
+
+
+ The amount (in milligrams) of a drug in a person's body following one dose is given by an exponential decay function. Let A(t) denote the amount of drug in the body at time t in hours after the dose was taken. In addition, suppose you know that A(3) = 22.7 and A(6) = 15.2.
+
+
+
+
+
+
+ Find a formula for A in the form A(t) = ab^t, where you determine the values of a and b exactly.
+
+
+
+
+ What is the size of the initial dose the person was given?
+
+
+
+
+ How much of the drug remains in the person's body 8 hours after the dose was taken?
+
+
+
+
+ Estimate how long it will take until there is less than 1 mg of the drug remaining in the body.
+
+
+
+
+ Compute the average rate of change of A on the intervals [3,5], [5,7], and [7,9]. Write at least one careful sentence to explain the meaning of the values you found, including appropriate units. Then write at least one additional sentence to explain any overall trend(s) you observe in the average rate of change.
+
+
+
+
+ Plot A(t) on an appropriate interval and label important points and features of the graph to highlight graphical interpretations of your answers in (b), (c), (d), and (e).
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
+
+
+
+ The value of b is about 0.875, or \sqrt[3]{15.2/22.7}. The value of a is about 33.9 milligrams.
+
+
+
+
+ The initial dose is equal to the coefficient a, about 33.9 milligrams.
+
+
+
+
+ Eight hours after the initial dose, the remaining amount is 33.9(0.875)^8 \approx 11.6 milligrams.
+
+
+
+
+ There will be less than 1 mg in the body after about 26.4 hours.
+
+
+
+
+ AV_{[3,5]} \approx -2.7 mg/hour, AV_{[5,7]} \approx -2.0 mg/hour, and AV_{[7,9]} \approx -1.6 mg/hour. On the interval [3,5], the amount of drug in the patient is decreasing at an average rate of 2.7 mg per hour. The rate of decrease is decreasing over time.
+
- For a population that is growing exponentially according to a model of the form P(t) = Ae^{kt}, the doubling timeexponential functiondoubling time is the amount of time that it takes the population to double. For each population described below, assume the function is growing exponentially according to a model P(t) = Ae^{kt}, where t is measured in years.
-
-
-
-
-
-
- Suppose that a certain population initially has 100 members and doubles after 3 years. What are the values of A and k in the model?
-
-
-
-
- A different population is observed to satisfy P(4) = 250 and P(11) = 500. What is the population's doubling time? When will 2000 members of the population be present?
-
-
-
-
- Another population is observed to have doubling time t = 21. What is the value of k in the model?
-
-
-
-
- How is k related to a population's doubling time, regardless of how long the doubling time is?
-
-
-
-
-
-
-
- Exercise Answer
-
-
-
-
-
-
-
- Since the population initially has 100 members, A = 100. Since the population doubles in 3 years, we have P(3) = 200, so 200 = 100e^{3k}, giving e^{3k} = 2 and thus k = \frac{1}{3}\ln 2 \approx 0.2310.
-
-
-
-
- Since P(4) = 250 and P(11) = 500, the population doubles every 11 - 4 = 7 years. So the doubling time is 7 years. The population will reach 2000 = 4 \cdot 500 members two doublings after t = 11, i.e., at t = 11 + 14 = 25 years.
-
-
-
-
- If the doubling time is 21 years, then 2A = Ae^{21k}, so e^{21k} = 2 and k = \frac{1}{21}\ln 2.
-
-
-
-
- If the doubling time is t_2, then k = \frac{1}{t_2}\ln 2 = \frac{\ln 2}{t_2}.
-
-
-
-
-
-
-
-
-
-
- A new car is purchased for $28000. Exactly 1 year later, the value of the car is $23200. Assume that the car's value in dollars, V, t years after purchase decays exponentially according to a model of form V(t) = Ae^{-kt}.
-
-
-
-
-
-
- Determine the exact values of A and k in the model.
-
-
-
-
- How many years will it take until the car's value is $10000?
-
-
-
-
- Suppose that rather than having the car's value decay all the way to $0, the lowest dollar amount its value ever approaches is $500. Explain why a model of the form V(t) = Ae^{-kt} + c is more appropriate.
-
-
-
-
- Under the original assumptions (V(0) = 28000 and V(1) = 23200) along with the condition in (c) that the car's value will approach $500 in the long-term, determine the exact values of A, k, and c in the model V(t) = Ae^{-kt} + c. Are the values of A and k the same or different from the model explored in (a)? Why?
-
-
-
-
-
-
-
- Exercise Answer
-
-
-
-
-
-
-
- At purchase (t = 0), V(0) = A = \$28000. Using V(1) = 23200, we get 23200 = 28000e^{-k}, so e^{-k} = \frac{23200}{28000} and k = -\ln\!\left(\frac{23200}{28000}\right) \approx 0.1880.
-
-
-
-
- We solve 10000 = 28000e^{-kt}, giving e^{-kt} = \frac{10000}{28000}, so
-
- t = \frac{\ln(10000/28000)}{-k} = \frac{\ln(10000) - \ln(28000)}{\ln(23200) - \ln(28000)} \approx 5.48 \text{ years}.
-
-
-
-
-
- As t \to \infty, the model V(t) = Ae^{-kt} approaches 0, but if the car's value approaches \$500, adding a positive constant c = 500 gives V(t) = Ae^{-kt} + 500, whose long-term behavior approaches \$500 as desired.
-
-
-
-
- With c = 500, from V(0) = 28000 we get A + 500 = 28000, so A = 27500. From V(1) = 23200 we get 27500e^{-k} + 500 = 23200, so e^{-k} = \frac{22700}{27500} and k = -\ln\!\left(\frac{22700}{27500}\right) \approx 0.1913. The values of A and k are different from part (a) because the model now incorporates the long-term value of \$500.
-
-
-
-
-
-
-
-
-
-
- In Exercise, we explored graphically how the function y = \log_b(x) can be thought of as a vertical stretch of the nautral logarithm, y = \ln(x). In this exercise, we determine the exact value of the vertical stretch that is needed.
-
-
-
- Recall that \log_b(x) is the power to which we raise b to get x.
-
-
-
-
-
-
- Write the equation y = \log_b(x) as an equivalent equation involving exponents with no logarithms present.
-
-
-
-
- Take the equation you found in (a) and take the natural logarithm of each side.
-
-
-
-
- Use rules and properties of logarithms appropriately to solve the equation from (b) for y. Your result here should express y in terms of \ln(x) and \ln(b).
-
-
-
-
- Recall that y = \log_b(x). Explain why the following equation (often called the Golden Rule for Logarithms) is true:
-
- \log_b(x) = \frac{\ln(x)}{\ln(b)}
- .
-
-
-
-
- What is the value of k that allows us to express the function y = \log_b(x) as a vertical stretch of the function y = \ln(x)?
-
-
-
-
-
-
-
- Exercise Answer
-
-
-
-
-
-
-
- The equation y = \log_b(x) is equivalent to b^y = x.
-
-
-
-
- Taking the natural log of both sides of b^y = x gives y\ln(b) = \ln(x).
-
-
-
-
- Dividing both sides by \ln(b) gives y = \dfrac{\ln(x)}{\ln(b)}.
-
-
-
-
- Since y = \log_b(x) and we showed y = \dfrac{\ln(x)}{\ln(b)}, it must be true that \log_b(x) = \dfrac{\ln(x)}{\ln(b)}.
-
-
-
-
- Since \log_b(x) = \dfrac{\ln(x)}{\ln(b)} = \dfrac{1}{\ln(b)} \cdot \ln(x), the value of k that gives a vertical stretch is k = \dfrac{1}{\ln(b)}.
-
+ For a population that is growing exponentially according to a model of the form P(t) = Ae^{kt}, the doubling timeexponential functiondoubling time is the amount of time that it takes the population to double. For each population described below, assume the function is growing exponentially according to a model P(t) = Ae^{kt}, where t is measured in years.
+
+
+
+
+
+
+ Suppose that a certain population initially has 100 members and doubles after 3 years. What are the values of A and k in the model?
+
+
+
+
+ A different population is observed to satisfy P(4) = 250 and P(11) = 500. What is the population's doubling time? When will 2000 members of the population be present?
+
+
+
+
+ Another population is observed to have doubling time t = 21. What is the value of k in the model?
+
+
+
+
+ How is k related to a population's doubling time, regardless of how long the doubling time is?
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
+
+
+
+ Since the population initially has 100 members, A = 100. Since the population doubles in 3 years, we have P(3) = 200, so 200 = 100e^{3k}, giving e^{3k} = 2 and thus k = \frac{1}{3}\ln 2 \approx 0.2310.
+
+
+
+
+ Since P(4) = 250 and P(11) = 500, the population doubles every 11 - 4 = 7 years. So the doubling time is 7 years. The population will reach 2000 = 4 \cdot 500 members two doublings after t = 11, i.e., at t = 11 + 14 = 25 years.
+
+
+
+
+ If the doubling time is 21 years, then 2A = Ae^{21k}, so e^{21k} = 2 and k = \frac{1}{21}\ln 2.
+
+
+
+
+ If the doubling time is t_2, then k = \frac{1}{t_2}\ln 2 = \frac{\ln 2}{t_2}.
+
+
+
+
+
+
+
+
+
+
+ A new car is purchased for $28000. Exactly 1 year later, the value of the car is $23200. Assume that the car's value in dollars, V, t years after purchase decays exponentially according to a model of form V(t) = Ae^{-kt}.
+
+
+
+
+
+
+ Determine the exact values of A and k in the model.
+
+
+
+
+ How many years will it take until the car's value is $10000?
+
+
+
+
+ Suppose that rather than having the car's value decay all the way to $0, the lowest dollar amount its value ever approaches is $500. Explain why a model of the form V(t) = Ae^{-kt} + c is more appropriate.
+
+
+
+
+ Under the original assumptions (V(0) = 28000 and V(1) = 23200) along with the condition in (c) that the car's value will approach $500 in the long-term, determine the exact values of A, k, and c in the model V(t) = Ae^{-kt} + c. Are the values of A and k the same or different from the model explored in (a)? Why?
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
+
+
+
+ At purchase (t = 0), V(0) = A = \$28000. Using V(1) = 23200, we get 23200 = 28000e^{-k}, so e^{-k} = \frac{23200}{28000} and k = -\ln\!\left(\frac{23200}{28000}\right) \approx 0.1880.
+
+
+
+
+ We solve 10000 = 28000e^{-kt}, giving e^{-kt} = \frac{10000}{28000}, so
+
+ t = \frac{\ln(10000/28000)}{-k} = \frac{\ln(10000) - \ln(28000)}{\ln(23200) - \ln(28000)} \approx 5.48 \text{ years}.
+
+
+
+
+
+ As t \to \infty, the model V(t) = Ae^{-kt} approaches 0, but if the car's value approaches \$500, adding a positive constant c = 500 gives V(t) = Ae^{-kt} + 500, whose long-term behavior approaches \$500 as desired.
+
+
+
+
+ With c = 500, from V(0) = 28000 we get A + 500 = 28000, so A = 27500. From V(1) = 23200 we get 27500e^{-k} + 500 = 23200, so e^{-k} = \frac{22700}{27500} and k = -\ln\!\left(\frac{22700}{27500}\right) \approx 0.1913. The values of A and k are different from part (a) because the model now incorporates the long-term value of \$500.
+
+
+
+
+
+
+
+
+
+
+ In Exercise, we explored graphically how the function y = \log_b(x) can be thought of as a vertical stretch of the nautral logarithm, y = \ln(x). In this exercise, we determine the exact value of the vertical stretch that is needed.
+
+
+
+ Recall that \log_b(x) is the power to which we raise b to get x.
+
+
+
+
+
+
+ Write the equation y = \log_b(x) as an equivalent equation involving exponents with no logarithms present.
+
+
+
+
+ Take the equation you found in (a) and take the natural logarithm of each side.
+
+
+
+
+ Use rules and properties of logarithms appropriately to solve the equation from (b) for y. Your result here should express y in terms of \ln(x) and \ln(b).
+
+
+
+
+ Recall that y = \log_b(x). Explain why the following equation (often called the Golden Rule for Logarithms) is true:
+
+ \log_b(x) = \frac{\ln(x)}{\ln(b)}
+ .
+
+
+
+
+ What is the value of k that allows us to express the function y = \log_b(x) as a vertical stretch of the function y = \ln(x)?
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
+
+
+
+ The equation y = \log_b(x) is equivalent to b^y = x.
+
+
+
+
+ Taking the natural log of both sides of b^y = x gives y\ln(b) = \ln(x).
+
+
+
+
+ Dividing both sides by \ln(b) gives y = \dfrac{\ln(x)}{\ln(b)}.
+
+
+
+
+ Since y = \log_b(x) and we showed y = \dfrac{\ln(x)}{\ln(b)}, it must be true that \log_b(x) = \dfrac{\ln(x)}{\ln(b)}.
+
+
+
+
+ Since \log_b(x) = \dfrac{\ln(x)}{\ln(b)} = \dfrac{1}{\ln(b)} \cdot \ln(x), the value of k that gives a vertical stretch is k = \dfrac{1}{\ln(b)}.
+
- Recall that when a function y = f(x) has an inverse function, the two equations y = f(x) and x = f^{-1}(y) say the same thing from different perspectives: the first equation expresses y in terms of x, while the second expresses x in terms of y. When y = f(x) = e^x, we know its inverse is x = f^{-1}(y) = \ln(y). Through logarithms, we now have the ability to find the inverse of many different exponential functions. In particular, because exponential functions and their transformations are either always increasing or always decreasing, any function of the form y = f(x) = ae^{-kx} + c will have an inverse function.
-
-
-
- Find the inverse function for each given function by solving algebraically for x as a function of y. In addition, state the domain and range of the given function and the domain and range of the inverse function.
-
-
-
-
-
-
- y = g(x) = e^{-0.25x}
-
-
-
-
-
- y = h(x) = 2e^{x} + 1
-
-
-
-
-
-
- y = r(x) = 21 + 15e^{-0.1x}
-
-
-
-
-
- y = s(x) = 72 - 40e^{-0.05x}
-
-
-
-
- y = u(x) = -5e^{3x-4} + 8
-
-
-
-
- y = w(x) = 3\ln(x) + 4
-
-
-
-
- y = z(x) = -0.2 \ln(2x - 5) + 1
-
-
-
-
-
-
-
- Exercise Answer
-
-
-
-
-
-
-
- The domain of y = g(x) = e^{-0.25x} is all real numbers and the range is all positive real numbers. Solving for x: x = g^{-1}(y) = \dfrac{\ln y}{-0.25}. The domain of the inverse is all positive real numbers and the range is all real numbers.
-
-
-
-
- The domain of y = h(x) = 2e^x + 1 is all real numbers and the range is all real numbers greater than 1. The inverse is x = h^{-1}(y) = \ln\!\dfrac{y-1}{2}. The domain of the inverse is all real numbers greater than 1 and the range is all real numbers.
-
-
-
-
- The domain of y = r(x) = 21 + 15e^{-0.1x} is all real numbers and the range is all real numbers greater than 21. The inverse is x = r^{-1}(y) = \dfrac{\ln\!\left(\frac{y-21}{15}\right)}{-0.1}. The domain of the inverse is all real numbers greater than 21 and the range is all real numbers.
-
-
-
-
- The domain of y = s(x) = 72 - 40e^{-0.05x} is all real numbers and the range is all real numbers less than 72. The inverse is x = s^{-1}(y) = \dfrac{\ln\!\left(\frac{y-72}{-40}\right)}{-0.05}. The domain of the inverse is all real numbers less than 72 and the range is all real numbers.
-
-
-
-
- The domain of y = u(x) = -5e^{3x-4} + 8 is all real numbers and the range is all real numbers less than 8. The inverse is x = u^{-1}(y) = \dfrac{\ln\!\left(\frac{y-8}{-5}\right) + 4}{3}. The domain of the inverse is all real numbers less than 8 and the range is all real numbers.
-
-
-
-
- The domain of y = w(x) = 3\ln(x) + 4 is all positive real numbers and the range is all real numbers. The inverse is x = w^{-1}(y) = e^{(y-4)/3}. The domain of the inverse is all real numbers and the range is all positive real numbers.
-
-
-
-
- The domain of y = z(x) = -0.2\ln(2x-5) + 1 is all real numbers greater than \frac{5}{2} and the range is all real numbers. The inverse is x = z^{-1}(y) = \dfrac{e^{(y-1)/(-0.2)} + 5}{2}. The domain of the inverse is all real numbers and the range is all real numbers greater than \frac{5}{2}.
-
-
-
-
-
-
-
-
-
-
- We've seen that any exponential function f(t) = b^t (b \gt 0, b \ne 1) can be written in the form f(t) = e^{kt} for some real number k, and this is because f(t) = b^t is a horizontal scaling of the function E(t) = e^{t}. In this exercise, we explore how the natural logarithm can be scaled to achieve a logarithm of any base.
-
-
-
- Let b \gt 1. Because the function y = f(t) = b^t has an inverse function, it makes sense to define its inverse like we did when b = 10 or b = e. The base-b logarithm, logarithmbase-bdefinition denoted \log_b(y) is defined to be the power to which we raise b to get y. Thus, writing y = f(t) = b^t is the same as writing t = f^{-1}(y) = \log_b(y).
-
-
-
- In Desmos, the natural logarithm function is given by ln(t), while the base-10 logarithm by log(t). To get a logarithm of a different base, such as a base-2 logarithm, type log_2(t) (the underscore will generate a subscript; then use the right arrow to get out of subscript mode).
-
-
-
- In a new Desmos worksheet, enter V(t) = k * ln(t) and accept the slider for k. Set the lower and upper bounds for the slider to 0.01 and 15, respectively.
-
-
-
-
-
-
- Define f(t) = \log_2(t) in Desmos. Can you find a value of k for which \log_2(t) = k\ln(t)? If yes, what is the value? If not, why not?
-
-
-
-
- Repeat (a) for the functions g(t) = \log_3(t), h(t) = \log_5(t), and p(t) = \log_{1.25}(t). What pattern(s) do you observe?
-
-
-
-
- True or false: for any value of b \gt 1, the function \log_b(t) can be viewed as a vertical scaling of \ln(t).
-
-
-
-
- Compute the following values: \frac{1}{\ln(2)}, \frac{1}{\ln(3)}, \frac{1}{\ln(5)}, and \frac{1}{\ln(1.25)}. What do you notice about these values compared to those of k you found in (a) and (b)?
-
-
-
-
-
-
-
- Exercise Answer
-
-
-
-
-
-
-
- In Desmos, entering V(t) = k * ln(t) and f(t) = log_2(t), we can adjust the slider until the graphs coincide. The value k = \frac{1}{\ln 2} \approx 1.443 makes \log_2(t) = k\ln(t).
-
-
-
-
- Similarly, the values of k for the other functions are: for g(t) = \log_3(t), k = \frac{1}{\ln 3} \approx 0.910; for h(t) = \log_5(t), k = \frac{1}{\ln 5} \approx 0.621; and for p(t) = \log_{1.25}(t), k = \frac{1}{\ln(1.25)} \approx 4.481. In each case, k = \frac{1}{\ln b}.
-
-
-
-
- True. For any b \gt 1, the function \log_b(t) is a vertical scaling of \ln(t) by a factor of \frac{1}{\ln b}.
-
-
-
-
- We compute \frac{1}{\ln 2} \approx 1.443, \frac{1}{\ln 3} \approx 0.910, \frac{1}{\ln 5} \approx 0.621, and \frac{1}{\ln(1.25)} \approx 4.481. These are exactly the values of k found in (a) and (b), confirming that \log_b(t) = \frac{1}{\ln b} \cdot \ln(t).
-
-
-
-
-
-
-
-
-
-
- A can of soda is removed from a refrigerator at time t = 0 (in minutes)
- and its temperature, F(t), in degrees Fahrenheit, is computed at regular intervals.
- Based on the data, a model is formulated for the object's temperature, given by
-
- F(t) = 74.4 - 38.8e^{-0.05t}
- .
-
-
-
-
-
-
- Determine the exact time when the soda's temperature is 50^\circ.
-
-
-
-
- Is there ever a time when the soda's temperature is 36^\circ? Why or why not?
-
-
-
-
- For the model, its domain is the set of all positive real numbers, t \gt 0. What is its range?
-
-
-
-
- Find a formula for the inverse of the function y = F(t). What is the meaning of this function?
-
-
-
-
-
-
-
- Exercise Answer
-
-
-
-
-
-
-
- We solve 74.4 - 38.8e^{-0.05t} = 50. Subtracting 74.4 and dividing by -38.8 gives e^{-0.05t} = \frac{50 - 74.4}{-38.8} = \frac{24.4}{38.8}. Taking the natural log, -0.05t = \ln\!\left(\frac{24.4}{38.8}\right), so
-
- t = \frac{\ln\!\left(\frac{24.4}{38.8}\right)}{-0.05} \approx 9.28 \text{ minutes}.
-
-
-
-
-
- At t = 0, F(0) = 74.4 - 38.8 = 35.6^\circ. Since F is an increasing function (the exponential term decreases as t increases, so F increases from its initial value), and F(0) = 35.6 \lt 36, yes, there is a time when the soda's temperature reaches 36^\circ.
-
-
-
-
- Since e^{-0.05t} \to 0 as t \to \infty, the temperature approaches 74.4^\circ but never reaches it. At t = 0, F(0) = 35.6^\circ. Thus the range of F on the domain t \gt 0 is (35.6, 74.4).
-
-
-
-
- Solving y = 74.4 - 38.8e^{-0.05t} for t: we get e^{-0.05t} = \frac{y - 74.4}{-38.8}, so
-
- t = F^{-1}(y) = \frac{\ln\!\left(\frac{y - 74.4}{-38.8}\right)}{-0.05}.
-
- This inverse function tells us the time (in minutes) at which the soda reaches a given temperature y (in degrees Fahrenheit).
-
+ Recall that when a function y = f(x) has an inverse function, the two equations y = f(x) and x = f^{-1}(y) say the same thing from different perspectives: the first equation expresses y in terms of x, while the second expresses x in terms of y. When y = f(x) = e^x, we know its inverse is x = f^{-1}(y) = \ln(y). Through logarithms, we now have the ability to find the inverse of many different exponential functions. In particular, because exponential functions and their transformations are either always increasing or always decreasing, any function of the form y = f(x) = ae^{-kx} + c will have an inverse function.
+
+
+
+ Find the inverse function for each given function by solving algebraically for x as a function of y. In addition, state the domain and range of the given function and the domain and range of the inverse function.
+
+
+
+
+
+
+ y = g(x) = e^{-0.25x}
+
+
+
+
+
+ y = h(x) = 2e^{x} + 1
+
+
+
+
+
+
+ y = r(x) = 21 + 15e^{-0.1x}
+
+
+
+
+
+ y = s(x) = 72 - 40e^{-0.05x}
+
+
+
+
+ y = u(x) = -5e^{3x-4} + 8
+
+
+
+
+ y = w(x) = 3\ln(x) + 4
+
+
+
+
+ y = z(x) = -0.2 \ln(2x - 5) + 1
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
+
+
+
+ The domain of y = g(x) = e^{-0.25x} is all real numbers and the range is all positive real numbers. Solving for x: x = g^{-1}(y) = \dfrac{\ln y}{-0.25}. The domain of the inverse is all positive real numbers and the range is all real numbers.
+
+
+
+
+ The domain of y = h(x) = 2e^x + 1 is all real numbers and the range is all real numbers greater than 1. The inverse is x = h^{-1}(y) = \ln\!\dfrac{y-1}{2}. The domain of the inverse is all real numbers greater than 1 and the range is all real numbers.
+
+
+
+
+ The domain of y = r(x) = 21 + 15e^{-0.1x} is all real numbers and the range is all real numbers greater than 21. The inverse is x = r^{-1}(y) = \dfrac{\ln\!\left(\frac{y-21}{15}\right)}{-0.1}. The domain of the inverse is all real numbers greater than 21 and the range is all real numbers.
+
+
+
+
+ The domain of y = s(x) = 72 - 40e^{-0.05x} is all real numbers and the range is all real numbers less than 72. The inverse is x = s^{-1}(y) = \dfrac{\ln\!\left(\frac{y-72}{-40}\right)}{-0.05}. The domain of the inverse is all real numbers less than 72 and the range is all real numbers.
+
+
+
+
+ The domain of y = u(x) = -5e^{3x-4} + 8 is all real numbers and the range is all real numbers less than 8. The inverse is x = u^{-1}(y) = \dfrac{\ln\!\left(\frac{y-8}{-5}\right) + 4}{3}. The domain of the inverse is all real numbers less than 8 and the range is all real numbers.
+
+
+
+
+ The domain of y = w(x) = 3\ln(x) + 4 is all positive real numbers and the range is all real numbers. The inverse is x = w^{-1}(y) = e^{(y-4)/3}. The domain of the inverse is all real numbers and the range is all positive real numbers.
+
+
+
+
+ The domain of y = z(x) = -0.2\ln(2x-5) + 1 is all real numbers greater than \frac{5}{2} and the range is all real numbers. The inverse is x = z^{-1}(y) = \dfrac{e^{(y-1)/(-0.2)} + 5}{2}. The domain of the inverse is all real numbers and the range is all real numbers greater than \frac{5}{2}.
+
+
+
+
+
+
+
+
+
+
+ We've seen that any exponential function f(t) = b^t (b \gt 0, b \ne 1) can be written in the form f(t) = e^{kt} for some real number k, and this is because f(t) = b^t is a horizontal scaling of the function E(t) = e^{t}. In this exercise, we explore how the natural logarithm can be scaled to achieve a logarithm of any base.
+
+
+
+ Let b \gt 1. Because the function y = f(t) = b^t has an inverse function, it makes sense to define its inverse like we did when b = 10 or b = e. The base-b logarithm, logarithmbase-bdefinition denoted \log_b(y) is defined to be the power to which we raise b to get y. Thus, writing y = f(t) = b^t is the same as writing t = f^{-1}(y) = \log_b(y).
+
+
+
+ In Desmos, the natural logarithm function is given by ln(t), while the base-10 logarithm by log(t). To get a logarithm of a different base, such as a base-2 logarithm, type log_2(t) (the underscore will generate a subscript; then use the right arrow to get out of subscript mode).
+
+
+
+ In a new Desmos worksheet, enter V(t) = k * ln(t) and accept the slider for k. Set the lower and upper bounds for the slider to 0.01 and 15, respectively.
+
+
+
+
+
+
+ Define f(t) = \log_2(t) in Desmos. Can you find a value of k for which \log_2(t) = k\ln(t)? If yes, what is the value? If not, why not?
+
+
+
+
+ Repeat (a) for the functions g(t) = \log_3(t), h(t) = \log_5(t), and p(t) = \log_{1.25}(t). What pattern(s) do you observe?
+
+
+
+
+ True or false: for any value of b \gt 1, the function \log_b(t) can be viewed as a vertical scaling of \ln(t).
+
+
+
+
+ Compute the following values: \frac{1}{\ln(2)}, \frac{1}{\ln(3)}, \frac{1}{\ln(5)}, and \frac{1}{\ln(1.25)}. What do you notice about these values compared to those of k you found in (a) and (b)?
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
+
+
+
+ In Desmos, entering V(t) = k * ln(t) and f(t) = log_2(t), we can adjust the slider until the graphs coincide. The value k = \frac{1}{\ln 2} \approx 1.443 makes \log_2(t) = k\ln(t).
+
+
+
+
+ Similarly, the values of k for the other functions are: for g(t) = \log_3(t), k = \frac{1}{\ln 3} \approx 0.910; for h(t) = \log_5(t), k = \frac{1}{\ln 5} \approx 0.621; and for p(t) = \log_{1.25}(t), k = \frac{1}{\ln(1.25)} \approx 4.481. In each case, k = \frac{1}{\ln b}.
+
+
+
+
+ True. For any b \gt 1, the function \log_b(t) is a vertical scaling of \ln(t) by a factor of \frac{1}{\ln b}.
+
+
+
+
+ We compute \frac{1}{\ln 2} \approx 1.443, \frac{1}{\ln 3} \approx 0.910, \frac{1}{\ln 5} \approx 0.621, and \frac{1}{\ln(1.25)} \approx 4.481. These are exactly the values of k found in (a) and (b), confirming that \log_b(t) = \frac{1}{\ln b} \cdot \ln(t).
+
+
+
+
+
+
+
+
+
+
+ A can of soda is removed from a refrigerator at time t = 0 (in minutes)
+ and its temperature, F(t), in degrees Fahrenheit, is computed at regular intervals.
+ Based on the data, a model is formulated for the object's temperature, given by
+
+ F(t) = 74.4 - 38.8e^{-0.05t}
+ .
+
+
+
+
+
+
+ Determine the exact time when the soda's temperature is 50^\circ.
+
+
+
+
+ Is there ever a time when the soda's temperature is 36^\circ? Why or why not?
+
+
+
+
+ For the model, its domain is the set of all positive real numbers, t \gt 0. What is its range?
+
+
+
+
+ Find a formula for the inverse of the function y = F(t). What is the meaning of this function?
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
+
+
+
+ We solve 74.4 - 38.8e^{-0.05t} = 50. Subtracting 74.4 and dividing by -38.8 gives e^{-0.05t} = \frac{50 - 74.4}{-38.8} = \frac{24.4}{38.8}. Taking the natural log, -0.05t = \ln\!\left(\frac{24.4}{38.8}\right), so
+
+ t = \frac{\ln\!\left(\frac{24.4}{38.8}\right)}{-0.05} \approx 9.28 \text{ minutes}.
+
+
+
+
+
+ At t = 0, F(0) = 74.4 - 38.8 = 35.6^\circ. Since F is an increasing function (the exponential term decreases as t increases, so F increases from its initial value), and F(0) = 35.6 \lt 36, yes, there is a time when the soda's temperature reaches 36^\circ.
+
+
+
+
+ Since e^{-0.05t} \to 0 as t \to \infty, the temperature approaches 74.4^\circ but never reaches it. At t = 0, F(0) = 35.6^\circ. Thus the range of F on the domain t \gt 0 is (35.6, 74.4).
+
+
+
+
+ Solving y = 74.4 - 38.8e^{-0.05t} for t: we get e^{-0.05t} = \frac{y - 74.4}{-38.8}, so
+
+ t = F^{-1}(y) = \frac{\ln\!\left(\frac{y - 74.4}{-38.8}\right)}{-0.05}.
+
+ This inverse function tells us the time (in minutes) at which the soda reaches a given temperature y (in degrees Fahrenheit).
+
- A can of soda has been in a refrigerator for several days; the refrigerator has temperature 41^\circ Fahrenheit. Upon removal, the soda is placed on a kitchen table in a room with surrounding temperature 72^\circ. Let F(t) represent the soda's temperature in degrees Fahrenheit at time t in minutes, where t = 0 corresponds to the time the can is removed from the refrigerator. We know from Newton's Law of Cooling that F has form F(t) = ab^t + c for some constants a, b, and c, where 0 \lt b \lt 1.
-
-
-
-
-
-
- What is the numerical value of the soda's initial temperature? What is the value of F(0) in terms of a, b, and c? What do these two observations tell us?
-
-
-
-
- What is the numerical value of the soda's long-term temperature? What is the long-term value of F(t) in terms of a, b, and c? What do these two observations tell us?
-
-
-
-
- Using your work in (a) and (b), determine the numerical values of a and c.
-
-
-
-
- Suppose it can be determined that b = 0.931. What is the soda's temperature after 10 minutes?
-
-
-
-
-
-
-
- Exercise Answer
-
-
-
-
-
-
-
- The soda's initial temperature is 41^\circ Fahrenheit. Since F(0) = ab^0 + c = a + c, we have a + c = 41.
-
-
-
-
- The soda's long-term temperature is the room temperature, 72^\circ. Since 0 \lt b \lt 1, the term ab^t \to 0 as t \to \infty, so F(t) \to c. Thus c = 72.
-
-
-
-
- Using a + c = 41 and c = 72, we find a = 41 - 72 = -31.
-
-
-
-
- With b = 0.931, the soda's temperature after 10 minutes is F(10) = -31(0.931)^{10} + 72 \approx 56.8^\circ Fahrenheit.
-
-
-
-
-
-
-
-
-
-
- Consider the graphs of the following four functions p, q, r, and s. Each is a shifted exponential function of the form ab^t + c.
-
-
-
-
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- For each function p, q, r, and s, determine
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- whether a \gt 0 or a \lt 0;
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- whether 0 \lt b \lt 1 or b \gt 1;
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- whether c \gt 0, c = 0, or c \lt 0; and
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- the range of the function in terms of c.
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- Exercise Answer
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- Each function has the form ab^t + c, where b \gt 0. For the function p(t) (upper left graph), a \gt 0, 0 \lt b \lt 1, and c \lt 0. For the function q(t) (upper right graph), a \lt 0, 0 \lt b \lt 1, and c appears to be close to zero. For the function r(t) (lower left graph), a \gt 0, b \gt 1, and c \gt 0. Finally, for s(t) (lower right graph), a \lt 0, b \gt 1, and c \gt 0.
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- A cup of coffee has its temperature, C(t), measured in degrees Celsius. When poured outdoors on a cold morning, its temperature is C(0) = 95. Ten minutes later, C(10) = 80. If the surrounding temperature outside is 0^\circ Celsius, find a formula for a function C(t) that models the coffee's temperature at time t.
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- In addition, recall that we can convert between Celsius and Fahrenheit according to the equations F = \frac{9}{5}C + 32 and C = \frac{5}{9}(F-32). Use this information to also find a formula for F(t), the coffee's Fahrenheit temperature at time t. What is similar and what is different regarding the functions C(t) and F(t)?
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- Exercise Answer
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- Assuming a cooling law of the form C(t) = ab^t + c, we use C(0) = 95, C(10) = 80, and end behavior C(t) \to 0 as t \to \infty (so c = 0). From C(0) = a = 95 and C(10) = 95b^{10} = 80, we find b = (80/95)^{1/10} \approx 0.983. So C(t) = 95(0.983)^t.
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- Converting to Fahrenheit: F(t) = \frac{9}{5}C(t) + 32 = \frac{9}{5}(95(0.983)^t) + 32 = 171(0.983)^t + 32. The cooling law has the same basic ab^t + c form in either temperature scale. The coefficient c happens to be zero in the Celsius scale and 32 in the Fahrenheit scale, reflecting the different zero points of the two scales.
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+ A can of soda has been in a refrigerator for several days; the refrigerator has temperature 41^\circ Fahrenheit. Upon removal, the soda is placed on a kitchen table in a room with surrounding temperature 72^\circ. Let F(t) represent the soda's temperature in degrees Fahrenheit at time t in minutes, where t = 0 corresponds to the time the can is removed from the refrigerator. We know from Newton's Law of Cooling that F has form F(t) = ab^t + c for some constants a, b, and c, where 0 \lt b \lt 1.
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+ What is the numerical value of the soda's initial temperature? What is the value of F(0) in terms of a, b, and c? What do these two observations tell us?
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+ What is the numerical value of the soda's long-term temperature? What is the long-term value of F(t) in terms of a, b, and c? What do these two observations tell us?
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+ Using your work in (a) and (b), determine the numerical values of a and c.
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+ Suppose it can be determined that b = 0.931. What is the soda's temperature after 10 minutes?
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+ Exercise Answer
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+ The soda's initial temperature is 41^\circ Fahrenheit. Since F(0) = ab^0 + c = a + c, we have a + c = 41.
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+ The soda's long-term temperature is the room temperature, 72^\circ. Since 0 \lt b \lt 1, the term ab^t \to 0 as t \to \infty, so F(t) \to c. Thus c = 72.
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+ Using a + c = 41 and c = 72, we find a = 41 - 72 = -31.
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+ With b = 0.931, the soda's temperature after 10 minutes is F(10) = -31(0.931)^{10} + 72 \approx 56.8^\circ Fahrenheit.
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+ Consider the graphs of the following four functions p, q, r, and s. Each is a shifted exponential function of the form ab^t + c.
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+ For each function p, q, r, and s, determine
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+ whether a \gt 0 or a \lt 0;
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+ whether 0 \lt b \lt 1 or b \gt 1;
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+ whether c \gt 0, c = 0, or c \lt 0; and
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+ the range of the function in terms of c.
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+ Exercise Answer
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+ Each function has the form ab^t + c, where b \gt 0. For the function p(t) (upper left graph), a \gt 0, 0 \lt b \lt 1, and c \lt 0. For the function q(t) (upper right graph), a \lt 0, 0 \lt b \lt 1, and c appears to be close to zero. For the function r(t) (lower left graph), a \gt 0, b \gt 1, and c \gt 0. Finally, for s(t) (lower right graph), a \lt 0, b \gt 1, and c \gt 0.
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+ A cup of coffee has its temperature, C(t), measured in degrees Celsius. When poured outdoors on a cold morning, its temperature is C(0) = 95. Ten minutes later, C(10) = 80. If the surrounding temperature outside is 0^\circ Celsius, find a formula for a function C(t) that models the coffee's temperature at time t.
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+ In addition, recall that we can convert between Celsius and Fahrenheit according to the equations F = \frac{9}{5}C + 32 and C = \frac{5}{9}(F-32). Use this information to also find a formula for F(t), the coffee's Fahrenheit temperature at time t. What is similar and what is different regarding the functions C(t) and F(t)?
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+ Exercise Answer
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+ Assuming a cooling law of the form C(t) = ab^t + c, we use C(0) = 95, C(10) = 80, and end behavior C(t) \to 0 as t \to \infty (so c = 0). From C(0) = a = 95 and C(10) = 95b^{10} = 80, we find b = (80/95)^{1/10} \approx 0.983. So C(t) = 95(0.983)^t.
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+ Converting to Fahrenheit: F(t) = \frac{9}{5}C(t) + 32 = \frac{9}{5}(95(0.983)^t) + 32 = 171(0.983)^t + 32. The cooling law has the same basic ab^t + c form in either temperature scale. The coefficient c happens to be zero in the Celsius scale and 32 in the Fahrenheit scale, reflecting the different zero points of the two scales.
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In Exercise below, use the following structure/formula for N(t): N(t)=\frac{L}{1+Ab^{-kt}}. In particular, note that when the instructions say find A, this use of A is not in reference to carrying capacity.
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- A glass filled with ice and water is set on a table in a climate-controlled room with constant temperature of 71^\circ Fahrenheit. A temperature probe is placed in the glass, and we find that the following temperatures are recorded (at time t in minutes).
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- Make a rough sketch of how you think the temperature graph should appear. Is the temperature function always increasing? always decreasing? always concave up? always concave down? what's its long-range behavior?
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- By describing F as a transformation of e^t, explain why a function of form F(t) = c - ae^{-kt}, where a, c, and k are positive constants is an appropriate model for how we expect the temperature function to behave.
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- Use the given information to determine the exact values of a, c, and k in the model F(t) = c - ae^{-kt}.
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- Determine the exact time when the water's temperature is 60^\circ.
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- Exercise Answer
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- The temperature is always increasing (the ice water warms toward room temperature), concave down, and approaches 71^\circF asymptotically in the long run.
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- The function F(t) = c - ae^{-kt} is a transformation of e^t that is reflected over the horizontal axis, vertically stretched by a, horizontally stretched by k, and shifted vertically by c. As t \to \infty, e^{-kt} \to 0 so F(t) \to c, giving the correct long-term behavior. The function increases at a decreasing rate (concave down) because -ae^{-kt} is increasing toward 0, and the rate of increase slows as the exponential term shrinks.
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- Since F(t) \to c = 71 as t \to \infty, we have c = 71. From F(0) = 71 - a = 34.2, we get a = 36.8. Using F(20) = 41.7: 71 - 36.8e^{-20k} = 41.7, so 36.8e^{-20k} = 29.3 and k = -\dfrac{1}{20}\ln\!\left(\dfrac{29.3}{36.8}\right) \approx 0.01140.
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- Setting F(t) = 60: 71 - 36.8e^{-kt} = 60, so 36.8e^{-kt} = 11 and t = -\dfrac{1}{k}\ln\!\left(\dfrac{11}{36.8}\right) \approx 106 minutes.
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- A popular cruise ship sets sail in the Gulf of Mexico with 5000 passengers and crew on board. Unfortunately, a five family members who board the ship are carrying a highly contagious virus. After interacting with many other passengers in the first few hours of the cruise, all five of them get very sick.
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- Let S(t) be the number of people who have acquired the virus t days after the ship has left port. It turns out that a logistic function is a good model for S, and thus we assume that
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- S(t) = \frac{A}{1 + Me^{-kt}}
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- for some positive constants A, M, and k. Suppose that after 1 day, 20 people have gotten the virus.
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- Recall we know that S(0) = 5 and S(1) = 20. In addition, assume that 5000 is the number of people who will eventually get sick. Use this information determine the exact values of A, M, and k in the logistic model.
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- How many days will it take for 4000 of the people on the cruise ship to have acquired the virus?
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- Compute the average rate of change of S on the intervals [1,2], [3,4], and [5,7]. What is the meaning of each of these values (with units) in the context of the question, and what trend(s) do you observe in these average rates of change?
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- Exercise Answer
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- Since all 5000 passengers will eventually get sick, A = 5000. From S(0) = \frac{5000}{1+M} = 5, we get M = 999. Using S(1) = 20: \frac{5000}{1 + 999e^{-k}} = 20, so 999e^{-k} = 249 and k = \ln\!\left(\frac{999}{249}\right) \approx 1.389.
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- Setting S(t) = 4000: \frac{5000}{1 + 999e^{-1.389t}} = 4000, so 999e^{-1.389t} = 0.25 and t = -\dfrac{\ln(0.25/999)}{1.389} \approx 5.97 days.
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- The average rates of change are approximately AV_{[1,2]} \approx 59.3, AV_{[3,4]} \approx 726.1, and AV_{[5,7]} \approx 1084.3 people per day. These represent the average number of people per day acquiring the virus on each time interval. The rate of spread is increasing through the first seven days, but must eventually decrease since the total number who can be infected is bounded by 5000.
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- A closed tank with an inflow and outflow contains a 100 liters of saltwater solution. Let the amount of salt in the tank at time t (in minutes) be given by the function A(t), whose output is measured in grams. At time t = 0 there is an initial amount of salt present in the tank, and the inflow line also carries a saltwater mixture to the tank at a fixed rate; the outflow occurs at the same rate and carries a perfectly mixed solution out of the tank. Because of these conditions, the volume of solution in the tank stays fixed over time, but the amount of salt possibly changes.
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- It turns out that the problem of determining the amount of salt in the tank at time t is similar to the problem of determining the temperature of a warming or cooling object, and that the function A(t) has form
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- A(t) = ae^{-kt} + c
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- for constants a, c, and k. Suppose that for a particular set of conditions, we know that
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- A(t) = -500e^{-0.25t} + 750
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- Again, A(t) measures the amount of salt in the tank after t minutes.
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- How much salt is in the tank initially?
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- In the long run, how much salt do we expect to eventually be in the tank?
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- At what exact time are there exactly 500 grams of salt present in the tank?
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- Can you determine the concentration of the solution that is being delivered by the inflow to the tank? If yes, explain why and determine this value. If not, explain why that information cannot be found without additional data.
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- Exercise Answer
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- At t = 0: A(0) = -500e^{0} + 750 = -500 + 750 = 250 grams of salt.
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- As t \to \infty, e^{-0.25t} \to 0, so A(t) \to 750 grams.
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- Setting A(t) = 500: -500e^{-0.25t} + 750 = 500, so e^{-0.25t} = \frac{1}{2} and t = -\dfrac{\ln(1/2)}{0.25} = 4\ln 2 \approx 2.773 minutes.
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- Yes. In the long run, the tank contains 750 grams of salt in 100 liters of solution, giving a concentration of \frac{750}{100} = 7.5 grams per liter. Since the salt concentration in the tank approaches that of the inflow over time, the inflow concentration is 7.5 grams per liter.
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In Exercise below, use the following structure/formula for N(t): N(t)=\frac{L}{1+Ab^{-kt}}. In particular, note that when the instructions say find A, this use of A is not in reference to carrying capacity.
+
+
+
+
+
+
+
+
+
+ A glass filled with ice and water is set on a table in a climate-controlled room with constant temperature of 71^\circ Fahrenheit. A temperature probe is placed in the glass, and we find that the following temperatures are recorded (at time t in minutes).
+
+ Make a rough sketch of how you think the temperature graph should appear. Is the temperature function always increasing? always decreasing? always concave up? always concave down? what's its long-range behavior?
+
+
+
+
+ By describing F as a transformation of e^t, explain why a function of form F(t) = c - ae^{-kt}, where a, c, and k are positive constants is an appropriate model for how we expect the temperature function to behave.
+
+
+
+
+ Use the given information to determine the exact values of a, c, and k in the model F(t) = c - ae^{-kt}.
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+
+
+
+ Determine the exact time when the water's temperature is 60^\circ.
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+
+
+
+
+
+
+ Exercise Answer
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+
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+ The temperature is always increasing (the ice water warms toward room temperature), concave down, and approaches 71^\circF asymptotically in the long run.
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+
+
+
+ The function F(t) = c - ae^{-kt} is a transformation of e^t that is reflected over the horizontal axis, vertically stretched by a, horizontally stretched by k, and shifted vertically by c. As t \to \infty, e^{-kt} \to 0 so F(t) \to c, giving the correct long-term behavior. The function increases at a decreasing rate (concave down) because -ae^{-kt} is increasing toward 0, and the rate of increase slows as the exponential term shrinks.
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+ Since F(t) \to c = 71 as t \to \infty, we have c = 71. From F(0) = 71 - a = 34.2, we get a = 36.8. Using F(20) = 41.7: 71 - 36.8e^{-20k} = 41.7, so 36.8e^{-20k} = 29.3 and k = -\dfrac{1}{20}\ln\!\left(\dfrac{29.3}{36.8}\right) \approx 0.01140.
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+ Setting F(t) = 60: 71 - 36.8e^{-kt} = 60, so 36.8e^{-kt} = 11 and t = -\dfrac{1}{k}\ln\!\left(\dfrac{11}{36.8}\right) \approx 106 minutes.
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+
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+
+
+ A popular cruise ship sets sail in the Gulf of Mexico with 5000 passengers and crew on board. Unfortunately, a five family members who board the ship are carrying a highly contagious virus. After interacting with many other passengers in the first few hours of the cruise, all five of them get very sick.
+
+
+
+ Let S(t) be the number of people who have acquired the virus t days after the ship has left port. It turns out that a logistic function is a good model for S, and thus we assume that
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+ S(t) = \frac{A}{1 + Me^{-kt}}
+
+ for some positive constants A, M, and k. Suppose that after 1 day, 20 people have gotten the virus.
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+
+
+
+
+
+ Recall we know that S(0) = 5 and S(1) = 20. In addition, assume that 5000 is the number of people who will eventually get sick. Use this information determine the exact values of A, M, and k in the logistic model.
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+
+
+
+ How many days will it take for 4000 of the people on the cruise ship to have acquired the virus?
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+
+
+
+ Compute the average rate of change of S on the intervals [1,2], [3,4], and [5,7]. What is the meaning of each of these values (with units) in the context of the question, and what trend(s) do you observe in these average rates of change?
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+
+
+
+
+
+
+ Exercise Answer
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+ Since all 5000 passengers will eventually get sick, A = 5000. From S(0) = \frac{5000}{1+M} = 5, we get M = 999. Using S(1) = 20: \frac{5000}{1 + 999e^{-k}} = 20, so 999e^{-k} = 249 and k = \ln\!\left(\frac{999}{249}\right) \approx 1.389.
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+ Setting S(t) = 4000: \frac{5000}{1 + 999e^{-1.389t}} = 4000, so 999e^{-1.389t} = 0.25 and t = -\dfrac{\ln(0.25/999)}{1.389} \approx 5.97 days.
+
+
+
+
+ The average rates of change are approximately AV_{[1,2]} \approx 59.3, AV_{[3,4]} \approx 726.1, and AV_{[5,7]} \approx 1084.3 people per day. These represent the average number of people per day acquiring the virus on each time interval. The rate of spread is increasing through the first seven days, but must eventually decrease since the total number who can be infected is bounded by 5000.
+
+
+
+
+
+
+
+
+
+
+ A closed tank with an inflow and outflow contains a 100 liters of saltwater solution. Let the amount of salt in the tank at time t (in minutes) be given by the function A(t), whose output is measured in grams. At time t = 0 there is an initial amount of salt present in the tank, and the inflow line also carries a saltwater mixture to the tank at a fixed rate; the outflow occurs at the same rate and carries a perfectly mixed solution out of the tank. Because of these conditions, the volume of solution in the tank stays fixed over time, but the amount of salt possibly changes.
+
+
+
+ It turns out that the problem of determining the amount of salt in the tank at time t is similar to the problem of determining the temperature of a warming or cooling object, and that the function A(t) has form
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+ A(t) = ae^{-kt} + c
+
+ for constants a, c, and k. Suppose that for a particular set of conditions, we know that
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+ A(t) = -500e^{-0.25t} + 750
+ .
+ Again, A(t) measures the amount of salt in the tank after t minutes.
+
+
+
+
+
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+ How much salt is in the tank initially?
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+
+
+
+ In the long run, how much salt do we expect to eventually be in the tank?
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+
+
+ At what exact time are there exactly 500 grams of salt present in the tank?
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+
+
+
+ Can you determine the concentration of the solution that is being delivered by the inflow to the tank? If yes, explain why and determine this value. If not, explain why that information cannot be found without additional data.
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+
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+ Exercise Answer
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+ At t = 0: A(0) = -500e^{0} + 750 = -500 + 750 = 250 grams of salt.
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+ As t \to \infty, e^{-0.25t} \to 0, so A(t) \to 750 grams.
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+ Setting A(t) = 500: -500e^{-0.25t} + 750 = 500, so e^{-0.25t} = \frac{1}{2} and t = -\dfrac{\ln(1/2)}{0.25} = 4\ln 2 \approx 2.773 minutes.
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+
+
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+ Yes. In the long run, the tank contains 750 grams of salt in 100 liters of solution, giving a concentration of \frac{750}{100} = 7.5 grams per liter. Since the salt concentration in the tank approaches that of the inflow over time, the inflow concentration is 7.5 grams per liter.
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- We've observed that several different familiar functions grow without bound as x \to \infty, including f(x) = \ln(x), g(x) = x^2, and h(x) = e^x. In this exercise, we compare and contrast how these three functions grow.
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- Use a computational device to compute decimal expressions for f(10), g(10), and h(10), as well as f(100), g(100), and h(100). What do you observe?
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- For each of f, g, and h, how large an input is needed in order to ensure that the function's output value is at least 10^{10}? What do these values tell us about how each function grows?
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- Consider the new function r(x) = \frac{g(x)}{h(x)} = \frac{x^2}{e^x}. Compute r(10), r(100), and r(1000). What do the results suggest about the long-range behavior of r? What is surprising about this, in light of the fact that both x^2 and e^x grow without bound?
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- Exercise Answer
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- f(10) = \ln(10) \approx 2.303, g(10) = 100, h(10) = e^{10} \approx 22026;
- f(100) \approx 4.605, g(100) = 10000, h(100) \approx 2.69 \times 10^{43}.
- The exponential h grows dramatically faster than the polynomial g, which in turn grows faster than the logarithm f.
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- f(x) = \ln(x) \ge 10^{10} requires x \ge e^{10^{10}} \approx 10^{4.3 \times 10^9} (an astronomically large value).
- g(x) = x^2 \ge 10^{10} requires x \ge 10^5 = 100000.
- h(x) = e^x \ge 10^{10} requires x \ge 10\ln(10) \approx 23.03.
- The exponential reaches 10^{10} at a tiny input compared to the others, confirming that e^x grows fastest.
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- r(10) = \frac{100}{e^{10}} \approx 0.00454, r(100) \approx 3.7 \times 10^{-40}, r(1000) \approx 10^{-432}.
- Despite both x^2 and e^x growing without bound, r(x) \to 0 because the exponential grows far faster than the polynomial, so the ratio shrinks toward zero.
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- Consider the familiar graph of f(x) = \frac{1}{x}, which has a vertical asypmtote at x = 0 and a horizontal asymptote at y = 0, as pictured in Figure. In addition, consider the similarly-shaped function g shown in Figure, which has vertical asymptote x = -1 and horizontal asymptote y = -2.
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A plot of y = f(x) = \frac{1}{x}.
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A plot of y = f(x) = \frac{1}{x}.
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A plot of a related function y = g(x).
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A plot of a related function y = g(x).
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- How can we view g as a transformation of f? Explain, and state how g can be expressed algebraically in terms of f.
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- Find a formula for g as a function of x. What is the domain of g?
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- Explain algebraically (using the form of g from (b)) why \lim_{x \to \infty} g(x) = -2 and \lim_{x \to -1^+} g(x) = \infty.
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- What if a function h (again of a similar shape as f) has vertical asymptote x = 5 and horizontal asymtote y = 10. What is a possible formula for h(x)?
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- Suppose that r(x) = \frac{1}{x+35} - 27. Without using a graphing utility, how do you expect the graph of r to appear? Does it have a horizontal asymptote? A vertical asymptote? What is its domain?
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- Exercise Answer
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- g is f shifted 1 unit to the left and 2 units down: g(x) = f(x+1) - 2.
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- g(x) = \dfrac{1}{x+1} - 2. The domain is all real numbers except x = -1.
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- As x \to \infty, \frac{1}{x+1} \to 0, so g(x) \to -2. As x \to -1^+, x+1 \to 0^+ so \frac{1}{x+1} \to +\infty, giving g(x) \to +\infty.
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- A possible formula is h(x) = \dfrac{1}{x-5} + 10.
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-
- r(x) = \frac{1}{x+35} - 27 has vertical asymptote x = -35, horizontal asymptote y = -27, and domain all real numbers except x = -35. The graph looks like y = 1/x shifted left 35 units and down 27 units.
-
-
-
-
-
-
-
-
-
-
-
-
- Power functions can have powers that are not whole numbers. For instance, we can consider such functions as
- f(x)=x^{2.4}, g(x)=x^{2.5}, and h(x)=x^{2.6}.
-
-
-
-
-
-
- Compare and contrast the graphs of f, g, and h. How are they similar? How are they different? (There is a lot you can discuss here.)
-
-
-
-
- Observe that we can think of f(x) = x^{2.4} as f(x) = x^{24/10} = x^{12/5}. In addition, recall by exponent rules that we can also view f as having the form f(x) = \sqrt[5]{x^{12}}. Write g and h in similar forms, and explain why g has a different domain than f and h.
-
-
-
-
- How do the graphs of f, g, and h compare to the graphs of y = x^2 and y = x^3? Why are these natural functions to use for comparison?
-
-
-
-
- Explore similar questions for the graphs of p(x) = x^{-2.4}, q(x) = x^{-2.5}, and r(x) = x^{-2.6}.
-
-
-
-
-
-
-
- Exercise Answer
-
-
-
-
-
-
-
- All three functions are positive for x \gt 0 and grow roughly like x^2. They differ for x \lt 0: f(x) = x^{2.4} is even-symmetric (defined and positive for all x); g(x) = x^{2.5} is only defined for x \ge 0; h(x) = x^{2.6} is odd-symmetric (defined for all x, negative for x \lt 0).
-
-
-
-
- g(x) = x^{5/2} = \sqrt{x^5}, defined only for x \ge 0 since square roots are undefined for negative inputs. h(x) = x^{13/5} = \sqrt[5]{x^{13}}, defined for all real x. g has a different domain (only x \ge 0) because it involves an even root.
-
-
-
-
- f and h resemble x^2 and x^3 respectively (since 2.4 and 2.6 are close to those integers). These are natural comparisons because they represent the even and odd symmetry templates for power functions with exponents near 2.4 and 2.6.
-
-
-
-
- p(x) = x^{-2.4} = \sqrt[5]{x^{-12}}, defined for all x \ne 0, even-symmetric, approaching 0 as x \to \pm\infty. q(x) = x^{-2.5} = \frac{1}{\sqrt{x^5}}, defined only for x \gt 0. r(x) = x^{-2.6} = \sqrt[5]{x^{-13}}, defined for all x \ne 0, odd-symmetric. All approach 0 as x \to \pm\infty.
-
+ We've observed that several different familiar functions grow without bound as x \to \infty, including f(x) = \ln(x), g(x) = x^2, and h(x) = e^x. In this exercise, we compare and contrast how these three functions grow.
+
+
+
+
+
+
+ Use a computational device to compute decimal expressions for f(10), g(10), and h(10), as well as f(100), g(100), and h(100). What do you observe?
+
+
+
+
+ For each of f, g, and h, how large an input is needed in order to ensure that the function's output value is at least 10^{10}? What do these values tell us about how each function grows?
+
+
+
+
+ Consider the new function r(x) = \frac{g(x)}{h(x)} = \frac{x^2}{e^x}. Compute r(10), r(100), and r(1000). What do the results suggest about the long-range behavior of r? What is surprising about this, in light of the fact that both x^2 and e^x grow without bound?
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
+
+
+
+ f(10) = \ln(10) \approx 2.303, g(10) = 100, h(10) = e^{10} \approx 22026;
+ f(100) \approx 4.605, g(100) = 10000, h(100) \approx 2.69 \times 10^{43}.
+ The exponential h grows dramatically faster than the polynomial g, which in turn grows faster than the logarithm f.
+
+
+
+
+ f(x) = \ln(x) \ge 10^{10} requires x \ge e^{10^{10}} \approx 10^{4.3 \times 10^9} (an astronomically large value).
+ g(x) = x^2 \ge 10^{10} requires x \ge 10^5 = 100000.
+ h(x) = e^x \ge 10^{10} requires x \ge 10\ln(10) \approx 23.03.
+ The exponential reaches 10^{10} at a tiny input compared to the others, confirming that e^x grows fastest.
+
+
+
+
+ r(10) = \frac{100}{e^{10}} \approx 0.00454, r(100) \approx 3.7 \times 10^{-40}, r(1000) \approx 10^{-432}.
+ Despite both x^2 and e^x growing without bound, r(x) \to 0 because the exponential grows far faster than the polynomial, so the ratio shrinks toward zero.
+
+
+
+
+
+
+
+
+
+
+ Consider the familiar graph of f(x) = \frac{1}{x}, which has a vertical asypmtote at x = 0 and a horizontal asymptote at y = 0, as pictured in Figure. In addition, consider the similarly-shaped function g shown in Figure, which has vertical asymptote x = -1 and horizontal asymptote y = -2.
+
+
+
+
+
A plot of y = f(x) = \frac{1}{x}.
+
A plot of y = f(x) = \frac{1}{x}.
+
+
+
+
A plot of a related function y = g(x).
+
A plot of a related function y = g(x).
+
+
+
+
+
+
+
+
+ How can we view g as a transformation of f? Explain, and state how g can be expressed algebraically in terms of f.
+
+
+
+
+ Find a formula for g as a function of x. What is the domain of g?
+
+
+
+
+ Explain algebraically (using the form of g from (b)) why \lim_{x \to \infty} g(x) = -2 and \lim_{x \to -1^+} g(x) = \infty.
+
+
+
+
+ What if a function h (again of a similar shape as f) has vertical asymptote x = 5 and horizontal asymtote y = 10. What is a possible formula for h(x)?
+
+
+
+
+ Suppose that r(x) = \frac{1}{x+35} - 27. Without using a graphing utility, how do you expect the graph of r to appear? Does it have a horizontal asymptote? A vertical asymptote? What is its domain?
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
+
+
+
+ g is f shifted 1 unit to the left and 2 units down: g(x) = f(x+1) - 2.
+
+
+
+
+ g(x) = \dfrac{1}{x+1} - 2. The domain is all real numbers except x = -1.
+
+
+
+
+ As x \to \infty, \frac{1}{x+1} \to 0, so g(x) \to -2. As x \to -1^+, x+1 \to 0^+ so \frac{1}{x+1} \to +\infty, giving g(x) \to +\infty.
+
+
+
+
+ A possible formula is h(x) = \dfrac{1}{x-5} + 10.
+
+
+
+
+ r(x) = \frac{1}{x+35} - 27 has vertical asymptote x = -35, horizontal asymptote y = -27, and domain all real numbers except x = -35. The graph looks like y = 1/x shifted left 35 units and down 27 units.
+
+
+
+
+
+
+
+
+
+
+
+
+ Power functions can have powers that are not whole numbers. For instance, we can consider such functions as
+ f(x)=x^{2.4}, g(x)=x^{2.5}, and h(x)=x^{2.6}.
+
+
+
+
+
+
+ Compare and contrast the graphs of f, g, and h. How are they similar? How are they different? (There is a lot you can discuss here.)
+
+
+
+
+ Observe that we can think of f(x) = x^{2.4} as f(x) = x^{24/10} = x^{12/5}. In addition, recall by exponent rules that we can also view f as having the form f(x) = \sqrt[5]{x^{12}}. Write g and h in similar forms, and explain why g has a different domain than f and h.
+
+
+
+
+ How do the graphs of f, g, and h compare to the graphs of y = x^2 and y = x^3? Why are these natural functions to use for comparison?
+
+
+
+
+ Explore similar questions for the graphs of p(x) = x^{-2.4}, q(x) = x^{-2.5}, and r(x) = x^{-2.6}.
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
+
+
+
+ All three functions are positive for x \gt 0 and grow roughly like x^2. They differ for x \lt 0: f(x) = x^{2.4} is even-symmetric (defined and positive for all x); g(x) = x^{2.5} is only defined for x \ge 0; h(x) = x^{2.6} is odd-symmetric (defined for all x, negative for x \lt 0).
+
+
+
+
+ g(x) = x^{5/2} = \sqrt{x^5}, defined only for x \ge 0 since square roots are undefined for negative inputs. h(x) = x^{13/5} = \sqrt[5]{x^{13}}, defined for all real x. g has a different domain (only x \ge 0) because it involves an even root.
+
+
+
+
+ f and h resemble x^2 and x^3 respectively (since 2.4 and 2.6 are close to those integers). These are natural comparisons because they represent the even and odd symmetry templates for power functions with exponents near 2.4 and 2.6.
+
+
+
+
+ p(x) = x^{-2.4} = \sqrt[5]{x^{-12}}, defined for all x \ne 0, even-symmetric, approaching 0 as x \to \pm\infty. q(x) = x^{-2.5} = \frac{1}{\sqrt{x^5}}, defined only for x \gt 0. r(x) = x^{-2.6} = \sqrt[5]{x^{-13}}, defined for all x \ne 0, odd-symmetric. All approach 0 as x \to \pm\infty.
+
- An open triangular trough, as pictured in Figure is being constructed from aluminum. The trough is to have equilateral triangular ends of side length s and a length of l. We want the trough to used a fixed 100 square feet of aluminum.
-
-
-
-
A triangular trough.
-
A triangular trough.
-
-
-
-
-
-
-
- What is the area of one of the equilateral triangle ends as a function of s?
-
-
-
-
- Recall that for an object with constant cross-sectional area, its volume is the area of one of those cross-sections times its height (or length). Hence determine a formula for the volume of the trough that depends on s and l.
-
-
-
-
- Find a formula involving s and l for the surface area of the trough.
-
-
-
-
- Use the constraint that we have 100 square feet of available aluminum to generate an equation that connects s and l and hence solve for l in terms of s.
-
-
-
-
- Use your work in (d) and (b) to express the volume of the trough, V, as a function of s only.
-
-
-
-
- What is the domain of the function V in the context of the situation being modeled? Why?
-
-
-
-
-
-
-
- Exercise Answer
-
-
-
-
-
-
-
- The area of an equilateral triangle with side length s is A_{\triangle} = \dfrac{\sqrt{3}}{4}s^2.
-
-
-
-
- The volume of the trough is cross-sectional area times length: V = \dfrac{\sqrt{3}}{4}s^2 l.
-
-
-
-
- The trough has two equilateral triangular ends and two rectangular side panels (the open top means there is no rectangular top panel). The surface area is
-
- A = 2 \cdot \frac{\sqrt{3}}{4}s^2 + 2sl = \frac{\sqrt{3}}{2}s^2 + 2sl.
-
-
-
-
-
- Setting the surface area equal to 100: \dfrac{\sqrt{3}}{2}s^2 + 2sl = 100, so
-
- l = \frac{100 - \frac{\sqrt{3}}{2}s^2}{2s} = \frac{50}{s} - \frac{\sqrt{3}}{4}s.
-
-
- We need s > 0 and l > 0. From l = \dfrac{50}{s} - \dfrac{\sqrt{3}}{4}s > 0, we get s^2 < \dfrac{200}{\sqrt{3}} = \dfrac{200\sqrt{3}}{3}, so s < \sqrt{\dfrac{200\sqrt{3}}{3}}. The domain is \left(0,\ \sqrt{\dfrac{200\sqrt{3}}{3}}\right).
-
-
-
-
-
-
-
-
-
-
- A rectangular box is being constructed so that its base is twice as long as it is wide. In addition, the base and top of the box cost $2 per square foot while the sides cost $1.50 per square foot. If we only want to spend $10 on materials for the box, how can we write the box's volume as a function of a single variable? What is the domain of this volume function? (Hint: first find the box's surface area in terms of two variables, and then find an expression for the cost of the box in terms of those same variables. Use the fact that cost is constrained to solve for one variable in terms of another.)
-
-
-
-
- Exercise Answer
-
-
-
-
- Let the box have width w, length 2w, and height h. The areas of the base and top are each 2w^2, and the combined area of the four sides is 2(wh) + 2(2wh) = 6wh. The cost constraint is
-
- 2(2w^2 + 2w^2) + 1.5(6wh) = 8w^2 + 9wh = 10.
-
- Solving for h: h = \dfrac{10 - 8w^2}{9w}. The volume is
-
- V(w) = w \cdot 2w \cdot h = 2w^2 \cdot \frac{10 - 8w^2}{9w} = \frac{20w - 16w^3}{9}.
-
- For the domain, we need h > 0, so 10 - 8w^2 > 0, giving w < \sqrt{\dfrac{10}{8}} = \dfrac{\sqrt{5}}{2}. With w > 0, the domain is \left(0,\ \dfrac{\sqrt{5}}{2}\right).
-
-
-
-
-
-
-
- Suppose that we want a cylindrical barrel to hold 8 cubic feet of volume. Let the barrel have radius r and height h, each measured in feet. How can we write the surface area, A, of the barrel solely as a function of r?
-
-
-
-
-
-
- Draw several possible pictures of how the barrel might look. For instance, what if the radius is very small? How will the height appear in comparison? Likewise, what happens if the height is very small?
-
-
-
-
- Use the fact that volume is fixed at 8 cubic feet to state a constraint equation and solve that equation for h in terms of r.
-
-
-
-
- Recall that the surface area of a cylinder is A = 2\pi r^2 + 2\pi rh. Use your work in (c) to write A as a function of only r.
-
-
-
-
- What is the domain of A? Why?
-
-
-
-
- Explain why A is not a polynomial function of r.
-
-
-
-
-
-
-
- Exercise Answer
-
-
-
-
-
-
-
- If the radius is very small, the cylinder must be very tall to hold 8 cubic feet; if the radius is large, the cylinder is short and flat.
-
-
-
-
- The volume constraint \pi r^2 h = 8 gives h = \dfrac{8}{\pi r^2}.
-
-
-
-
- Substituting into the surface area formula:
-
- A(r) = 2\pi r^2 + 2\pi r \cdot \frac{8}{\pi r^2} = 2\pi r^2 + \frac{16}{r}.
-
-
-
-
-
- The domain is r > 0, since the radius must be positive.
-
-
-
-
- A(r) = 2\pi r^2 + \dfrac{16}{r} is not a polynomial because of the term \dfrac{16}{r} = 16r^{-1}, which has a negative exponent. Polynomial functions have only non-negative integer exponents.
-
+ An open triangular trough, as pictured in Figure is being constructed from aluminum. The trough is to have equilateral triangular ends of side length s and a length of l. We want the trough to used a fixed 100 square feet of aluminum.
+
+
+
+
A triangular trough.
+
A triangular trough.
+
+
+
+
+
+
+
+ What is the area of one of the equilateral triangle ends as a function of s?
+
+
+
+
+ Recall that for an object with constant cross-sectional area, its volume is the area of one of those cross-sections times its height (or length). Hence determine a formula for the volume of the trough that depends on s and l.
+
+
+
+
+ Find a formula involving s and l for the surface area of the trough.
+
+
+
+
+ Use the constraint that we have 100 square feet of available aluminum to generate an equation that connects s and l and hence solve for l in terms of s.
+
+
+
+
+ Use your work in (d) and (b) to express the volume of the trough, V, as a function of s only.
+
+
+
+
+ What is the domain of the function V in the context of the situation being modeled? Why?
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
+
+
+
+ The area of an equilateral triangle with side length s is A_{\triangle} = \dfrac{\sqrt{3}}{4}s^2.
+
+
+
+
+ The volume of the trough is cross-sectional area times length: V = \dfrac{\sqrt{3}}{4}s^2 l.
+
+
+
+
+ The trough has two equilateral triangular ends and two rectangular side panels (the open top means there is no rectangular top panel). The surface area is
+
+ A = 2 \cdot \frac{\sqrt{3}}{4}s^2 + 2sl = \frac{\sqrt{3}}{2}s^2 + 2sl.
+
+
+
+
+
+ Setting the surface area equal to 100: \dfrac{\sqrt{3}}{2}s^2 + 2sl = 100, so
+
+ l = \frac{100 - \frac{\sqrt{3}}{2}s^2}{2s} = \frac{50}{s} - \frac{\sqrt{3}}{4}s.
+
+
+ We need s > 0 and l > 0. From l = \dfrac{50}{s} - \dfrac{\sqrt{3}}{4}s > 0, we get s^2 < \dfrac{200}{\sqrt{3}} = \dfrac{200\sqrt{3}}{3}, so s < \sqrt{\dfrac{200\sqrt{3}}{3}}. The domain is \left(0,\ \sqrt{\dfrac{200\sqrt{3}}{3}}\right).
+
+
+
+
+
+
+
+
+
+
+ A rectangular box is being constructed so that its base is twice as long as it is wide. In addition, the base and top of the box cost $2 per square foot while the sides cost $1.50 per square foot. If we only want to spend $10 on materials for the box, how can we write the box's volume as a function of a single variable? What is the domain of this volume function? (Hint: first find the box's surface area in terms of two variables, and then find an expression for the cost of the box in terms of those same variables. Use the fact that cost is constrained to solve for one variable in terms of another.)
+
+
+
+
+ Exercise Answer
+
+
+
+
+ Let the box have width w, length 2w, and height h. The areas of the base and top are each 2w^2, and the combined area of the four sides is 2(wh) + 2(2wh) = 6wh. The cost constraint is
+
+ 2(2w^2 + 2w^2) + 1.5(6wh) = 8w^2 + 9wh = 10.
+
+ Solving for h: h = \dfrac{10 - 8w^2}{9w}. The volume is
+
+ V(w) = w \cdot 2w \cdot h = 2w^2 \cdot \frac{10 - 8w^2}{9w} = \frac{20w - 16w^3}{9}.
+
+ For the domain, we need h > 0, so 10 - 8w^2 > 0, giving w < \sqrt{\dfrac{10}{8}} = \dfrac{\sqrt{5}}{2}. With w > 0, the domain is \left(0,\ \dfrac{\sqrt{5}}{2}\right).
+
+
+
+
+
+
+
+ Suppose that we want a cylindrical barrel to hold 8 cubic feet of volume. Let the barrel have radius r and height h, each measured in feet. How can we write the surface area, A, of the barrel solely as a function of r?
+
+
+
+
+
+
+ Draw several possible pictures of how the barrel might look. For instance, what if the radius is very small? How will the height appear in comparison? Likewise, what happens if the height is very small?
+
+
+
+
+ Use the fact that volume is fixed at 8 cubic feet to state a constraint equation and solve that equation for h in terms of r.
+
+
+
+
+ Recall that the surface area of a cylinder is A = 2\pi r^2 + 2\pi rh. Use your work in (c) to write A as a function of only r.
+
+
+
+
+ What is the domain of A? Why?
+
+
+
+
+ Explain why A is not a polynomial function of r.
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
+
+
+
+ If the radius is very small, the cylinder must be very tall to hold 8 cubic feet; if the radius is large, the cylinder is short and flat.
+
+
+
+
+ The volume constraint \pi r^2 h = 8 gives h = \dfrac{8}{\pi r^2}.
+
+
+
+
+ Substituting into the surface area formula:
+
+ A(r) = 2\pi r^2 + 2\pi r \cdot \frac{8}{\pi r^2} = 2\pi r^2 + \frac{16}{r}.
+
+
+
+
+
+ The domain is r > 0, since the radius must be positive.
+
+
+
+
+ A(r) = 2\pi r^2 + \dfrac{16}{r} is not a polynomial because of the term \dfrac{16}{r} = 16r^{-1}, which has a negative exponent. Polynomial functions have only non-negative integer exponents.
+
- Consider the polynomial function given by
-
- p(x) = 0.0005(x+21.7)^3 (x-20.9)^2 (x-31.4)(x^2+100)
- .
-
-
-
-
-
-
- What is the degree of p?
-
-
-
-
-
- What are the real zeros of p?
- State them with multiplicity.
-
-
-
-
-
- Construct a carefully labeled sign chart for p(x).
-
-
-
-
-
- Plot the function p in Desmos.
- Are the zeros obvious from the graph?
- How do you have to adjust the window in order to tell?
- Even in an adjusted window, can you tell them exactly from the graph?
-
-
-
-
-
- Now consider the related but different polynomial
-
- q(x) = -0.0005(x+21.7)^3 (x-20.9)^2 (x-31.4)(x^2+100)(x-92.3)
- .
- What is the degree of q?
- What are the zeros of q?
- What is obvious from its graph and what is not?
-
-
-
-
-
-
-
- Exercise Answer
-
-
-
-
-
-
-
- The degree of p is 3+2+1+2 = 8.
-
-
-
-
- The real zeros are x=-21.7 (multiplicity 3), x=20.9 (multiplicity 2), and x=31.4 (multiplicity 1). (The factor x^2+100 has no real zeros.)
-
-
-
-
- The sign of p is determined by (x+21.7)^3 and (x-31.4) (since the other factors are always non-negative and (x-20.9)^2 doesn't change sign): p \gt 0 for x \lt -21.7; p \lt 0 for -21.7 \lt x \lt 31.4 (with p=0 at x=20.9, a bounce); p \gt 0 for x \gt 31.4.
-
-
-
-
- The zeros span a large range (-21.7 to 31.4), so a viewing window such as -40 \le x \le 40 and -3 \times 10^8 \le y \le 2 \times 10^8 is needed. Even then, the bounce at x=20.9 is subtle.
-
-
-
-
- q(x) = -0.0005(x+21.7)^3(x-20.9)^2(x-31.4)(x^2+100)(x-92.3) has degree 9. Its real zeros are x=-21.7 (mult 3), x=20.9 (mult 2), x=31.4 (mult 1), and x=92.3 (mult 1). With such widely-spaced zeros, no single window shows all features clearly.
-
-
-
-
-
-
-
-
-
-
- Consider the (non-polynomial) function r(x) = e^{-x^2}(x^2+1)(x-2)(x-3).
-
-
-
-
-
-
- What are the zeros of r(x)? (Hint:
- is e^{\Box} ever equal to zero?)
-
-
-
-
-
- Construct a sign chart for r(x).
-
-
-
-
-
- Plot r(x) in Desmos.
- Is the sign and overall behavior of r obvious from the plot?
- Why or why not?
-
-
-
-
-
- From the graph,
- what appears to be the value of \lim_{x \to \infty} r(x)?
- Why is this surprising in light of the behavior of
- f(x)=(x^2+1)(x-2)(x-3) as x \to \infty?
-
-
-
-
-
-
-
- Exercise Answer
-
-
-
-
-
-
-
- Since e^{-x^2} \gt 0 always and x^2+1 \gt 0 always, the only real zeros come from (x-2)(x-3). The real zeros are x = 2 and x = 3.
-
-
-
-
- The sign of r equals the sign of (x-2)(x-3): positive for x \lt 2, negative for 2 \lt x \lt 3, positive for x \gt 3.
-
-
-
-
- In Desmos, the zeros at x=2 and x=3 are visible, but the graph decays very rapidly to 0 for large |x| due to the e^{-x^2} factor.
-
-
-
-
- \lim_{x \to \infty} r(x) = 0. This is surprising because (x^2+1)(x-2)(x-3) \to \infty, but the factor e^{-x^2} decays to zero much faster than any polynomial grows, so the product is driven to zero.
-
-
-
-
-
-
-
-
-
-
- In each following question, find a formula for a polynomial with certain properties, generate a plot that demonstrates you’ve found a function with the given specifications, and write several sentences to explain your thinking.
-
-
-
-
-
-
- A quadratic function q has zeros at x = −7 and x = 11 and its y-value at its vertex is 42.
-
-
-
-
- A polynomial r of degree 4 has zeros at x = −3 and x = 5, both of multiplicity 2, and the function has a y-intercept at the point (0, 28).
-
-
-
-
- A polynomial f has degree 11 and the following zeros: zeros of multiplicity 1 at x = −3 and x = 5, zeros of multiplicity 2 at x = −2 and x = 3, and a zero of multiplicity 3 at x = 1. In addition, \lim_{x \to \infty} f(x) = -\infty.
-
-
-
-
- A polynomial g has its graph given in Figure below. Determine a possible formula for g(x) where the polynomial you find has the lowest possible degree to match the graph. What is the degree of the function you find?
-
-
-
-
A polynomial function g.
-
A polynomial function g.
-
-
-
-
-
-
-
-
- Exercise Answer
-
-
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-
- q(x) = -\dfrac{14}{27}(x+7)(x-11). The vertex is at x = \frac{-7+11}{2} = 2. Setting q(2) = -\frac{14}{27}(9)(-9) = 42 confirms the vertex value.
-
-
-
-
- r(x) = \dfrac{28}{225}(x+3)^2(x-5)^2. We have r(0) = \frac{28}{225}(9)(25) = 28 ✓.
-
-
-
-
- Since the stated multiplicities sum to 1+1+2+2+3=9 but the degree must be 11, we need two additional roots (a complex conjugate pair). One example: f(x) = -(x+3)(x-5)(x+2)^2(x-3)^2(x-1)^3(x^2+1), which has degree 11 and \lim_{x \to \infty} f(x) = -\infty.
-
-
-
-
- Reading from the graph (which shows zeros at approximately x = -3, 1, 4 with the zero at x=-3 being a touch/bounce and those at x=1, 4 being crossings): one possible formula is g(x) = -\dfrac{27}{50}\!\left(\dfrac{x+3}{3}\right)^3(x-1)^2\!\left(\dfrac{x-4}{4}\right), a degree-6 polynomial.
-
-
-
-
-
-
-
-
-
-
- Like we have worked to understand families of functions that involve parameters such as
- p(t) = a\cos(k(t-b)) + c and F(t) = a + be^{-kt},
- we are often interested in polynomials that involve one or more parameters and understanding how those parameters affect the function's behavior.
-
-
-
- For example, let a \gt 0 be a positive constant,
- and consider p(x) = x^3 - a^2x.
-
-
-
-
-
-
- What is the degree of p?
-
-
-
-
- What is the long-term behavior of p? State your responses using limit notation.
-
-
-
-
- In terms of the constant a, what are the zeros of p?
-
-
-
-
- Construct a carefully labeled sign chart for p.
-
-
-
-
- How does changing the value of a affect the graph of p?
-
-
-
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-
-
- Exercise Answer
-
-
-
-
-
-
-
- p has degree 3.
-
-
-
-
- \lim_{x \to -\infty} p(x) = -\infty and \lim_{x \to \infty} p(x) = +\infty (since the leading term is x^3, positive coefficient, odd degree).
-
-
-
-
- Factor: p(x) = x(x^2-a^2) = x(x-a)(x+a). The zeros are x = -a, 0, a.
-
-
-
-
- Sign chart (with a \gt 0): p \lt 0 for x \lt -a; p \gt 0 for -a \lt x \lt 0; p \lt 0 for 0 \lt x \lt a; p \gt 0 for x \gt a.
-
-
-
-
- As a increases, the zeros at \pm a spread further apart and the local maximum and minimum values grow in magnitude. The overall shape remains an S-curve, but wider and taller.
-
+ Consider the polynomial function given by
+
+ p(x) = 0.0005(x+21.7)^3 (x-20.9)^2 (x-31.4)(x^2+100)
+ .
+
+
+
+
+
+
+ What is the degree of p?
+
+
+
+
+
+ What are the real zeros of p?
+ State them with multiplicity.
+
+
+
+
+
+ Construct a carefully labeled sign chart for p(x).
+
+
+
+
+
+ Plot the function p in Desmos.
+ Are the zeros obvious from the graph?
+ How do you have to adjust the window in order to tell?
+ Even in an adjusted window, can you tell them exactly from the graph?
+
+
+
+
+
+ Now consider the related but different polynomial
+
+ q(x) = -0.0005(x+21.7)^3 (x-20.9)^2 (x-31.4)(x^2+100)(x-92.3)
+ .
+ What is the degree of q?
+ What are the zeros of q?
+ What is obvious from its graph and what is not?
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
+
+
+
+ The degree of p is 3+2+1+2 = 8.
+
+
+
+
+ The real zeros are x=-21.7 (multiplicity 3), x=20.9 (multiplicity 2), and x=31.4 (multiplicity 1). (The factor x^2+100 has no real zeros.)
+
+
+
+
+ The sign of p is determined by (x+21.7)^3 and (x-31.4) (since the other factors are always non-negative and (x-20.9)^2 doesn't change sign): p \gt 0 for x \lt -21.7; p \lt 0 for -21.7 \lt x \lt 31.4 (with p=0 at x=20.9, a bounce); p \gt 0 for x \gt 31.4.
+
+
+
+
+ The zeros span a large range (-21.7 to 31.4), so a viewing window such as -40 \le x \le 40 and -3 \times 10^8 \le y \le 2 \times 10^8 is needed. Even then, the bounce at x=20.9 is subtle.
+
+
+
+
+ q(x) = -0.0005(x+21.7)^3(x-20.9)^2(x-31.4)(x^2+100)(x-92.3) has degree 9. Its real zeros are x=-21.7 (mult 3), x=20.9 (mult 2), x=31.4 (mult 1), and x=92.3 (mult 1). With such widely-spaced zeros, no single window shows all features clearly.
+
+
+
+
+
+
+
+
+
+
+ Consider the (non-polynomial) function r(x) = e^{-x^2}(x^2+1)(x-2)(x-3).
+
+
+
+
+
+
+ What are the zeros of r(x)? (Hint:
+ is e^{\Box} ever equal to zero?)
+
+
+
+
+
+ Construct a sign chart for r(x).
+
+
+
+
+
+ Plot r(x) in Desmos.
+ Is the sign and overall behavior of r obvious from the plot?
+ Why or why not?
+
+
+
+
+
+ From the graph,
+ what appears to be the value of \lim_{x \to \infty} r(x)?
+ Why is this surprising in light of the behavior of
+ f(x)=(x^2+1)(x-2)(x-3) as x \to \infty?
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
+
+
+
+ Since e^{-x^2} \gt 0 always and x^2+1 \gt 0 always, the only real zeros come from (x-2)(x-3). The real zeros are x = 2 and x = 3.
+
+
+
+
+ The sign of r equals the sign of (x-2)(x-3): positive for x \lt 2, negative for 2 \lt x \lt 3, positive for x \gt 3.
+
+
+
+
+ In Desmos, the zeros at x=2 and x=3 are visible, but the graph decays very rapidly to 0 for large |x| due to the e^{-x^2} factor.
+
+
+
+
+ \lim_{x \to \infty} r(x) = 0. This is surprising because (x^2+1)(x-2)(x-3) \to \infty, but the factor e^{-x^2} decays to zero much faster than any polynomial grows, so the product is driven to zero.
+
+
+
+
+
+
+
+
+
+
+ In each following question, find a formula for a polynomial with certain properties, generate a plot that demonstrates you’ve found a function with the given specifications, and write several sentences to explain your thinking.
+
+
+
+
+
+
+ A quadratic function q has zeros at x = −7 and x = 11 and its y-value at its vertex is 42.
+
+
+
+
+ A polynomial r of degree 4 has zeros at x = −3 and x = 5, both of multiplicity 2, and the function has a y-intercept at the point (0, 28).
+
+
+
+
+ A polynomial f has degree 11 and the following zeros: zeros of multiplicity 1 at x = −3 and x = 5, zeros of multiplicity 2 at x = −2 and x = 3, and a zero of multiplicity 3 at x = 1. In addition, \lim_{x \to \infty} f(x) = -\infty.
+
+
+
+
+ A polynomial g has its graph given in Figure below. Determine a possible formula for g(x) where the polynomial you find has the lowest possible degree to match the graph. What is the degree of the function you find?
+
+
+
+
A polynomial function g.
+
A polynomial function g.
+
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
+
+
+
+ q(x) = -\dfrac{14}{27}(x+7)(x-11). The vertex is at x = \frac{-7+11}{2} = 2. Setting q(2) = -\frac{14}{27}(9)(-9) = 42 confirms the vertex value.
+
+
+
+
+ r(x) = \dfrac{28}{225}(x+3)^2(x-5)^2. We have r(0) = \frac{28}{225}(9)(25) = 28 ✓.
+
+
+
+
+ Since the stated multiplicities sum to 1+1+2+2+3=9 but the degree must be 11, we need two additional roots (a complex conjugate pair). One example: f(x) = -(x+3)(x-5)(x+2)^2(x-3)^2(x-1)^3(x^2+1), which has degree 11 and \lim_{x \to \infty} f(x) = -\infty.
+
+
+
+
+ Reading from the graph (which shows zeros at approximately x = -3, 1, 4 with the zero at x=-3 being a touch/bounce and those at x=1, 4 being crossings): one possible formula is g(x) = -\dfrac{27}{50}\!\left(\dfrac{x+3}{3}\right)^3(x-1)^2\!\left(\dfrac{x-4}{4}\right), a degree-6 polynomial.
+
+
+
+
+
+
+
+
+
+
+ Like we have worked to understand families of functions that involve parameters such as
+ p(t) = a\cos(k(t-b)) + c and F(t) = a + be^{-kt},
+ we are often interested in polynomials that involve one or more parameters and understanding how those parameters affect the function's behavior.
+
+
+
+ For example, let a \gt 0 be a positive constant,
+ and consider p(x) = x^3 - a^2x.
+
+
+
+
+
+
+ What is the degree of p?
+
+
+
+
+ What is the long-term behavior of p? State your responses using limit notation.
+
+
+
+
+ In terms of the constant a, what are the zeros of p?
+
+
+
+
+ Construct a carefully labeled sign chart for p.
+
+
+
+
+ How does changing the value of a affect the graph of p?
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
+
+
+
+ p has degree 3.
+
+
+
+
+ \lim_{x \to -\infty} p(x) = -\infty and \lim_{x \to \infty} p(x) = +\infty (since the leading term is x^3, positive coefficient, odd degree).
+
+
+
+
+ Factor: p(x) = x(x^2-a^2) = x(x-a)(x+a). The zeros are x = -a, 0, a.
+
+
+
+
+ Sign chart (with a \gt 0): p \lt 0 for x \lt -a; p \gt 0 for -a \lt x \lt 0; p \lt 0 for 0 \lt x \lt a; p \gt 0 for x \gt a.
+
+
+
+
+ As a increases, the zeros at \pm a spread further apart and the local maximum and minimum values grow in magnitude. The overall shape remains an S-curve, but wider and taller.
+
- For each of the following rational functions, determine, with justification, the exact locations of all (i) horizontal asymptotes, (ii) vertical asymptotes, (iii) zeros, and (iv) holes of the function. Clearly show your work and thinking.
-
- r(x) = \dfrac{-19(x+11.3)^2(x-15.1)(x-17.3)}{41(x+5.7)(x+11.3)(x-8.4)(x-15.1)}. Both (x+11.3) and (x-15.1) appear in both numerator and denominator, so they cancel. The reduced form is \dfrac{-19(x+11.3)(x-17.3)}{41(x+5.7)(x-8.4)}.
-
-
Horizontal asymptote:y = -\dfrac{19}{41} (same degree after cancellation).
-
Vertical asymptotes:x = -5.7 and x = 8.4.
-
Zero:x = 17.3 (the factor x+11.3 in reduced numerator gives x=-11.3, but that is a hole).
-
Holes: at x = -11.3 (hole value 0, since x+11.3=0 makes reduced numerator zero) and at x = 15.1.
-
-
-
-
-
- s(x) = \dfrac{-29(x^2-16)(x^2+99)(x-53)}{101(x^2-4)(x-13)^2(x+104)}. Factor: numerator = -29(x-4)(x+4)(x^2+99)(x-53); denominator = 101(x-2)(x+2)(x-13)^2(x+104). No common factors.
-
-
Horizontal asymptote:y = -\dfrac{29}{101} (degree 5 over degree 5).
-
Vertical asymptotes:x = 2, x = -2, x = 13, x = -104.
-
Zeros:x = 4, x = -4, x = 53 (note x^2+99 > 0 always).
-
Holes: none.
-
-
-
-
-
- u(x) = \dfrac{-71(x^2-13x+36)(x-58.4)(x+78.2)}{83(x+58.4)(x-78.2)(x^2-12x+27)}. Factor: x^2-13x+36 = (x-4)(x-9) and x^2-12x+27 = (x-3)(x-9). The factor (x-9) cancels.
-
-
Horizontal asymptote:y = -\dfrac{71}{83} (degree 4 over degree 4).
-
Vertical asymptotes:x = -58.4, x = 78.2, x = 3.
-
Zeros:x = 4, x = 58.4, x = -78.2.
-
Hole: at x = 9.
-
-
-
-
-
-
-
-
-
-
-
- Find a formula for a rational function that meets the stated criteria, with justification. If no such formula is possible, explain why.
-
-
-
-
-
-
- A rational function r(x) in the form r(x) = \frac{k}{x-a} + b so that r has a horizontal asymptote of y = -\frac{3}{7}, a vertical asymptote of x = \frac{5}{2}, and r(0) = 4.
-
-
-
-
- A rational function s(x) that has no horizontal asymptote, has zeros at x = -5 and x = 3, has a single vertical asymptote at x = -1, and satisfies \lim_{x \to \infty} s(x) = -\infty and \lim_{x \to -\infty} s(x) = +\infty.
-
-
-
-
- A rational function u(x) that is positive for x \lt -4, negative for -4 \lt x \lt -2, negative for -2 \lt x \lt 1, positive for 1 \lt x \lt 5, and negative for x \gt 5. The only zeros of u are located at x = -4 and x = -2. In addition, u has a hole at x = 4.
-
-
-
-
- A rational function w(x) whose graph is shown in Figure.
-
-
-
-
A plot of the rational function w.
-
A plot of the rational function w.
-
-
-
-
-
-
-
-
- Exercise Answer
-
-
-
-
-
-
-
- We need b = -\dfrac{3}{7} (horizontal asymptote) and a = \dfrac{5}{2} (vertical asymptote). Then r(0) = \dfrac{k}{0 - \frac{5}{2}} - \dfrac{3}{7} = -\dfrac{2k}{5} - \dfrac{3}{7} = 4, so -\dfrac{2k}{5} = \dfrac{31}{7} and k = -\dfrac{155}{14}. Thus r(x) = -\dfrac{155/14}{x - 5/2} - \dfrac{3}{7}.
-
-
-
-
- We need: numerator degree greater than denominator degree (no horizontal asymptote); zeros at x=-5 and x=3; single vertical asymptote at x=-1; and the correct end behavior. The function
-
- s(x) = -\frac{(x+5)(x-3)}{x+1}
-
- satisfies all criteria: as x \to \infty, s(x) \to -\infty (since -(+)(+)/(+) \to -\infty); as x \to -\infty, s(x) \to +\infty.
-
-
-
-
- Zeros only at x=-4 and x=-2, with x=-2 having even multiplicity (no sign change), VAs at x=1 and x=5, hole at x=4. One formula:
-
- u(x) = \frac{(x+4)(x+2)^2(x-4)}{(x-1)(x-5)(x-4)}.
-
-
-
-
-
- Reading from the graph: zeros near x=-1 and x=5, vertical asymptotes near x=3 and x=8, a hole near x=10, and a horizontal asymptote near y=0. One formula consistent with the graph is
-
- w(x) = -\frac{(x+2)(x-5)(x-10)}{5(x-3)(x-8)(x-10)}.
-
-
-
-
-
-
-
-
-
-
-
- Graph each of the following rational functions and decide whether or not each function has an inverse function. If an inverse function exists, find its formula. In addition, state the domain and range of each function you consider (the original function as well as its inverse function, if the inverse function exists).
-
-
-
-
-
-
- \displaystyle r(x) = -\frac{3}{x-4} + 5
-
-
-
-
- \displaystyle s(x) = \frac{4 - 3x}{7x - 2}
-
-
-
-
- \displaystyle u(x) = \frac{2x - 1}{(x-1)^2}
-
-
-
-
- \displaystyle w(x) = \frac{11}{(x+4)^3} - 7
-
-
-
-
-
-
-
- Exercise Answer
-
-
-
-
-
-
-
- r(x) = -\dfrac{3}{x-4} + 5. Domain: x \neq 4. Range: r \neq 5. The graph passes the horizontal line test, so an inverse exists. Solving y = -\dfrac{3}{x-4}+5 for x: y-5 = -\dfrac{3}{x-4}, so x-4 = -\dfrac{3}{y-5} and x = 4 - \dfrac{3}{y-5}. Thus
-
- r^{-1}(x) = 4 - \frac{3}{x-5}.
-
- Domain of r^{-1}: x \neq 5. Range of r^{-1}: r^{-1} \neq 4.
-
-
-
-
- s(x) = \dfrac{4-3x}{7x-2}. Domain: x \neq \dfrac{2}{7}. Range: s \neq -\dfrac{3}{7}. The graph passes the horizontal line test. Solving y = \dfrac{4-3x}{7x-2} for x: y(7x-2) = 4-3x, so 7xy + 3x = 4+2y, giving x = \dfrac{2y+4}{7y+3}. Thus
-
- s^{-1}(x) = \frac{2x+4}{7x+3}.
-
- Domain of s^{-1}: x \neq -\dfrac{3}{7}. Range of s^{-1}: s^{-1} \neq \dfrac{2}{7}.
-
-
-
-
- u(x) = \dfrac{2x-1}{(x-1)^2}. Domain: x \neq 1. This function fails the horizontal line test (not monotone), so it has no inverse function.
-
-
-
-
- w(x) = \dfrac{11}{(x+4)^3} - 7. Domain: x \neq -4. Range: w \neq -7. The graph passes the horizontal line test. Solving for x: y+7 = \dfrac{11}{(x+4)^3}, so (x+4)^3 = \dfrac{11}{y+7} and x = -4 + \sqrt[3]{\dfrac{11}{y+7}}. Thus
-
- w^{-1}(x) = -4 + \sqrt[3]{\frac{11}{x+7}}.
-
- Domain of w^{-1}: x \neq -7. Range of w^{-1}: w^{-1} \neq -4.
-
-
-
-
-
-
-
-
-
-
- For each of the following rational functions, identify the location of any potential hole in the graph. Then, create a table of function values for input values near where the hole should be located. Use your work to decide whether or not the graph indeed has a hole, with written justification.
-
- True or false: given r(x) = \frac{p(x)}{q(x)}, if p(a) = 0 and q(a) = 0, then r has a hole at x = a.
-
-
-
-
-
-
-
- Exercise Answer
-
-
-
-
-
-
-
- r(x) = \dfrac{x^2-16}{x+4} = \dfrac{(x-4)(x+4)}{x+4}. At x = -4, both numerator and denominator are zero. After cancellation, r(x) = x-4 for x \neq -4. Near x = -4, the values approach -4-4 = -8 (e.g., r(-3.9) \approx -7.9). So the graph has a hole at (-4, -8).
-
-
-
-
- s(x) = \dfrac{(x-2)^2(x+3)}{x^2-5x-6} = \dfrac{(x-2)^2(x+3)}{(x-6)(x+1)}. The denominator zeros are x = 6 and x = -1; neither equals a numerator zero. There are no holes.
-
-
-
-
- u(x) = \dfrac{(x-2)^3(x+3)}{(x^2-5x-6)(x-7)} = \dfrac{(x-2)^3(x+3)}{(x-6)(x+1)(x-7)}. The denominator zeros x=6, x=-1, x=7 share no factors with the numerator. There are no holes.
-
-
-
-
- w(x) = \dfrac{x^2+x-6}{(x^2+5x+6)(x+3)} = \dfrac{(x+3)(x-2)}{(x+2)(x+3)^2}. One factor of (x+3) cancels, leaving \dfrac{x-2}{(x+2)(x+3)}. At x=-3, the original is 0/0 and the reduced form still has (x+3) in the denominator, so x=-3 is a vertical asymptote, not a hole. There are no holes.
-
-
-
-
- False. Part (d) provides a counterexample: at x=-3, both p(-3)=0 and q(-3)=0, yet the graph has a vertical asymptote rather than a hole. A hole occurs only when the factor cancels completely in the reduced form.
-
-
-
-
-
-
-
-
-
-
- In the questions that follow, we explore the average rate of change of power functions on the interval [1,x]. To begin, let f(x) = x^2 and let A(x) be the average rate of change of f on [1,x].
-
-
-
-
-
-
- Explain why A is a rational function of x.
-
-
-
-
- What is the domain of A?
-
-
-
-
- At the point where A is undefined, does A have a vertical asymptote or a hole? Justify your thinking clearly.
-
-
-
-
- What can you say about the average rate of change of f on [1,x] as x gets closer and closer (but not equal) to 1?
-
-
-
-
- Now let g(x) = x^3 and B(x) be the average rate of change of B on [1,x]. Respond to prompts (a) - (d) but this time for the function B instead of A.
-
-
-
-
- Finally, let h(x) = x^4 and C(x) be the average rate of change of C on [1,x]. Respond to prompts (a) - (d) but this time for the function C instead of A.
-
-
-
-
-
-
-
- Exercise Answer
-
-
-
-
-
-
-
- A(x) = \dfrac{f(x)-f(1)}{x-1} = \dfrac{x^2-1}{x-1}, which is a ratio of polynomials and hence a rational function.
-
-
-
-
- A is undefined at x = 1 (divides by zero). Domain: all real numbers except x = 1.
-
-
-
-
- Factoring: A(x) = \dfrac{(x+1)(x-1)}{x-1} = x+1 for x \neq 1. Near x=1, A(x) \to 2. So A has a hole at (1, 2), not a vertical asymptote.
-
-
-
-
- As x \to 1, A(x) = x+1 \to 2. The average rate of change of f(x)=x^2 on [1,x] approaches 2.
-
-
-
-
- For g(x) = x^3: B(x) = \dfrac{x^3-1}{x-1} = \dfrac{(x-1)(x^2+x+1)}{x-1} = x^2+x+1 for x \neq 1. This is a rational function. Domain: x \neq 1. Hole at (1, 3) (since 1^2+1+1=3). As x \to 1, B(x) \to 3.
-
-
-
-
- For h(x) = x^4: C(x) = \dfrac{x^4-1}{x-1} = \dfrac{(x-1)(x^3+x^2+x+1)}{x-1} = x^3+x^2+x+1 for x \neq 1. This is a rational function. Domain: x \neq 1. Hole at (1, 4). As x \to 1, C(x) \to 4.
-
+ For each of the following rational functions, determine, with justification, the exact locations of all (i) horizontal asymptotes, (ii) vertical asymptotes, (iii) zeros, and (iv) holes of the function. Clearly show your work and thinking.
+
+ r(x) = \dfrac{-19(x+11.3)^2(x-15.1)(x-17.3)}{41(x+5.7)(x+11.3)(x-8.4)(x-15.1)}. Both (x+11.3) and (x-15.1) appear in both numerator and denominator, so they cancel. The reduced form is \dfrac{-19(x+11.3)(x-17.3)}{41(x+5.7)(x-8.4)}.
+
+
Horizontal asymptote:y = -\dfrac{19}{41} (same degree after cancellation).
+
Vertical asymptotes:x = -5.7 and x = 8.4.
+
Zero:x = 17.3 (the factor x+11.3 in reduced numerator gives x=-11.3, but that is a hole).
+
Holes: at x = -11.3 (hole value 0, since x+11.3=0 makes reduced numerator zero) and at x = 15.1.
+
+
+
+
+
+ s(x) = \dfrac{-29(x^2-16)(x^2+99)(x-53)}{101(x^2-4)(x-13)^2(x+104)}. Factor: numerator = -29(x-4)(x+4)(x^2+99)(x-53); denominator = 101(x-2)(x+2)(x-13)^2(x+104). No common factors.
+
+
Horizontal asymptote:y = -\dfrac{29}{101} (degree 5 over degree 5).
+
Vertical asymptotes:x = 2, x = -2, x = 13, x = -104.
+
Zeros:x = 4, x = -4, x = 53 (note x^2+99 > 0 always).
+
Holes: none.
+
+
+
+
+
+ u(x) = \dfrac{-71(x^2-13x+36)(x-58.4)(x+78.2)}{83(x+58.4)(x-78.2)(x^2-12x+27)}. Factor: x^2-13x+36 = (x-4)(x-9) and x^2-12x+27 = (x-3)(x-9). The factor (x-9) cancels.
+
+
Horizontal asymptote:y = -\dfrac{71}{83} (degree 4 over degree 4).
+
Vertical asymptotes:x = -58.4, x = 78.2, x = 3.
+
Zeros:x = 4, x = 58.4, x = -78.2.
+
Hole: at x = 9.
+
+
+
+
+
+
+
+
+
+
+
+ Find a formula for a rational function that meets the stated criteria, with justification. If no such formula is possible, explain why.
+
+
+
+
+
+
+ A rational function r(x) in the form r(x) = \frac{k}{x-a} + b so that r has a horizontal asymptote of y = -\frac{3}{7}, a vertical asymptote of x = \frac{5}{2}, and r(0) = 4.
+
+
+
+
+ A rational function s(x) that has no horizontal asymptote, has zeros at x = -5 and x = 3, has a single vertical asymptote at x = -1, and satisfies \lim_{x \to \infty} s(x) = -\infty and \lim_{x \to -\infty} s(x) = +\infty.
+
+
+
+
+ A rational function u(x) that is positive for x \lt -4, negative for -4 \lt x \lt -2, negative for -2 \lt x \lt 1, positive for 1 \lt x \lt 5, and negative for x \gt 5. The only zeros of u are located at x = -4 and x = -2. In addition, u has a hole at x = 4.
+
+
+
+
+ A rational function w(x) whose graph is shown in Figure.
+
+
+
+
A plot of the rational function w.
+
A plot of the rational function w.
+
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
+
+
+
+ We need b = -\dfrac{3}{7} (horizontal asymptote) and a = \dfrac{5}{2} (vertical asymptote). Then r(0) = \dfrac{k}{0 - \frac{5}{2}} - \dfrac{3}{7} = -\dfrac{2k}{5} - \dfrac{3}{7} = 4, so -\dfrac{2k}{5} = \dfrac{31}{7} and k = -\dfrac{155}{14}. Thus r(x) = -\dfrac{155/14}{x - 5/2} - \dfrac{3}{7}.
+
+
+
+
+ We need: numerator degree greater than denominator degree (no horizontal asymptote); zeros at x=-5 and x=3; single vertical asymptote at x=-1; and the correct end behavior. The function
+
+ s(x) = -\frac{(x+5)(x-3)}{x+1}
+
+ satisfies all criteria: as x \to \infty, s(x) \to -\infty (since -(+)(+)/(+) \to -\infty); as x \to -\infty, s(x) \to +\infty.
+
+
+
+
+ Zeros only at x=-4 and x=-2, with x=-2 having even multiplicity (no sign change), VAs at x=1 and x=5, hole at x=4. One formula:
+
+ u(x) = \frac{(x+4)(x+2)^2(x-4)}{(x-1)(x-5)(x-4)}.
+
+
+
+
+
+ Reading from the graph: zeros near x=-1 and x=5, vertical asymptotes near x=3 and x=8, a hole near x=10, and a horizontal asymptote near y=0. One formula consistent with the graph is
+
+ w(x) = -\frac{(x+2)(x-5)(x-10)}{5(x-3)(x-8)(x-10)}.
+
+
+
+
+
+
+
+
+
+
+
+ Graph each of the following rational functions and decide whether or not each function has an inverse function. If an inverse function exists, find its formula. In addition, state the domain and range of each function you consider (the original function as well as its inverse function, if the inverse function exists).
+
+
+
+
+
+
+ \displaystyle r(x) = -\frac{3}{x-4} + 5
+
+
+
+
+ \displaystyle s(x) = \frac{4 - 3x}{7x - 2}
+
+
+
+
+ \displaystyle u(x) = \frac{2x - 1}{(x-1)^2}
+
+
+
+
+ \displaystyle w(x) = \frac{11}{(x+4)^3} - 7
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
+
+
+
+ r(x) = -\dfrac{3}{x-4} + 5. Domain: x \neq 4. Range: r \neq 5. The graph passes the horizontal line test, so an inverse exists. Solving y = -\dfrac{3}{x-4}+5 for x: y-5 = -\dfrac{3}{x-4}, so x-4 = -\dfrac{3}{y-5} and x = 4 - \dfrac{3}{y-5}. Thus
+
+ r^{-1}(x) = 4 - \frac{3}{x-5}.
+
+ Domain of r^{-1}: x \neq 5. Range of r^{-1}: r^{-1} \neq 4.
+
+
+
+
+ s(x) = \dfrac{4-3x}{7x-2}. Domain: x \neq \dfrac{2}{7}. Range: s \neq -\dfrac{3}{7}. The graph passes the horizontal line test. Solving y = \dfrac{4-3x}{7x-2} for x: y(7x-2) = 4-3x, so 7xy + 3x = 4+2y, giving x = \dfrac{2y+4}{7y+3}. Thus
+
+ s^{-1}(x) = \frac{2x+4}{7x+3}.
+
+ Domain of s^{-1}: x \neq -\dfrac{3}{7}. Range of s^{-1}: s^{-1} \neq \dfrac{2}{7}.
+
+
+
+
+ u(x) = \dfrac{2x-1}{(x-1)^2}. Domain: x \neq 1. This function fails the horizontal line test (not monotone), so it has no inverse function.
+
+
+
+
+ w(x) = \dfrac{11}{(x+4)^3} - 7. Domain: x \neq -4. Range: w \neq -7. The graph passes the horizontal line test. Solving for x: y+7 = \dfrac{11}{(x+4)^3}, so (x+4)^3 = \dfrac{11}{y+7} and x = -4 + \sqrt[3]{\dfrac{11}{y+7}}. Thus
+
+ w^{-1}(x) = -4 + \sqrt[3]{\frac{11}{x+7}}.
+
+ Domain of w^{-1}: x \neq -7. Range of w^{-1}: w^{-1} \neq -4.
+
+
+
+
+
+
+
+
+
+
+ For each of the following rational functions, identify the location of any potential hole in the graph. Then, create a table of function values for input values near where the hole should be located. Use your work to decide whether or not the graph indeed has a hole, with written justification.
+
+ True or false: given r(x) = \frac{p(x)}{q(x)}, if p(a) = 0 and q(a) = 0, then r has a hole at x = a.
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
+
+
+
+ r(x) = \dfrac{x^2-16}{x+4} = \dfrac{(x-4)(x+4)}{x+4}. At x = -4, both numerator and denominator are zero. After cancellation, r(x) = x-4 for x \neq -4. Near x = -4, the values approach -4-4 = -8 (e.g., r(-3.9) \approx -7.9). So the graph has a hole at (-4, -8).
+
+
+
+
+ s(x) = \dfrac{(x-2)^2(x+3)}{x^2-5x-6} = \dfrac{(x-2)^2(x+3)}{(x-6)(x+1)}. The denominator zeros are x = 6 and x = -1; neither equals a numerator zero. There are no holes.
+
+
+
+
+ u(x) = \dfrac{(x-2)^3(x+3)}{(x^2-5x-6)(x-7)} = \dfrac{(x-2)^3(x+3)}{(x-6)(x+1)(x-7)}. The denominator zeros x=6, x=-1, x=7 share no factors with the numerator. There are no holes.
+
+
+
+
+ w(x) = \dfrac{x^2+x-6}{(x^2+5x+6)(x+3)} = \dfrac{(x+3)(x-2)}{(x+2)(x+3)^2}. One factor of (x+3) cancels, leaving \dfrac{x-2}{(x+2)(x+3)}. At x=-3, the original is 0/0 and the reduced form still has (x+3) in the denominator, so x=-3 is a vertical asymptote, not a hole. There are no holes.
+
+
+
+
+ False. Part (d) provides a counterexample: at x=-3, both p(-3)=0 and q(-3)=0, yet the graph has a vertical asymptote rather than a hole. A hole occurs only when the factor cancels completely in the reduced form.
+
+
+
+
+
+
+
+
+
+
+ In the questions that follow, we explore the average rate of change of power functions on the interval [1,x]. To begin, let f(x) = x^2 and let A(x) be the average rate of change of f on [1,x].
+
+
+
+
+
+
+ Explain why A is a rational function of x.
+
+
+
+
+ What is the domain of A?
+
+
+
+
+ At the point where A is undefined, does A have a vertical asymptote or a hole? Justify your thinking clearly.
+
+
+
+
+ What can you say about the average rate of change of f on [1,x] as x gets closer and closer (but not equal) to 1?
+
+
+
+
+ Now let g(x) = x^3 and B(x) be the average rate of change of B on [1,x]. Respond to prompts (a) - (d) but this time for the function B instead of A.
+
+
+
+
+ Finally, let h(x) = x^4 and C(x) be the average rate of change of C on [1,x]. Respond to prompts (a) - (d) but this time for the function C instead of A.
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
+
+
+
+ A(x) = \dfrac{f(x)-f(1)}{x-1} = \dfrac{x^2-1}{x-1}, which is a ratio of polynomials and hence a rational function.
+
+
+
+
+ A is undefined at x = 1 (divides by zero). Domain: all real numbers except x = 1.
+
+
+
+
+ Factoring: A(x) = \dfrac{(x+1)(x-1)}{x-1} = x+1 for x \neq 1. Near x=1, A(x) \to 2. So A has a hole at (1, 2), not a vertical asymptote.
+
+
+
+
+ As x \to 1, A(x) = x+1 \to 2. The average rate of change of f(x)=x^2 on [1,x] approaches 2.
+
+
+
+
+ For g(x) = x^3: B(x) = \dfrac{x^3-1}{x-1} = \dfrac{(x-1)(x^2+x+1)}{x-1} = x^2+x+1 for x \neq 1. This is a rational function. Domain: x \neq 1. Hole at (1, 3) (since 1^2+1+1=3). As x \to 1, B(x) \to 3.
+
+
+
+
+ For h(x) = x^4: C(x) = \dfrac{x^4-1}{x-1} = \dfrac{(x-1)(x^3+x^2+x+1)}{x-1} = x^3+x^2+x+1 for x \neq 1. This is a rational function. Domain: x \neq 1. Hole at (1, 4). As x \to 1, C(x) \to 4.
+
- f(x) = \dfrac{17x^2+34}{19x^2-76} = \dfrac{17(x^2+2)}{19(x-2)(x+2)}. Domain: all real numbers except x = \pm 2. Horizontal asymptote: y = \dfrac{17}{19} (ratio of leading coefficients, same degree).
-
-
-
-
- g(x) = \dfrac{29}{53} + \dfrac{1}{x-2} is undefined at x = 2. Domain: all real numbers except x = 2. Horizontal asymptote: y = \dfrac{29}{53} (as x \to \pm\infty, the term \frac{1}{x-2} \to 0).
-
-
-
-
- h(x) = \dfrac{4-31x}{11x-7} is undefined at x = \dfrac{7}{11}. Domain: all real numbers except x = \dfrac{7}{11}. Horizontal asymptote: y = \dfrac{-31}{11} (ratio of leading coefficients).
-
-
-
-
- r(x) = \dfrac{151(x-4)(x+5)^2(x-2)}{537(x+5)(x+1)(x^2+1)(x-15)}. The factor (x+5) cancels, leaving a hole at x = -5. The reduced denominator zeros are x = -1 and x = 15 (note x^2+1 > 0 always). Domain: all real numbers except x = -5, x = -1, and x = 15. Since the numerator has degree 4 and denominator has degree 5, the horizontal asymptote is y = 0.
-
-
-
-
-
-
-
-
-
-
- A rectangular box is being constructed so that its base is 1.5 times as long as it is wide. In addition, suppose that material for the base and top of the box costs $3.75 per square foot, while material for the sides costs $2.50 per square foot. Finally, we want the box to hold 8 cubic feet of volume.
-
-
-
-
-
-
- Draw a labeled picture of the box with x as the length of the shorter side of the box's base and h as its height.
-
-
-
-
- Determine a formula involving x and h for the total surface area, S, of the box.
-
-
-
-
- Use your work from (b) along with the given information about cost to determine a formula for the total cost, C, oif the box in terms of x and h.
-
-
-
-
- Use the volume constraint given in the problem to write an equation that relates x and h, and solve that equation for h in terms of x.
-
-
-
-
- Combine your work in (c) and (d) to write the cost, C, of the box as a function solely of x.
-
-
-
-
- What is the domain of the cost function? How does a graph of the cost function appear? What does this suggest about the ideal box for the given constraints?
-
-
-
-
-
-
-
- Exercise Answer
-
-
-
-
-
-
-
- The box has width x, length 1.5x, and height h.
-
-
-
-
- The surface area is: base and top each 1.5x^2, two short sides each xh, two long sides each 1.5xh. So S = 3x^2 + 5xh.
-
- The domain is x > 0. As x \to 0^+, C \to \infty; as x \to \infty, C \to \infty. A graph shows a minimum at x \approx 1.44 ft, suggesting the ideal box has approximate dimensions 1.44 \times 2.15 \times 2.58 feet.
-
-
-
-
-
-
-
-
-
-
- A cylindrical can is being constructed so that its volume is 16 cubic inches. Suppose that material for the lids (the top and bottom) cost $0.11 per square inch and material for the side of the can costs $0.07 per square inch. Determine a formula for the total cost of the can as a function of the can's radius. What is the domain of the function and why?
-
-
-
-
- You may find it helpful to ask yourself a sequence of questions like those stated in Exercise).
-
-
-
-
- Exercise Answer
-
-
-
-
- With radius r and height h, the volume constraint \pi r^2 h = 16 gives h = \dfrac{16}{\pi r^2}. The cost is
-
- C(r) &= 0.11 \cdot 2\pi r^2 + 0.07 \cdot 2\pi r h
- &= 0.22\pi r^2 + 0.14\pi r \cdot \frac{16}{\pi r^2}
- &= 0.22\pi r^2 + \frac{2.24}{r}.
-
- Numerically, C(r) \approx 0.691 r^2 + \dfrac{2.240}{r}. The domain is r > 0 (radius must be positive). A graph shows a minimum at r \approx 1.48 inches, giving a height of about h \approx 2.33 inches and cost of about \$3.03.
-
+ f(x) = \dfrac{17x^2+34}{19x^2-76} = \dfrac{17(x^2+2)}{19(x-2)(x+2)}. Domain: all real numbers except x = \pm 2. Horizontal asymptote: y = \dfrac{17}{19} (ratio of leading coefficients, same degree).
+
+
+
+
+ g(x) = \dfrac{29}{53} + \dfrac{1}{x-2} is undefined at x = 2. Domain: all real numbers except x = 2. Horizontal asymptote: y = \dfrac{29}{53} (as x \to \pm\infty, the term \frac{1}{x-2} \to 0).
+
+
+
+
+ h(x) = \dfrac{4-31x}{11x-7} is undefined at x = \dfrac{7}{11}. Domain: all real numbers except x = \dfrac{7}{11}. Horizontal asymptote: y = \dfrac{-31}{11} (ratio of leading coefficients).
+
+
+
+
+ r(x) = \dfrac{151(x-4)(x+5)^2(x-2)}{537(x+5)(x+1)(x^2+1)(x-15)}. The factor (x+5) cancels, leaving a hole at x = -5. The reduced denominator zeros are x = -1 and x = 15 (note x^2+1 > 0 always). Domain: all real numbers except x = -5, x = -1, and x = 15. Since the numerator has degree 4 and denominator has degree 5, the horizontal asymptote is y = 0.
+
+
+
+
+
+
+
+
+
+
+ A rectangular box is being constructed so that its base is 1.5 times as long as it is wide. In addition, suppose that material for the base and top of the box costs $3.75 per square foot, while material for the sides costs $2.50 per square foot. Finally, we want the box to hold 8 cubic feet of volume.
+
+
+
+
+
+
+ Draw a labeled picture of the box with x as the length of the shorter side of the box's base and h as its height.
+
+
+
+
+ Determine a formula involving x and h for the total surface area, S, of the box.
+
+
+
+
+ Use your work from (b) along with the given information about cost to determine a formula for the total cost, C, oif the box in terms of x and h.
+
+
+
+
+ Use the volume constraint given in the problem to write an equation that relates x and h, and solve that equation for h in terms of x.
+
+
+
+
+ Combine your work in (c) and (d) to write the cost, C, of the box as a function solely of x.
+
+
+
+
+ What is the domain of the cost function? How does a graph of the cost function appear? What does this suggest about the ideal box for the given constraints?
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
+
+
+
+ The box has width x, length 1.5x, and height h.
+
+
+
+
+ The surface area is: base and top each 1.5x^2, two short sides each xh, two long sides each 1.5xh. So S = 3x^2 + 5xh.
+
+ The domain is x > 0. As x \to 0^+, C \to \infty; as x \to \infty, C \to \infty. A graph shows a minimum at x \approx 1.44 ft, suggesting the ideal box has approximate dimensions 1.44 \times 2.15 \times 2.58 feet.
+
+
+
+
+
+
+
+
+
+
+ A cylindrical can is being constructed so that its volume is 16 cubic inches. Suppose that material for the lids (the top and bottom) cost $0.11 per square inch and material for the side of the can costs $0.07 per square inch. Determine a formula for the total cost of the can as a function of the can's radius. What is the domain of the function and why?
+
+
+
+
+ You may find it helpful to ask yourself a sequence of questions like those stated in Exercise).
+
+
+
+
+ Exercise Answer
+
+
+
+
+ With radius r and height h, the volume constraint \pi r^2 h = 16 gives h = \dfrac{16}{\pi r^2}. The cost is
+
+ C(r) &= 0.11 \cdot 2\pi r^2 + 0.07 \cdot 2\pi r h
+ &= 0.22\pi r^2 + 0.14\pi r \cdot \frac{16}{\pi r^2}
+ &= 0.22\pi r^2 + \frac{2.24}{r}.
+
+ Numerically, C(r) \approx 0.691 r^2 + \dfrac{2.240}{r}. The domain is r > 0 (radius must be positive). A graph shows a minimum at r \approx 1.48 inches, giving a height of about h \approx 2.33 inches and cost of about \$3.03.
+
- At an airshow, a pilot is flying low over a runway while maintaining a constant altitude of 2000 feet and a constant speed. On a straight path over the runway, the pilot observes on her laser range-finder that the distance from the plane to a fixed building adjacent to the runway is 7500 feet. Five seconds later, she observes that distance to the same building is now 6000 feet.
-
-
-
-
-
-
- What is the angle of depression from the plane to the building when the plane is 7500 feet away from the building? (The angle of depression is the angle that the pilot's line of sight makes with the horizontal.)
-
-
-
-
- What is the angle of depression when the plane is 6000 feet from the building?
-
-
-
-
- How far did the plane travel during the time between the two different observations?
-
-
-
-
- What is the plane's velocity (in miles per hour)?
-
-
-
-
-
-
-
- Exercise Answer
-
-
-
-
-
-
-
- The plane is at altitude 2000 feet and 7500 feet from the building (slant distance). The angle of depression is \arcsin\!\left(\dfrac{2000}{7500}\right) = \arcsin\!\left(\dfrac{4}{15}\right) \approx 0.2699 radians, or about 15.47^{\circ}.
-
-
-
-
- When the slant distance is 6000 feet, the angle of depression is \arcsin\!\left(\dfrac{2000}{6000}\right) = \arcsin\!\left(\dfrac{1}{3}\right) \approx 0.3398 radians, or about 19.47^{\circ}.
-
-
-
-
- The horizontal distance at 7500 feet is \sqrt{7500^2 - 2000^2} = 500\sqrt{209} \approx 7228 feet, and at 6000 feet it is \sqrt{6000^2 - 2000^2} = 4000\sqrt{2} \approx 5657 feet. The plane traveled 500\sqrt{209} - 4000\sqrt{2} = 500(\sqrt{209} - 8\sqrt{2}) \approx 1572 feet.
-
-
-
-
- Traveling 500(\sqrt{209}-8\sqrt{2}) feet in 5 seconds gives a speed of 100(\sqrt{209}-8\sqrt{2}) \approx 314 feet per second, which is approximately 214 miles per hour.
-
-
-
-
-
-
-
-
-
-
- On a calm day, a photographer is filming a hot air balloon. When the balloon launches, the photographer is stationed 850 feet away from the balloon.
-
-
-
-
-
-
- When the balloon is 200 feet off the ground, what is the angle of elevation of the camera?
-
-
-
-
- When the balloon is 275 feet off the ground, what is the angle of elevation of the camera?
-
-
-
-
- Let \theta represent the camera's angle of elevation when the balloon is at an arbitrary height h above the ground. Express \theta as a function of h.
-
-
-
-
- Determine AV_{[200,275]} for \theta (as a function of h) and write at least one sentence to carefully explain the meaning of the value you find, including units.
-
-
-
-
-
-
-
- Exercise Answer
-
-
-
-
-
-
-
- When the balloon is 200 feet off the ground, the angle of elevation is \arctan\!\left(\dfrac{200}{850}\right) \approx 0.2311 radians (about 13.24^{\circ}).
-
-
-
-
- When the balloon is 275 feet off the ground, the angle of elevation is \arctan\!\left(\dfrac{275}{850}\right) \approx 0.3129 radians (about 17.93^{\circ}).
-
-
-
-
- The angle of elevation as a function of height is \theta(h) = \arctan\!\left(\dfrac{h}{850}\right).
-
-
-
-
- The average rate of change on [200,275] is AV_{[200,275]} = \dfrac{\theta(275) - \theta(200)}{75} \approx \dfrac{0.3129 - 0.2311}{75} \approx 0.001091 radians per foot. This means that as the balloon rises from 200 to 275 feet, the camera angle must increase at an average rate of about 0.001091 radians for each foot of altitude gained.
-
-
-
-
-
-
-
-
-
-
- Consider a right triangle where the two legs measure 5 and 12 respectively and \alpha is the angle opposite the shorter leg and \beta is the angle opposite the longer leg.
-
-
-
-
-
-
- What is the exact value of \cos(\alpha)?
-
-
-
-
- What is the exact value of \sin(\beta)?
-
-
-
-
- What is the exact value of \tan(\beta)? of \tan(\alpha)?
-
-
-
-
- What is the exact radian measure of \alpha? approximate measure?
-
-
-
-
- What is the exact radian measure of \beta? approximate measure?
-
-
-
-
- True or false: for any two angles \theta and \gamma such that \theta + \gamma = \frac{\pi}{2} (radians), it follows that \cos(\theta) = \sin(\gamma).
-
-
-
-
-
-
-
- Exercise Answer
-
-
-
-
-
-
-
- The hypotenuse has length \sqrt{5^2 + 12^2} = 13. Since \alpha is opposite the leg of length 5, we have \cos(\alpha) = \dfrac{12}{13}.
-
-
-
-
- \sin(\beta) = \dfrac{12}{13}.
-
-
-
-
- \tan(\beta) = \dfrac{12}{5} and \tan(\alpha) = \dfrac{5}{12}.
-
- True. If \theta + \gamma = \frac{\pi}{2}, then \theta and \gamma are complementary angles in a right triangle. Labeling the side opposite \theta as B and the side opposite \gamma as C, with hypotenuse H, we get \cos(\theta) = \frac{C}{H} = \sin(\gamma).
-
+ At an airshow, a pilot is flying low over a runway while maintaining a constant altitude of 2000 feet and a constant speed. On a straight path over the runway, the pilot observes on her laser range-finder that the distance from the plane to a fixed building adjacent to the runway is 7500 feet. Five seconds later, she observes that distance to the same building is now 6000 feet.
+
+
+
+
+
+
+ What is the angle of depression from the plane to the building when the plane is 7500 feet away from the building? (The angle of depression is the angle that the pilot's line of sight makes with the horizontal.)
+
+
+
+
+ What is the angle of depression when the plane is 6000 feet from the building?
+
+
+
+
+ How far did the plane travel during the time between the two different observations?
+
+
+
+
+ What is the plane's velocity (in miles per hour)?
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
+
+
+
+ The plane is at altitude 2000 feet and 7500 feet from the building (slant distance). The angle of depression is \arcsin\!\left(\dfrac{2000}{7500}\right) = \arcsin\!\left(\dfrac{4}{15}\right) \approx 0.2699 radians, or about 15.47^{\circ}.
+
+
+
+
+ When the slant distance is 6000 feet, the angle of depression is \arcsin\!\left(\dfrac{2000}{6000}\right) = \arcsin\!\left(\dfrac{1}{3}\right) \approx 0.3398 radians, or about 19.47^{\circ}.
+
+
+
+
+ The horizontal distance at 7500 feet is \sqrt{7500^2 - 2000^2} = 500\sqrt{209} \approx 7228 feet, and at 6000 feet it is \sqrt{6000^2 - 2000^2} = 4000\sqrt{2} \approx 5657 feet. The plane traveled 500\sqrt{209} - 4000\sqrt{2} = 500(\sqrt{209} - 8\sqrt{2}) \approx 1572 feet.
+
+
+
+
+ Traveling 500(\sqrt{209}-8\sqrt{2}) feet in 5 seconds gives a speed of 100(\sqrt{209}-8\sqrt{2}) \approx 314 feet per second, which is approximately 214 miles per hour.
+
+
+
+
+
+
+
+
+
+
+ On a calm day, a photographer is filming a hot air balloon. When the balloon launches, the photographer is stationed 850 feet away from the balloon.
+
+
+
+
+
+
+ When the balloon is 200 feet off the ground, what is the angle of elevation of the camera?
+
+
+
+
+ When the balloon is 275 feet off the ground, what is the angle of elevation of the camera?
+
+
+
+
+ Let \theta represent the camera's angle of elevation when the balloon is at an arbitrary height h above the ground. Express \theta as a function of h.
+
+
+
+
+ Determine AV_{[200,275]} for \theta (as a function of h) and write at least one sentence to carefully explain the meaning of the value you find, including units.
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
+
+
+
+ When the balloon is 200 feet off the ground, the angle of elevation is \arctan\!\left(\dfrac{200}{850}\right) \approx 0.2311 radians (about 13.24^{\circ}).
+
+
+
+
+ When the balloon is 275 feet off the ground, the angle of elevation is \arctan\!\left(\dfrac{275}{850}\right) \approx 0.3129 radians (about 17.93^{\circ}).
+
+
+
+
+ The angle of elevation as a function of height is \theta(h) = \arctan\!\left(\dfrac{h}{850}\right).
+
+
+
+
+ The average rate of change on [200,275] is AV_{[200,275]} = \dfrac{\theta(275) - \theta(200)}{75} \approx \dfrac{0.3129 - 0.2311}{75} \approx 0.001091 radians per foot. This means that as the balloon rises from 200 to 275 feet, the camera angle must increase at an average rate of about 0.001091 radians for each foot of altitude gained.
+
+
+
+
+
+
+
+
+
+
+ Consider a right triangle where the two legs measure 5 and 12 respectively and \alpha is the angle opposite the shorter leg and \beta is the angle opposite the longer leg.
+
+
+
+
+
+
+ What is the exact value of \cos(\alpha)?
+
+
+
+
+ What is the exact value of \sin(\beta)?
+
+
+
+
+ What is the exact value of \tan(\beta)? of \tan(\alpha)?
+
+
+
+
+ What is the exact radian measure of \alpha? approximate measure?
+
+
+
+
+ What is the exact radian measure of \beta? approximate measure?
+
+
+
+
+ True or false: for any two angles \theta and \gamma such that \theta + \gamma = \frac{\pi}{2} (radians), it follows that \cos(\theta) = \sin(\gamma).
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
+
+
+
+ The hypotenuse has length \sqrt{5^2 + 12^2} = 13. Since \alpha is opposite the leg of length 5, we have \cos(\alpha) = \dfrac{12}{13}.
+
+
+
+
+ \sin(\beta) = \dfrac{12}{13}.
+
+
+
+
+ \tan(\beta) = \dfrac{12}{5} and \tan(\alpha) = \dfrac{5}{12}.
+
+ True. If \theta + \gamma = \frac{\pi}{2}, then \theta and \gamma are complementary angles in a right triangle. Labeling the side opposite \theta as B and the side opposite \gamma as C, with hypotenuse H, we get \cos(\theta) = \frac{C}{H} = \sin(\gamma).
+
- Use the special points on the unit circle (see, for instance, Figure) to determine the exact values of each of the following numerical expressions. Do so without using a computational device.
-
- For each of the following claims, determine whether the statement is true or false. If true, write one sentence to justify your reasoning. If false, give an example of a value that shows the claim fails.
-
-
-
-
-
-
- For any y such that -1 \le y \le 1, \sin(\arcsin(y)) = y.
-
-
-
-
- For any real number t, \arcsin(\sin(t)) = t.
-
-
-
-
- For any real number t, \arccos(\cos(t)) = t.
-
-
-
-
- For any y such that -1 \le y \le 1, \cos(\arccos(y)) = y.
-
-
-
-
- For any real number y, \tan(\arctan(y)) = y.
-
-
-
-
- For any real number t, \arctan(\tan(t)) = t.
-
-
-
-
-
-
-
- Exercise Answer
-
-
-
-
-
-
-
- True. Sine is the inverse function of arcsine, and therefore \sin(\arcsin(y)) = y for any y with -1 \le y \le 1.
-
-
-
-
- False. For example, \arcsin\!\left(\sin\!\left(\frac{3\pi}{4}\right)\right) = \arcsin\!\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4} \ne \frac{3\pi}{4}.
-
-
-
-
- False. For example, \arccos\!\left(\cos\!\left(-\frac{\pi}{4}\right)\right) = \arccos\!\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4} \ne -\frac{\pi}{4}.
-
-
-
-
- True. Cosine is the inverse function of arccosine, and therefore \cos(\arccos(y)) = y for any y with -1 \le y \le 1.
-
-
-
-
- True. Tangent is the inverse function of arctangent, and therefore \tan(\arctan(y)) = y for any real number y.
-
-
-
-
- False. For example, \arctan\!\left(\tan\!\left(\frac{3\pi}{4}\right)\right) = \arctan(-1) = -\frac{\pi}{4} \ne \frac{3\pi}{4}.
-
-
-
-
-
-
-
-
-
-
- Let's consider the composite function h(x) = \cos(\arcsin(x)). This function makes sense to consider since the arcsine function produces an angle, at which the cosine function can then be evaluated. In the questions that follow, we investigate how to express h without using trigonometric functions at all.
-
-
-
-
-
-
- What is the domain of h? The range of h?
-
-
-
-
- Since the arcsine function produces an angle, let's say that \theta = \arcsin(x), so that \theta is the angle whose sine is x. By definition, we can picture \theta as an angle in a right triangle with hypotenuse 1 and a vertical leg of length x, as shown in Figure. Use the Pythagorean Theorem to determine the length of the horizontal leg as a function of x.
-
-
-
-
-
-
The right triangle that corresponds to the angle \theta = \arcsin(x).
-
The right triangle that corresponds to the angle \theta = \arcsin(x).
-
-
-
-
-
The right triangle that corresponds to the angle \alpha = \arctan(x).
-
The right triangle that corresponds to the angle \alpha = \arctan(x).
-
-
-
-
-
-
-
- What is the value of \cos(\theta) as a function of x? What have we shown about h(x) = \cos(\arcsin(x))?
-
-
-
-
-
- How about the function p(x) = \cos(\arctan(x))? How can you reason similarly to write p in a way that doesn't involve any trigonometric functions at all? (Hint: let \alpha = \arctan(x) and consider the right triangle in Figure.)
-
-
-
-
-
-
-
- Exercise Answer
-
-
-
-
-
-
-
- The domain of h is [-1, 1] (the domain of arcsine). The range of h is [0, 1], since arcsin returns values in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] and cosine is non-negative on that interval.
-
-
-
-
- With hypotenuse 1 and vertical leg x, the Pythagorean Theorem gives the horizontal leg length as \sqrt{1 - x^2}.
-
-
-
-
- We have \cos(\theta) = \dfrac{\sqrt{1-x^2}}{1} = \sqrt{1-x^2}. Therefore h(x) = \cos(\arcsin(x)) = \sqrt{1 - x^2}.
-
-
-
-
- Let \alpha = \arctan(x), so \tan(\alpha) = x. In the corresponding right triangle, the opposite leg has length x, the adjacent leg has length 1, and by the Pythagorean Theorem the hypotenuse has length \sqrt{1 + x^2}. Therefore p(x) = \cos(\arctan(x)) = \dfrac{1}{\sqrt{1+x^2}}.
-
+ Use the special points on the unit circle (see, for instance, Figure) to determine the exact values of each of the following numerical expressions. Do so without using a computational device.
+
+ For each of the following claims, determine whether the statement is true or false. If true, write one sentence to justify your reasoning. If false, give an example of a value that shows the claim fails.
+
+
+
+
+
+
+ For any y such that -1 \le y \le 1, \sin(\arcsin(y)) = y.
+
+
+
+
+ For any real number t, \arcsin(\sin(t)) = t.
+
+
+
+
+ For any real number t, \arccos(\cos(t)) = t.
+
+
+
+
+ For any y such that -1 \le y \le 1, \cos(\arccos(y)) = y.
+
+
+
+
+ For any real number y, \tan(\arctan(y)) = y.
+
+
+
+
+ For any real number t, \arctan(\tan(t)) = t.
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
+
+
+
+ True. Sine is the inverse function of arcsine, and therefore \sin(\arcsin(y)) = y for any y with -1 \le y \le 1.
+
+
+
+
+ False. For example, \arcsin\!\left(\sin\!\left(\frac{3\pi}{4}\right)\right) = \arcsin\!\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4} \ne \frac{3\pi}{4}.
+
+
+
+
+ False. For example, \arccos\!\left(\cos\!\left(-\frac{\pi}{4}\right)\right) = \arccos\!\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4} \ne -\frac{\pi}{4}.
+
+
+
+
+ True. Cosine is the inverse function of arccosine, and therefore \cos(\arccos(y)) = y for any y with -1 \le y \le 1.
+
+
+
+
+ True. Tangent is the inverse function of arctangent, and therefore \tan(\arctan(y)) = y for any real number y.
+
+
+
+
+ False. For example, \arctan\!\left(\tan\!\left(\frac{3\pi}{4}\right)\right) = \arctan(-1) = -\frac{\pi}{4} \ne \frac{3\pi}{4}.
+
+
+
+
+
+
+
+
+
+
+ Let's consider the composite function h(x) = \cos(\arcsin(x)). This function makes sense to consider since the arcsine function produces an angle, at which the cosine function can then be evaluated. In the questions that follow, we investigate how to express h without using trigonometric functions at all.
+
+
+
+
+
+
+ What is the domain of h? The range of h?
+
+
+
+
+ Since the arcsine function produces an angle, let's say that \theta = \arcsin(x), so that \theta is the angle whose sine is x. By definition, we can picture \theta as an angle in a right triangle with hypotenuse 1 and a vertical leg of length x, as shown in Figure. Use the Pythagorean Theorem to determine the length of the horizontal leg as a function of x.
+
+
+
+
+
+
The right triangle that corresponds to the angle \theta = \arcsin(x).
+
The right triangle that corresponds to the angle \theta = \arcsin(x).
+
+
+
+
+
The right triangle that corresponds to the angle \alpha = \arctan(x).
+
The right triangle that corresponds to the angle \alpha = \arctan(x).
+
+
+
+
+
+
+
+ What is the value of \cos(\theta) as a function of x? What have we shown about h(x) = \cos(\arcsin(x))?
+
+
+
+
+
+ How about the function p(x) = \cos(\arctan(x))? How can you reason similarly to write p in a way that doesn't involve any trigonometric functions at all? (Hint: let \alpha = \arctan(x) and consider the right triangle in Figure.)
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
+
+
+
+ The domain of h is [-1, 1] (the domain of arcsine). The range of h is [0, 1], since arcsin returns values in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] and cosine is non-negative on that interval.
+
+
+
+
+ With hypotenuse 1 and vertical leg x, the Pythagorean Theorem gives the horizontal leg length as \sqrt{1 - x^2}.
+
+
+
+
+ We have \cos(\theta) = \dfrac{\sqrt{1-x^2}}{1} = \sqrt{1-x^2}. Therefore h(x) = \cos(\arcsin(x)) = \sqrt{1 - x^2}.
+
+
+
+
+ Let \alpha = \arctan(x), so \tan(\alpha) = x. In the corresponding right triangle, the opposite leg has length x, the adjacent leg has length 1, and by the Pythagorean Theorem the hypotenuse has length \sqrt{1 + x^2}. Therefore p(x) = \cos(\arctan(x)) = \dfrac{1}{\sqrt{1+x^2}}.
+
- Let \beta be an angle in quadrant II that satisfies \cos(\beta) = -\frac{12}{13}.
- Determine the values of the other five trigonometric functions evaluated at \beta exactly and without evaluating any trigonometric function on a computational device.
-
-
-
- How do your answers change if \beta lies in quadrant III?
-
-
-
-
- Exercise Answer
-
-
-
-
- With \cos(\beta) = -\frac{12}{13} and \beta in quadrant II, the Pythagorean identity gives \sin(\beta) = \frac{5}{13} (positive in QII). The remaining values are:
-
- \sec(\beta) = -\frac{13}{12}, \quad \csc(\beta) = \frac{13}{5}, \quad \tan(\beta) = -\frac{5}{12}, \quad \cot(\beta) = -\frac{12}{5}.
-
-
-
- If \beta lies in quadrant III, then \cos(\beta) is still -\frac{12}{13} but now \sin(\beta) = -\frac{5}{13} (negative in QIII). So \sec(\beta) = -\frac{13}{12} (unchanged), \csc(\beta) = -\frac{13}{5}, \tan(\beta) = \frac{5}{12}, and \cot(\beta) = \frac{12}{5}.
-
-
-
-
-
-
-
-
- For each of the following transformations of standard trigonometric functions, use your understanding of transformations to determine the domain, range, asymptotes, and period of the function, with careful justification. Then, check your results using Desmos or another graphing utility.
-
-
-
-
-
-
- f(t) = 5\sec(t-\frac{\pi}{2}) + 3
-
-
-
-
- g(t) = -\frac{1}{3}\csc(2t) - 4
-
-
-
-
- h(t) = -7\tan(t+\frac{\pi}{4}) + 1
-
-
-
-
- j(t) = \frac{1}{2}\cot(4t) - 2
-
-
-
-
-
-
-
- Exercise Answer
-
-
-
-
-
-
-
- f(t) = 5\sec\!\left(t - \frac{\pi}{2}\right) + 3: The parent \sec(t) has period 2\pi, domain all reals except \frac{\pi}{2} + j\pi, range (-\infty,-1]\cup[1,\infty). Shifting by \frac{\pi}{2} moves the asymptotes to t = j\pi. The vertical stretch by 5 and shift by 3 give range (-\infty,-2]\cup[8,\infty). Period remains 2\pi.
-
-
-
-
- g(t) = -\frac{1}{3}\csc(2t) - 4: The horizontal compression by 2 gives period \pi and moves asymptotes to all multiples of \frac{\pi}{2} (i.e., t = \frac{j\pi}{2} for any integer j), so the domain is all reals except multiples of \frac{\pi}{2}. The factor -\frac{1}{3} and vertical shift -4 give range (-\infty, -\frac{13}{3}]\cup[-\frac{11}{3},\infty).
-
-
-
-
- h(t) = -7\tan\!\left(t + \frac{\pi}{4}\right) + 1: The parent \tan(t) has period \pi, asymptotes at \frac{\pi}{2}+j\pi, range all reals. The horizontal shift moves asymptotes to t = \frac{\pi}{4} + j\pi. Domain is all reals except \frac{\pi}{4}+j\pi. Range remains all reals. Period remains \pi.
-
-
-
-
- j(t) = \frac{1}{2}\cot(4t) - 2: The horizontal compression by 4 gives period \frac{\pi}{4} and moves asymptotes to t = \frac{j\pi}{4}. Domain is all reals except \frac{j\pi}{4}. Range remains all reals.
-
-
-
-
-
-
-
-
-
-
- In a right triangle with hypotenuse 1 and vertical leg x,
- with angle \theta opposite x,
- determine the simplest expression you can for each of the following quantities in terms of x.
-
-
-
-
-
-
- \sin(\theta)
-
-
-
-
-
- \sec(\theta)
-
-
-
-
-
- \csc(\theta)
-
-
-
-
-
- \tan(\theta)
-
-
-
-
-
- \cos(\arcsin(x))
-
-
-
-
-
- \cot(\arcsin(x))
-
-
-
-
-
-
-
- Exercise Answer
-
-
-
-
- In the right triangle, the horizontal leg has length \sqrt{1-x^2} by the Pythagorean Theorem.
-
+ Let \beta be an angle in quadrant II that satisfies \cos(\beta) = -\frac{12}{13}.
+ Determine the values of the other five trigonometric functions evaluated at \beta exactly and without evaluating any trigonometric function on a computational device.
+
+
+
+ How do your answers change if \beta lies in quadrant III?
+
+
+
+
+ Exercise Answer
+
+
+
+
+ With \cos(\beta) = -\frac{12}{13} and \beta in quadrant II, the Pythagorean identity gives \sin(\beta) = \frac{5}{13} (positive in QII). The remaining values are:
+
+ \sec(\beta) = -\frac{13}{12}, \quad \csc(\beta) = \frac{13}{5}, \quad \tan(\beta) = -\frac{5}{12}, \quad \cot(\beta) = -\frac{12}{5}.
+
+
+
+ If \beta lies in quadrant III, then \cos(\beta) is still -\frac{12}{13} but now \sin(\beta) = -\frac{5}{13} (negative in QIII). So \sec(\beta) = -\frac{13}{12} (unchanged), \csc(\beta) = -\frac{13}{5}, \tan(\beta) = \frac{5}{12}, and \cot(\beta) = \frac{12}{5}.
+
+
+
+
+
+
+
+
+ For each of the following transformations of standard trigonometric functions, use your understanding of transformations to determine the domain, range, asymptotes, and period of the function, with careful justification. Then, check your results using Desmos or another graphing utility.
+
+
+
+
+
+
+ f(t) = 5\sec(t-\frac{\pi}{2}) + 3
+
+
+
+
+ g(t) = -\frac{1}{3}\csc(2t) - 4
+
+
+
+
+ h(t) = -7\tan(t+\frac{\pi}{4}) + 1
+
+
+
+
+ j(t) = \frac{1}{2}\cot(4t) - 2
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
+
+
+
+ f(t) = 5\sec\!\left(t - \frac{\pi}{2}\right) + 3: The parent \sec(t) has period 2\pi, domain all reals except \frac{\pi}{2} + j\pi, range (-\infty,-1]\cup[1,\infty). Shifting by \frac{\pi}{2} moves the asymptotes to t = j\pi. The vertical stretch by 5 and shift by 3 give range (-\infty,-2]\cup[8,\infty). Period remains 2\pi.
+
+
+
+
+ g(t) = -\frac{1}{3}\csc(2t) - 4: The horizontal compression by 2 gives period \pi and moves asymptotes to all multiples of \frac{\pi}{2} (i.e., t = \frac{j\pi}{2} for any integer j), so the domain is all reals except multiples of \frac{\pi}{2}. The factor -\frac{1}{3} and vertical shift -4 give range (-\infty, -\frac{13}{3}]\cup[-\frac{11}{3},\infty).
+
+
+
+
+ h(t) = -7\tan\!\left(t + \frac{\pi}{4}\right) + 1: The parent \tan(t) has period \pi, asymptotes at \frac{\pi}{2}+j\pi, range all reals. The horizontal shift moves asymptotes to t = \frac{\pi}{4} + j\pi. Domain is all reals except \frac{\pi}{4}+j\pi. Range remains all reals. Period remains \pi.
+
+
+
+
+ j(t) = \frac{1}{2}\cot(4t) - 2: The horizontal compression by 4 gives period \frac{\pi}{4} and moves asymptotes to t = \frac{j\pi}{4}. Domain is all reals except \frac{j\pi}{4}. Range remains all reals.
+
+
+
+
+
+
+
+
+
+
+ In a right triangle with hypotenuse 1 and vertical leg x,
+ with angle \theta opposite x,
+ determine the simplest expression you can for each of the following quantities in terms of x.
+
+
+
+
+
+
+ \sin(\theta)
+
+
+
+
+
+ \sec(\theta)
+
+
+
+
+
+ \csc(\theta)
+
+
+
+
+
+ \tan(\theta)
+
+
+
+
+
+ \cos(\arcsin(x))
+
+
+
+
+
+ \cot(\arcsin(x))
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
+ In the right triangle, the horizontal leg has length \sqrt{1-x^2} by the Pythagorean Theorem.
+
- A person standing 50 feet away from a streetlight observes that they cast a shadow that is 14 feet long. If a ray of light from the streetlight to the tip of the person's shadow forms an angle of 27.5^\circ with the ground, how tall is the person and how tall is the streetlight? What other information about the situation can you determine?
-
-
-
-
- Exercise Answer
-
-
-
-
- Let the angle of elevation be 27.5^\circ. The line from the streetlight to the tip of the shadow is the hypotenuse of a large right triangle with base 50 + 14 = 64 feet, so this hypotenuse is \frac{64}{\cos(27.5^\circ)} \approx 72.1 feet. The line from the person's head to the tip of the shadow is the hypotenuse of a smaller right triangle with base 14 feet, so it has length \frac{14}{\cos(27.5^\circ)} \approx 15.7 feet. The person's height is \sqrt{15.7^2 - 14^2} \approx 7.1 feet, and the streetlight's height is \sqrt{72.1^2 - 64^2} \approx 33.2 feet.
-
-
-
-
-
-
-
- A person watching a rocket launch uses a laser range-finder to measure the distance from themselves to the rocket. The range-finder also reports the angle at which the finder is being elevated from horizontal. At a certain instant, the range-finder reports that it is elevated at an angle of 17.4^\circ from horizontal and that the distance to the rocket is 1650 meters. How high off the ground is the rocket? Assuming a straight-line vertical path for the rocket that is perpendicular to the earth, how far away was the rocket from the range-finder at the moment it was launched?
-
-
-
-
- Exercise Answer
-
-
-
-
- With the range-finder elevated at 17.4^\circ and at distance 1650 meters from the rocket, the height of the rocket is 1650\sin(17.4^\circ) \approx 493.4 meters. The horizontal distance from the range-finder to the launch point is 1650\cos(17.4^\circ) \approx 1574 meters.
-
-
-
-
-
-
-
- A trough is constructed by bending a 4' \times 24' rectangular sheet of metal. Two symmetric folds 2 feet apart are made parallel to the longest side of the rectangle so that the trough has cross-sections in the shape of a trapezoid, as pictured in Figure. Determine a formula for V(\theta), the volume of the trough as a function of \theta.
-
-
-
-
A cross-section of the trough.
-
A cross-section of the trough.
-
-
-
-
-
- The volume of the trough is the area of a cross-section times the length of the trough.
-
-
-
-
- Exercise Answer
-
-
-
-
- Each side panel is a 2' \times 24' strip folded at angle \theta from the base. The depth of the trough is \sin(\theta) feet, and the two slanted sides each add \cos(\theta) feet to the top width, giving a total top width of 2 + 2\cos(\theta) feet. The cross-sectional area of the trapezoidal cross-section is \frac{1}{2}(2 + 2 + 2\cos(\theta))\sin(\theta) = \sin(\theta)(2 + \cos(\theta)) square feet. Multiplying by the length of 24 feet gives
-
- V(\theta) = 24\sin(\theta)(2 + \cos(\theta)).
-
-
+ A person standing 50 feet away from a streetlight observes that they cast a shadow that is 14 feet long. If a ray of light from the streetlight to the tip of the person's shadow forms an angle of 27.5^\circ with the ground, how tall is the person and how tall is the streetlight? What other information about the situation can you determine?
+
+
+
+
+ Exercise Answer
+
+
+
+
+ Let the angle of elevation be 27.5^\circ. The line from the streetlight to the tip of the shadow is the hypotenuse of a large right triangle with base 50 + 14 = 64 feet, so this hypotenuse is \frac{64}{\cos(27.5^\circ)} \approx 72.1 feet. The line from the person's head to the tip of the shadow is the hypotenuse of a smaller right triangle with base 14 feet, so it has length \frac{14}{\cos(27.5^\circ)} \approx 15.7 feet. The person's height is \sqrt{15.7^2 - 14^2} \approx 7.1 feet, and the streetlight's height is \sqrt{72.1^2 - 64^2} \approx 33.2 feet.
+
+
+
+
+
+
+
+ A person watching a rocket launch uses a laser range-finder to measure the distance from themselves to the rocket. The range-finder also reports the angle at which the finder is being elevated from horizontal. At a certain instant, the range-finder reports that it is elevated at an angle of 17.4^\circ from horizontal and that the distance to the rocket is 1650 meters. How high off the ground is the rocket? Assuming a straight-line vertical path for the rocket that is perpendicular to the earth, how far away was the rocket from the range-finder at the moment it was launched?
+
+
+
+
+ Exercise Answer
+
+
+
+
+ With the range-finder elevated at 17.4^\circ and at distance 1650 meters from the rocket, the height of the rocket is 1650\sin(17.4^\circ) \approx 493.4 meters. The horizontal distance from the range-finder to the launch point is 1650\cos(17.4^\circ) \approx 1574 meters.
+
+
+
+
+
+
+
+ A trough is constructed by bending a 4' \times 24' rectangular sheet of metal. Two symmetric folds 2 feet apart are made parallel to the longest side of the rectangle so that the trough has cross-sections in the shape of a trapezoid, as pictured in Figure. Determine a formula for V(\theta), the volume of the trough as a function of \theta.
+
+
+
+
A cross-section of the trough.
+
A cross-section of the trough.
+
+
+
+
+
+ The volume of the trough is the area of a cross-section times the length of the trough.
+
+
+
+
+ Exercise Answer
+
+
+
+
+ Each side panel is a 2' \times 24' strip folded at angle \theta from the base. The depth of the trough is \sin(\theta) feet, and the two slanted sides each add \cos(\theta) feet to the top width, giving a total top width of 2 + 2\cos(\theta) feet. The cross-sectional area of the trapezoidal cross-section is \frac{1}{2}(2 + 2 + 2\cos(\theta))\sin(\theta) = \sin(\theta)(2 + \cos(\theta)) square feet. Multiplying by the length of 24 feet gives
+
+ V(\theta) = 24\sin(\theta)(2 + \cos(\theta)).
+
+
- A wheelchair ramp is to be built so that the angle it forms with level ground is 4^{\circ}.
- If the ramp is going to rise from a level sidewalk up to a front porch that is 3 feet above the ground,
- how long does the ramp have to be?
- How far from the front porch will it meet the sidewalk?
- What is the slope of the ramp?
-
-
-
-
- Exercise Answer
-
-
-
-
- The ramp makes a 4^\circ angle with the ground and rises 3 feet. Its length is \frac{3}{\sin(4^\circ)} \approx 43.0 feet. It meets the sidewalk approximately \frac{3}{\tan(4^\circ)} \approx 42.9 feet from the porch. The slope is \frac{3}{42.9} \approx 0.0699.
-
-
-
-
-
-
-
- A person is flying a kite and at the end of a fixed length of string. Assume there is no slack in the string.
-
-
-
- At a certain moment, the kite is 170 feet off the ground, and the angle of elevation the string makes with the ground is 40^\circ.
-
-
-
-
-
-
- How far is it from the person flying the kite to another person who is standing directly beneath the kite?
-
-
-
-
- How much string is out between the person flying the kite and the kite itself?
-
-
-
-
- With the same amount of string out, the angle of elevation increases to 50^\circ. How high is the kite at this time?
-
-
-
-
-
-
-
- Exercise Answer
-
-
-
-
-
-
-
- With the kite at 170 feet elevation and elevation angle 40^\circ, the horizontal distance is \frac{170}{\tan(40^\circ)} \approx 202.6 feet.
-
-
-
-
- The string length is \frac{170}{\sin(40^\circ)} \approx 264.5 feet.
-
-
-
-
- With the same string length and elevation angle 50^\circ, the kite height is 264.5\sin(50^\circ) \approx 202.6 feet.
-
-
-
-
-
-
-
-
-
-
- An airplane is flying at a constant speed along a straight path above a straight road at a constant elevation of 2400 feet. A person on the road observes the plane flying directly at them and uses a sextant to measure the angle of elevation from them to the plane. The first measurement they take records an angle of 36^\circ; a second measurement taken 2 seconds later is 41^\circ.
-
-
-
- How far did the plane travel during the two seconds between the two angle measurements? How fast was the plane flying?
-
-
-
-
- Exercise Answer
-
-
-
-
- At angle 36^\circ, the plane's horizontal distance is \frac{2400}{\tan(36^\circ)} \approx 3301 feet. Two seconds later at 41^\circ, the distance is \frac{2400}{\tan(41^\circ)} \approx 2759 feet. The plane traveled 3301 - 2759 \approx 542 feet in 2 seconds, giving a speed of approximately 271 feet per second, or about 185 miles per hour.
-
+ A wheelchair ramp is to be built so that the angle it forms with level ground is 4^{\circ}.
+ If the ramp is going to rise from a level sidewalk up to a front porch that is 3 feet above the ground,
+ how long does the ramp have to be?
+ How far from the front porch will it meet the sidewalk?
+ What is the slope of the ramp?
+
+
+
+
+ Exercise Answer
+
+
+
+
+ The ramp makes a 4^\circ angle with the ground and rises 3 feet. Its length is \frac{3}{\sin(4^\circ)} \approx 43.0 feet. It meets the sidewalk approximately \frac{3}{\tan(4^\circ)} \approx 42.9 feet from the porch. The slope is \frac{3}{42.9} \approx 0.0699.
+
+
+
+
+
+
+
+ A person is flying a kite and at the end of a fixed length of string. Assume there is no slack in the string.
+
+
+
+ At a certain moment, the kite is 170 feet off the ground, and the angle of elevation the string makes with the ground is 40^\circ.
+
+
+
+
+
+
+ How far is it from the person flying the kite to another person who is standing directly beneath the kite?
+
+
+
+
+ How much string is out between the person flying the kite and the kite itself?
+
+
+
+
+ With the same amount of string out, the angle of elevation increases to 50^\circ. How high is the kite at this time?
+
+
+
+
+
+
+
+ Exercise Answer
+
+
+
+
+
+
+
+ With the kite at 170 feet elevation and elevation angle 40^\circ, the horizontal distance is \frac{170}{\tan(40^\circ)} \approx 202.6 feet.
+
+
+
+
+ The string length is \frac{170}{\sin(40^\circ)} \approx 264.5 feet.
+
+
+
+
+ With the same string length and elevation angle 50^\circ, the kite height is 264.5\sin(50^\circ) \approx 202.6 feet.
+
+
+
+
+
+
+
+
+
+
+ An airplane is flying at a constant speed along a straight path above a straight road at a constant elevation of 2400 feet. A person on the road observes the plane flying directly at them and uses a sextant to measure the angle of elevation from them to the plane. The first measurement they take records an angle of 36^\circ; a second measurement taken 2 seconds later is 41^\circ.
+
+
+
+ How far did the plane travel during the two seconds between the two angle measurements? How fast was the plane flying?
+
+
+
+
+ Exercise Answer
+
+
+
+
+ At angle 36^\circ, the plane's horizontal distance is \frac{2400}{\tan(36^\circ)} \approx 3301 feet. Two seconds later at 41^\circ, the distance is \frac{2400}{\tan(41^\circ)} \approx 2759 feet. The plane traveled 3301 - 2759 \approx 542 feet in 2 seconds, giving a speed of approximately 271 feet per second, or about 185 miles per hour.
+
- This text began as my sabbatical project in the fall semester of 2018,
- during which I wrote most of the material.
- For the sabbatical leave, I express my deep gratitude to Grand Valley State University for its support of the project,
- as well as to my colleagues in the Department of Mathematics and the College of Liberal Arts and Sciences for their endorsement of the project.
-
-
-
-
-
- The beautiful full-color .eps graphics, as well as the occasional interactive JavaScript graphics, use David Austin's Python library that employs Bill Casselman's PiScript.
- The .html version of the text is the result Rob Beezer's amazing work to develop the publishing language (formerly known as Mathbook XML); learn more at pretextbook.org.
- I'm grateful to the American Institute of Mathematics for hosting and funding
- a weeklong workshop in San Jose, CA, in April 2016, which enabled me to get started in . The ongoing support of the user group is invaluable, and David Farmer of AIM is has also been a source of major support and advocacy.
- Mitch Keller of the University of Wisconsin is the production editor of both Active Calculus: Single Variable and this text; his technical expertise is a gift.
-
-
-
-
-
-
-
- Contributors
-
- Users of the text contribute important insight: they find errors, suggest improvements, and offer feedback and impressions. I'm grateful for all of it. As you use the text, I hope you'll contact me to share anything you think could make the book better.
-
-
-
- The following contributing editors have offered feedback that includes information about typographical errors or suggestions to improve the exposition.
-
- This text is designed for college students who aspire to take calculus and who either need to take a course to prepare them for calculus or want to do some additional self-study. Many of the core topics of the course will be familiar to students who have completed high school. At the same time, we take a perspective on every topic that emphasizes how it is important in calculus. This text is written in the spirit of Active Calculus and is especially ideal for students who will eventually study calculus from that text. The reader will find that the text requires them to engage actively with the material, to view topics from multiple perspectives, and to develop deep conceptual undersanding of ideas.
-
-
-
- Many courses at the high school and college level with titles such as college algebra, precalculus, and trigonometry serve other disciplines and courses other than calculus. As such, these prerequisite classes frequently contain wide-ranging material that, while mathematically interesting and important, isn't necessary for calculus. Perhaps because of these additional topics, certain ideas that are essential in calculus are under-emphasized or ignored. In Active Prelude to Calculus, one of our top goals is to keep the focus narrow on the following most important ideas.
-
-
-
-
-
-
- Functions as processes. The mathematical concept of function is sophisticated. Understanding how a function is a special mathematical process that converts a collection of inputs to a collection of outputs is crucial for success in calculus, as calculus is the study of how functions change.
-
-
-
-
- Average rate of change. The central idea in differential calculus is the instantaneous rate of change of a function, which measures how fast a function's output changes with respect to changes in the input at a particular location. Because instantaneous rate of change is defined in terms of average rate of change, it's essential that students are comfortable and familiar with the idea, meaning, and applications of average rate of change.
-
-
-
-
- Library of basic functions. The vast majority of functions in calculus come from an algebraic combination of a collection of familiar basic functions that include power, circular, exponential, and logarithmic functions. By developing understanding of a relatively small family of basic functions and using these along with transformations to consider larger collections of functions, we work to make the central objects of calculus more intuitive and accessible.
-
-
-
-
- Families of functions that model important phenomena. Mathematics is the language of science, and it's remarkable how effective mathematics is at representing observable physical phenomena. From quadratic functions that model how an object falls under the influence of gravity, to shifted exponential functions that model how coffee cools, to sinusoidal functions that model how a spring-mass system oscillates, familiar basic functions find many important applications in the world around us. We regularly use these physical situations to help us see the importance of functions and to understand how families of functions that depend on different parameters are needed to represent these situations.
-
-
-
-
- The sine and cosine are circular functions. Many students are first introduced to the sine and cosine functions through right triangles. While this perspective is important, it is more important in calculus and other advanced courses to understand how the sine and cosine functions arise from a point traversing a circle. We take this circular function perspective early and first, and do so in order to develop deep understanding of how the familiar sine and cosine waves are generated.
-
-
-
-
- Inverses of functions. When a function has an inverse function, the inverse function affords us the opportunity to view an idea from a new perspective. Inverses also play a crucial role in solving algebraic equations and in determining unknown parameters in models. We emphasize the perspective that an inverse function is a process that reverses the process of the original function, as well as important basic functions that arise as inverses of other functions, especially logarithms and inverse trigonometric functions.
-
-
-
-
- Exact values versus approximate ones. The ability to represent numbers exactly is a powerful tool in mathematics. We regularly and consistently distinguish between a number's exact value, such as \sqrt{2}, and its approximation, say 1.414. This idea is also closely tied to functions and function notation: e^{-1}, \cos(2), and \ln(7) are all symbolic representations of exact numbers that can only be approximated by a computer.
-
-
-
-
- Finding function formulas in applied settings. In applied settings with unknown variables, it's especially useful to be able to represent relationships among variables, since such relationships often lead to functions whose behavior we can study. We work throughout Active Prelude to Calculus to ready students for problems in calculus that ask them to develop function formulas by observing relationships.
-
-
-
-
- Long-term trends, unbounded behavior, and limits. By working to study functions as objects themselves, we often focus on trends and overall behavior. In addition to introducing the ideas of a function being increasing or decreasing, or concave up or concave down, we also focus on using algebraic approaces to comprehend function behavior where the input and/or output increase without bound. In anticipation of calculus, we use limit notation and work to understand how this shorthand summarizes key features of functions.
-
-
-
-
-
-
-
-
- Features of the Text
-
- Instructors and students alike will find several consistent features in the presentation,
- including:
-
-
- Motivating Questions
-
- At the start of each section,
- we list 23 motivating questions
- that provide motivation for why the following material is of interest to us.
- One goal of each section is to answer each of the motivating questions.
-
-
-
-
- Preview Activities
-
- Each section of the text begins with a short introduction,
- followed by a preview activity.
- This brief reading and preview activity are designed to foreshadow the upcoming ideas in the remainder of the section;
- both the reading and preview activity are intended to be accessible to students
- in advance of class,
- and indeed to be completed by students before the particular section is to be considered
- in class.
-
-
-
-
- Activities
-
- A typical section in the text has at least three activities.
- These are designed to engage students in an inquiry-based style that encourages them to construct solutions to key examples on their own,
- working in small groups or individually.
-
-
-
-
- Exercises
-
- There are dozens of college algebra and trignometry texts with (collectively) tens of thousands of exercises.
- Rather than repeat standard and routine exercises in this text,
- we recommend the use of
-
- with its access to the Open Problem Library (OPL) and many thousands of relevant problems.
- In this text,
- each section includes a small collection of anonymous exercises that offer
- students immediate feedback without penalty,
- as well as 34 additional challenging exercises per section.
- Each of the non- exercises has multiple parts,
- requires the student to connect several key ideas,
- and expects that the student will do at least a modest amount of writing to answer the questions and explain their findings.
-
-
-
-
- Graphics
-
- As much as possible,
- we strive to demonstrate key fundamental ideas visually,
- and to encourage students to do the same.
- Throughout the text, we use full-colorTo keep cost low,
- the graphics in the print-on-demand version are in black and white.
- When the text itself refers to color in images,
- one needs to view the .html or .pdf electronically. graphics to exemplify and magnify key ideas,
- and to use this graphical perspective alongside both numerical and algebraic representations of calculus.
-
-
-
-
- Interactive graphics
-
- Many of the ideas of how functions behave are best understood dynamically;
- applets offer an often ideal format for investigations and demonstrations.
- Desmos provides a free and easy-to-use online graphing utility that we occasionally link to and often direct students to use. Thanks to David Austin, there are also select interactive javascript figures within the text itself.
-
-
-
-
- Summary of Key Ideas
-
- Each section concludes with a summary of the key ideas encountered in the preceding section;
- this summary normally reflects responses to the motivating questions that began the section.
-
-
-
-
-
-
-
-
- Students! Read this!
-
- This book is different.
-
-
-
- The text is available in three different formats: HTML, PDF, and print, each of which is available via links on the landing page at https://activecalculus.org/. The first two formats are free. If you are going to use the book electronically, the best mode is the HTML. The HTML version looks great in any browser, including on a smartphone, and the links are much easier to navigate in HTML than in PDF. Some particular direct suggestions about using the HTML follow among the next few paragraphs; alternatively, you can watch this short video from the author (based on using the text Active Calculus, which is similar). It is also wise to download and save the PDF, since you can use the PDF offline, while the HTML version requires an internet connection. An inexpensive print copy is available on Amazon.
-
-
-
- This book is intended to be read sequentially and engaged with, much more than to be used as a lookup reference. For example, each section begins with a short introduction and a Preview Activity; you should read the short introduction and complete the Preview Activity prior to class. Your instructor may require you to do this. Most Preview Activities can be completed in 15-20 minutes and are intended to be accessible based on the understanding you have from preceding sections.
-
-
-
- As you use the book, think of it as a workbook, not a worked-book. There is a great deal of scholarship that shows people learn better when they interactively engage and struggle with ideas themselves, rather than passively watch others. Thus, instead of reading worked examples or watching an instructor complete examples, you will engage with Activities that prompt you to grapple with concepts and develop deep understanding. You should expect to spend time in class working with peers on Activities and getting feedback from them and from your instructor. You can purchase a separate Activities Workbook from Amazon in which to record your work on the activities, or you can find a free PDF at the home page for the textbooks (scroll down in the left panel and look for the link to download PDF in the student workbook section) that has only the activities along with room to record your work. Your goal should be to do all of the activities in the relevant sections of the text and keep a careful record of your work.
-
-
-
- Each section concludes with a Summary. Reading the Summary after you have read the section and worked the Activities is a good way to find a short list of key ideas that are most essential to take from the section. A good study habit is to write similar summaries in your own words.
-
-
-
- At the end of each section, you'll find two types of Exercises. First, there are several anonymous exercises. These are online, interactive exercises that allow you to submit answers for immediate feedback with unlimited attempts without penalty; to submit answers, you have to be using the HTML version of the text (see this short video on the HTML version that includes a demonstration). You should use these exercises as a way to test your understanding of basic ideas in the preceding section. If your institution uses , you may also need to log in to a server as directed by your instructor to complete assigned sets as part of your course grade. The exercises included in this text are ungraded and not connected to any individual account. Following the exercises there are 3-4 additional challenging exercises that are designed to encourage you to connect ideas, investigate new situations, and write about your understanding.
-
-
-
-
-
- The best way to be successful in mathematics generally and calculus specifically is to strive to make sense of the main ideas. We make sense of ideas by asking questions, interacting with others, attempting to solve problems, making mistakes, revising attempts, and writing and speaking about our understanding. This text has been designed to help you make sense of key ideas that are needed in calculus and to help you be well-prepared for success in calculus; we wish you the very best as you undertake the large and challenging task of doing so.
-
-
-
-
-
- Instructors! Read this!
-
- This book is different. Before you read further, first read Students! Read this! as well as Our Goals. More information for instructors can also be found at the home page for the text, at activecalculus.org generally, and on the instructors page.
-
-
-
- Among the three formats (HTML, PDF, print), the HTML is optimal for display in class if you have a suitable projector. The HTML is also best for navigation, as links to internal and external references are much more obvious. We recommend saving a downloaded version of the PDF format as a backup in the event you don't have internet access. It's a good idea for each student to have a printed version of the Activities Workbook, which can be purchased from Amazon, or you can find a free PDF at the home page for the textbooks (scroll down in the left panel and look for the link to download PDF in the student workbook section) that has only the activities along with room to work; many instructors use the PDF to have coursepacks printed for students to purchase from their local bookstore.
-
-
-
- The text is written so that, on average, one section corresponds to two hours of class meeting time. A typical instructional sequence when starting a new section might look like the following:
-
-
-
- Students complete a Preview Activity in advance of class. Class begins with a short debrief among peers followed by all class discussion. (5-10 minutes)
-
-
-
-
- Brief lecture and discussion to build on the preview activity and set the stage for the next activity. (5-10 minutes)
-
-
-
-
- Students engage with peers to work on and discuss the first activity in the section. (15-20 minutes)
-
-
-
-
- Brief discussion and possibly lecture to reach closure on the preceding activity, followed by transition to new ideas. (Varies, but 5-15 minutes)
-
-
-
-
- Possibly begin next activity.
-
-
-
- The next hour of class would be similar, but without the Preview Activity to complete prior to class: the principal focus of class will be completing 2 activities. Then rinse and repeat.
-
-
-
- We recommend that instructors use appropriate incentives to encourage students to complete Preview Activities prior to class. Having these be part of completion-based assignments that count 5% of the semester grade usually results in the vast majority of students completing the vast majority of the previews. If you'd like to see a sample syllabus for how to organize a course and weight various assignments, you can request one via email to the author.
-
-
-
- Note that the exercises in the HTML version are anonymous and there's not a way to track students' engagement with them. These are intended to be formative for students and provide them with immediate feedback without penalty. If your institution is a user, in the near future we will have sets of .def files that correspond to the sections in the text; these will be available upon request to the author.
-
-
-
-
-
-
-
- The source code for the text can be found on GitHub. If you find errors in the text or have other suggestions, you can file an issue on GitHub or email the author directly. To engage with instructors who use the text, we maintain a Google group and a blog; you can request to join the Google group via the link at the instructors page. Finally, if you're interested in a video presentation on using the similar Active Calculus text, you can see this online video presentation to the MIT Electronic Seminar on Mathematics Education; at about the 17-minute mark, the portion begins where we demonstrate features of and how to use the text.
-
-
-
- Thank you for considering Active Prelude to Calculus as a resource to help your students develop deep understanding of the subject. I wish you the very best in your work and hope to hear from you.
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ This text began as my sabbatical project in the fall semester of 2018,
+ during which I wrote most of the material.
+ For the sabbatical leave, I express my deep gratitude to Grand Valley State University for its support of the project,
+ as well as to my colleagues in the Department of Mathematics and the College of Liberal Arts and Sciences for their endorsement of the project.
+
+
+
+
+
+ The beautiful full-color .eps graphics, as well as the occasional interactive JavaScript graphics, use David Austin's Python library that employs Bill Casselman's PiScript.
+ The .html version of the text is the result Rob Beezer's amazing work to develop the publishing language (formerly known as Mathbook XML); learn more at pretextbook.org.
+ I'm grateful to the American Institute of Mathematics for hosting and funding
+ a weeklong workshop in San Jose, CA, in April 2016, which enabled me to get started in . The ongoing support of the user group is invaluable, and David Farmer of AIM is has also been a source of major support and advocacy.
+ Mitch Keller of the University of Wisconsin is the production editor of both Active Calculus: Single Variable and this text; his technical expertise is a gift.
+
+
+
+
+
+
+
+ Contributors
+
+ Users of the text contribute important insight: they find errors, suggest improvements, and offer feedback and impressions. I'm grateful for all of it. As you use the text, I hope you'll contact me to share anything you think could make the book better.
+
+
+
+ The following contributing editors have offered feedback that includes information about typographical errors or suggestions to improve the exposition.
+
+ This text is designed for college students who aspire to take calculus and who either need to take a course to prepare them for calculus or want to do some additional self-study. Many of the core topics of the course will be familiar to students who have completed high school. At the same time, we take a perspective on every topic that emphasizes how it is important in calculus. This text is written in the spirit of Active Calculus and is especially ideal for students who will eventually study calculus from that text. The reader will find that the text requires them to engage actively with the material, to view topics from multiple perspectives, and to develop deep conceptual undersanding of ideas.
+
+
+
+ Many courses at the high school and college level with titles such as college algebra, precalculus, and trigonometry serve other disciplines and courses other than calculus. As such, these prerequisite classes frequently contain wide-ranging material that, while mathematically interesting and important, isn't necessary for calculus. Perhaps because of these additional topics, certain ideas that are essential in calculus are under-emphasized or ignored. In Active Prelude to Calculus, one of our top goals is to keep the focus narrow on the following most important ideas.
+
+
+
+
+
+
+ Functions as processes. The mathematical concept of function is sophisticated. Understanding how a function is a special mathematical process that converts a collection of inputs to a collection of outputs is crucial for success in calculus, as calculus is the study of how functions change.
+
+
+
+
+ Average rate of change. The central idea in differential calculus is the instantaneous rate of change of a function, which measures how fast a function's output changes with respect to changes in the input at a particular location. Because instantaneous rate of change is defined in terms of average rate of change, it's essential that students are comfortable and familiar with the idea, meaning, and applications of average rate of change.
+
+
+
+
+ Library of basic functions. The vast majority of functions in calculus come from an algebraic combination of a collection of familiar basic functions that include power, circular, exponential, and logarithmic functions. By developing understanding of a relatively small family of basic functions and using these along with transformations to consider larger collections of functions, we work to make the central objects of calculus more intuitive and accessible.
+
+
+
+
+ Families of functions that model important phenomena. Mathematics is the language of science, and it's remarkable how effective mathematics is at representing observable physical phenomena. From quadratic functions that model how an object falls under the influence of gravity, to shifted exponential functions that model how coffee cools, to sinusoidal functions that model how a spring-mass system oscillates, familiar basic functions find many important applications in the world around us. We regularly use these physical situations to help us see the importance of functions and to understand how families of functions that depend on different parameters are needed to represent these situations.
+
+
+
+
+ The sine and cosine are circular functions. Many students are first introduced to the sine and cosine functions through right triangles. While this perspective is important, it is more important in calculus and other advanced courses to understand how the sine and cosine functions arise from a point traversing a circle. We take this circular function perspective early and first, and do so in order to develop deep understanding of how the familiar sine and cosine waves are generated.
+
+
+
+
+ Inverses of functions. When a function has an inverse function, the inverse function affords us the opportunity to view an idea from a new perspective. Inverses also play a crucial role in solving algebraic equations and in determining unknown parameters in models. We emphasize the perspective that an inverse function is a process that reverses the process of the original function, as well as important basic functions that arise as inverses of other functions, especially logarithms and inverse trigonometric functions.
+
+
+
+
+ Exact values versus approximate ones. The ability to represent numbers exactly is a powerful tool in mathematics. We regularly and consistently distinguish between a number's exact value, such as \sqrt{2}, and its approximation, say 1.414. This idea is also closely tied to functions and function notation: e^{-1}, \cos(2), and \ln(7) are all symbolic representations of exact numbers that can only be approximated by a computer.
+
+
+
+
+ Finding function formulas in applied settings. In applied settings with unknown variables, it's especially useful to be able to represent relationships among variables, since such relationships often lead to functions whose behavior we can study. We work throughout Active Prelude to Calculus to ready students for problems in calculus that ask them to develop function formulas by observing relationships.
+
+
+
+
+ Long-term trends, unbounded behavior, and limits. By working to study functions as objects themselves, we often focus on trends and overall behavior. In addition to introducing the ideas of a function being increasing or decreasing, or concave up or concave down, we also focus on using algebraic approaches to comprehend function behavior where the input and/or output increase without bound. In anticipation of calculus, we use limit notation and work to understand how this shorthand summarizes key features of functions.
+
+
+
+
+
+
+
+
+ Features of the Text
+
+ Instructors and students alike will find several consistent features in the presentation,
+ including:
+
+
+ Motivating Questions
+
+ At the start of each section,
+ we list 23 motivating questions
+ that provide motivation for why the following material is of interest to us.
+ One goal of each section is to answer each of the motivating questions.
+
+
+
+
+ Preview Activities
+
+ Each section of the text begins with a short introduction,
+ followed by a preview activity.
+ This brief reading and preview activity are designed to foreshadow the upcoming ideas in the remainder of the section;
+ both the reading and preview activity are intended to be accessible to students
+ in advance of class,
+ and indeed to be completed by students before the particular section is to be considered
+ in class.
+
+
+
+
+ Activities
+
+ A typical section in the text has at least three activities.
+ These are designed to engage students in an inquiry-based style that encourages them to construct solutions to key examples on their own,
+ working in small groups or individually.
+
+
+
+
+ Exercises
+
+ There are dozens of college algebra and trignometry texts with (collectively) tens of thousands of exercises.
+ Rather than repeat standard and routine exercises in this text,
+ we recommend the use of
+
+ with its access to the Open Problem Library (OPL) and many thousands of relevant problems.
+ In this text,
+ each section includes a small collection of anonymous exercises that offer
+ students immediate feedback without penalty,
+ as well as 34 additional challenging exercises per section.
+ Each of the non- exercises has multiple parts,
+ requires the student to connect several key ideas,
+ and expects that the student will do at least a modest amount of writing to answer the questions and explain their findings.
+
+
+
+
+ Graphics
+
+ As much as possible,
+ we strive to demonstrate key fundamental ideas visually,
+ and to encourage students to do the same.
+ Throughout the text, we use full-colorTo keep cost low,
+ the graphics in the print-on-demand version are in black and white.
+ When the text itself refers to color in images,
+ one needs to view the .html or .pdf electronically. graphics to exemplify and magnify key ideas,
+ and to use this graphical perspective alongside both numerical and algebraic representations of calculus.
+
+
+
+
+ Interactive graphics
+
+ Many of the ideas of how functions behave are best understood dynamically;
+ applets offer an often ideal format for investigations and demonstrations.
+ Desmos provides a free and easy-to-use online graphing utility that we occasionally link to and often direct students to use. Thanks to David Austin, there are also select interactive javascript figures within the text itself.
+
+
+
+
+ Summary of Key Ideas
+
+ Each section concludes with a summary of the key ideas encountered in the preceding section;
+ this summary normally reflects responses to the motivating questions that began the section.
+
+
+
+
+
+
+
+
+ Students! Read this!
+
+ This book is different.
+
+
+
+ The text is available in three different formats: HTML, PDF, and print, each of which is available via links on the landing page at https://activecalculus.org/. The first two formats are free. If you are going to use the book electronically, the best mode is the HTML. The HTML version looks great in any browser, including on a smartphone, and the links are much easier to navigate in HTML than in PDF. Some particular direct suggestions about using the HTML follow among the next few paragraphs; alternatively, you can watch this short video from the author (based on using the text Active Calculus, which is similar). It is also wise to download and save the PDF, since you can use the PDF offline, while the HTML version requires an internet connection. An inexpensive print copy is available on Amazon.
+
+
+
+ This book is intended to be read sequentially and engaged with, much more than to be used as a lookup reference. For example, each section begins with a short introduction and a Preview Activity; you should read the short introduction and complete the Preview Activity prior to class. Your instructor may require you to do this. Most Preview Activities can be completed in 15-20 minutes and are intended to be accessible based on the understanding you have from preceding sections.
+
+
+
+ As you use the book, think of it as a workbook, not a worked-book. There is a great deal of scholarship that shows people learn better when they interactively engage and struggle with ideas themselves, rather than passively watch others. Thus, instead of reading worked examples or watching an instructor complete examples, you will engage with Activities that prompt you to grapple with concepts and develop deep understanding. You should expect to spend time in class working with peers on Activities and getting feedback from them and from your instructor. You can purchase a separate Activities Workbook from Amazon in which to record your work on the activities, or you can find a free PDF at the home page for the textbooks (scroll down in the left panel and look for the link to download PDF in the student workbook section) that has only the activities along with room to record your work. Your goal should be to do all of the activities in the relevant sections of the text and keep a careful record of your work.
+
+
+
+ Each section concludes with a Summary. Reading the Summary after you have read the section and worked the Activities is a good way to find a short list of key ideas that are most essential to take from the section. A good study habit is to write similar summaries in your own words.
+
+
+
+ At the end of each section, you'll find two types of Exercises. First, there are several anonymous exercises. These are online, interactive exercises that allow you to submit answers for immediate feedback with unlimited attempts without penalty; to submit answers, you have to be using the HTML version of the text (see this short video on the HTML version that includes a demonstration). You should use these exercises as a way to test your understanding of basic ideas in the preceding section. If your institution uses , you may also need to log in to a server as directed by your instructor to complete assigned sets as part of your course grade. The exercises included in this text are ungraded and not connected to any individual account. Following the exercises there are 3-4 additional challenging exercises that are designed to encourage you to connect ideas, investigate new situations, and write about your understanding.
+
+
+
+
+
+ The best way to be successful in mathematics generally and calculus specifically is to strive to make sense of the main ideas. We make sense of ideas by asking questions, interacting with others, attempting to solve problems, making mistakes, revising attempts, and writing and speaking about our understanding. This text has been designed to help you make sense of key ideas that are needed in calculus and to help you be well-prepared for success in calculus; we wish you the very best as you undertake the large and challenging task of doing so.
+
+
+
+
+
+ Instructors! Read this!
+
+ This book is different. Before you read further, first read Students! Read this! as well as Our Goals. More information for instructors can also be found at the home page for the text, at activecalculus.org generally, and on the instructors page.
+
+
+
+ Among the three formats (HTML, PDF, print), the HTML is optimal for display in class if you have a suitable projector. The HTML is also best for navigation, as links to internal and external references are much more obvious. We recommend saving a downloaded version of the PDF format as a backup in the event you don't have internet access. It's a good idea for each student to have a printed version of the Activities Workbook, which can be purchased from Amazon, or you can find a free PDF at the home page for the textbooks (scroll down in the left panel and look for the link to download PDF in the student workbook section) that has only the activities along with room to work; many instructors use the PDF to have coursepacks printed for students to purchase from their local bookstore.
+
+
+
+ The text is written so that, on average, one section corresponds to two hours of class meeting time. A typical instructional sequence when starting a new section might look like the following:
+
+
+
+ Students complete a Preview Activity in advance of class. Class begins with a short debrief among peers followed by all class discussion. (5-10 minutes)
+
+
+
+
+ Brief lecture and discussion to build on the preview activity and set the stage for the next activity. (5-10 minutes)
+
+
+
+
+ Students engage with peers to work on and discuss the first activity in the section. (15-20 minutes)
+
+
+
+
+ Brief discussion and possibly lecture to reach closure on the preceding activity, followed by transition to new ideas. (Varies, but 5-15 minutes)
+
+
+
+
+ Possibly begin next activity.
+
+
+
+ The next hour of class would be similar, but without the Preview Activity to complete prior to class: the principal focus of class will be completing 2 activities. Then rinse and repeat.
+
+
+
+ We recommend that instructors use appropriate incentives to encourage students to complete Preview Activities prior to class. Having these be part of completion-based assignments that count 5% of the semester grade usually results in the vast majority of students completing the vast majority of the previews. If you'd like to see a sample syllabus for how to organize a course and weight various assignments, you can request one via email to the author.
+
+
+
+ Note that the exercises in the HTML version are anonymous and there's not a way to track students' engagement with them. These are intended to be formative for students and provide them with immediate feedback without penalty. If your institution is a user, in the near future we will have sets of .def files that correspond to the sections in the text; these will be available upon request to the author.
+
+
+
+
+
+
+
+ The source code for the text can be found on GitHub. If you find errors in the text or have other suggestions, you can file an issue on GitHub or email the author directly. To engage with instructors who use the text, we maintain a Google group and a blog; you can request to join the Google group via the link at the instructors page. Finally, if you're interested in a video presentation on using the similar Active Calculus text, you can see this online video presentation to the MIT Electronic Seminar on Mathematics Education; at about the 17-minute mark, the portion begins where we demonstrate features of and how to use the text.
+
+
+
+ Thank you for considering Active Prelude to Calculus as a resource to help your students develop deep understanding of the subject. I wish you the very best in your work and hope to hear from you.
+
- Let the height function for a ball tossed vertically be given by s(t) = 64 - 16(t-1)^2,
- where t is measured in seconds and s is measured in feet above the ground.
-
- What are the units on the quantity AV_{[1.5,2.5]}?
- What is the meaning of this number in the context of the rising/falling ball?
-
-
-
-
-
-
- The units are feet per second. The value AV_{[1.5,2.5]} = -32 means that
- between t = 1.5 and t = 2.5 seconds, the ball's height decreases
- at an average rate of 32 feet per second.
-
-
-
-
-
-
- In Desmos, plot the function
- s(t) = 64 - 16(t-1)^2 along with the points
- (1.5,s(1.5)) and (2.5, s(2.5)).
- Make a copy of your plot on the axes in the figure provided,
- labeling key points as well as the scale on your axes.
- What is the domain of the model?
- The range?
- Why?
-
-
-
ADD ALT TEXT TO THIS IMAGE
-
-
-
-
-
-
- Without additional context specifying when the ball is launched or lands, the
- domain and range of the model are not fully determined by the equation alone.
- The formula s(t) = 64 - 16(t-1)^2 is a classical free-fall equation,
- and negative values of t or s are not necessarily unphysical.
-
-
-
-
-
-
-
-
- Work by hand to find the equation of the line through the points
- (1.5,s(1.5)) and (2.5, s(2.5)).
- Write the line in the form
- y = mt + b and plot the line in Desmos,
- as well as on the axes above.
-
-
-
-
-
-
- The slope is m = AV_{[1.5,2.5]} = -32. Using the point (1.5, 60)
- to find the vertical intercept:
-
- 60 &= -32(1.5) + b
- b &= 108.
-
- The equation of the line is y = -32t + 108.
-
-
-
-
-
-
-
-
- What is a geometric interpretation of the value
- AV_{[1.5,2.5]} in light of your work in the preceding questions?
-
-
-
-
-
-
- AV_{[1.5,2.5]} is the slope of the secant line connecting the points
- (1.5, s(1.5)) and (2.5, s(2.5)) on the graph of s.
-
-
-
-
-
-
- How do your answers in the preceding questions change if we instead consider the interval
- [0.25, 0.75]? [0.5, 1.5]? [1,3]?
-
-
-
-
-
-
- Interval [0.25, 0.75]:
- s(0.25) = 55, s(0.75) = 63, so
- AV_{[0.25,0.75]} = \frac{63-55}{0.5} = 16 \text{ ft/sec.}
- The secant line through (0.25, 55) with slope 16 is y = 16t + 51.
-
-
- Interval [0.5, 1.5]:
- s(0.5) = 60 and s(1.5) = 60, so
- AV_{[0.5,1.5]} = \frac{60-60}{1} = 0 \text{ ft/sec.}
- The secant line is the horizontal line y = 60.
-
-
- Interval [1, 3]:
- s(1) = 64, s(3) = 0, so
- AV_{[1,3]} = \frac{0-64}{2} = -32 \text{ ft/sec.}
- The secant line through (3, 0) with slope -32 is y = -32t + 96.
-
+ Let the height function for a ball tossed vertically be given by s(t) = 64 - 16(t-1)^2,
+ where t is measured in seconds and s is measured in feet above the ground.
+
+ What are the units on the quantity AV_{[1.5,2.5]}?
+ What is the meaning of this number in the context of the rising/falling ball?
+
+
+
+
+
+ The units are feet per second. The value AV_{[1.5,2.5]} = -32 means that
+ between t = 1.5 and t = 2.5 seconds, the ball's height decreases
+ at an average rate of 32 feet per second.
+
+
+
+
+
+
+ In Desmos, plot the function
+ s(t) = 64 - 16(t-1)^2 along with the points
+ (1.5,s(1.5)) and (2.5, s(2.5)).
+ Make a copy of your plot on the axes in the figure provided,
+ labeling key points as well as the scale on your axes.
+ What is the domain of the model?
+ The range?
+ Why?
+
+
+
ADD ALT TEXT TO THIS IMAGE
+
+
+
+
+
+ Without additional context specifying when the ball is launched or lands, the
+ domain and range of the model are not fully determined by the equation alone.
+ The formula s(t) = 64 - 16(t-1)^2 is a classical free-fall equation,
+ and negative values of t or s are not necessarily unphysical.
+
+
+
+
+
+
+
+
+ Work by hand to find the equation of the line through the points
+ (1.5,s(1.5)) and (2.5, s(2.5)).
+ Write the line in the form
+ y = mt + b and plot the line in Desmos,
+ as well as on the axes above.
+
+
+
+
+
+ The slope is m = AV_{[1.5,2.5]} = -32. Using the point (1.5, 60)
+ to find the vertical intercept:
+
+ 60 &= -32(1.5) + b
+ b &= 108.
+
+ The equation of the line is y = -32t + 108.
+
+
+
+
+
+
+
+
+ What is a geometric interpretation of the value
+ AV_{[1.5,2.5]} in light of your work in the preceding questions?
+
+
+
+
+
+ AV_{[1.5,2.5]} is the slope of the secant line connecting the points
+ (1.5, s(1.5)) and (2.5, s(2.5)) on the graph of s.
+
+
+
+
+
+
+ How do your answers in the preceding questions change if we instead consider the interval
+ [0.25, 0.75]? [0.5, 1.5]? [1,3]?
+
+
+
+
+
+ Interval [0.25, 0.75]:
+ s(0.25) = 55, s(0.75) = 63, so
+ AV_{[0.25,0.75]} = \frac{63-55}{0.5} = 16 \text{ ft/sec.}
+ The secant line through (0.25, 55) with slope 16 is y = 16t + 51.
+
+
+ Interval [0.5, 1.5]:
+ s(0.5) = 60 and s(1.5) = 60, so
+ AV_{[0.5,1.5]} = \frac{60-60}{1} = 0 \text{ ft/sec.}
+ The secant line is the horizontal line y = 60.
+
+
+ Interval [1, 3]:
+ s(1) = 64, s(3) = 0, so
+ AV_{[1,3]} = \frac{0-64}{2} = -32 \text{ ft/sec.}
+ The secant line through (3, 0) with slope -32 is y = -32t + 96.
+
- Consider the functions f and g defined by the following table and the piecewise linear functions p and q defined by the following figure. Assume that the lines in the figure pass through whole number coordinates where they appear to do so; for example, (2,2) lies on the graph of q, and (3,-3) lies on the graph of p.
-
- From the graph, p(-1) = \frac{1}{5} and q(-1) = \frac{7}{2}, so
- r(-1) = p(-1) - q(-1) = \frac{1}{5} - \frac{7}{2} = \frac{2}{10} - \frac{35}{10} = -\frac{33}{10}.
-
-
-
-
-
-
- Are there any values of x for which r(x) = 0? If not, explain why; if so, determine all such values, with justification.
-
-
-
-
-
-
- Yes. Setting r(x) = 0 requires p(x) = q(x).
- On the relevant interval, p(x) = -\frac{2}{5}x - \frac{1}{5} and
- q(x) = 4x + 12, so:
-
- -\frac{2}{5}x - \frac{1}{5} &= 4x + 12
- -\frac{22}{5}x &= \frac{61}{5}
- x &= -\frac{61}{22}.
-
-
-
-
-
-
-
- Let k(x) = f(x) \cdot g(x). Determine k(0).
-
-
-
-
-
-
- From the table, f(0) = 5 and g(0) = 9, so
- k(0) = f(0) \cdot g(0) = 5 \cdot 9 = 45.
-
-
-
-
-
-
- Let s(x) = \frac{p(x)}{q(x)}. Determine s(1) exactly.
-
-
-
-
-
-
- From the graph, p(1) = -\frac{3}{5} and q(1) = \frac{5}{2}, so
- s(1) = \frac{p(1)}{q(1)} = \frac{-3/5}{5/2} = -\frac{3}{5} \cdot \frac{2}{5} = -\frac{6}{25}.
-
-
-
-
-
-
- Are there any values of x in the interval -4 \le x \le 4 for which s(x) is not defined? If not, explain why; if so, determine all such values, with justification.
-
-
-
-
-
-
- Yes: s(x) is undefined wherever q(x) = 0.
- From the graph, q(-3) = 0, so s(-3) is undefined.
-
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ Consider the functions f and g defined by the following table and the piecewise linear functions p and q defined by the following figure. Assume that the lines in the figure pass through whole number coordinates where they appear to do so; for example, (2,2) lies on the graph of q, and (3,-3) lies on the graph of p.
+
+ From the graph, p(-1) = \frac{1}{5} and q(-1) = \frac{7}{2}, so
+ r(-1) = p(-1) - q(-1) = \frac{1}{5} - \frac{7}{2} = \frac{2}{10} - \frac{35}{10} = -\frac{33}{10}.
+
+
+
+
+
+
+ Are there any values of x for which r(x) = 0? If not, explain why; if so, determine all such values, with justification.
+
+
+
+
+
+ Yes. Setting r(x) = 0 requires p(x) = q(x).
+ On the relevant interval, p(x) = -\frac{2}{5}x - \frac{1}{5} and
+ q(x) = 4x + 12, so:
+
+ -\frac{2}{5}x - \frac{1}{5} &= 4x + 12
+ -\frac{22}{5}x &= \frac{61}{5}
+ x &= -\frac{61}{22}.
+
+
+
+
+
+
+
+ Let k(x) = f(x) \cdot g(x). Determine k(0).
+
+
+
+
+
+ From the table, f(0) = 5 and g(0) = 9, so
+ k(0) = f(0) \cdot g(0) = 5 \cdot 9 = 45.
+
+
+
+
+
+
+ Let s(x) = \frac{p(x)}{q(x)}. Determine s(1) exactly.
+
+
+
+
+
+ From the graph, p(1) = -\frac{3}{5} and q(1) = \frac{5}{2}, so
+ s(1) = \frac{p(1)}{q(1)} = \frac{-3/5}{5/2} = -\frac{3}{5} \cdot \frac{2}{5} = -\frac{6}{25}.
+
+
+
+
+
+
+ Are there any values of x in the interval -4 \le x \le 4 for which s(x) is not defined? If not, explain why; if so, determine all such values, with justification.
+
+
+
+
+
+ Yes: s(x) is undefined wherever q(x) = 0.
+ From the graph, q(-3) = 0, so s(-3) is undefined.
+
- Review the introductory example with f(x) = x^2 - 1 and g(t) = 3t - 4, which involved functions similar to p and q in part (a). What is the biggest difference between your work in (a) above and in the introductory example?
-
-
-
-
-
-
- In the introductory example, the inner function g(t) = 3t-4 is linear
- and the outer function f(x) = x^2 - 1 is quadratic. In part (a), the
- roles are reversed: the inner function q(t) = t^2-1 is quadratic and the
- outer function p(x) = 3x-4 is linear.
-
-
-
-
-
-
- Let t = s(z) = \frac{1}{z+4} and recall that x = q(t) = t^2 - 1. Determine a formula for x = q(s(z)) that depends only on z.
-
+ Review the introductory example with f(x) = x^2 - 1 and g(t) = 3t - 4, which involved functions similar to p and q in part (a). What is the biggest difference between your work in (a) above and in the introductory example?
+
+
+
+
+
+ In the introductory example, the inner function g(t) = 3t-4 is linear
+ and the outer function f(x) = x^2 - 1 is quadratic. In part (a), the
+ roles are reversed: the inner function q(t) = t^2-1 is quadratic and the
+ outer function p(x) = 3x-4 is linear.
+
+
+
+
+
+
+ Let t = s(z) = \frac{1}{z+4} and recall that x = q(t) = t^2 - 1. Determine a formula for x = q(s(z)) that depends only on z.
+
- Use the equation T = 40 + 0.25N that relates the temperature, T, to the number of chirps per minute, N, to respond to the questions below. The equation is also called Dolbear's Law.
-
-
-
-
-
-
- If we hear snowy tree crickets chirping at a rate of 92 chirps per minute, what does Dolbear's Law suggest should be the outside temperature?
-
-
-
-
-
-
We seek T when N = 92: T = 40 + 0.25(92) = 63. So the temperature should be 63^\circ F.
-
-
-
-
-
- If the outside temperature is 77^\circ F, how many chirps per minute should we expect to hear?
-
-
-
-
-
-
We seek N when T = 77. Solving 77 = 40 + 0.25N gives N = \frac{77 - 40}{0.25} = 148 chirps per minute.
-
-
-
-
-
- Is the model valid for determining the number of chirps one should hear when the outside temperature is 35^\circ F? Why or why not?
-
-
-
-
-
-
The model is known to be accurate only for temperatures from 50^\circ to 85^\circ F, so 35^\circ F lies outside the valid range and we should not expect reasonable results at this temperature.
-
-
-
-
-
- Suppose that in the morning an observer hears 65 chirps per minute, and several hours later hears 75 chirps per minute. How much has the temperature risen between observations?
-
-
-
-
-
-
At 65 chirps per minute the temperature is T = 40 + 0.25(65) = 56.25^\circ F, and at 75 chirps per minute the temperature is T = 40 + 0.25(75) = 58.75^\circ F. The temperature has risen by 58.75 - 56.25 = 2.5 degrees Fahrenheit.
-
-
-
-
-
- Dolbear's Law is known to be accurate for temperatures from 50^\circ to 85^\circ. What is the fewest number of chirps per minute an observer could expect to hear? the greatest number of chirps per minute?
-
-
-
-
-
-
At 50^\circ F we get N = \frac{50 - 40}{0.25} = 40 chirps per minute; at 85^\circ F we get N = \frac{85 - 40}{0.25} = 180 chirps per minute. The fewest expected is 40 and the greatest is 180 chirps per minute.
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ Use the equation T = 40 + 0.25N that relates the temperature, T, to the number of chirps per minute, N, to respond to the questions below. The equation is also called Dolbear's Law.
+
+
+
+
+
+
+ If we hear snowy tree crickets chirping at a rate of 92 chirps per minute, what does Dolbear's Law suggest should be the outside temperature?
+
+
+
+
+
We seek T when N = 92: T = 40 + 0.25(92) = 63. So the temperature should be 63^\circ F.
+
+
+
+
+
+ If the outside temperature is 77^\circ F, how many chirps per minute should we expect to hear?
+
+
+
+
+
We seek N when T = 77. Solving 77 = 40 + 0.25N gives N = \frac{77 - 40}{0.25} = 148 chirps per minute.
+
+
+
+
+
+ Is the model valid for determining the number of chirps one should hear when the outside temperature is 35^\circ F? Why or why not?
+
+
+
+
+
The model is known to be accurate only for temperatures from 50^\circ to 85^\circ F, so 35^\circ F lies outside the valid range and we should not expect reasonable results at this temperature.
+
+
+
+
+
+ Suppose that in the morning an observer hears 65 chirps per minute, and several hours later hears 75 chirps per minute. How much has the temperature risen between observations?
+
+
+
+
+
At 65 chirps per minute the temperature is T = 40 + 0.25(65) = 56.25^\circ F, and at 75 chirps per minute the temperature is T = 40 + 0.25(75) = 58.75^\circ F. The temperature has risen by 58.75 - 56.25 = 2.5 degrees Fahrenheit.
+
+
+
+
+
+ Dolbear's Law is known to be accurate for temperatures from 50^\circ to 85^\circ. What is the fewest number of chirps per minute an observer could expect to hear? the greatest number of chirps per minute?
+
+
+
+
+
At 50^\circ F we get N = \frac{50 - 40}{0.25} = 40 chirps per minute; at 85^\circ F we get N = \frac{85 - 40}{0.25} = 180 chirps per minute. The fewest expected is 40 and the greatest is 180 chirps per minute.
- Recall that F = g(C) = \frac{9}{5}C + 32 is the function that takes Celsius temperature inputs and produces the corresponding Fahrenheit temperature outputs.
-
-
-
-
-
-
- Show that it is possible to solve the equation F = \frac{9}{5}C + 32 for C in terms of F and that doing so results in the equation C = \frac{5}{9}(F-32).
-
-
-
-
-
-
- Subtracting 32 from both sides: F - 32 = \frac{9}{5}C.
- Multiplying both sides by \frac{5}{9}:
- C = \frac{5}{9}(F-32).
-
-
-
-
-
-
- Note that the equation C = \frac{5}{9}(F-32) expresses C as a function of F. Call this function h so that C = h(F) = \frac{5}{9}(F-32).
-
-
- Find the simplest expression that you can for the composite function j(C) = h(g(C)).
-
- Why are the functions j and k so simple? Explain by discussing how the functions g and h process inputs to generate outputs and what happens when we first execute one followed by the other.
-
-
-
-
-
-
- The function g converts Celsius to Fahrenheit, and h converts
- Fahrenheit back to Celsius. Composing them in either order simply undoes the
- conversion: applying both functions in sequence returns the original input.
- This is why j(C) = C and k(F) = F — the two functions are
- inverses of each other.
-
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ Recall that F = g(C) = \frac{9}{5}C + 32 is the function that takes Celsius temperature inputs and produces the corresponding Fahrenheit temperature outputs.
+
+
+
+
+
+
+ Show that it is possible to solve the equation F = \frac{9}{5}C + 32 for C in terms of F and that doing so results in the equation C = \frac{5}{9}(F-32).
+
+
+
+
+
+ Subtracting 32 from both sides: F - 32 = \frac{9}{5}C.
+ Multiplying both sides by \frac{5}{9}:
+ C = \frac{5}{9}(F-32).
+
+
+
+
+
+
+ Note that the equation C = \frac{5}{9}(F-32) expresses C as a function of F. Call this function h so that C = h(F) = \frac{5}{9}(F-32).
+
+
+ Find the simplest expression that you can for the composite function j(C) = h(g(C)).
+
+ Why are the functions j and k so simple? Explain by discussing how the functions g and h process inputs to generate outputs and what happens when we first execute one followed by the other.
+
+
+
+
+
+ The function g converts Celsius to Fahrenheit, and h converts
+ Fahrenheit back to Celsius. Composing them in either order simply undoes the
+ conversion: applying both functions in sequence returns the original input.
+ This is why j(C) = C and k(F) = F — the two functions are
+ inverses of each other.
+
- Determine AV_{[-5,-2]}, AV_{[-1,1]}, and AV_{[0,4]} for the function g.
-
-
-
-
-
-
- Reading values from the table:
- AV_{[-5,-2]} = \frac{g(-2)-g(-5)}{-2-(-5)} = \frac{-1.25-(-2.75)}{3} = \frac{1.5}{3} = 0.5,
- AV_{[-1,1]} = \frac{g(1)-g(-1)}{1-(-1)} = \frac{0.25-(-0.75)}{2} = \frac{1}{2} = 0.5,
- AV_{[0,4]} = \frac{g(4)-g(0)}{4-0} = \frac{1.75-(-0.25)}{4} = \frac{2}{4} = 0.5.
- All three average rates of change equal 0.5.
-
-
-
-
-
-
- Consider the function y = h(x) defined by the graph in the following figure.
-
-
-
ADD ALT TEXT TO THIS IMAGE
-
-
- Determine AV_{[-5,-2]}, AV_{[-1,1]}, and AV_{[0,4]} for the function h.
-
-
-
-
-
-
- Reading values from the graph:
- AV_{[-5,-2]} = \frac{h(-2)-h(-5)}{-2-(-5)} = \frac{3-4}{3} = -\frac{1}{3},
- AV_{[-1,1]} = \frac{h(1)-h(-1)}{1-(-1)} = \frac{2-\frac{8}{3}}{2} = -\frac{1}{3},
- AV_{[0,4]} = \frac{h(4)-h(0)}{4-0} = \frac{1-\frac{7}{3}}{4} = -\frac{1}{3}.
- All three average rates of change equal -\frac{1}{3}.
-
-
-
-
-
-
- What do all three examples above have in common? How do they differ?
-
-
-
-
-
-
- All three functions have a constant average rate of change: the average rate of
- change is the same regardless of which interval is chosen. They differ in their
- specific rate of change (-3, 0.5, and -\frac{1}{3} respectively)
- and in how they were presented (formula, table, graph).
-
-
-
-
-
-
- For the function y = f(x) = 7 - 3x from (a), find the simplest expression you can for
-
- AV_{[a,b]} = \frac{f(b)-f(a)}{b-a}
-
- where a \ne b.
-
-
-
-
-
-
-
- AV_{[a,b]} &= \frac{f(b)-f(a)}{b-a} = \frac{(7-3b)-(7-3a)}{b-a}
- &= \frac{3a-3b}{b-a} = \frac{3(a-b)}{b-a} = -3.
-
- For any a \ne b, AV_{[a,b]} = -3.
-
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ Let y = f(x) = 7 - 3x. Determine AV_{[-3,-1]}, AV_{[2,5]}, and AV_{[4,10]} for the function f.
+
+
+
+
+
+ AV_{[-3,-1]} = \frac{f(-1)-f(-3)}{-1-(-3)} = \frac{10-16}{2} = -3,
+ AV_{[2,5]} = \frac{f(5)-f(2)}{5-2} = \frac{-8-1}{3} = -3,
+ AV_{[4,10]} = \frac{f(10)-f(4)}{10-4} = \frac{-23-(-5)}{6} = -3.
+ All three average rates of change equal -3.
+
+
+
+
+
+
+ Let y = g(x) be given by the data in the following table.
+
+ Determine AV_{[-5,-2]}, AV_{[-1,1]}, and AV_{[0,4]} for the function g.
+
+
+
+
+
+ Reading values from the table:
+ AV_{[-5,-2]} = \frac{g(-2)-g(-5)}{-2-(-5)} = \frac{-1.25-(-2.75)}{3} = \frac{1.5}{3} = 0.5,
+ AV_{[-1,1]} = \frac{g(1)-g(-1)}{1-(-1)} = \frac{0.25-(-0.75)}{2} = \frac{1}{2} = 0.5,
+ AV_{[0,4]} = \frac{g(4)-g(0)}{4-0} = \frac{1.75-(-0.25)}{4} = \frac{2}{4} = 0.5.
+ All three average rates of change equal 0.5.
+
+
+
+
+
+
+ Consider the function y = h(x) defined by the graph in the following figure.
+
+
+
ADD ALT TEXT TO THIS IMAGE
+
+
+ Determine AV_{[-5,-2]}, AV_{[-1,1]}, and AV_{[0,4]} for the function h.
+
+
+
+
+
+ Reading values from the graph:
+ AV_{[-5,-2]} = \frac{h(-2)-h(-5)}{-2-(-5)} = \frac{3-4}{3} = -\frac{1}{3},
+ AV_{[-1,1]} = \frac{h(1)-h(-1)}{1-(-1)} = \frac{2-\frac{8}{3}}{2} = -\frac{1}{3},
+ AV_{[0,4]} = \frac{h(4)-h(0)}{4-0} = \frac{1-\frac{7}{3}}{4} = -\frac{1}{3}.
+ All three average rates of change equal -\frac{1}{3}.
+
+
+
+
+
+
+ What do all three examples above have in common? How do they differ?
+
+
+
+
+
+ All three functions have a constant average rate of change: the average rate of
+ change is the same regardless of which interval is chosen. They differ in their
+ specific rate of change (-3, 0.5, and -\frac{1}{3} respectively)
+ and in how they were presented (formula, table, graph).
+
+
+
+
+
+
+ For the function y = f(x) = 7 - 3x from (a), find the simplest expression you can for
+
+ AV_{[a,b]} = \frac{f(b)-f(a)}{b-a}
+
+ where a \ne b.
+
+
+
+
+
+
+ AV_{[a,b]} &= \frac{f(b)-f(a)}{b-a} = \frac{(7-3b)-(7-3a)}{b-a}
+ &= \frac{3a-3b}{b-a} = \frac{3(a-b)}{b-a} = -3.
+
+ For any a \ne b, AV_{[a,b]} = -3.
+
- A water balloon is tossed vertically from a fifth story window. Its height, h, in meters, at time t, in seconds, is modeled by the function
-
- h = q(t) = -5t^2 + 20t + 25
- .
-
-
-
-
-
-
- Execute appropriate computations to complete both of the following tables: values of the function h on the left, average rates of change for h on the right.
-
- What pattern(s) do you observe in the table of function values and in the table of average rates of change?
-
-
-
-
-
-
- The function values increase from 25 to a peak of 45 at t=2, then decrease
- back to 0 symmetrically. The average rates of change decrease by exactly 10 m/s
- on each successive interval: 15, 5, -5, -15, -25.
-
-
-
-
-
-
- Explain why h = q(t) is not a linear function. Use the definition of a linear function (that is, referencing average rate of change) in your response.
-
-
-
-
-
-
- A linear function must have the same average rate of change on every interval.
- But AV_{[0,1]} = 15 \ne 5 = AV_{[1,2]}, so q is not linear.
-
-
-
-
-
-
- What is the average velocity of the water balloon in the final second before it lands? How does this value compare to the average velocity on the time interval [4.9, 5]?
-
-
-
-
-
-
- The average velocity on [4,5] is AV_{[4,5]} = -25 m/s.
- On the shorter interval, q(4.9) = -5(4.9)^2 + 20(4.9) + 25 = 2.95, so
- AV_{[4.9,5]} = \frac{0 - 2.95}{5 - 4.9} = -29.5 \text{ m/s.}
- The balloon is falling faster (greater speed) on [4.9, 5] than over the
- full final second [4,5].
-
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ A water balloon is tossed vertically from a fifth story window. Its height, h, in meters, at time t, in seconds, is modeled by the function
+
+ h = q(t) = -5t^2 + 20t + 25
+ .
+
+
+
+
+
+
+ Execute appropriate computations to complete both of the following tables: values of the function h on the left, average rates of change for h on the right.
+
+ What pattern(s) do you observe in the table of function values and in the table of average rates of change?
+
+
+
+
+
+ The function values increase from 25 to a peak of 45 at t=2, then decrease
+ back to 0 symmetrically. The average rates of change decrease by exactly 10 m/s
+ on each successive interval: 15, 5, -5, -15, -25.
+
+
+
+
+
+
+ Explain why h = q(t) is not a linear function. Use the definition of a linear function (that is, referencing average rate of change) in your response.
+
+
+
+
+
+ A linear function must have the same average rate of change on every interval.
+ But AV_{[0,1]} = 15 \ne 5 = AV_{[1,2]}, so q is not linear.
+
+
+
+
+
+
+ What is the average velocity of the water balloon in the final second before it lands? How does this value compare to the average velocity on the time interval [4.9, 5]?
+
+
+
+
+
+ The average velocity on [4,5] is AV_{[4,5]} = -25 m/s.
+ On the shorter interval, q(4.9) = -5(4.9)^2 + 20(4.9) + 25 = 2.95, so
+ AV_{[4.9,5]} = \frac{0 - 2.95}{5 - 4.9} = -29.5 \text{ m/s.}
+ The balloon is falling faster (greater speed) on [4.9, 5] than over the
+ full final second [4,5].
+
- Suppose that a rectangular aquarium is being filled with water. The tank is 4 feet long by 2 feet wide by 3 feet high, and the hose that is filling the tank is delivering water at a rate of 0.5 cubic feet per minute.
-
-
-
-
The empty aquarium.
-
The empty aquarium.
-
-
-
-
The aquarium, partially filled.
-
The aquarium, partially filled.
-
-
-
-
-
-
-
-
- What are some different quantities that are changing in this scenario?
-
-
-
-
-
-
The depth of the water in the tank, the amount of water in the tank, and time are all changing.
-
-
-
-
-
- After 1 minute has elapsed, how much water is in the tank? At this moment, how deep is the water?
-
-
-
-
-
-
Since water is entering at a rate of 0.5 cubic feet per minute, after 1 minute the tank contains 0.5 cubic feet of water. If d is the depth at this time, then 4 \times 2 \times d = 0.5, so d = \frac{1}{16} feet.
-
-
-
-
-
- How much water is in the tank and how deep is the water after 2 minutes? After 3 minutes?
-
-
-
-
-
-
After 2 minutes the tank contains 1 cubic foot of water, and the depth satisfies 4 \times 2 \times d = 1, giving d = \frac{1}{8} feet. After 3 minutes the tank contains 1.5 cubic feet of water and the depth satisfies 4 \times 2 \times d = 1.5, giving d = \frac{3}{16} feet.
-
-
-
-
-
- How long will it take for the tank to be completely full? Why?
-
-
-
-
-
-
The tank holds 4 \times 2 \times 3 = 24 cubic feet. Since water enters at 0.5 cubic feet per minute, the volume at time t is 0.5t cubic feet. The tank is completely full when 0.5t = 24, so t = 48 minutes.
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ Suppose that a rectangular aquarium is being filled with water. The tank is 4 feet long by 2 feet wide by 3 feet high, and the hose that is filling the tank is delivering water at a rate of 0.5 cubic feet per minute.
+
+
+
+
The empty aquarium.
+
The empty aquarium.
+
+
+
+
The aquarium, partially filled.
+
The aquarium, partially filled.
+
+
+
+
+
+
+
+
+ What are some different quantities that are changing in this scenario?
+
+
+
+
+
The depth of the water in the tank, the amount of water in the tank, and time are all changing.
+
+
+
+
+
+ After 1 minute has elapsed, how much water is in the tank? At this moment, how deep is the water?
+
+
+
+
+
Since water is entering at a rate of 0.5 cubic feet per minute, after 1 minute the tank contains 0.5 cubic feet of water. If d is the depth at this time, then 4 \times 2 \times d = 0.5, so d = \frac{1}{16} feet.
+
+
+
+
+
+ How much water is in the tank and how deep is the water after 2 minutes? After 3 minutes?
+
+
+
+
+
After 2 minutes the tank contains 1 cubic foot of water, and the depth satisfies 4 \times 2 \times d = 1, giving d = \frac{1}{8} feet. After 3 minutes the tank contains 1.5 cubic feet of water and the depth satisfies 4 \times 2 \times d = 1.5, giving d = \frac{3}{16} feet.
+
+
+
+
+
+ How long will it take for the tank to be completely full? Why?
+
+
+
+
+
The tank holds 4 \times 2 \times 3 = 24 cubic feet. Since water enters at 0.5 cubic feet per minute, the volume at time t is 0.5t cubic feet. The tank is completely full when 0.5t = 24, so t = 48 minutes.
- Open a new Desmos graph and define the function f(x) = x^2.
- Adjust the window so that the range is for -4 \le x \le 4 and -10 \le y \le 10.
-
-
-
-
-
-
- In Desmos,
- define the function g(x) = f(x) + a. (That is,
- in Desmos on line 2, enter
- g(x) = f(x) + a.) You will get prompted to add a slider for a.
- Do so.
-
-
- Explore by moving the slider for a and write at least one sentence to describe
- the effect that changing the value of a has on the graph of g.
-
-
-
-
-
-
- Changing a shifts the graph of f vertically: positive values of
- a shift it up, and negative values shift it down.
-
-
-
-
-
-
- Next,
- define the function h(x) = f(x-b). (That is,
- in Desmos on line 4, enter
- h(x) = f(x-b) and add the slider for b.)
-
-
- Move the slider for b and write at least one sentence to describe
- the effect that changing the value of b has on the graph of h.
-
-
-
-
-
-
- Changing b shifts the graph of f horizontally: positive values of
- b shift it to the right, and negative values shift it to the left.
-
-
-
-
-
-
- Now define the function p(x) = cf(x). (That is,
- in Desmos on line 6, enter
- p(x) = cf(x) and add the slider for c.)
-
-
- Move the slider for c and write at least one sentence to describe
- the effect that changing the value of c has on the graph of p.
- In particular, when c = -1,
- how is the graph of p related to the graph of f?
-
-
-
-
-
-
- The value c vertically stretches or compresses the graph of f:
- when |c| \gt 1 the graph is stretched away from the x-axis,
- and when 0 \lt |c| \lt 1 it is compressed toward the x-axis.
- When c = -1, the graph of p is a perfect reflection of the graph
- of f across the x-axis.
-
-
-
-
-
-
- Finally, click on the icons next to g, h, and p to temporarily hide them, and go back to Line 1 and change your formula for f.
- You can make it whatever you'd like,
- but try something like f(x) = x^2 + 2x + 3 or f(x) = x^3 - 1.
- Then, investigate with the sliders a, b,
- and c to see the effects on g, h, and p (unhiding them appropriately).
- Write a couple of sentences to describe your observations of your explorations.
-
-
-
-
-
-
- The same three effects hold for any choice of f: adding a shifts
- the graph vertically, replacing x with x - b shifts it horizontally,
- and multiplying by c stretches, compresses, or reflects it vertically.
- The shape of the particular function f changes, but the roles of
- a, b, and c remain the same.
-
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ Open a new Desmos graph and define the function f(x) = x^2.
+ Adjust the window so that the range is for -4 \le x \le 4 and -10 \le y \le 10.
+
+
+
+
+
+
+ In Desmos,
+ define the function g(x) = f(x) + a. (That is,
+ in Desmos on line 2, enter
+ g(x) = f(x) + a.) You will get prompted to add a slider for a.
+ Do so.
+
+
+ Explore by moving the slider for a and write at least one sentence to describe
+ the effect that changing the value of a has on the graph of g.
+
+
+
+
+
+ Changing a shifts the graph of f vertically: positive values of
+ a shift it up, and negative values shift it down.
+
+
+
+
+
+
+ Next,
+ define the function h(x) = f(x-b). (That is,
+ in Desmos on line 4, enter
+ h(x) = f(x-b) and add the slider for b.)
+
+
+ Move the slider for b and write at least one sentence to describe
+ the effect that changing the value of b has on the graph of h.
+
+
+
+
+
+ Changing b shifts the graph of f horizontally: positive values of
+ b shift it to the right, and negative values shift it to the left.
+
+
+
+
+
+
+ Now define the function p(x) = cf(x). (That is,
+ in Desmos on line 6, enter
+ p(x) = cf(x) and add the slider for c.)
+
+
+ Move the slider for c and write at least one sentence to describe
+ the effect that changing the value of c has on the graph of p.
+ In particular, when c = -1,
+ how is the graph of p related to the graph of f?
+
+
+
+
+
+ The value c vertically stretches or compresses the graph of f:
+ when |c| \gt 1 the graph is stretched away from the x-axis,
+ and when 0 \lt |c| \lt 1 it is compressed toward the x-axis.
+ When c = -1, the graph of p is a perfect reflection of the graph
+ of f across the x-axis.
+
+
+
+
+
+
+ Finally, click on the icons next to g, h, and p to temporarily hide them, and go back to Line 1 and change your formula for f.
+ You can make it whatever you'd like,
+ but try something like f(x) = x^2 + 2x + 3 or f(x) = x^3 - 1.
+ Then, investigate with the sliders a, b,
+ and c to see the effects on g, h, and p (unhiding them appropriately).
+ Write a couple of sentences to describe your observations of your explorations.
+
+
+
+
+
+ The same three effects hold for any choice of f: adding a shifts
+ the graph vertically, replacing x with x - b shifts it horizontally,
+ and multiplying by c stretches, compresses, or reflects it vertically.
+ The shape of the particular function f changes, but the roles of
+ a, b, and c remain the same.
+
- If we consider the unit circle with 16 labeled special points in Figure 2.3.1, start at t = 0, and traverse the circle counterclockwise, we may view the height, h, of the traversing point as a function of the angle, t, in radians. From there, we can plot the resulting (t,h) ordered pairs and connect them to generate the circular function pictured in the following figure, which tracks the height of a point traversing the unit circle.
-
-
-
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-
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-
-
-
- What is the exact value of f( \frac{\pi}{4} )? of f( \frac{\pi}{3} )?
-
-
-
-
-
-
- The exact value of f\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}, and the exact value of f\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}.
-
-
-
-
-
-
- Complete the following table with the exact values of h that correspond to the stated inputs.
-
- What is the exact value of f( \frac{11\pi}{4} )? of f( \frac{14\pi}{3} )?
-
-
-
-
-
-
- The exact value of f\left(\frac{11\pi}{4}\right) is the same as f\left(\frac{3\pi}{4}\right) = \frac{\sqrt{2}}{2}, because of the periodicity of 2\pi. Likewise, f\left(\frac{14\pi}{3}\right) = f\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2}.
-
-
-
-
-
-
- Give four different values of t for which f(t) = -\frac{\sqrt{3}}{2}.
-
-
-
-
-
-
- f(t) = -\frac{\sqrt{3}}{2} for values t = n(2\pi) + \frac{4\pi}{3} and t = n(2\pi) + \frac{5\pi}{3}, where n is a non-negative integer. Four such values are: t = \frac{4\pi}{3},\ \frac{5\pi}{3},\ \frac{10\pi}{3},\ \frac{11\pi}{3}.
-
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ If we consider the unit circle with 16 labeled special points in Figure 2.3.1, start at t = 0, and traverse the circle counterclockwise, we may view the height, h, of the traversing point as a function of the angle, t, in radians. From there, we can plot the resulting (t,h) ordered pairs and connect them to generate the circular function pictured in the following figure, which tracks the height of a point traversing the unit circle.
+
+
+
ADD ALT TEXT TO THIS IMAGE
+
+
+
+
+
+
+ What is the exact value of f( \frac{\pi}{4} )? of f( \frac{\pi}{3} )?
+
+
+
+
+
+ The exact value of f\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}, and the exact value of f\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}.
+
+
+
+
+
+
+ Complete the following table with the exact values of h that correspond to the stated inputs.
+
+ What is the exact value of f( \frac{11\pi}{4} )? of f( \frac{14\pi}{3} )?
+
+
+
+
+
+ The exact value of f\left(\frac{11\pi}{4}\right) is the same as f\left(\frac{3\pi}{4}\right) = \frac{\sqrt{2}}{2}, because of the periodicity of 2\pi. Likewise, f\left(\frac{14\pi}{3}\right) = f\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2}.
+
+
+
+
+
+
+ Give four different values of t for which f(t) = -\frac{\sqrt{3}}{2}.
+
+
+
+
+
+ f(t) = -\frac{\sqrt{3}}{2} for values t = n(2\pi) + \frac{4\pi}{3} and t = n(2\pi) + \frac{5\pi}{3}, where n is a non-negative integer. Four such values are: t = \frac{4\pi}{3},\ \frac{5\pi}{3},\ \frac{10\pi}{3},\ \frac{11\pi}{3}.
+
- Let f(t) = \cos(t). First, answer all of the questions below without using Desmos;
- then use Desmos to confirm your conjectures. For each prompt, describe the graphs of g and h as transformations of f and, in addition, state the amplitude, midline, and period of both g and h.
-
-
-
-
-
-
g(t) = 3\cos(t) and h(t) = -\frac{1}{4}\cos(t)
-
-
-
-
-
- g(t) is a vertical stretch of f by a factor of 3, with amplitude 3, midline y=0, and period 2\pi. h(t) is a vertical shrink by a factor of \frac{1}{4} combined with a reflection over the x-axis, with amplitude \frac{1}{4}, midline y=0, and period 2\pi.
-
-
-
-
-
-
g(t) = \cos(t-\pi) and h(t) = \cos\left(t+ \frac{\pi}{2}\right)
-
-
-
-
-
- g(t) is a horizontal translation of f by \pi units to the right (equivalently, a reflection over the x-axis due to periodicity), with amplitude 1, midline y=0, and period 2\pi. h(t) is a horizontal translation \frac{\pi}{2} units to the left, with amplitude 1, midline y=0, and period 2\pi.
-
-
-
-
-
-
g(t) = \cos(t)+4 and h(t) = \cos\left(t\right)-2
-
-
-
-
-
- g(t) is a vertical translation of f by 4 units up, with amplitude 1, midline y=4, and period 2\pi. h(t) is a vertical translation 2 units down, with amplitude 1, midline y=-2, and period 2\pi.
-
-
-
-
-
-
g(t) = 3\cos(t-\pi)+4 and h(t) = -\frac{1}{4}\cos\left(t+ \frac{\pi}{2}\right)-2
-
-
-
-
-
- g(t) is a vertical stretch by a factor of 3, followed by a horizontal shift \pi units to the right, and a vertical shift 4 units up. Due to the periodicity, it is also accurate to say that g(t) is reflected over the x-axis, stretched by a factor of 3, and shifted vertically up 4 units. g(t) has amplitude 3, midline y=4, and period 2\pi.
-
-
- h(t) is a reflection over the x-axis, followed by a vertical shrink by a factor of \frac{1}{4}, then a horizontal shift \frac{\pi}{2} units to the left, and a vertical shift 2 units down. Due to the periodicity, it is also accurate to say that h(t) is shifted \frac{\pi}{2} units to the right, shrunk by a factor of \frac{1}{4}, and shifted 2 units down. h(t) has amplitude \frac{1}{4}, midline y=-2, and period 2\pi.
-
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ Let f(t) = \cos(t). First, answer all of the questions below without using Desmos;
+ then use Desmos to confirm your conjectures. For each prompt, describe the graphs of g and h as transformations of f and, in addition, state the amplitude, midline, and period of both g and h.
+
+
+
+
+
+
g(t) = 3\cos(t) and h(t) = -\frac{1}{4}\cos(t)
+
+
+
+
+ g(t) is a vertical stretch of f by a factor of 3, with amplitude 3, midline y=0, and period 2\pi. h(t) is a vertical shrink by a factor of \frac{1}{4} combined with a reflection over the x-axis, with amplitude \frac{1}{4}, midline y=0, and period 2\pi.
+
+
+
+
+
+
g(t) = \cos(t-\pi) and h(t) = \cos\left(t+ \frac{\pi}{2}\right)
+
+
+
+
+ g(t) is a horizontal translation of f by \pi units to the right (equivalently, a reflection over the x-axis due to periodicity), with amplitude 1, midline y=0, and period 2\pi. h(t) is a horizontal translation \frac{\pi}{2} units to the left, with amplitude 1, midline y=0, and period 2\pi.
+
+
+
+
+
+
g(t) = \cos(t)+4 and h(t) = \cos\left(t\right)-2
+
+
+
+
+ g(t) is a vertical translation of f by 4 units up, with amplitude 1, midline y=4, and period 2\pi. h(t) is a vertical translation 2 units down, with amplitude 1, midline y=-2, and period 2\pi.
+
+
+
+
+
+
g(t) = 3\cos(t-\pi)+4 and h(t) = -\frac{1}{4}\cos\left(t+ \frac{\pi}{2}\right)-2
+
+
+
+
+ g(t) is a vertical stretch by a factor of 3, followed by a horizontal shift \pi units to the right, and a vertical shift 4 units up. Due to the periodicity, it is also accurate to say that g(t) is reflected over the x-axis, stretched by a factor of 3, and shifted vertically up 4 units. g(t) has amplitude 3, midline y=4, and period 2\pi.
+
+
+ h(t) is a reflection over the x-axis, followed by a vertical shrink by a factor of \frac{1}{4}, then a horizontal shift \frac{\pi}{2} units to the left, and a vertical shift 2 units down. Due to the periodicity, it is also accurate to say that h(t) is shifted \frac{\pi}{2} units to the right, shrunk by a factor of \frac{1}{4}, and shifted 2 units down. h(t) has amplitude \frac{1}{4}, midline y=-2, and period 2\pi.
+
- In the context of the ferris wheel pictured in Figure 2.1.1 in the text, assume that the height, h, of the moving point (the cab in which you are riding), and the distance, d, that the point has traveled around the circumference of the ferris wheel are both measured in meters.
-
-
- Further, assume that the circumference of the ferris wheel is 150 meters. In addition, suppose that after getting in your cab at the lowest point on the wheel, you traverse the full circle several times.
-
-
-
-
-
-
- Recall that the circumference, C, of a circle is connected to the circle's radius, r, by the formula C = 2\pi r. What is the radius of the ferris wheel? How high is the highest point on the ferris wheel?
-
-
-
-
-
-
- The circumference 2\pi r of the Ferris wheel is 150 meters, so the radius is \frac{150}{2\pi} = \frac{75}{\pi} meters, or about 23.9 meters. The highest point on the Ferris wheel is twice the radius plus however far the wheel begins off the ground at the bottom. If we neglect this starting height, the height of this particular Ferris wheel is \frac{150}{\pi} meters, or about 47.7 meters.
-
-
-
-
-
-
- How high is the cab after it has traveled 1/4 of the circumference of the circle?
-
-
-
-
-
-
- The cab starts at the bottom, at a height of zero. At 1/4 of the distance around, the cab is at the same height as the hub of the wheel, so it is one radius up: \frac{75}{\pi} meters.
-
-
-
-
-
-
- How much distance along the circle has the cab traversed at the moment it first reaches a height of \frac{150}{\pi} \approx 47.75 meters?
-
-
-
-
-
-
- The highest point of the Ferris wheel is at \frac{150}{\pi} meters. There is only one point on the Ferris wheel that is this high. The first time the cab reaches it is when the cab has traveled one-half the way around the wheel, a distance of 75 meters.
-
-
-
-
-
-
- Can h be thought of as a function of d? Why or why not?
-
-
-
-
-
-
- Yes. In this scenario, the height h may be thought of as a function of distance traveled d because given a distance it is possible to determine the (singular) height of the cab.
-
-
-
-
-
-
- Can d be thought of as a function of h? Why or why not?
-
-
-
-
-
-
- No. The distance d traveled is not a function of height h because the Ferris wheel may make multiple turns so that there are multiple distances associated with a given height.
-
-
-
-
-
-
- Why do you think the curve shown at right in Figure 2.1.1 has the shape that it does? Write several sentences to explain.
-
-
-
-
-
-
- The curve has the shape that it does because the height of the Ferris wheel changes slowly at the bottom and at the top, where much of the distance travelled is horizontal instead of vertical. The function describing height in terms of distance is a sine function.
-
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ In the context of the ferris wheel pictured in Figure 2.1.1 in the text, assume that the height, h, of the moving point (the cab in which you are riding), and the distance, d, that the point has traveled around the circumference of the ferris wheel are both measured in meters.
+
+
+ Further, assume that the circumference of the ferris wheel is 150 meters. In addition, suppose that after getting in your cab at the lowest point on the wheel, you traverse the full circle several times.
+
+
+
+
+
+
+ Recall that the circumference, C, of a circle is connected to the circle's radius, r, by the formula C = 2\pi r. What is the radius of the ferris wheel? How high is the highest point on the ferris wheel?
+
+
+
+
+
+ The circumference 2\pi r of the Ferris wheel is 150 meters, so the radius is \frac{150}{2\pi} = \frac{75}{\pi} meters, or about 23.9 meters. The highest point on the Ferris wheel is twice the radius plus however far the wheel begins off the ground at the bottom. If we neglect this starting height, the height of this particular Ferris wheel is \frac{150}{\pi} meters, or about 47.7 meters.
+
+
+
+
+
+
+ How high is the cab after it has traveled 1/4 of the circumference of the circle?
+
+
+
+
+
+ The cab starts at the bottom, at a height of zero. At 1/4 of the distance around, the cab is at the same height as the hub of the wheel, so it is one radius up: \frac{75}{\pi} meters.
+
+
+
+
+
+
+ How much distance along the circle has the cab traversed at the moment it first reaches a height of \frac{150}{\pi} \approx 47.75 meters?
+
+
+
+
+
+ The highest point of the Ferris wheel is at \frac{150}{\pi} meters. There is only one point on the Ferris wheel that is this high. The first time the cab reaches it is when the cab has traveled one-half the way around the wheel, a distance of 75 meters.
+
+
+
+
+
+
+ Can h be thought of as a function of d? Why or why not?
+
+
+
+
+
+ Yes. In this scenario, the height h may be thought of as a function of distance traveled d because given a distance it is possible to determine the (singular) height of the cab.
+
+
+
+
+
+
+ Can d be thought of as a function of h? Why or why not?
+
+
+
+
+
+ No. The distance d traveled is not a function of height h because the Ferris wheel may make multiple turns so that there are multiple distances associated with a given height.
+
+
+
+
+
+
+ Why do you think the curve shown at right in Figure 2.1.1 has the shape that it does? Write several sentences to explain.
+
+
+
+
+
+ The curve has the shape that it does because the height of the Ferris wheel changes slowly at the bottom and at the top, where much of the distance travelled is horizontal instead of vertical. The function describing height in terms of distance is a sine function.
+
- In the following figure there are 24 equally spaced points on the unit circle. Since the circumference of the unit circle is 2\pi, each of the points is \frac{1}{24} \cdot 2\pi = \frac{\pi}{12} units apart (traveled along the circle). Thus,
- the first point counterclockwise from (1,0) corresponds to the distance
- t = \frac{\pi}{12} traveled along the unit circle. The second point is twice as far, and thus t = 2 \cdot \frac{\pi}{12} = \frac{\pi}{6} units along the circle away from (1,0).
-
-
-
ADD ALT TEXT TO THIS IMAGE
-
-
-
-
-
-
- Label each of the subsequent points on the unit circle with the exact distance they lie counter-clockwise away from (1,0); write each fraction in lowest terms.
-
-
-
-
-
-
-
-
-
-
-
- Which distance along the unit circle corresponds to \frac{1}{4} of a full rotation around? to \frac{5}{8} of a full rotation?
-
-
-
-
-
-
- One-quarter of a full rotation corresponds to a distance of \frac{\pi}{2}. Five-eighths of a full rotation corresponds to a distance of \frac{5\pi}{4}.
-
-
-
-
-
-
- One way to measure angles is connected to the arc length along a circle. For an angle whose vertex is at (0,0) in the unit circle, we say the angle's measure is 1 radianradian provided that the angle intercepts an arc of the circle that is 1 unit in length, as pictured in the following figure. Note particularly that an angle measuring 1 radian intercepts an arc of the same length as the circle's radius.
-
-
-
-
ADD ALT TEXT TO THIS IMAGE
-
-
-
- Suppose that \alpha and \beta are angles with respective radian measures \alpha = \frac{\pi}{3} and \beta = \frac{3\pi}{4}. Assuming that we view \alpha and \beta as having their vertex at (0,0) and one side along the positive x-axis, sketch the angles \alpha and \beta on the unit circle in part (a).
-
-
-
-
-
-
-
-
-
-
-
- What is the radian measure that corresponds to a 90^\circ angle?
-
-
-
-
-
-
- 90^\circ corresponds to \frac{\pi}{2} radians.
-
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ In the following figure there are 24 equally spaced points on the unit circle. Since the circumference of the unit circle is 2\pi, each of the points is \frac{1}{24} \cdot 2\pi = \frac{\pi}{12} units apart (traveled along the circle). Thus,
+ the first point counterclockwise from (1,0) corresponds to the distance
+ t = \frac{\pi}{12} traveled along the unit circle. The second point is twice as far, and thus t = 2 \cdot \frac{\pi}{12} = \frac{\pi}{6} units along the circle away from (1,0).
+
+
+
ADD ALT TEXT TO THIS IMAGE
+
+
+
+
+
+
+ Label each of the subsequent points on the unit circle with the exact distance they lie counter-clockwise away from (1,0); write each fraction in lowest terms.
+
+
+
+
+
+
+
+
+
+
+ Which distance along the unit circle corresponds to \frac{1}{4} of a full rotation around? to \frac{5}{8} of a full rotation?
+
+
+
+
+
+ One-quarter of a full rotation corresponds to a distance of \frac{\pi}{2}. Five-eighths of a full rotation corresponds to a distance of \frac{5\pi}{4}.
+
+
+
+
+
+
+ One way to measure angles is connected to the arc length along a circle. For an angle whose vertex is at (0,0) in the unit circle, we say the angle's measure is 1 radianradian provided that the angle intercepts an arc of the circle that is 1 unit in length, as pictured in the following figure. Note particularly that an angle measuring 1 radian intercepts an arc of the same length as the circle's radius.
+
+
+
+
ADD ALT TEXT TO THIS IMAGE
+
+
+
+ Suppose that \alpha and \beta are angles with respective radian measures \alpha = \frac{\pi}{3} and \beta = \frac{3\pi}{4}. Assuming that we view \alpha and \beta as having their vertex at (0,0) and one side along the positive x-axis, sketch the angles \alpha and \beta on the unit circle in part (a).
+
+
+
+
+
+
+
+
+
+
+ What is the radian measure that corresponds to a 90^\circ angle?
+
+
+
+
+
+ 90^\circ corresponds to \frac{\pi}{2} radians.
+
- Open a new Desmos worksheet and define the following functions: f(t) = 2^t, g(t) = 3^t, h(t) = (\frac{1}{3})^t, and p(t) = f(kt). After you define p, accept the slider for k, and set the range of the slider to be -2 \le k \le 2.
-
-
-
-
-
-
- By experimenting with the value of k, find a value of k so that the graph of p(t) = f(kt) = 2^{kt} appears to align with the graph of g(t) = 3^t. What is the value of k?
-
-
-
-
-
-
- The value of k is approximately k = \ln 3 / \ln 2 \approx 1.585.
-
-
-
-
-
-
- Similarly, experiment to find a value of k so that the graph of p(t) = f(kt) = 2^{kt} appears to align with the graph of h(t) = (\frac{1}{3})^t. What is the value of k?
-
-
-
-
-
-
- The value of k is approximately k = -\ln 3 / \ln 2 \approx -1.585.
-
-
-
-
-
-
- For the value of k you determined in (a), compute 2^k. What do you observe?
-
-
-
-
-
-
- 2^k = 2^{\ln 3 / \ln 2} = 3. This makes sense because the equation 2^{kt} = 3^t requires 2^k = 3.
-
-
-
-
-
-
- For the value of k you determined in (b), compute 2^k. What do you observe?
-
-
-
-
-
-
- 2^k = 2^{-\ln 3 / \ln 2} = \frac{1}{3}. This makes sense because the equation 2^{kt} = (1/3)^t requires 2^k = 1/3.
-
-
-
-
-
-
- Given any exponential function of the form b^t, do you think it's possible to find a value of k to that p(t) = f(kt) = 2^{kt} is the same function as b^t? Why or why not?
-
-
-
-
-
-
- Yes, it is possible to find a value of k for any exponential function of the form b^t, provided b is a positive number. Setting 2^{kt} = b^t and taking logarithms gives k \ln 2 = \ln b, so k = \dfrac{\ln b}{\ln 2}.
-
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ Open a new Desmos worksheet and define the following functions: f(t) = 2^t, g(t) = 3^t, h(t) = (\frac{1}{3})^t, and p(t) = f(kt). After you define p, accept the slider for k, and set the range of the slider to be -2 \le k \le 2.
+
+
+
+
+
+
+ By experimenting with the value of k, find a value of k so that the graph of p(t) = f(kt) = 2^{kt} appears to align with the graph of g(t) = 3^t. What is the value of k?
+
+
+
+
+
+ The value of k is approximately k = \ln 3 / \ln 2 \approx 1.585.
+
+
+
+
+
+
+ Similarly, experiment to find a value of k so that the graph of p(t) = f(kt) = 2^{kt} appears to align with the graph of h(t) = (\frac{1}{3})^t. What is the value of k?
+
+
+
+
+
+ The value of k is approximately k = -\ln 3 / \ln 2 \approx -1.585.
+
+
+
+
+
+
+ For the value of k you determined in (a), compute 2^k. What do you observe?
+
+
+
+
+
+ 2^k = 2^{\ln 3 / \ln 2} = 3. This makes sense because the equation 2^{kt} = 3^t requires 2^k = 3.
+
+
+
+
+
+
+ For the value of k you determined in (b), compute 2^k. What do you observe?
+
+
+
+
+
+ 2^k = 2^{-\ln 3 / \ln 2} = \frac{1}{3}. This makes sense because the equation 2^{kt} = (1/3)^t requires 2^k = 1/3.
+
+
+
+
+
+
+ Given any exponential function of the form b^t, do you think it's possible to find a value of k to that p(t) = f(kt) = 2^{kt} is the same function as b^t? Why or why not?
+
+
+
+
+
+ Yes, it is possible to find a value of k for any exponential function of the form b^t, provided b is a positive number. Setting 2^{kt} = b^t and taking logarithms gives k \ln 2 = \ln b, so k = \dfrac{\ln b}{\ln 2}.
+
- Suppose that at age 20 you have $20000 and you can choose between one of two ways to use the money: you can invest it in a mutual fund that will, on average, earn 8% interest annually, or you can purchase a new automobile that will, on average, depreciate 12% annually. Let's explore how the $20000 changes over time.
-
-
- Let I(t) denote the value of the $20000 after t years if it is invested in the mutual fund, and let V(t) denote the value of the automobile t years after it is purchased.
-
- Based on the patterns in your computations in (a) and (b), determine formulas for I(t) and V(t).
-
-
-
-
-
-
- For the appreciating investment of $20,000: I(t) = \$20{,}000 \times 1.08^t, and for the depreciating $20,000 vehicle: V(t) = \$20{,}000 \times 0.88^t.
-
-
-
-
-
-
- Use Desmos to define I(t) and V(t). Plot each function on the interval 0 \le t \le 20 and record your results on the axes in the figure below, being sure to label the scale on the axes. What trends do you observe in the graphs? How do I(20) and V(20) compare?
-
-
-
ADD ALT TEXT TO THIS IMAGE
-
-
-
-
-
-
- The red curve shows the growing value of a $20,000 investment earning 8% interest per year. The blue curve shows the decreasing value of a $20,000 vehicle losing 12% of its value each year.
-
-
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ Suppose that at age 20 you have $20000 and you can choose between one of two ways to use the money: you can invest it in a mutual fund that will, on average, earn 8% interest annually, or you can purchase a new automobile that will, on average, depreciate 12% annually. Let's explore how the $20000 changes over time.
+
+
+ Let I(t) denote the value of the $20000 after t years if it is invested in the mutual fund, and let V(t) denote the value of the automobile t years after it is purchased.
+
+ Based on the patterns in your computations in (a) and (b), determine formulas for I(t) and V(t).
+
+
+
+
+
+ For the appreciating investment of $20,000: I(t) = \$20{,}000 \times 1.08^t, and for the depreciating $20,000 vehicle: V(t) = \$20{,}000 \times 0.88^t.
+
+
+
+
+
+
+ Use Desmos to define I(t) and V(t). Plot each function on the interval 0 \le t \le 20 and record your results on the axes in the figure below, being sure to label the scale on the axes. What trends do you observe in the graphs? How do I(20) and V(20) compare?
+
+
+
ADD ALT TEXT TO THIS IMAGE
+
+
+
+
+
+ The red curve shows the growing value of a $20,000 investment earning 8% interest per year. The blue curve shows the decreasing value of a $20,000 vehicle losing 12% of its value each year.
+
- In the following questions, we investigate how \log_{10}(a \cdot b) can be equivalently written in terms of \log_{10}(a) and \log_{10}(b).
-
-
-
-
-
-
- Write 10^x \cdot 10^y as 10 raised to a single power. That is, complete the equation
-
- 10^x \cdot 10^y = 10^{\Box}
-
- by filling in the box with an appropriate expression involving x and y.
-
-
-
-
-
-
- 10^x \cdot 10^y = 10^{x+y}
-
-
-
-
-
-
- What is the simplest possible way to write \log_{10}10^x? What about the simplest equivalent expression for \log_{10}10^y?
-
-
-
-
-
-
- \log_{10}10^x = x and \log_{10}10^y = y.
-
-
-
-
-
-
- Explain why each of the following three equal signs is valid in the sequence of equalities:
- \log_{10}(10^x \cdot 10^y) &= \log_{10}(10^{x+y}) &= x+y &= \log_{10}(10^x) + \log_{10}(10^y).
-
-
-
-
-
-
- The first equal sign holds because 10^x \cdot 10^y = 10^{x+y} by the rule for multiplying exponentials with the same base. The second equal sign holds because \log_{10}(10^A) = A for any real number A. The third equal sign holds because x = \log_{10}(10^x) and y = \log_{10}(10^y).
-
-
-
-
-
-
- Suppose that a and b are positive real numbers so we can think of a as 10^x for some real number x and
- b as 10^y for some real number y. That is, say that a = 10^x and b = 10^y. What does our work in (c) tell us about
- \log_{10}(ab)?
-
-
-
-
-
-
- Since a = 10^x and b = 10^y, we have ab = 10^x \cdot 10^y. Our work in (c) shows that \log_{10}(10^x \cdot 10^y) = \log_{10}(10^x) + \log_{10}(10^y). With \log_{10}(10^x) = \log_{10}(a) and \log_{10}(10^y) = \log_{10}(b), it follows that \log_{10}(ab) = \log_{10}(a) + \log_{10}(b).
-
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ In the following questions, we investigate how \log_{10}(a \cdot b) can be equivalently written in terms of \log_{10}(a) and \log_{10}(b).
+
+
+
+
+
+
+ Write 10^x \cdot 10^y as 10 raised to a single power. That is, complete the equation
+
+ 10^x \cdot 10^y = 10^{\Box}
+
+ by filling in the box with an appropriate expression involving x and y.
+
+
+
+
+
+ 10^x \cdot 10^y = 10^{x+y}
+
+
+
+
+
+
+ What is the simplest possible way to write \log_{10}10^x? What about the simplest equivalent expression for \log_{10}10^y?
+
+
+
+
+
+ \log_{10}10^x = x and \log_{10}10^y = y.
+
+
+
+
+
+
+ Explain why each of the following three equal signs is valid in the sequence of equalities:
+ \log_{10}(10^x \cdot 10^y) &= \log_{10}(10^{x+y}) &= x+y &= \log_{10}(10^x) + \log_{10}(10^y).
+
+
+
+
+
+ The first equal sign holds because 10^x \cdot 10^y = 10^{x+y} by the rule for multiplying exponentials with the same base. The second equal sign holds because \log_{10}(10^A) = A for any real number A. The third equal sign holds because x = \log_{10}(10^x) and y = \log_{10}(10^y).
+
+
+
+
+
+
+ Suppose that a and b are positive real numbers so we can think of a as 10^x for some real number x and
+ b as 10^y for some real number y. That is, say that a = 10^x and b = 10^y. What does our work in (c) tell us about
+ \log_{10}(ab)?
+
+
+
+
+
+ Since a = 10^x and b = 10^y, we have ab = 10^x \cdot 10^y. Our work in (c) shows that \log_{10}(10^x \cdot 10^y) = \log_{10}(10^x) + \log_{10}(10^y). With \log_{10}(10^x) = \log_{10}(a) and \log_{10}(10^y) = \log_{10}(b), it follows that \log_{10}(ab) = \log_{10}(a) + \log_{10}(b).
+
- The function P(t) has an inverse because it is a one-to-one function, where each value of y has only one corresponding value of t.
-
-
-
-
-
-
- Since P has an inverse function, we know there exists some other function, say L,
- such that writing y = P(t)
- says the exact same thing as writing t = L(y). In words, where P produces the result of raising 10 to a given power, the function L reverses this process and instead tells us the power to which we need to raise 10, given a desired result. Complete the following table to generate a collection of values of L.
-
- What are the domain and range of the function P? What are the domain and range of the function L?
-
-
-
-
-
-
- The domain of y = P(t) = 10^t is all real numbers, and the range is all positive real numbers. The domain of t = L(y) is all positive real numbers, and the range is all real numbers.
-
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ Let P(t) be the powers of 10 function, which is given by P(t) = 10^t.
+
+
+
+
+
+
+ Complete the following table to generate certain values of P.
+
+ The function P(t) has an inverse because it is a one-to-one function, where each value of y has only one corresponding value of t.
+
+
+
+
+
+
+ Since P has an inverse function, we know there exists some other function, say L,
+ such that writing y = P(t)
+ says the exact same thing as writing t = L(y). In words, where P produces the result of raising 10 to a given power, the function L reverses this process and instead tells us the power to which we need to raise 10, given a desired result. Complete the following table to generate a collection of values of L.
+
+ What are the domain and range of the function P? What are the domain and range of the function L?
+
+
+
+
+
+ The domain of y = P(t) = 10^t is all real numbers, and the range is all positive real numbers. The domain of t = L(y) is all positive real numbers, and the range is all real numbers.
+
- In Desmos, define
- g(t) = ab^t+c and accept the prompt for sliders for a, b, and c.
- Edit the sliders so that a has values from a = 5 to a = 50,
- b has values from b = 0.7 to b = 1.3, and
- c has values from c = -5 to c = 5
- (each with a step-size of 0.01).
- In addition, in Desmos
- let P = (0, g(0)) and check the box to show the label.
- Finally, zoom out so that the window shows an interval of t-values from -30 \le t \le 30.
-
-
-
-
-
-
- Set b = 1.1 and explore the effects of changing the values of a and c. Write several sentences to summarize your observations.
-
-
-
-
-
-
- When b = 1.1, changing the value of a where 5 \le a \le 50 causes a vertical stretch effect. The coefficient c causes a vertical shift.
-
-
-
-
-
-
- Follow the directions for (a) again, this time with b = 0.9
-
-
-
-
-
- The same vertical stretch and vertical shift behavior is apparent when b = 0.9.
-
-
-
-
-
-
- Set a = 5 and c = 4.
- Explore the effects of changing the value of b;
- be sure to include values of b both less than and greater than 1.
- Write several sentences to summarize your observations.
-
-
-
-
-
-
- When a = 5 and c = 4, the line rises to the left and flattens to the right when b \lt 1, the line is flat when b = 1, and the line flattens to the left and rises to the right when b \gt 1.
-
-
-
-
-
-
- When 0 \lt b \lt 1,
- what happens to the graph of g when we consider positive t-values that get larger and larger?
-
-
-
-
-
-
- When 0 \lt b \lt 1, the graph gets closer and closer to c at greater and greater values of t.
-
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ In Desmos, define
+ g(t) = ab^t+c and accept the prompt for sliders for a, b, and c.
+ Edit the sliders so that a has values from a = 5 to a = 50,
+ b has values from b = 0.7 to b = 1.3, and
+ c has values from c = -5 to c = 5
+ (each with a step-size of 0.01).
+ In addition, in Desmos
+ let P = (0, g(0)) and check the box to show the label.
+ Finally, zoom out so that the window shows an interval of t-values from -30 \le t \le 30.
+
+
+
+
+
+
+ Set b = 1.1 and explore the effects of changing the values of a and c. Write several sentences to summarize your observations.
+
+
+
+
+
+ When b = 1.1, changing the value of a where 5 \le a \le 50 causes a vertical stretch effect. The coefficient c causes a vertical shift.
+
+
+
+
+
+
+ Follow the directions for (a) again, this time with b = 0.9
+
+
+
+
+ The same vertical stretch and vertical shift behavior is apparent when b = 0.9.
+
+
+
+
+
+
+ Set a = 5 and c = 4.
+ Explore the effects of changing the value of b;
+ be sure to include values of b both less than and greater than 1.
+ Write several sentences to summarize your observations.
+
+
+
+
+
+ When a = 5 and c = 4, the line rises to the left and flattens to the right when b \lt 1, the line is flat when b = 1, and the line flattens to the left and rises to the right when b \gt 1.
+
+
+
+
+
+
+ When 0 \lt b \lt 1,
+ what happens to the graph of g when we consider positive t-values that get larger and larger?
+
+
+
+
+
+ When 0 \lt b \lt 1, the graph gets closer and closer to c at greater and greater values of t.
+
- In each of the following situations, determine the exact value of the unknown quantity that is identified.
-
-
-
-
-
-
- The temperature of a warming object in an oven is given by F(t) = 275 - 203e^{-kt}, and we know that the object's temperature after 20 minutes is F(20) = 101. Determine the exact value of k.
-
-
-
-
-
-
- Setting F(20) = 101, we have 275 - 203e^{-20k} = 101, so 203e^{-20k} = 174 and e^{-20k} = \frac{174}{203}. Thus k = -\dfrac{1}{20}\ln\!\left(\frac{174}{203}\right) \approx 7.708 \times 10^{-3}.
-
-
-
-
-
-
- The temperature of a cooling object in a refrigerator is modeled by F(t) = a + 37.4e^{-0.05t}, and the temperature of the refrigerator is 39.8^\circ. By thinking about the long-term behavior of e^{-0.05t} and the long-term behavior of the object's temperature, determine the exact value of a.
-
-
-
-
-
-
- As t \to \infty, e^{-0.05t} \to 0, so F(t) \to a. Since the object cools toward the refrigerator temperature of 39.8^\circ, the long-term temperature is 39.8^\circ, and therefore a = 39.8.
-
-
-
-
-
-
- Later in this section, we'll learn that one model for how a population grows over time can be given by a function of the form
-
- P(t) = \frac{A}{1 + Me^{-kt}}
- . Models of this form lead naturally to equations that have structure like
-
- 3 = \frac{10}{1+x}.
-
- Solve the equation 3 = \frac{10}{1+x} for the exact value of x.
-
-
-
-
-
-
- Multiplying both sides by (1+x) gives 3(1+x) = 10, so 1+x = \frac{10}{3} and x = \frac{10}{3} - 1 = \frac{7}{3}.
-
-
-
-
-
-
- Suppose that y = a + be^{-kt}. Solve for t in terms of a, b, k, and y. What does this new equation represent?
-
-
-
-
-
-
- Starting from y = a + be^{-kt}, we subtract a: y - a = be^{-kt}. Dividing by b: \frac{y-a}{b} = e^{-kt}. Taking the natural log: -kt = \ln\!\left(\frac{y-a}{b}\right). Therefore t = -\dfrac{1}{k}\ln\!\left(\frac{y-a}{b}\right). This is the inverse of the original function and gives the time at which the object reaches temperature y.
-
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ In each of the following situations, determine the exact value of the unknown quantity that is identified.
+
+
+
+
+
+
+ The temperature of a warming object in an oven is given by F(t) = 275 - 203e^{-kt}, and we know that the object's temperature after 20 minutes is F(20) = 101. Determine the exact value of k.
+
+
+
+
+
+ Setting F(20) = 101, we have 275 - 203e^{-20k} = 101, so 203e^{-20k} = 174 and e^{-20k} = \frac{174}{203}. Thus k = -\dfrac{1}{20}\ln\!\left(\frac{174}{203}\right) \approx 7.708 \times 10^{-3}.
+
+
+
+
+
+
+ The temperature of a cooling object in a refrigerator is modeled by F(t) = a + 37.4e^{-0.05t}, and the temperature of the refrigerator is 39.8^\circ. By thinking about the long-term behavior of e^{-0.05t} and the long-term behavior of the object's temperature, determine the exact value of a.
+
+
+
+
+
+ As t \to \infty, e^{-0.05t} \to 0, so F(t) \to a. Since the object cools toward the refrigerator temperature of 39.8^\circ, the long-term temperature is 39.8^\circ, and therefore a = 39.8.
+
+
+
+
+
+
+ Later in this section, we'll learn that one model for how a population grows over time can be given by a function of the form
+
+ P(t) = \frac{A}{1 + Me^{-kt}}
+ . Models of this form lead naturally to equations that have structure like
+
+ 3 = \frac{10}{1+x}.
+
+ Solve the equation 3 = \frac{10}{1+x} for the exact value of x.
+
+
+
+
+
+ Multiplying both sides by (1+x) gives 3(1+x) = 10, so 1+x = \frac{10}{3} and x = \frac{10}{3} - 1 = \frac{7}{3}.
+
+
+
+
+
+
+ Suppose that y = a + be^{-kt}. Solve for t in terms of a, b, k, and y. What does this new equation represent?
+
+
+
+
+
+ Starting from y = a + be^{-kt}, we subtract a: y - a = be^{-kt}. Dividing by b: \frac{y-a}{b} = e^{-kt}. Taking the natural log: -kt = \ln\!\left(\frac{y-a}{b}\right). Therefore t = -\dfrac{1}{k}\ln\!\left(\frac{y-a}{b}\right). This is the inverse of the original function and gives the time at which the object reaches temperature y.
+
- Complete each of the following statements with an appropriate number or the symbols \infty or -\infty. Do your best to do so without using a graphing utility; instead use your understanding of the function's graph.
-
-
-
-
-
-
- As t \to \infty, e^{-t} \to .
-
-
-
-
-
-
0
-
-
-
-
-
- As t \to \infty, \ln(t) \to .
-
-
-
-
-
-
\infty
-
-
-
-
-
- As t \to \infty, e^{t} \to .
-
-
-
-
-
-
\infty
-
-
-
-
-
- As t \to 0^+, e^{-t} \to . (When we write t \to 0^+, this means that we are letting t get closer and closer to 0, but only allowing t to take on positive values.)
-
-
-
-
-
-
1. Since e^{-t} \to e^0 = 1 as t \to 0^+.
-
-
-
-
-
- As t \to \infty, 35 + 53e^{-0.025t} \to.
-
-
-
-
-
-
35. As t \to \infty, e^{-0.025t} \to 0, so the expression approaches 35 + 53 \cdot 0 = 35.
-
-
-
-
-
- As t \to \frac{\pi}{2}^-, \tan(t) \to . (When we write t \to \frac{\pi}{2}^-, this means that we are letting t get closer and closer to \frac{\pi}{2}^-, but only allowing t to take on values that lie to the left of \frac{\pi}{2}.)
-
-
-
-
-
-
\infty. As t \to \frac{\pi}{2}^-, \cos(t) \to 0^+ while \sin(t) \to 1, so \tan(t) = \frac{\sin(t)}{\cos(t)} \to +\infty.
-
-
-
-
-
- As t \to \frac{\pi}{2}^+, \tan(t) \to . (When we write t \to \frac{\pi}{2}^+, this means that we are letting t get closer and closer to \frac{\pi}{2}^+, but only allowing t to take on values that lie to the right of \frac{\pi}{2}.)
-
-
-
-
-
-
-\infty. As t \to \frac{\pi}{2}^+, \cos(t) \to 0^- while \sin(t) \to 1, so \tan(t) = \frac{\sin(t)}{\cos(t)} \to -\infty.
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ Complete each of the following statements with an appropriate number or the symbols \infty or -\infty. Do your best to do so without using a graphing utility; instead use your understanding of the function's graph.
+
+
+
+
+
+
+ As t \to \infty, e^{-t} \to .
+
+
+
+
+
0
+
+
+
+
+
+ As t \to \infty, \ln(t) \to .
+
+
+
+
+
\infty
+
+
+
+
+
+ As t \to \infty, e^{t} \to .
+
+
+
+
+
\infty
+
+
+
+
+
+ As t \to 0^+, e^{-t} \to . (When we write t \to 0^+, this means that we are letting t get closer and closer to 0, but only allowing t to take on positive values.)
+
+
+
+
+
1. Since e^{-t} \to e^0 = 1 as t \to 0^+.
+
+
+
+
+
+ As t \to \infty, 35 + 53e^{-0.025t} \to.
+
+
+
+
+
35. As t \to \infty, e^{-0.025t} \to 0, so the expression approaches 35 + 53 \cdot 0 = 35.
+
+
+
+
+
+ As t \to \frac{\pi}{2}^-, \tan(t) \to . (When we write t \to \frac{\pi}{2}^-, this means that we are letting t get closer and closer to \frac{\pi}{2}^-, but only allowing t to take on values that lie to the left of \frac{\pi}{2}.)
+
+
+
+
+
\infty. As t \to \frac{\pi}{2}^-, \cos(t) \to 0^+ while \sin(t) \to 1, so \tan(t) = \frac{\sin(t)}{\cos(t)} \to +\infty.
+
+
+
+
+
+ As t \to \frac{\pi}{2}^+, \tan(t) \to . (When we write t \to \frac{\pi}{2}^+, this means that we are letting t get closer and closer to \frac{\pi}{2}^+, but only allowing t to take on values that lie to the right of \frac{\pi}{2}.)
+
+
+
+
+
-\infty. As t \to \frac{\pi}{2}^+, \cos(t) \to 0^- while \sin(t) \to 1, so \tan(t) = \frac{\sin(t)}{\cos(t)} \to -\infty.
- A piece of cardboard that is 12 \times 18 (each measured in inches) is being made into a box without a top. To do so, squares are cut from each corner of the cardboard and the remaining sides are folded up.
-
-
-
-
-
-
- Let x be the side length of the squares being cut from the corners of the cardboard. Draw a labeled diagram that shows the given information and the variable being used.
-
-
-
-
-
-
-
-
-
-
-
- Determine a formula for the function V whose output is the volume of the box that results from a square of size x \times x being cut from each corner of the cardboard.
-
-
-
-
-
-
After cutting x \times x squares from each corner, the base of the box has dimensions (12 - 2x) by (18 - 2x), and the height of the box is x. Thus
-
- V(x) = (12-2x)(18-2x)x = 4(6-x)(9-x)x.
-
-
-
-
-
-
-
- What familiar kind of function is V?
-
-
-
-
-
-
V is a cubic (degree 3) polynomial.
-
-
-
-
-
- If we start with a small positive value for x and let that value get larger and larger, what is the first value of xwe encounter that makes it impossible to remove x \times x squares from the cardboard and still form a box?
-
-
-
-
-
-
The shorter side of the cardboard is 12 inches. Cutting squares of size x from both ends of the short side removes 2x inches, so we need 12 - 2x > 0, i.e., x < 6. The first value of x that makes it impossible to form a box is x = 6.
-
-
-
-
-
- What are the zeros of V? What is the domain of the model V in the context of the rectangular box?
-
-
-
-
-
-
From V(x) = 4x(6-x)(9-x), the zeros are x = 0, x = 6, and x = 9. In the context of the physical problem, we need x > 0 and x < 6 (so that both pairs of opposite sides of the base remain positive), giving domain (0, 6).
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ A piece of cardboard that is 12 \times 18 (each measured in inches) is being made into a box without a top. To do so, squares are cut from each corner of the cardboard and the remaining sides are folded up.
+
+
+
+
+
+
+ Let x be the side length of the squares being cut from the corners of the cardboard. Draw a labeled diagram that shows the given information and the variable being used.
+
+
+
+
+
+
+
+
+
+
+ Determine a formula for the function V(x) whose output is the volume of the box that results from a square of size x \times x being cut from each corner of the cardboard.
+
+
+
+
+
After cutting x \times x squares from each corner, the base of the box has dimensions (12 - 2x) by (18 - 2x), and the height of the box is x. Thus
+
+ V(x) = (12-2x)(18-2x)x = 4(6-x)(9-x)x.
+
+
+
+
+
+
+
+ What familiar kind of function is V?
+
+
+
+
+
V is a cubic (degree 3) polynomial.
+
+
+
+
+
+ If we start with a small positive value for x and let that value get larger and larger, what is the first value of x we encounter that makes it impossible to remove x \times x squares from the cardboard and still form a box?
+
+
+
+
+
The shorter side of the cardboard is 12 inches. Cutting squares of size x from both ends of the short side removes 2x inches, so we need 12 - 2x > 0, i.e., x < 6. The first value of x that makes it impossible to form a box is x = 6.
+
+
+
+
+
+ What are the zeros of V? What is the domain of the model V in the context of the rectangular box?
+
+
+
+
+
From V(x) = 4x(6-x)(9-x), the zeros are x = 0, x = 6, and x = 9. In the context of the physical problem, we need x > 0 and x < 6 (so that both pairs of opposite sides of the base remain positive), giving domain (0, 6).
- Point your browser to the Desmos worksheet at http://gvsu.edu/s/0zy. There you'll find a degree 4 polynomial of the form p(x) = a_0 + a_1x + a_2x^2 + a_3x^3 + a_4x^4, where a_0, \ldots, a_4 are set up as sliders. In the questions that follow, you'll experiment with different values of a_0, \ldots, a_4 to investigate different possible behaviors in a degree 4 polynomial. Note that we require a_4 \ne 0 in order to ensure p is a degree 4 polynomial.
-
-
-
-
-
-
- What is the largest number of distinct points at which p(x) can cross the x-axis?
-
-
- Recall from the definition of a polynoimal function what we mean by a zero of the polynomial. Give examples of values for a_0, \ldots, a_4 that lead to that largest number of zeros for p(x).
-
-
-
-
-
-
- A degree 4 polynomial can cross the x-axis at most 4 times. For example, a_0=-24, a_1=26, a_2=-1, a_3=-4, a_4=1 gives a polynomial with 4 real zeros near x=-2,-1,3,4.
-
-
-
-
-
-
- What other numbers of zeros are possible for p(x)? Said differently, can you get each possible number of fewer zeros than the largest number that you found in (a)? Why or why not?
-
-
-
-
-
-
- A degree 4 polynomial can have 0, 1, 2, 3, or 4 real zeros. Yes, each number from 0 to 4 is achievable (though complex zeros always come in conjugate pairs, so the count of real zeros must have the same parity as the degree; a degree 4 polynomial can have 0, 2, or 4 real zeros if it has no repeated zeros, but with repeated zeros 1 or 3 are also possible).
-
-
-
-
-
-
- We say that a function has a turning pointturning point if the function changes from decreasing to increasing or increasing to decreasing at the point. For example, any quadratic function has a turning point at its vertex.
-
-
- What is the largest number of turning points that p(x) (the function in the Desmos worksheet) can have? Experiment with the sliders, and give examples of values for a_0, \ldots, a_4 that lead to that largest number of turning points for p(x).
-
-
-
-
-
-
- A degree 4 polynomial can have at most 3 turning points (one fewer than the degree). For example, p(x) = x^4 - 3x^2 has turning points at approximately x \approx -1.22, 0, 1.22.
-
-
-
-
-
-
- What other numbers of turning points are possible for p(x)? Can it have no turning points? Just one? Exactly two? Experiment and explain.
-
-
-
-
-
-
- A degree 4 polynomial can have 1 or 3 turning points (or 0 is technically possible, but not for most configurations). With 0 real zeros, it can have just 1 turning point (like x^4 + 1). It can also have 2 turning points, for example when one zero has multiplicity 2. So 1, 2, or 3 turning points are all possible.
-
-
-
-
-
-
- What long-range behavior is possible for p(x)? Said differently, what are the possible results for \displaystyle \lim_{x \to -\infty} p(x) and \displaystyle \lim_{x \to \infty} p(x)?
-
-
-
-
-
-
- Since the leading term is a_4 x^4 with a_4 \ne 0: if a_4 \gt 0, then both \lim_{x \to \pm\infty} p(x) = +\infty; if a_4 \lt 0, then both limits equal -\infty. The two limits are always equal to each other (both +\infty or both -\infty).
-
-
-
-
-
-
- What happens when we plot y = a_4 x^4 in Desmos and compare p(x) and a_4 x^4? How do they look when we zoom out? (Experiment with different values of each of the sliders, too.)
-
-
-
-
-
-
- When zoomed out, p(x) and a_4 x^4 look nearly identical: the lower-degree terms become insignificant compared to the dominant a_4 x^4 term. On a large scale, all degree-4 polynomials with the same leading coefficient look alike.
-
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ Point your browser to the Desmos worksheet at http://gvsu.edu/s/0zy. There you'll find a degree 4 polynomial of the form p(x) = a_0 + a_1x + a_2x^2 + a_3x^3 + a_4x^4, where a_0, \ldots, a_4 are set up as sliders. In the questions that follow, you'll experiment with different values of a_0, \ldots, a_4 to investigate different possible behaviors in a degree 4 polynomial. Note that we require a_4 \ne 0 in order to ensure p is a degree 4 polynomial.
+
+
+
+
+
+
+ What is the largest number of distinct points at which p(x) can cross the x-axis?
+
+
+ Recall from the definition of a polynomial function what we mean by a zero of the polynomial. Give examples of values for a_0, \ldots, a_4 that lead to that largest number of zeros for p(x).
+
+
+
+
+
+ A degree 4 polynomial can cross the x-axis at most 4 times. For example, a_0=-24, a_1=26, a_2=-1, a_3=-4, a_4=1 gives a polynomial with 4 real zeros near x=-2,-1,3,4.
+
+
+
+
+
+
+ What other numbers of zeros are possible for p(x)? Said differently, can you get each possible number of fewer zeros than the largest number that you found in (a)? Why or why not?
+
+
+
+
+
+ A degree 4 polynomial can have 0, 1, 2, 3, or 4 real zeros. Yes, each number from 0 to 4 is achievable (though complex zeros always come in conjugate pairs, so the count of real zeros must have the same parity as the degree; a degree 4 polynomial can have 0, 2, or 4 real zeros if it has no repeated zeros, but with repeated zeros 1 or 3 are also possible).
+
+
+
+
+
+
+ We say that a function has a turning pointturning point if the function changes from decreasing to increasing or increasing to decreasing at the point. For example, any quadratic function has a turning point at its vertex.
+
+
+ What is the largest number of turning points that p(x) (the function in the Desmos worksheet) can have? Experiment with the sliders, and give examples of values for a_0, \ldots, a_4 that lead to that largest number of turning points for p(x).
+
+
+
+
+
+ A degree 4 polynomial can have at most 3 turning points (one fewer than the degree). For example, p(x) = x^4 - 3x^2 has turning points at approximately x \approx -1.22, 0, 1.22.
+
+
+
+
+
+
+ What other numbers of turning points are possible for p(x)? Can it have no turning points? Just one? Exactly two? Experiment and explain.
+
+
+
+
+
+ A degree 4 polynomial can have 1 or 3 turning points. When it has 0 real zeros, it can have just 1 turning point (like x^4 + 1).
+
+
+
+
+
+
+ What long-range behavior is possible for p(x)? Said differently, what are the possible results for \displaystyle \lim_{x \to -\infty} p(x) and \displaystyle \lim_{x \to \infty} p(x)?
+
+
+
+
+
+ Since the leading term is a_4 x^4 with a_4 \ne 0: if a_4 \gt 0, then both \lim_{x \to \pm\infty} p(x) = +\infty; if a_4 \lt 0, then both limits equal -\infty. The two limits are always equal to each other (both +\infty or both -\infty).
+
+
+
+
+
+
+ What happens when we plot y = a_4 x^4 in Desmos and compare p(x) and a_4 x^4? How do they look when we zoom out? (Experiment with different values of each of the sliders, too.)
+
+
+
+
+
+ When zoomed out, p(x) and a_4 x^4 look nearly identical: the lower-degree terms become insignificant compared to the dominant a_4 x^4 term. On a large scale, all degree-4 polynomials with the same leading coefficient look alike.
+
- Consider the rational function r(x) = \frac{x^2 - 1}{x^2 - 3x - 4}, and let p(x) = x^2 - 1 (the numerator of r(x)) and q(x) = x^2 - 3x - 4 (the denominator of r(x)).
-
-
-
-
-
-
-
- Reasoning algebraically, for what values of x is p(x) = 0?
-
-
-
-
-
-
Factoring: p(x) = x^2 - 1 = (x-1)(x+1). So p(x) = 0 when x = 1 or x = -1.
-
-
-
-
-
- Again reasoning algebraically, for what values of x is q(x) = 0?
-
-
-
-
-
-
Factoring: q(x) = x^2 - 3x - 4 = (x-4)(x+1). So q(x) = 0 when x = 4 or x = -1.
-
-
-
-
-
- Define r(x) in Desmos, and evaluate the function appropriately to find numerical values for the output of r and hence complete the following tables.
-
Using technology to evaluate r(x) = \dfrac{x^2-1}{x^2-3x-4}: near x=4, the values grow very large in magnitude; near x=1, the values are small and negative then positive; near x=-1, the values are close to \frac{2}{5} = 0.4. Approximate values:
-
Near x=4: r(4.1) \approx 31, r(3.9) \approx -29. Near x=1: r(1.1) \approx -0.035, r(0.9) \approx 0.032. Near x=-1: r(-1.1) \approx 0.412, r(-0.9) \approx 0.387.
-
-
-
-
-
- Why does r behave the way it does near x = 4? Explain by describing the behavior of the numerator and denominator.
-
-
-
-
-
-
Near x = 4, the denominator q(x) = (x-4)(x+1) approaches 0 (since x-4 \to 0), while the numerator p(4) = 15 \neq 0. So r(x) increases or decreases without bound near x = 4: this is a vertical asymptote. For x just above 4, (x-4) > 0 and (x+1) > 0, so r \to +\infty; just below 4, (x-4) < 0 so r \to -\infty.
-
-
-
-
-
- Why does r behave the way it does near x = 1? Explain by describing the behavior of the numerator and denominator.
-
-
-
-
-
-
Near x = 1, the numerator p(x) \to 0 while the denominator q(1) = 1 - 3 - 4 = -6 \neq 0. So r(x) \to 0 near x = 1, meaning x = 1 is an ordinary zero of r.
-
-
-
-
-
- Why does r behave the way it does near x = -1? Explain by describing the behavior of the numerator and denominator.
-
-
-
-
-
-
Near x = -1, both the numerator and denominator approach 0 (since p(-1) = 0 and q(-1) = 0). The factor (x+1) cancels: r(x) = \dfrac{(x-1)(x+1)}{(x-4)(x+1)} = \dfrac{x-1}{x-4} (for x \neq -1). Near x = -1, r(x) \approx \dfrac{-1-1}{-1-4} = \dfrac{-2}{-5} = \dfrac{2}{5}. So x = -1 is a hole at the point \left(-1, \dfrac{2}{5}\right).
-
-
-
-
-
- Plot r in Desmos. Is there anything surprising or misleading about the graph that Desmos generates?
-
-
-
-
-
-
The Desmos graph of r appears continuous near x = -1; the hole is not visible because Desmos plots points at closely-spaced x-values and connects them, so the single missing point at x=-1 is not apparent. One must know to look for the hole based on the algebra.
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ Consider the rational function r(x) = \frac{x^2 - 1}{x^2 - 3x - 4}, and let p(x) = x^2 - 1 (the numerator of r(x)) and q(x) = x^2 - 3x - 4 (the denominator of r(x)).
+
+
+
+
+
+
+
+ Reasoning algebraically, for what values of x is p(x) = 0?
+
+
+
+
+
Factoring: p(x) = x^2 - 1 = (x-1)(x+1). So p(x) = 0 when x = 1 or x = -1.
+
+
+
+
+
+ Again reasoning algebraically, for what values of x is q(x) = 0?
+
+
+
+
+
Factoring: q(x) = x^2 - 3x - 4 = (x-4)(x+1). So q(x) = 0 when x = 4 or x = -1.
+
+
+
+
+
+ Define r(x) in Desmos, and evaluate the function appropriately to find numerical values for the output of r and hence complete the following tables.
+
Using technology to evaluate r(x) = \dfrac{x^2-1}{x^2-3x-4}: near x=4, the values grow very large in magnitude; near x=1, the values are small and negative then positive; near x=-1, the values are close to \frac{2}{5} = 0.4. Approximate values:
+
Near x=4: r(4.1) \approx 31, r(3.9) \approx -29. Near x=1: r(1.1) \approx -0.035, r(0.9) \approx 0.032. Near x=-1: r(-1.1) \approx 0.412, r(-0.9) \approx 0.387.
+
+
+
+
+
+ Why does r behave the way it does near x = 4? Explain by describing the behavior of the numerator and denominator.
+
+
+
+
+
Near x = 4, the denominator q(x) = (x-4)(x+1) approaches 0 (since x-4 \to 0), while the numerator p(4) = 15 \neq 0. So r(x) increases or decreases without bound near x = 4: this is a vertical asymptote. For x just above 4, (x-4) > 0 and (x+1) > 0, so r \to +\infty; just below 4, (x-4) < 0 so r \to -\infty.
+
+
+
+
+
+ Why does r behave the way it does near x = 1? Explain by describing the behavior of the numerator and denominator.
+
+
+
+
+
Near x = 1, the numerator p(x) \to 0 while the denominator q(1) = 1 - 3 - 4 = -6 \neq 0. So r(x) \to 0 near x = 1, meaning x = 1 is an ordinary zero of r.
+
+
+
+
+
+ Why does r behave the way it does near x = -1? Explain by describing the behavior of the numerator and denominator.
+
+
+
+
+
Near x = -1, both the numerator and denominator approach 0 (since p(-1) = 0 and q(-1) = 0). The factor (x+1) cancels: r(x) = \dfrac{(x-1)(x+1)}{(x-4)(x+1)} = \dfrac{x-1}{x-4} (for x \neq -1). Near x = -1, r(x) \approx \dfrac{-1-1}{-1-4} = \dfrac{-2}{-5} = \dfrac{2}{5}. So x = -1 is a hole at the point \left(-1, \dfrac{2}{5}\right).
+
+
+
+
+
+ Plot r in Desmos. Is there anything surprising or misleading about the graph that Desmos generates?
+
+
+
+
+
The Desmos graph of r appears continuous near x = -1; the hole is not visible because Desmos plots points at closely-spaced x-values and connects them, so the single missing point at x=-1 is not apparent. One must know to look for the hole based on the algebra.
- A drug company estimates that to produce a new drug,
- it will cost $5 million in startup resources, and that once they reach production, each gram of the drug will cost $2500 to make.
-
-
-
-
-
-
- Determine a formula for a function C(q) that models the cost of producing q grams of the drug. What familiar kind of function is C?
-
-
-
-
-
-
C(q) = 5{,}000{,}000 + 2500q gives the total cost in dollars to produce q grams. C is a linear function whose slope is the per-gram manufacturing cost and whose y-intercept is the startup cost.
-
-
-
-
-
- The drug company needs to sell the drug at a price of more than $2500 per gram in order to at least break even. To investigate how they might set prices, they first consider what their average cost per gram is. What is the total cost of producing 1000 grams? What is the average cost per gram to produce 1000 grams?
-
-
-
-
-
-
The total cost of producing 1000 grams is C(1000) = 5{,}000{,}000 + 2500(1000) = \$7{,}500{,}000. The average cost per gram is \frac{7{,}500{,}000}{1000} = \$7{,}500 per gram.
-
-
-
-
-
- What is the total cost of producing 10000 grams? What is the average cost per gram to produce 10000 grams?
-
-
-
-
-
-
The total cost of producing 10{,}000 grams is C(10000) = 5{,}000{,}000 + 2500(10000) = \$30{,}000{,}000. The average cost per gram is \frac{30{,}000{,}000}{10{,}000} = \$3{,}000 per gram.
-
-
-
-
-
- Our computations in (b) and (c) naturally lead us to define the average cost per gram function, A(q), whose output is the average cost of producing q grams of the drug. What is a formula for A(q)?
-
-
-
-
-
-
The average cost per gram to produce q grams is
-
- A(q) = \frac{C(q)}{q} = \frac{5{,}000{,}000 + 2500q}{q}.
-
-
-
-
-
-
-
- Explain why another formula for A is A(q) = 2500 + \frac{5000000}{q}.
-
-
-
-
-
-
Dividing both terms in the numerator of A(q) by q:
-
- A(q) = \frac{5{,}000{,}000 + 2500q}{q} = \frac{5{,}000{,}000}{q} + \frac{2500q}{q} = 2500 + \frac{5{,}000{,}000}{q}.
-
-
-
-
-
-
-
- What can you say about the long-range behavior of A? What does this behavior mean in the context of the problem?
-
-
-
-
-
-
As q \to \infty, the term \frac{5{,}000{,}000}{q} \to 0, so A(q) \to 2500. In context, this means that as production volume grows very large, the startup cost is spread over so many grams that the average cost per gram approaches the per-gram manufacturing cost of \$2500.
-
-
-
-
- This activity is based on p. 457ff in Functions Modeling Change, by Connally et al.
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ A drug company estimates that to produce a new drug,
+ it will cost $5 million in startup resources, and that once they reach production, each gram of the drug will cost $2500 to make.
+
+
+
+
+
+
+ Determine a formula for a function C(q) that models the cost of producing q grams of the drug. What familiar kind of function is C?
+
+
+
+
+
C(q) = 5{,}000{,}000 + 2500q gives the total cost in dollars to produce q grams. C is a linear function whose slope is the per-gram manufacturing cost and whose y-intercept is the startup cost.
+
+
+
+
+
+ The drug company needs to sell the drug at a price of more than $2500 per gram in order to at least break even. To investigate how they might set prices, they first consider what their average cost per gram is. What is the total cost of producing 1000 grams? What is the average cost per gram to produce 1000 grams?
+
+
+
+
+
The total cost of producing 1000 grams is C(1000) = 5{,}000{,}000 + 2500(1000) = \$7{,}500{,}000. The average cost per gram is \frac{7{,}500{,}000}{1000} = \$7{,}500 per gram.
+
+
+
+
+
+ What is the total cost of producing 10000 grams? What is the average cost per gram to produce 10000 grams?
+
+
+
+
+
The total cost of producing 10{,}000 grams is C(10000) = 5{,}000{,}000 + 2500(10000) = \$30{,}000{,}000. The average cost per gram is \frac{30{,}000{,}000}{10{,}000} = \$3{,}000 per gram.
+
+
+
+
+
+ Our computations in (b) and (c) naturally lead us to define the average cost per gram function, A(q), whose output is the average cost of producing q grams of the drug. What is a formula for A(q)?
+
+
+
+
+
The average cost per gram to produce q grams is
+
+ A(q) = \frac{C(q)}{q} = \frac{5{,}000{,}000 + 2500q}{q}.
+
+
+
+
+
+
+
+ Explain why another formula for A is A(q) = 2500 + \frac{5000000}{q}.
+
+
+
+
+
Dividing both terms in the numerator of A(q) by q:
+
+ A(q) = \frac{5{,}000{,}000 + 2500q}{q} = \frac{5{,}000{,}000}{q} + \frac{2500q}{q} = 2500 + \frac{5{,}000{,}000}{q}.
+
+
+
+
+
+
+
+ What can you say about the long-range behavior of A? What does this behavior mean in the context of the problem?
+
+
+
+
+
As q \to \infty, the term \frac{5{,}000{,}000}{q} \to 0, so A(q) \to 2500. In context, this means that as production volume grows very large, the startup cost is spread over so many grams that the average cost per gram approaches the per-gram manufacturing cost of \$2500.
+
+
+
+
+ This activity is based on p. 457ff in Functions Modeling Change, 5th edition, by Connally et al.
+
- Consider a right triangle that has one leg of length 3 and another leg of length \sqrt{3}. Let \theta be the angle that lies opposite the shorter leg.
-
-
-
-
-
-
- Sketch a labeled picture of the triangle.
-
-
-
-
-
-
-
-
-
-
-
- What is the exact length of the triangle's hypotenuse?
-
-
-
-
-
-
- The hypotenuse has length \sqrt{(\sqrt{3})^2 + 3^2} = \sqrt{12} = 2\sqrt{3}.
-
-
-
-
-
-
- What is the exact value of \sin(\theta)?
-
-
-
-
-
-
- Since \theta is opposite the shorter leg of length \sqrt{3}, we have \sin(\theta) = \dfrac{\sqrt{3}}{2\sqrt{3}} = \dfrac{1}{2}.
-
-
-
-
-
-
- Rewrite your equation from (c) using the arcsine function in the form \arcsin(\Box) = \Delta, where \Box and \Delta are numerical values.
-
-
-
-
-
-
- Since \sin(\theta) = \dfrac{1}{2}, we can write \arcsin\!\left(\dfrac{1}{2}\right) = \theta.
-
-
-
-
-
-
- What special angle from the unit circle is \theta?
-
-
-
-
-
-
- Since \sin\!\left(\dfrac{\pi}{6}\right) = \dfrac{1}{2}, we have \theta = \dfrac{\pi}{6}.
-
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ Consider a right triangle that has one leg of length 3 and another leg of length \sqrt{3}. Let \theta be the angle that lies opposite the shorter leg.
+
+
+
+
+
+
+ Sketch a labeled picture of the triangle.
+
+
+
+
+
+
+
+
+
+
+ What is the exact length of the triangle's hypotenuse?
+
+
+
+
+
+ The hypotenuse has length \sqrt{(\sqrt{3})^2 + 3^2} = \sqrt{12} = 2\sqrt{3}.
+
+
+
+
+
+
+ What is the exact value of \sin(\theta)?
+
+
+
+
+
+ Since \theta is opposite the shorter leg of length \sqrt{3}, we have \sin(\theta) = \dfrac{\sqrt{3}}{2\sqrt{3}} = \dfrac{1}{2}.
+
+
+
+
+
+
+ Rewrite your equation from (c) using the arcsine function in the form \arcsin(\Box) = \Delta, where \Box and \Delta are numerical values.
+
+
+
+
+
+ Since \sin(\theta) = \dfrac{1}{2}, we can write \arcsin\!\left(\dfrac{1}{2}\right) = \theta.
+
+
+
+
+
+
+ What special angle from the unit circle is \theta?
+
+
+
+
+
+ Since \sin\!\left(\dfrac{\pi}{6}\right) = \dfrac{1}{2}, we have \theta = \dfrac{\pi}{6}.
+
- Consider the plot of the standard cosine function in the following figure along with the emphasized portion of the graph on [0,\pi].
-
-
-
ADD ALT TEXT TO THIS IMAGE
-
-
- Let g be the function whose domain is 0 \le t \le \pi and whose outputs are determined by the rule g(t) = \cos(t). Note well: g is defined in terms of the cosine function, but because it has a different domain, it is not the cosine function.
-
-
-
-
-
-
- What is the domain of g?
-
-
-
-
-
-
The domain of g is the interval [0, \pi].
-
-
-
-
-
- What is the range of g?
-
-
-
-
-
-
The range of g is the interval [-1, 1].
-
-
-
-
-
- Does g pass the horizontal line test?
- Why or why not?
-
-
-
-
-
-
Yes, g passes the horizontal line test. On its restricted domain [0,\pi], the cosine function is strictly decreasing, so no horizontal line can intersect its graph more than once.
-
-
-
-
-
- Explain why g has an inverse function,
- g^{-1}, and state the domain and range of g^{-1}.
-
-
-
-
-
-
Since g passes the horizontal line test, it is one-to-one and has an inverse function g^{-1}. The domain of g^{-1} is the range of g, which is [-1, 1], and the range of g^{-1} is the domain of g, which is [0, \pi].
-
-
-
-
-
- We know that g(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}.
- What is the exact value of g^{-1}(\frac{\sqrt{2}}{2})?
- How about the exact value of g^{-1}(-\frac{\sqrt{2}}{2})?
-
-
-
-
-
-
Since g\!\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}, we have g^{-1}\!\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4}. Since g\!\left(\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2}, we have g^{-1}\!\left(-\frac{\sqrt{2}}{2}\right) = \frac{3\pi}{4}.
-
-
-
-
-
- Determine the exact values of g^{-1}(-\frac{1}{2}),
- g^{-1}(\frac{\sqrt{3}}{2}),
- g^{-1}(0), and g^{-1}(-1).
- Use proper notation to label your results.
-
+ Consider the plot of the standard cosine function in the following figure along with the emphasized portion of the graph on [0,\pi].
+
+
+
ADD ALT TEXT TO THIS IMAGE
+
+
+ Let g be the function whose domain is 0 \le t \le \pi and whose outputs are determined by the rule g(t) = \cos(t). Note well: g is defined in terms of the cosine function, but because it has a different domain, it is not the cosine function.
+
+
+
+
+
+
+ What is the domain of g?
+
+
+
+
+
The domain of g is the interval [0, \pi].
+
+
+
+
+
+ What is the range of g?
+
+
+
+
+
The range of g is the interval [-1, 1].
+
+
+
+
+
+ Does g pass the horizontal line test?
+ Why or why not?
+
+
+
+
+
Yes, g passes the horizontal line test. On its restricted domain [0,\pi], the cosine function is strictly decreasing, so no horizontal line can intersect its graph more than once.
+
+
+
+
+
+ Explain why g has an inverse function,
+ g^{-1}, and state the domain and range of g^{-1}.
+
+
+
+
+
Since g passes the horizontal line test, it is one-to-one and has an inverse function g^{-1}. The domain of g^{-1} is the range of g, which is [-1, 1], and the range of g^{-1} is the domain of g, which is [0, \pi].
+
+
+
+
+
+ We know that g(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}.
+ What is the exact value of g^{-1}(\frac{\sqrt{2}}{2})?
+ How about the exact value of g^{-1}(-\frac{\sqrt{2}}{2})?
+
+
+
+
+
Since g\!\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}, we have g^{-1}\!\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4}. Since g\!\left(\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2}, we have g^{-1}\!\left(-\frac{\sqrt{2}}{2}\right) = \frac{3\pi}{4}.
+
+
+
+
+
+ Determine the exact values of g^{-1}(-\frac{1}{2}),
+ g^{-1}(\frac{\sqrt{3}}{2}),
+ g^{-1}(0), and g^{-1}(-1).
+ Use proper notation to label your results.
+
- Consider a right triangle with hypotenuse of length 61 and one leg of length 11.
- Let \alpha be the angle opposite the side of length 11.
- Find the exact length of the other leg and then determine the value of each of the six trigonometric functions evaluated at \alpha.
- In addition, what are the exact and approximate measures of the two non-right angles in the triangle?
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ Consider a right triangle with hypotenuse of length 61 and one leg of length 11.
+ Let \alpha be the angle opposite the side of length 11.
+ Find the exact length of the other leg and then determine the value of each of the six trigonometric functions evaluated at \alpha.
+ In addition, what are the exact and approximate measures of the two non-right angles in the triangle?
+
- For each of the following situations, sketch a right triangle that satisfies the given conditions, and then either determine the requested missing information in the triangle or explain why you don't have enough information to determine it. Assume that all angles are being considered in radian measure.
-
-
-
-
-
-
- The length of the other leg of a right triangle with hypotenuse of length 1 and one leg of length \frac{3}{5}.
-
-
-
-
-
-
- By the Pythagorean theorem, the other leg has length \sqrt{1^2 - (3/5)^2} = \sqrt{1 - 9/25} = \sqrt{16/25} = \frac{4}{5}.
-
-
-
-
-
-
- The lengths of the two legs in a right triangle with hypotenuse of length 1 where one of the non-right angles measures \frac{\pi}{3}.
-
-
-
-
-
-
- In a right triangle with hypotenuse 1 and one non-right angle of measure \frac{\pi}{3}, the adjacent leg has length \cos\!\left(\frac{\pi}{3}\right) = \frac{1}{2} and the opposite leg has length \sin\!\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}.
-
-
-
-
-
-
- The length of the other leg of a right triangle with hypotenuse of length 7 and one leg of length 6.
-
-
-
-
-
-
- By the Pythagorean theorem, the other leg has length \sqrt{7^2 - 6^2} = \sqrt{49 - 36} = \sqrt{13}.
-
-
-
-
-
-
- The lengths of the two legs in a right triangle with hypotenuse 5 and where one of the non-right angles measures \frac{\pi}{4}.
-
-
-
-
-
-
- In a right triangle with hypotenuse 5 and one non-right angle of measure \frac{\pi}{4}, both legs have the same length: 5\cos\!\left(\frac{\pi}{4}\right) = 5 \cdot \frac{\sqrt{2}}{2} = \frac{5\sqrt{2}}{2}.
-
-
-
-
-
-
- The length of the other leg of a right triangle with hypotenuse of length 1 and one leg of length \cos(0.7).
-
-
-
-
-
-
- By the Pythagorean theorem, the other leg has length \sqrt{1 - \cos^2(0.7)} = \sin(0.7), using the Pythagorean identity \sin^2(t) + \cos^2(t) = 1.
-
-
-
-
-
-
- The measures of the two angles in a right triangle with hypotenuse of length 1 where the two legs have lengths \cos(1.1) and \sin(1.1), respectively.
-
-
-
-
-
-
- Since the hypotenuse is 1 and the adjacent leg to angle \theta has length \cos(1.1), we have \cos(\theta) = \cos(1.1), so \theta = 1.1 radians. The other non-right angle is \frac{\pi}{2} - 1.1 \approx 0.471 radians.
-
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ For each of the following situations, sketch a right triangle that satisfies the given conditions, and then either determine the requested missing information in the triangle or explain why you don't have enough information to determine it. Assume that all angles are being considered in radian measure.
+
+
+
+
+
+
+ The length of the other leg of a right triangle with hypotenuse of length 1 and one leg of length \frac{3}{5}.
+
+
+
+
+
+ By the Pythagorean theorem, the other leg has length \sqrt{1^2 - (3/5)^2} = \sqrt{1 - 9/25} = \sqrt{16/25} = \frac{4}{5}.
+
+
+
+
+
+
+ The lengths of the two legs in a right triangle with hypotenuse of length 1 where one of the non-right angles measures \frac{\pi}{3}.
+
+
+
+
+
+ In a right triangle with hypotenuse 1 and one non-right angle of measure \frac{\pi}{3}, the adjacent leg has length \cos\!\left(\frac{\pi}{3}\right) = \frac{1}{2} and the opposite leg has length \sin\!\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}.
+
+
+
+
+
+
+ The length of the other leg of a right triangle with hypotenuse of length 7 and one leg of length 6.
+
+
+
+
+
+ By the Pythagorean theorem, the other leg has length \sqrt{7^2 - 6^2} = \sqrt{49 - 36} = \sqrt{13}.
+
+
+
+
+
+
+ The lengths of the two legs in a right triangle with hypotenuse 5 and where one of the non-right angles measures \frac{\pi}{4}.
+
+
+
+
+
+ In a right triangle with hypotenuse 5 and one non-right angle of measure \frac{\pi}{4}, both legs have the same length: 5\cos\!\left(\frac{\pi}{4}\right) = 5 \cdot \frac{\sqrt{2}}{2} = \frac{5\sqrt{2}}{2}.
+
+
+
+
+
+
+ The length of the other leg of a right triangle with hypotenuse of length 1 and one leg of length \cos(0.7).
+
+
+
+
+
+ By the Pythagorean theorem, the other leg has length \sqrt{1 - \cos^2(0.7)} = \sin(0.7), using the Pythagorean identity \sin^2(t) + \cos^2(t) = 1.
+
+
+
+
+
+
+ The measures of the two angles in a right triangle with hypotenuse of length 1 where the two legs have lengths \cos(1.1) and \sin(1.1), respectively.
+
+
+
+
+
+ Since the hypotenuse is 1 and the adjacent leg to angle \theta has length \cos(1.1), we have \cos(\theta) = \cos(1.1), so \theta = 1.1 radians. The other non-right angle is \frac{\pi}{2} - 1.1 \approx 0.471 radians.
+
Using \tan(t) = \frac{\sin(t)}{\cos(t)}:
@@ -73,8 +73,8 @@
What are three other input values x for which \tan(x) is not defined?
-
-
+
Since \tan(t) = \frac{\sin(t)}{\cos(t)} and \cos\!\left(\frac{\pi}{2}\right) = 0, \tan\!\left(\frac{\pi}{2}\right) is undefined because division by zero is not allowed. Three other values where \tan is undefined are -\frac{\pi}{2}, \frac{3\pi}{2}, and -\frac{3\pi}{2} (any value of the form \frac{\pi}{2} + k\pi for integer k).
@@ -99,8 +99,8 @@
Why is the value of \tan(\frac{11\pi}{24}) so large relative to the other values of \tan(x) in the table?
-
-
+
At x = \frac{11\pi}{24}, the point is approximately \left(\frac{11\pi}{24},\ 7.596\right). We have \sin\!\left(\frac{11\pi}{24}\right) \approx 0.9914 and \cos\!\left(\frac{11\pi}{24}\right) \approx 0.1305. The value of \tan\!\left(\frac{11\pi}{24}\right) is so large because \frac{11\pi}{24} is very close to \frac{\pi}{2}, making \cos\!\left(\frac{11\pi}{24}\right) very small; since we divide by cosine, the tangent value becomes very large.
@@ -133,8 +133,8 @@
-->
+ Suppose that a rectangular aquarium is being filled with water. The tank is 4 feet long by 2 feet wide by 3 feet high, and the hose that is filling the tank is delivering water at a rate of 0.5 cubic feet per minute.
+
+
+
+ The empty aquarium, with length 4 feet, width 2 feet, and height 3 feet.
+
+ The empty aquarium
+
+
+
+ The aquarium partially filled to some unmarked height
+
+ The aquarium partially filled
+
+
+
+
+
+
+
+ What are some different quantities that are changing in this scenario?
+
+
+ Both the
+ first changing quantity
+ amount of water in the tank
+ capacity of the tank
+ length of the water
+ depth (height) of the water
+ and the
+ second changing quantity
+ amount of water in the tank
+ capacity of the tank
+ length of the water
+ depth (height) of the water
+ are changing as the hose delivers water into the tank.
+ correct changing quantities
+ $c1.selectedIndices=1 and $c2.selectedIndices=4
+ $c2.selectedIndices=1 and $c1.selectedIndices=4
+
+
+
+
+
+
+ After 1 minute has elapsed, how much water is in the tank?
+
+
+ amount of water after 1 minute cubic feet $amount1 = 0.5
+
+
+ At this moment, how deep is the water?
+
+
+ depth of water after 1 minute feet $depth1 = 0.5/8$depth1-.0625 < .01
+
+ Your answer is close to the correct answer. Make sure to either enter the exact fraction or use all the decimal places and don't round.
+
+
+
+ How much water is in the tank and how deep is the water after 2 minutes?
+
+
+ 1 cubic feet and 1/8 feet
+
+
+ After 3 minutes?
+
+
+ 1.5 cubic feet and 1.5/8 feet
+
+
+
+
+ How long will it take for the tank to be completely full?
+
+
+ $time=48
+
+
+
+ How much water is in a full tank? amount of water in a full tank24cubic feet
+
+
+
+
\ No newline at end of file
diff --git a/source/previews/doenet/PA-changing-aroc-D.doenetml b/source/previews/doenet/PA-changing-aroc-D.doenetml
new file mode 100644
index 00000000..c0949003
--- /dev/null
+++ b/source/previews/doenet/PA-changing-aroc-D.doenetml
@@ -0,0 +1,168 @@
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ 64-16(x-1)^2
+ 64-16(x-1)^2
+ x=0
+ x=1/2
+ x=1
+ x=3/2
+ x=2
+ x=3
+
+
+
+ domain of s of t[0,3][0,64](-\infty,64](-\infty,\infty)
+
+
+
+
+
+
+ Let the height function for a ball tossed vertically be given by s(t) = 64 - 16(t-1)^2,
+ where t is measured in seconds and s is measured in feet above the ground.
+
+ Let h(x)=f(x)+g(x). Then
+ \hspace{.3in}h(3)=f\big(input of f3\big)+g\big(input of g3\big)=value of h of 3$f[4]+$g[4]
+
+
+ Let k(x)=f(x)\cdot g(x). Then
+ \hspace{.3in}k(0)=f\big(input of f0\big)\cdot g\big(input of g0\big)=value of k of 0$f[1]*$g[1]
+
+
+
+
+ Now consider the functions p(x) and q(x) shown in the graph below.
+
+
+ graph of 2 piecewise linear functions and a moveable point
+
The first piece of p(x) is a line starting at the point (-4,4) and ending at the point (-3,1). The second piece of p(x) is a line starting at the point (-3,1) and ending at the point (2,-1). The third piece of p(x) is a line starting at the point (2,-1) going through the point (3,-3) and ending where x=4.
+
The first piece of q(x) is a line starting at the point (-4,-4) and ending at the point (-2,4). The second piece of p(x) is a line starting at the point (-2,4) and ending at the point (4,1).
+ $p
+
+ $q
+
+ (-1,-3)
+
+
+ Calculate exactly: You can use the slope and the change in x to see exactly how much the y-value has changed from a known y-value. Or, you can find the equation of a line from the slope and a point on the line, and use the equation to exactly calculate the y-value for p(-1).$$p(-1).
+ Calculate exactly: You can use the slope and the change in x to see exactly how much the y-value has changed from a known y-value. Or, you can find the equation of a line from the slope and a point on the line, and use the equation to exactly calculate the y-value for q(-1).$$q(-1).
+
+
+ Let r(x) = p(x) - q(x). Then
+ $$p(-1)-$$q(-1)
+
+
+ Note that r(x)=0 means that p(x)-q(x)=0, which means that p(x)=q(x). Drag the point on the above graph to a location where r(x)=0. abs($r0.x + 61/22)< 0.03 and abs($r0.y + 4*61/22-12)< 0.03
+
+
+ Calculate exactly: You can use the slope and the change in x to see exactly how much the y-value has changed from a known y-value. Or, you can find the equation of a line from the slope and a point on the line, and use the equation to exactly calculate the y-value for p(1).$$p(1).
+ Calculate exactly: You can use the slope and the change in x to see exactly how much the y-value has changed from a known y-value. Or, you can find the equation of a line from the slope and a point on the line, and use the equation to exactly calculate the y-value for q(1).$$q(1).
+
+
+ Let s(x) = \frac{p(x)}{q(x)}. Then
+ $$p(1)/$$q(1)
+
+
+ We have reproduced the graph of p(x) and q(x) below. Note that a fraction is undefined when the denominator is 0. Drag the point on the below graph to a location where s(x) is undefined. abs($sundef.x + 3)< 0.03 and abs($sundef.y - 0)< 0.03
+
+
+ graph of 2 piecewise linear functions and a moveable point
+
The first piece of p(x) is a line starting at the point (-4,4) and ending at the point (-3,1). The second piece of p(x) is a line starting at the point (-3,1) and ending at the point (2,-1). The third piece of p(x) is a line starting at the point (2,-1) going through the point (3,-3) and ending where x=4.
+
The first piece of q(x) is a line starting at the point (-4,-4) and ending at the point (-2,4). The second piece of p(x) is a line starting at the point (-2,4) and ending at the point (4,1).
+ Let r(t) = p(q(t)). Determine a formula for r that depends only on t and not on p or q.
+
+
+ t^2-1\bigg)
+
+
+ \hspace{.65in}=3\bigg(q of t\bigg)-4=p of q of t3t^2-73t^2-7
+
+
+ Your process is correct, but your answer isn't simplified enough, yet.
+
+
+
+ In the introductory example with y=f(x) = x^2 - 1 and x=g(t) = 3t - 4, we substituted 3t-4 in for the x and calculated f(g(t)).
+
+ y \amp = f(x)
+ \amp = f(g(t))
+ \amp = f(3t-4)
+ \amp = (3t-4)^2-1.
+
+
+ In both part a. and the example above, we had similar functions
+
+ p(x) \amp = 3x-4 \amp q(t)\amp=t^2-1
+ g(t)\amp = 3t-4 \amp f(x)\amp=x^2-1
+
+ but we composed them in order of compositionin the same order;in the opposite order; in part a., we substituted the type of functionlinear functionquadratic function into the type of functionlinear functionquadratic function while in the introductory example, we substituted the type of functionlinear functionquadratic function into the type of functionlinear function.quadratic function.
+
+
+
+
+
+ Let t = s(z) = \frac{1}{z+4} and recall that x = q(t) = t^2 - 1. Determine a formula for x = q(s(z)) that depends only on z.
+
+
+ 1/(z+4)\bigg)
+
+ \hspace{.65in}=q of s of z(1/(z+4))^2-1(1/z+4)^2-1(1/(z+4))^2-1
+
+ Your answer is correct, but you could have left it as \displaystyle \left(\frac{1}{z+4}\right)^2-1 or \displaystyle \frac{1}{(z+4)^2}-1.
+
+
+
+ Suppose that h(t) = \sqrt{2t^2 + 5}. Determine formulas for two related functions, y = f(x) and x = g(t), so that h(t) = f(g(t)).
+
+
+
+
+
+
+ The result of composing f(g(t)) is .
+
+
+ $gfunc=0 and
+ $ffunc=0 and
+ $gfunc!=1 and
+ $ffunc!=1 and
+ $$f($g) = sqrt(2t^2+5)
+
+
+ The composition f(g(t)) is not equal to h(t)=\sqrt{2t^2+5} as required.
+ The function g(t) cannot contain the variable x.
+ The function f(x) cannot contain the variable t.
+ We are looking for two functions that each are related but not the same as h(t); neither f(x) nor g(t) should be just a single variable.
+ You might try setting f(x)=\sqrt{x}
+
+
+
+
\ No newline at end of file
diff --git a/source/previews/doenet/PA-changing-functions-crickets-D.doenetml b/source/previews/doenet/PA-changing-functions-crickets-D.doenetml
new file mode 100644
index 00000000..9f4935a7
--- /dev/null
+++ b/source/previews/doenet/PA-changing-functions-crickets-D.doenetml
@@ -0,0 +1,94 @@
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ Use the equation T = 40 + 0.25N that relates the temperature, T, to the number of chirps per minute, N, to respond to the questions below. The equation is also called Dolbear's Law.
+
+
+
+
+ If we hear snowy tree crickets chirping at a rate of 92 chirps per minute, what does Dolbear's Law suggest should be the outside temperature?
+
+
+ temperature at 92 chirps per minute40+0.25*92^\circ F \,
+
+
+
+
+ If the outside temperature is 77^\circ F, how many chirps per minute should we expect to hear?
+
+
+ chirps per minute at 77 degrees Fahrenheit 4*(77-40)chirps per minute
+
+
+
+
+ Is the model valid for determining the number of chirps one should hear when the outside temperature is 35^\circ F? Why or why not?
+
+
+ yes or no
+ Yes
+ No
+ because
+ reason part 1
+ we can find a value of N which makes T equal 35,
+ we can't find a value of N which makes T equal 35,
+ T adds a quarter of the number of chirps per minute to 40,
+ and
+ reason part 2
+ N=-20.
+ N=20.
+ crickets can't chirp a negative number of times per minute.
+
+ correct explanation
+ $c1.selectedIndex=2 and $c2.selectedIndices=3 and $c3.selectedIndices=3
+
+
+
+
+
+
+ Suppose that in the morning an observer hears 65 chirps per minute, and several hours later hears 75 chirps per minute. How much has the temperature risen between observations?
+
+
+ $changetemp=2.5
+
+
+
+
+ What temperature corresponds to 65 chirps per minute? temperature at 65 chirps per minute56.25^\circ F
+
+
+ What temperature corresponds to 75 chirps per minute? temperature at 75 chirps per minute58.75^\circ F
+
+
+
+
+
+
+
+
+ Dolbear's Law is known to be accurate for temperatures from 50^\circ to 85^\circ.
+
+
+ What is the fewest number of chirps per minute an observer could expect to hear? fewest chirps per minute40 chirps per minute
+
+
+ What is the greatest number of chirps per minute an observer could expect to hear? greatest chirps per minute180 chirps per minute
+
+
+
+
\ No newline at end of file
diff --git a/source/previews/doenet/PA-changing-inverse-F-C-D.doenetml b/source/previews/doenet/PA-changing-inverse-F-C-D.doenetml
new file mode 100644
index 00000000..4256ffb3
--- /dev/null
+++ b/source/previews/doenet/PA-changing-inverse-F-C-D.doenetml
@@ -0,0 +1,88 @@
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ Recall that F = g(C) = \frac{9}{5}C + 32 is the function that takes Celsius temperature inputs and produces the corresponding Fahrenheit temperature outputs.
+
+
+
+
+ Solve the equation F = \frac{9}{5}C + 32 for C in terms of F.
+
+
+ The first step in solving the equation F = \frac{9}{5}C + 32 for C in terms of F is possible first operation
+ Multiply both sides by \frac{9}{5}.
+ Multiply both sides by \frac{5}{9}.
+ Subtract 32 from both sides.
+ Add 32 to both sides.
+
+ \amp F\amp \amp = \frac{9}{5}C\amp \amp + \amp 32
+ \amp \amp-32\amp= \amp \amp \amp-32
+ \amp\amp F-32\amp = \frac{9}{5}C\amp \amp \amp
+
+
+
+ The second step in solving the equation F = \frac{9}{5}C + 32 for C in terms of F is possible second operation
+ Multiply both sides by \frac{9}{5}.
+ Multiply both sides by \frac{5}{9}.
+ Subtract 32 from both sides.
+ Add 32 to both sides.
+
+ F-32\amp = \frac{9}{5}C
+ \frac{5}{9}(F-32)\amp = \frac{5}{9}\cdot\frac{9}{5}C
+ \frac{5}{9}(F-32)\amp = C
+
+
+
+
+
+ Note that the equation C = \frac{5}{9}(F-32) expresses C as a function of F. Call this function h so that C = h(F) = \frac{5}{9}(F-32).
+
+
+ Find the simplest expression that you can for the composite function j(C) = h(g(C)).
+
+
+ j(C) = h(g(C))=\frac{5}{9}\big(g(C)-32\big)
+ Enter the expression for g(C).9/5 C +32-32\bigg)
+ Calculate and simplify the line above to find the simplest expression for h(g(C)).C
+
+
+
+
+ Find the simplest expression that you can for the composite function k(F) = g(h(F)).
+
+
+ k(F) = g(h(F))=\frac{9}{5}h(F)+32
+ Enter the expression for h(F).5/9(F-32)\bigg)+32
+ Calculate and simplify the line above to find the simplest expression for g(h(F)).F
+
+
+
+
+
+ Complete the following sentences to explain why the functions j and k are so simple.
+
+
+ The function g(C) takes in a units on the input to gCelsiusFahrenheittemperature and produces the same temperature converted to units on the output of gCelsiusFahrenheit while the function h(F) takes in a units on the input to hCelsiusFahrenheittemperature and produces the same temperature coverted to units on the output of gCelsius.Fahrenheit.
+
+
+ Then the function j(C)=h(g(C)) starts by taking in a units on the input to gCelsiusFahrenheittemperature and produces the same temperature converted to units on the output of gCelsiusFahrenheit and then h is applied to that output and converts it back to the equivalent temperature in units on the output of hCelsius.Fahrenheit. Whatever temperature we started with is also the one we ended with.
+
+
+ Similarly, the function k(F)=g(h(F)) starts by taking in a units on the input to hCelsiusFahrenheittemperature and produces the same temperature converted to units on the output of hCelsiusFahrenheit and then g is applied to that output and converts it back to the equivalent temperature in units on the output of hCelsius.Fahrenheit. Whatever temperature we started with is again the same as the one we ended with.
+
+
+
\ No newline at end of file
diff --git a/source/previews/doenet/PA-changing-linear-3-ex-D.doenetml b/source/previews/doenet/PA-changing-linear-3-ex-D.doenetml
new file mode 100644
index 00000000..e212cc30
--- /dev/null
+++ b/source/previews/doenet/PA-changing-linear-3-ex-D.doenetml
@@ -0,0 +1,159 @@
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ 7-3x
+ 0.5x-0.25
+ -1/3*x+7/3
+
+
+
+
+
+ Let y = f(x) = 7 - 3x. Determine AV_{[-3,-1]}, AV_{[2,5]}, and AV_{[4,10]} for the function f.
+
+ $AVneg3neg1.numerator=$$f(-1)-$$f(-3) and $AVneg3neg1.denominator=2=single answer for the average rate of change from negative 3 to negative 1($$f(-1)-$$f(-3))/(-1+3)($$f(-1)-$$f(-3))/(-1+3)
+
+ Your process is correct, but you need to enter your answer as a single number.
+
+ $AV25.numerator=$$f(5)-$$f(2) and $AV25.denominator=3=single answer for the average rate of change from 2 to 5($$f(5)-$$f(2))/(5-2)($$f(5)-$$f(2))/(5-2)
+
+ Your process is correct, but you need to enter your answer as a single number.
+
+ $AV410.numerator=$$f(10)-$$f(4) and $AV410.denominator=6=single answer for the average rate of change from 4 to 10($$f(10)-$$f(4))/(10-4)($$f(10)-$$f(4))/(10-4)
+
+ Your process is correct, but you need to enter your answer as a single number.
+
+
+
+ Let y = g(x) be given by the data in the following table.
+
+ Determine AV_{[-5,-2]}, AV_{[-1,1]}, and AV_{[0,4]} for the function g.
+
+
+ $AVgneg5neg2.numerator=$$g(-2)-$$g(-5) and $AVgneg5neg2.denominator=3=single answer for the average rate of change of g from -5 to -2($$g(-2)-$$g(-5))/(-2+5)1/2($$g(-2)-$$g(-5))/(-2+5)
+
+ Your process is correct, but you need to enter your answer as a single number in reduced form.
+
+ $AVgneg11.numerator=$$g(1)-$$g(-1) and $AVgneg11.denominator=2=single answer for the average rate of change of g from -1 to 1($$g(1)-$$g(-1))/(2)1/2($$g(1)-$$g(-1))/(2)
+
+ Your process is correct, but you need to enter your answer as a single number in reduced form.
+
+ $AVg04.numerator=$$g(4)-$$g(0) and $AVg04.denominator=4=single answer for the average rate of change of g from 0 to 4($$g(4)-$$g(0))/(4)1/2($$g(4)-$$g(0))/(4)
+
+ Your process is correct, but you need to enter your answer as a single number in reduced form.
+
+
+
+
+ Consider the function y = h(x) defined by the graph in the following figure.
+
+
+ graph of a line through the points (-5,4) and (-2,3)
+
+ Graph of a line going through the points (-5,4), (-2,3), (1,2) and (4,1)
+
+
+
+
+
+
+
+ $h
+
+
+ Determine AV_{[-5,-2]}, AV_{[-1,1]}, and AV_{[0,4]} for the function h. Use exact values, not a decimal approximation.
+
+
+ -1/3-1/3
+
+ Your answer is close, but is not exact or is not simplified. Enter your answer as a fraction using exact values from the graph.
+
+ -1/3-1/3
+
+ Your answer is close, but is not exact or is not simplified. How does the answer to this question relate to the answer to AV_{[-5,-2]}?
+
+ -1/3-1/3
+
+ Your answer is close, but is not exact or is not simplified. How does the answer to this question relate to the answer to AV_{[-5,-2]}?
+
+
+
+ Note that in each of the three examples above, the functions involved were relation of the functionsthe same functiondifferent functions having relation of the valuesthe samedifferenty-values and relation of the rates of changethe samedifferent rates of change, but within each example, the value of AV_{[a,b]} was relation of the AVthe samedifferent for different values of a and b.
+
+
+
+
+ For the function y = f(x) = 7 - 3x from (a), find the simplest expression you can for
+
+ AV_{[a,b]} = \frac{f(b)-f(a)}{b-a}
+
+ where a \ne b.
+
+
+ 7-3b
+
+
+ 7-3a
+
+
+ -3b+3a-3(b-a)3(a-b)-3b+3a
+
+ Your process is correct, but your answer isn't simplified enough, yet.
+
+ -3-3
+
+ Your process is correct, but your answer isn't simplified enough, yet.
+
+ Does your answer to \displaystyle AV_{[a,b]} for this function depend on a or b? yes or noYesNo Does that surprise you? What do you wonder?
+
+
+
\ No newline at end of file
diff --git a/source/previews/doenet/PA-changing-quadratic-D.doenetml b/source/previews/doenet/PA-changing-quadratic-D.doenetml
new file mode 100644
index 00000000..bf293b3f
--- /dev/null
+++ b/source/previews/doenet/PA-changing-quadratic-D.doenetml
@@ -0,0 +1,100 @@
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ -5*x^2+20x+25
+
+
+
+ A water balloon is tossed vertically from a fifth story window. Its height, h, in meters, at time t, in seconds, is modeled by the function
+
+ h = q(t) = -5t^2 + 20t + 25.
+
+
+
+
+
+ Complete the table of function values below. For example, q(0)=-5(0)^2+20 \cdot 0 + 25 =25, so we've entered that in the table below under the input value 0. Enter each answer as a single number.
+
+
+
+
+ t
+ 0
+ 1
+ 2
+ 3
+ 4
+ 5
+
+
+ q(t)
+ 25
+ q of 1$$q(1)
+ q of 2$$q(2)
+ q of 3$$q(3)
+ q of 4$$q(4)
+ q of 5$$q(5)
+
+
+
+
+
+ Now use the values you computed to calculate AV_{[a,b]} for different intervals [a,b]. For example, AV_{[0,1]}=\frac{q(1)-q(0)}{1-0}=\frac{40-25}{1}=15. Enter each answer as a single number.
+
+
+
+
+
+ AV_{[0,1]}
+ AV_{[1,2]}
+ AV_{[2,3]}
+ AV_{[3,4]}
+ AV_{[4,5]}
+
+
+ 15
+ average rate of change of q from 1 to 2$$q(2)-$$q(1)
+ q of 2$$q(3)-$$q(2)
+ q of 3$$q(4)-$$q(3)
+ q of 4$$q(5)-$$q(4)
+
+
+
+
+
+
+ Complete the following sentences to record some observations about the function q(t) and its rate of change AV, the average velocity of the water balloon.
+
+
+ The function h=q(t) has values which change how q changesby the same amountin a symmetric patternseemingly at random as the input changes by the same amount, meaning that the average rate of change is behavior of the values of AVis constantis not constant and so the function q(t)description of qisis not a linear function. However, the values of AV itself change how AV changesby the same amountin a symmetric pattern as the input changes by the same amount, meaning that the average rate of change of AV is how AV changesis constantis not constant and so the function AVdescription of AVisis not a linear function.
+
+
+
+
+ When does the water balloon land on the ground?
+ 5units of tfeetmeterssecondsminutesfeet per minutemeters per second
+
+
+ What is the average velocity of the water balloon in the final second before it lands?
+ -25units of AVfeetmeterssecondsminutesfeet per minutemeters per second
+
+
+ What is the average velocity of the water balloon on the interval [4.9,5]?
+ -29.5units of AVfeetmeterssecondsminutesfeet per minutemeters per second
+
+ We are going to explore transformations of a familiar quadratic function, y=x^2.
+
+
+
+
+
+ First, move the slider for a back and forth and observe the effect(s) on the graph of f(x)+a compared to the graph of f(x). You can also drag the point around to see how the value of a changes.
+
+
+
+ graph of a function a plus f of x for different values of a
+ $f
+ $f+$a
+ ($zero,$$f($zero)+$a)
+
+
+
+ Set the value of the slider to a=9.
+ $a=9
+ You can move it again afterwards.
+
+
+ Which of the following observations are true for y=f(x)+a?
+ properties of a plus f of xThe graph of f(x) is deformed (stretched or shrunk) vertically by a.The graph of f(x) is shifted vertically by a.The graph of f(x) is shifted horizontally by a.The graph of f(x) is reflected across the x-axis when a is negative.The graph is the same as f(x) when a=0.The graph is the same as f(x) when a=1.
+
+
+
+
+ Next, move the slider for b back and forth and observe the effect(s) on the graph of f(x-b) compared to the graph of f(x). You can also drag the point around to see how the value of b changes.
+
+
+
+ graph of a function f of the quantity x minus b for different values of b
+ $f
+ $$f(x-$b)
+ ($b,$$fxminusb($b))
+
+
+
+ Set the value of the slider to b=6.
+ $b=6
+ You can move it again afterwards.
+
+
+ Which of the following observations are true for y=f(x-b)?
+ properties of f of the quantity x minus bThe graph of f(x) is deformed (stretched or shrunk) vertically by b.The graph of f(x) is shifted vertically by b.The graph of f(x) is shifted horizontally by b.The graph of f(x) is reflected across the x-axis when b is negative.The graph is the same as f(x) when b=0.The graph is the same as f(x) when b=1.
+
+
+
+
+ Third, move the slider for c back and forth and observe the effect(s) on the graph of cf(x) compared to the graph of f(x). You can also drag the point around to see how the value of c changes.
+
+
+
+ graph of a function c times f of x for different values of c
+ $f
+ $c*$f
+ ($one,$c*$$f($one))
+ ($zero,$c*$$f($zero))
+
+
+
+ Set the value of the slider to c=6.
+ $c=6
+ You can move it again afterwards.
+
+
+ Which of the following observations are true for y=cf(x)?
+ properties of c times f of x The graph of f(x) is deformed (stretched or shrunk) vertically by a.The graph of f(x) is shifted vertically by c.The graph of f(x) is shifted horizontally by c.The graph of f(x) is reflected across the x-axis when c is negative.The graph is the same as f(x) when c=0.The graph is the same as f(x) when c=1.
+
+
+
+
+ Finally, change the function entered below and explore the sliders above again for this new function. Do any of your selected choices change when the function we are comparing to has changed? You can come up with your own functions to try, but some good ones to try that will fit in the window nicely could be f(x)=x+1, f(x) = x^2 + 2x + 3, or f(x) = x^3 - 1.
+
+
+
+
+
+
\ No newline at end of file
diff --git a/source/previews/doenet/PA-circular-sine-D.doenetml b/source/previews/doenet/PA-circular-sine-D.doenetml
new file mode 100644
index 00000000..ad3ef6ab
--- /dev/null
+++ b/source/previews/doenet/PA-circular-sine-D.doenetml
@@ -0,0 +1,245 @@
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ pi/6 pi/4 pi/3 pi/2 2pi/3 3pi/4 5pi/6 pi 7pi/6 5pi/4 4pi/3 3pi/2 5pi/3 7pi/4 11pi/6 2pi
+ (sqrt(3)/2,1/2) (sqrt(2)/2, sqrt(2)/2) (1/2,sqrt(3)/2) (0,1) (-1/2,sqrt(3)/2) (-sqrt(2)/2, sqrt(2)/2) (-sqrt(3)/2,1/2) (-1,0) (-sqrt(3)/2,-1/2) (-sqrt(2)/2, -sqrt(2)/2) (-1/2,-sqrt(3)/2) (0,-1) (1/2,-sqrt(3)/2) (sqrt(2)/2, -sqrt(2)/2) (sqrt(3)/2,-1/2) (1,0)
+
+ (cos($anglelist[$k]),sin($anglelist[$k]))
+
+
+ $listEnt
+ -sqrt(3)/2
+
+
+ $P.coords=(cos($anglelist[$n]),sin($anglelist[$n]))
+
+
+
+
+ $anglelist[1] $plist[1]
+
+
+ $anglelist[2] $plist[2]
+
+
+ $anglelist[3] $plist[3]
+
+
+ $anglelist[4] $plist[4]
+
+
+ $anglelist[5] $plist[5]
+
+
+ $anglelist[6] $plist[6]
+
+
+ $anglelist[7] $plist[7]
+
+
+ $anglelist[8] $plist[8]
+
+
+ $anglelist[9] $plist[9]
+
+
+ $anglelist[10] $plist[10]
+
+
+ $anglelist[11] $plist[11]
+
+
+ $anglelist[12] $plist[12]
+
+
+ $anglelist[13] $plist[13]
+
+
+ $anglelist[14] $plist[14]
+
+
+ $anglelist[15] $plist[15]
+
+
+ $anglelist[16] $plist[16]
+
+
+
+
+ arccos($P.x)
+
+
+ 2pi-arccos($P.x)
+
+ $P.coords
+
+
+
+
+
+
+
+ graph of a unit circle and moveable point on the circle
+ x
+ y
+
+ $unitcir$points(.9,0.1)
+
+
+
+
+
+ (1/5cos(t), 1/5sin(t))
+
+
+ (1/5cos(t), 1/5sin(t))
+
+
+
+
+
+
+
+ graph of a function whose height is the y value of the point on the unit circle
+ t
+ h=f(t)
+
+
+
+ (t, sin(t))
+
+ (arccos($P.x),$P.y)
+
+ (2pi-arccos($P.x),$P.y)
+
+
+
+
+
+
+
+
+
+ When t=$decangle[1][1], the coordinates of the point on the unit circle are: $decangle[1][2]
+
+
+
+
+
+ Drag the point on the unit circle to the location where t=\frac{\pi}{4}. $P=$plist[2]
+
+
+ What is the exact value of f\left(\frac{\pi}{4}\right)?
+ sqrt(2)/2
+
+
+ What is the exact value of f\left(\frac{\pi}{3}\right)?
+ sqrt(3)/2
+
+
+
+
+ Complete the following table with the exact values of h that correspond to the stated inputs.
+
+
+
+ t
+ 0
+ $anglelist[1]
+ $anglelist[2]
+ $anglelist[3]
+ $anglelist[4]
+
+
+ h
+ value of h0
+ value of h$plist[1][2]
+ value of h$plist[2][2]
+ value of h$plist[3][2]
+ value of h$plist[4][2]
+
+
+ t
+ $anglelist[5]
+ $anglelist[6]
+ $anglelist[7]
+ $anglelist[8]
+ $anglelist[9]
+
+
+ h
+ value of h$plist[5][2]
+ value of h$plist[6][2]
+ value of h$plist[7][2]
+ value of h$plist[8][2]
+ value of h$plist[9][2]
+
+
+ t
+ $anglelist[10]
+ $anglelist[11]
+ $anglelist[12]
+ $anglelist[13]
+ $anglelist[14]
+
+
+ h
+ value of h$plist[10][2]
+ value of h$plist[11][2]
+ value of h$plist[12][2]
+ value of h$plist[13][2]
+ value of h$plist[14][2]
+
+
+ t
+ $anglelist[15]
+ $anglelist[16]
+
+
+
+
+
+ h
+ value of h$plist[15][2]
+ value of h$plist[16][2]
+
+
+
+
+
+
+
+
+ Use the patterns you observe above to answer the following questions.
+
+
+ What is the exact value of f\left(\frac{11\pi}{4}\right)?
+ -sqrt(2)/2
+
+
+ What is the exact value of f\left(\frac{14\pi}{3}\right)?
+ -sqrt(3)/2
+
+
+
+
+ Give four different values of t for which f(t) = -\frac{\sqrt{3}}{2}. Separate your answers with commas.
+ minWidth="200">$tsEnt.numComponents=4 and sin($tsEnt[1])=$ans and sin($tsEnt[2])=$ans and sin($tsEnt[3])=$ans and sin($tsEnt[4])=$ans
+
+
+
+
\ No newline at end of file
diff --git a/source/previews/doenet/PA-circular-sinusoidal-D.doenetml b/source/previews/doenet/PA-circular-sinusoidal-D.doenetml
new file mode 100644
index 00000000..4245d649
--- /dev/null
+++ b/source/previews/doenet/PA-circular-sinusoidal-D.doenetml
@@ -0,0 +1,503 @@
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ 0
+ -2.5
+ -6
+
+ -2.5
+ -6
+
+ -2.5
+ -6
+
+
+ (0,1.5)
+
+ (0,$horiz)
+ (2,$horiz)
+ ($vert,$zero)
+ ($vert,-1.5)
+ $ampa1.creditAchieved=1 and $ampa2.creditAchieved=1 and $mida1.creditAchieved=1 and $mida2.creditAchieved=1 and $perioda1.creditAchieved=1 and $perioda2.creditAchieved=1
+
+ (0,1.5)
+
+ (0,$horizb)
+ (2,$horizb)
+ ($vertb,$zero)
+ ($vertb,-1.5)
+ $ampb1.creditAchieved=1 and $ampb2.creditAchieved=1 and $midb1.creditAchieved=1 and $midb2.creditAchieved=1 and $periodb1.creditAchieved=1 and $periodb2.creditAchieved=1
+
+ (0,1.5)
+
+ (0,$horizc)
+ (2,$horizc)
+ ($vertc,$midgc.yintercept)
+ ($vertc,-1.5)
+ $ampc1.creditAchieved=1 and $ampc2.creditAchieved=1 and $midc1.creditAchieved=1 and $midc2.creditAchieved=1 and $periodc1.creditAchieved=1 and $periodc2.creditAchieved=1
+
+
+
+ Let f(t) = \cos(t), which has a midline of y=0, an amplitude of a=1, and a period of P=2\pi.
+
+
+ Answer all of the questions below without using a graph; after answering, a graph will appear and you'll either confirm or reflect on your answers using a graph. Note that there could be more than one correct answer to some questions.
+
+ A simplified version of the ferris wheel scenario pictured in Figure 2.1.1 has been reproduced below.
+
+
+
+a point going around a circle and a curve showing its height and distance
+
+ $htvcir
+
+
+
+
+
+
+ $cirpt
+ $height
+ $height2
+ $circum
+ $htvcirpt
+ $lencircum
+
+
+
+
+
+
+
+
+ Assume that the height, h, of the moving point (the cab in which you are riding), and the distance, d, that the point has traveled around the circumference of the ferris wheel are both measured in meters. Assume also that the circumference of the ferris wheel is 150 meters.
+
+
+
+
+
+
+ Recall that the circumference, C, of a circle is connected to the circle's radius, r, by the formula C = 2\pi r.
+
+
+ What is the radius of the ferris wheel? $r
+
+ How high is the highest point on the ferris wheel? 2*$r2*$rEnt
+
+
+
+
+ How high is the cab after it has traveled 1/4 of the circumference of the circle?
+ $r
+
+ In the animation above, the cab goes all the way around the circumference of the circle in 60 steps, and so the cab will be 1/4 of the way around when the slider is set to 15.
+
+
+
+ How much distance along the circle has the cab traversed at the moment it first reaches a height of \frac{150}{\pi} \approx 47.75 meters?
+
+ 7575
+
+
+ Your answer is close, but we can be exact. What fraction of the circle has the cab traversed at this moment, and what is the total distance around the circle?
+
+
+
+ Complete the following sentences.
+
+
+ The cab's height h at any time can or can'tcancan't be thought of as a function of the cab's distance traveled d because given any input or outputinput value of dinput value of houtput value of doutput value of h there is or is notisis not exactly one possible input or outputinput value of d.input value of h.output value of d.output value of h.
+
+
+ The cab's distance traveled d at any time can or can'tcancan't be thought of as a function of the cab's height h because for at least one input or outputvalue of d, value of h,number of inputs or outputsthere are multiple values of hthere are multiple values of dthere is exactly one value of hthere is exactly one value of d corresponding to that height or distancedistance.height.
+
+ In the following figure there are 24 equally spaced points on the unit circle. Since the circumference of the unit circle is 2\pi, each of the points is \frac{1}{24} \cdot 2\pi = \frac{\pi}{12} units apart (traveled along the circle). Thus, the first point counterclockwise from (1,0) corresponds to the distance d = \frac{\pi}{12} traveled along the unit circle. The second point is twice as far, and thus d = 2 \cdot \frac{\pi}{12} = \frac{\pi}{6} units along the circle away from (1,0).
+
+
+
+ Move the large square point to each of the points in the remainder of the top half of the circle and enter the distance traveled along the circle counterclockwise from the point (1,0) to that location.
+
+
+ When you have entered the correct distance for the selected point and hit enter or clicked outside the box, the distance traveled will appear on the graph near that point.
+
+ Hit Check Work when all of the correct distances appear for the top half of the circle and (-1,0), but not (1,0).
+ $H3=false and $H4=false and $H5=false and $H6=false and $H7=false and $H8=false and $H9=false and $H10=false and $H11=false and $H12=false
+
+
+
+
+
+Continuing around the circle counterclockwise after (-1,0), enter the remaining distances in order, stopping just before you get back to the point (1,0). Use commas to separate your answers.
+
+
+values for the distances covered in the bottom half of the unit circle$dists$dists2
+
+You have the correct distances, but they're not going in order counterclockwise starting with the first point after (-1,0).
+
+ Which distance along the unit circle corresponds to \frac{1}{4} of a full rotation around? distance for a quarter of the way aroundpi/2
+
+
+ Which distance along the unit circle corresponds to \frac{5}{8} of a full rotation around? distance for a quarter of the way around5pi/4
+
+
+ You can divide the circle up into 8 equal pieces and go around 5 of them to answer this question.
+ Or, you can recall that the total distance around a circle with radius 1 is 2\pi, and note that \frac{1}{4}\cdot 2\pi=\frac{\pi}{2}. You can use this same method to calculate the distance along the unit circle which corresponds to \frac{5}{8} of a full rotation.
+
+
+ One way to measure angles is connected to the arc length along a circle. For an angle whose vertex is at (0,0) in the unit circle, we say the angle's measure is 1 radian provided that the angle intercepts an arc of the circle that is 1 unit in length, as pictured in the following figure. Note particularly that an angle measuring 1 radian intercepts an arc of the same length as the circle's radius.
+
+ The functions 2^t, 3^t, \left(\frac{1}{3}\right)^t, and 2^{kt} appear in the graph below, along with a slider for the value of k and the ability to hide various of the functions.
+
+
+
+
+
+
+
+
+
+
+ graph of four exponential functions, one with a changing parameter
+ t
+
+
+
+
+
+
+
+
+ 2^x
+ 3^x
+ (1/3)^x
+ 2^($k x)
+
+
+
+
+
+
+
+
+
+ Use the slider to find a value of k so that the graph of 2^{kt} appears to align with the graph of 2^t. $k=1
+
+
+ When k=1, the graph of 2^{kt} looks like the graph of 2^t. This makes sense because 2^{1t}=2^{t}.
+
+ Enter a number, and use at least 3 digits after the decimal place.2^1.6
+
+
+
+
+ Use the slider to find a value of k so that the graph of 2^{kt} appears to align with the graph of 3^t. $k=1.6
+
+
+ When k=1.6, the graph of 2^{kt} \approx 3^t.
+
+ Enter a number, and use at least 3 digits after the decimal place.2^1.6
+
+
+
+
+ Use the slider to find a value of k so that the graph of 2^{kt} appears to align with the graph of \left(\frac{1}{3}\right)^t. $k=-1.6
+
+
+ When k=-1.6, the graph of 2^{kt} \approx \left(\frac{1}{3}\right)^t.
+
+ Enter a number, and use at least 3 digits after the decimal place.2^(-1.6)
+
+
+
+
+ Below is another graph and slider, where now you are the one choosing the target function.
+
+
+
+
+ Enter a number between .1 and 30. You are welcome to try other values and see what happens!
+
+
+ graph of four exponential functions, one with a changing parameter
+ t
+
+
+
+
+
+
+
+
+ $b^x
+ 2^($k2 x)
+
+
+
+
+
+
+ Enter a number for b such that 0.1 \le b \le 30, then use the slider to find a value of k so that the graph of 2^{kt} appears to align with the graph of b^t for your choice of b. Do this at least three times, once for each of the conditions below, and do not reuse any of the functions from above.
+
+
+ When .1 \le b \lt 1: .1<=$b and $b< 1 and ln($b)/ln(2)=$k2 and $differentfcns
+
+
+ When 1 \le b \lt 10 : 1<=$b and $b< 10 and ln($b)/ln(2)=$k2 and $differentfcns
+
+
+ When 11 \le b \le 30 : 11<=$b and $b<= 30 and ln($b)/ln(2)=$k2 and $differentfcns
+
+
+ Do you think you would be able to find a value of k so that 2^{kt} aligns with b^tfor any positive number b if you had freedom to pick any value of k and weren't restricted to a certain range like you are with a slider?
+
+
+
\ No newline at end of file
diff --git a/source/previews/doenet/PA-exp-growth-D.doenetml b/source/previews/doenet/PA-exp-growth-D.doenetml
new file mode 100644
index 00000000..372d3621
--- /dev/null
+++ b/source/previews/doenet/PA-exp-growth-D.doenetml
@@ -0,0 +1,103 @@
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ Suppose that at age 20 you have $20000 and you can choose between one of two ways to use the money: you can invest it in a mutual fund that will, on average, earn 8% interest annually, or you can purchase a new automobile that will, on average, depreciate 12% annually. Let's explore how the $20000 changes over time.
+
+
+ Let I(t) denote the value of the $20000 after t years if it is invested in the mutual fund, and let V(t) denote the value of the automobile t years after it is purchased.
+
+
+
+
+
+ Determine I(0), I(1), I(2), and I(3). Enter each answer as a single number.
+
+
+ The amount in the account after 0 years; in other words, at the beginning.20000
+
+
+ The amount in the account after 1 year earning 8 \% interest. Recall that this can be calculated by 1.08\cdot 20000.1.08*20000
+
+
+ The amount in the account after 2 years earning 8 \% interest. Recall that this can be calculated by 1.08\cdot \big(1.08\cdot 20000\big)=1.08^2\cdot 20000.1.08^2*20000
+
+
+ The amount in the account after 3 years earning 8 \% interest. Recall that this can be calculated by 1.08\cdot \big(1.08^2\cdot 20000\big)=1.08^3\cdot 20000.1.08^3*20000
+
+
+
+
+ Note that if a quantity depreciates12 \% annually then since 100-12=88, this means that after a given year 88 \% of the quantity remains. Compute V(0), V(1), V(2), and V(3). Enter each answer as a single number.
+
+
+ The value of the car after 0 years; in other words, at the beginning.20000
+
+
+ The value of the car after 1 year depreciating 12 \%. Recall that this can be calculated by .88\cdot 20000.0.88*20000
+
+
+ The value of the car after 2 years depreciating 12 \%. Recall that this can be calculated by 0.88\cdot \big(0.88\cdot 20000\big)=0.88^2\cdot 20000.0.88^2*20000
+
+
+ The value of the car after 3 years depreciating 12 \%. Recall that this can be calculated by 0.88\cdot \big(0.88^2\cdot 20000\big)=0.88^3\cdot 20000.0.88^3*20000
+
+
+
+
+ Based on the patterns in your computations in parts a. and b., determine formulas for I(t) and V(t).
+
+
+ 1.08^t*20000
+
+
+ 0.88^t*20000
+
+
+
+
+ The graphs of I(t) and V(t) appear below.
+
+
+ graph of the functions I and V
+ t
+ Value in dollars
+
+
+
+
+
+
+
+
+ 2*1.08^x
+ 2* 0.88^x
+
+
+
+
+
+ Complete the sentence with some observations about the behavior of the two functions.
+
+
+ Even though both functions had the same form, form of the functions20000t+b20000t^2+bt+cb^t\cdot 2000020000\cos(bt)+c, the function I(t) is behavior of I of tgetting largergetting smallerstaying the same year after year and V(t) is behavior of V of tgetting largergetting smallerstaying the same. At the end of 20 years, the account earning interest is worth amount in I of t around \$ 50000over \$ 90000a couple thousand dollars, while the car that depreciates is worth maybe amount in I of t around \$ 50000over \$ 90000a couple thousand dollars.
+
+
+ What else do you notice and wonder?
+
+
+
+
\ No newline at end of file
diff --git a/source/previews/doenet/PA-exp-log-D.doenetml b/source/previews/doenet/PA-exp-log-D.doenetml
new file mode 100644
index 00000000..10b178a1
--- /dev/null
+++ b/source/previews/doenet/PA-exp-log-D.doenetml
@@ -0,0 +1,124 @@
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ Let P(t) be the powers of 10 function, which is given by P(t) = 10^t.
+
+
+
+
+
+ Complete the following table to generate certain values of P.
+
+
+
+ t
+ y = 10^t
+
+
+
+ -3
+ 10^{-3}
+ =decimal value of P of t10^(-3)
+
+
+ -2
+ value of P of t10^(-2)
+ =decimal value of P of tEnter something in the first box that has the same form as in the first row, 10^{-3}, and the decimal value will fill in this second box automatically.$Pneg2
+
+
+ -1
+ value of P of t10^(-1)
+ =decimal value of P of tEnter something in the first box that has the same form as in the first row, 10^{-3}, and the decimal value will fill in this second box automatically.$Pneg1
+
+
+ 0
+ value of P of t10^(0)
+ =decimal value of P of tEnter something in the first box that has the same form as in the first row, 10^{-3}, and the decimal value will fill in this second box automatically.$P0
+
+
+ 1
+ value of P of t10^(1)
+ =decimal value of P of tEnter something in the first box that has the same form as in the first row, 10^{-3}, and the decimal value will fill in this second box automatically.$P1
+
+
+ 2
+ value of P of t10^(2)
+ =decimal value of P of tEnter something in the first box that has the same form as in the first row, 10^{-3}, and the decimal value will fill in this second box automatically.$P2
+
+
+ 0
+ value of P of t10^(3)
+ =decimal value of P of tEnter something in the first box that has the same form as in the first row, 10^{-3}, and the decimal value will fill in this second box automatically.$P3
+
+
+
+
+
+
+ We can see that the function P(t) has an inverse function because each output or inputinputoutput of P(t) can be reversed to find a singular output or inputinputoutput; the function P(t) passes the test for inversesVertical Line TestHorizontal Line Test. We know this happens because P(t) is behavior of Psometimes increasingsometimes decreasingalways increasingalways decreasing and so P(t) can't return to the same output value for a different input of t.
+
+
+
+
+ Since P has an inverse function, we know there exists some other function, say L, such that writing y = P(t) says the exact same thing as writing t = L(y). In words, where P produces the result of raising 10 to a given power, the function L reverses this process and instead tells us the power to which we need to raise 10, given a desired result. Complete the following table to generate a collection of values of L.
+
+
+
+ y
+ t=L(y)
+
+
+ 10^{-3}
+ value of L of yThe input to P(t) which resulted in the output y that appears on the left.-3
+
+
+ 10^{-2}
+ value of L of yThe input to P(t) which resulted in the output y that appears on the left.-2
+
+
+ 10^{-1}
+ value of L of yThe input to P(t) which resulted in the output y that appears on the left.-1
+
+
+ 10^{0}
+ value of L of yThe input to P(t) which resulted in the output y that appears on the left.0
+
+
+ 10^{1}
+ value of L of yThe input to P(t) which resulted in the output y that appears on the left.1
+
+
+ 10^{2}
+ value of L of yThe input to P(t) which resulted in the output y that appears on the left.2
+
+
+ 10^{3}
+ value of L of yThe input to P(t) which resulted in the output y that appears on the left.3
+
+
+
+
+
+
+ [0,\infty)(0,\infty)(-\infty,\infty) because any number 0 and greaterany positive numberany number. [0,\infty)(0,\infty)(-\infty,\infty) because a number 0 and greatera positive numbersome number.
+
+
+ [0,\infty)(0,\infty)(-\infty,\infty) because a number 0 and greatera positive numbersome number, so those are the inputs vs outpursinputs to L(y)inputs to P(t) which then outputs the power needed. [0,\infty)(0,\infty)(-\infty,\infty).
+
+
+
\ No newline at end of file
diff --git a/source/previews/doenet/PA-exp-log-properties-D.doenetml b/source/previews/doenet/PA-exp-log-properties-D.doenetml
new file mode 100644
index 00000000..05f23de5
--- /dev/null
+++ b/source/previews/doenet/PA-exp-log-properties-D.doenetml
@@ -0,0 +1,75 @@
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ In the following questions, we investigate how \log_{10}(a \cdot b) can be equivalently written in terms of \log_{10}(a) and \log_{10}(b).
+
+
+
+
+
+ Write 10^x \cdot 10^y as 10 raised to a single power.
+
+ 10^x \cdot 10^y = 10^{^{\underline{\hspace{.3in}}}}
+
+Enter the correct expression involving x and y.x+y
+
+
+ 10^x \cdot 10^y = 10^{x+y}
+
+
+
+
+ Enter the simplest possible way to write each of the expressions below.
+
+ x
+
+
+ y
+
+
+
+
+ Select the reason why each of the following equalities is true.
+
+
+ inside the parentheses 10^x\cdot 10^y=10^{x+y}.we know \displaystyle \log_{10}10^{z}=z, no matter what z is.we know that \displaystyle x= \log_{10}10^{x} and \displaystyle y= \log_{10}10^{y}.
+
+
+ inside the parentheses 10^x\cdot 10^y=10^{x+y}.we know \displaystyle \log_{10}10^{z}=z, no matter what z is.we know that \displaystyle x= \log_{10}10^{x} and \displaystyle y= \log_{10}10^{y}.
+
+
+ inside the parentheses 10^x\cdot 10^y=10^{x+y}.we know \displaystyle \log_{10}10^{z}=z, no matter what z is.we know that \displaystyle x= \log_{10}10^{x} and \displaystyle y= \log_{10}10^{y}.
+
+
+ Putting it all together,
+ \log_{10}(10^x \cdot 10^y) &= \log_{10}(10^{x+y}) &= x+y &= \log_{10}(10^x) + \log_{10}(10^y).
+
+
+
+
+ Suppose that a and b are positive real numbers, which means that we can think of a as 10^x for some real number x and
+ b as 10^y for some real number y. That is, say that a = 10^x and b = 10^y. Then we can add those equalities to both ends of the above list of equalities to get
+ \log_{10}(a \cdot b)\amp= \log_{10}(10^x \cdot 10^y) &= \log_{10}(10^{x+y}) &= x+y &= \log_{10}(10^x) + \log_{10}(10^y)\amp=\log_{10}(\underline{\hspace{.3in}})+ \log_{10}(\underline{\hspace{.3in}})
+
+
+ where the missing entries are: missing element 1a and missing element 1b
+
+
+
+
+ In conclusion, \log_{10}(ab)=\log_{10}(a)+\log_{10}(b)
+
+ The graph below shows the function g(t) = ab^t+c, along with sliders for the values of a, b, and c. We will explore the effects of a, b, and c on the graph.
+
+ All of the following situations contain expressions which can be used to model temperature or population growth. In each situation, we will solve for the exact value of various unknown quantities.
+
+
+
+
+
+
+
+
\ No newline at end of file
diff --git a/source/previews/doenet/PA-poly-infty-D.doenetml b/source/previews/doenet/PA-poly-infty-D.doenetml
new file mode 100644
index 00000000..c6583f40
--- /dev/null
+++ b/source/previews/doenet/PA-poly-infty-D.doenetml
@@ -0,0 +1,60 @@
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ Complete each of the following statements with an appropriate number or the symbols \infty or -\infty. Do your best to do so without using a graphing utility; instead use your understanding of the function's graph. To enter \infty, simply type the word infinity.
+
+
+
+
+
+ 0
+
+
+
+
+ infinity
+
+
+
+
+ infinity
+
+
+
+
+ 1
+ Note that when we write t \to 0^+, this means that we are letting t get closer and closer to 0, but only allowing t to take on positive values.
+
+
+
+
+ 35
+
+
+
+
+ infinity
+ Note that when we write t \to \frac{\pi}{2}^-, this means that we are letting t get closer and closer to \frac{\pi}{2}, but only allowing t to take on values that lie to the left of \frac{\pi}{2}.
+
+
+
+
+ -infinity
+ Note that when we write t \to \frac{\pi}{2}^+, this means that we are letting t get closer and closer to \frac{\pi}{2}, but only allowing t to take on values that lie to the right of \frac{\pi}{2}.
+
+ A piece of cardboard that is 18 \times 12 (each measured in inches) is being made into a box without a top. To do so, squares are cut from each corner of the cardboard and the remaining sides are folded up, as shown in the diagrams below.
+
+
+
+
+ a labeled rectangle showing squares cut out from the corners
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ a box coming from folding up the sides of a piece of paper with squares cut out from the corners
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ As shown in the diagram, let x be the side length of the squares being cut from the corners of the cardboard, and let \ell, w and h be the length, width and height of the box that results. Write each of the quantities below in terms of x.
+
+
+ x
+
+
+ 12-2x
+
+
+ 18-2x
+
+
+
+
+ We know that the volume of a rectangular box is given by V=\ell w h. Determine a formula for the function V(x) whose output is the volume of the box that results from a square of size x \times x being cut from each corner of the cardboard.
+
+ If any of the sides have length 0, we don't have a three-dimensional box.
+
+
+
+ You can enter your answer using interval notation like [0,\pi), or using inequalities such as 0\leq x < \pi.$mi1=(0,6)[0,6) subset $mi1 or (0,6] subset $mi1
+
+ Neither x=0 nor x=6 can be part of the domain, since in both cases one side of the "box" would have length 0 and so it would be flat and wouldn't be a box at all.
+
+
+ Separate your answers by commas.$mi2=0 6 9
+
+
+
\ No newline at end of file
diff --git a/source/previews/doenet/PA-poly-polynomials-D.doenetml b/source/previews/doenet/PA-poly-polynomials-D.doenetml
new file mode 100644
index 00000000..495ae9e5
--- /dev/null
+++ b/source/previews/doenet/PA-poly-polynomials-D.doenetml
@@ -0,0 +1,154 @@
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ $f.numMinima=2
+
+ $f.minimumValues[1]< 0
+ $f.minimumValues[2]< 0
+ $f.maximumValues[1]> 0
+
+
+
+ $f.numMaxima=2
+
+ $f.maximumValues[1]> 0
+ $f.maximumValues[2]> 0
+ $f.minimumValues[1]< 0
+
+
+
+ $xi> 0
+ $xi< 0
+
+
+
+ $a4> 0
+ $allextremapos
+
+
+ $a4< 0
+ $allextremaneg
+
+
+
+
+ $f.numExtrema=1
+ abs($f.extremumValues)< .15
+
+
+
+
+ $a4< 0
+ $allextremapos
+
+
+ $a4> 0
+ $allextremaneg
+
+
+
+Below you'll find a degree 4 polynomial of the form p(x) = a_0 + a_1x + a_2x^2 + a_3x^3 + a_4x^4 where a_0, \ldots, a_4 are set up as sliders. In the questions that follow, you'll experiment with different values of a_0, \ldots, a_4 to investigate different possible behaviors in a degree 4 polynomial. Note that a_4 \ne 0 is required for p to be a degree 4 polynomial.
+
+
+
+
+
+
+
+
+
+
+
+ graph of a degree 4 polynomial whose coefficients are controlled by sliders
+ $a4 x^4 + $a3 x^3 + $a2 x^2 + $a1 x + $a0
+
+
+ $a4 x^4
+
+
+
+
+
+
+
+
\ No newline at end of file
diff --git a/source/previews/doenet/PA-poly-rational-D.doenetml b/source/previews/doenet/PA-poly-rational-D.doenetml
new file mode 100644
index 00000000..ab2161ba
--- /dev/null
+++ b/source/previews/doenet/PA-poly-rational-D.doenetml
@@ -0,0 +1,75 @@
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ A drug company estimates that to produce a new drug, it will cost $5 million in startup resources, and that once they reach production, each gram of the drug will cost $2500 to make.
+
+
+
+
+
+
+ Determine a formula for a function C(q) that models the cost of producing q grams of the drug.
+ 5000000 + 2500q
+
+ The drug company needs to sell the drug at a price of more than $2500 per gram in order to at least break even. To investigate how they might set prices, they first consider what their average cost per gram is.
+
+ 5000000+2500*1000
+
+
+ (5000000+2500*1000)/1000
+
+
+
+
+ 5000000+2500*10000
+
+
+ (5000000+2500*10000)/10000
+
+
+
+
+ Our computations in b. and c. naturally lead us to define the average cost per gram function, A(q), whose output is the average cost of producing q grams of the drug.
+
+ The average cost per gram of producing q grams is the numerator of A of qcost per gramtotal costnumber of grams to produce q grams divided by the denominator of A of qcost per gram A(q)total cost C(q)number of grams q, which means that formula for A of qA(q)=C(q)A(q)=q\displaystyle A(q)=\frac{q}{C(q)}\displaystyle A(q)=\frac{C(q)}{q}, and from our work in part a, that is the same as A(q)=formula for A of q using C of q\frac{5000000+2500q}{q}. Dividing both terms in the numerator by q results in an equivalent form, A(q)=another formula for A of q using C of q2500+\frac{5000000}{q}.
+
+
+
+
+ What can you say about the long-range behavior of A?
+
+
+ 2500
+
+
+ In the context of this scenario, this means 1000 grams10000 gramsas many grams as possibleas few grams as possible so that their average cost per gram is as small as possible.
+
+
+
+
+
+ This activity is based on p. 457ff in Functions Modeling Change, 5th edition, by Connally et al.
+
+
diff --git a/source/previews/doenet/PA-poly-rational-features-D.doenetml b/source/previews/doenet/PA-poly-rational-features-D.doenetml
new file mode 100644
index 00000000..c78d26d5
--- /dev/null
+++ b/source/previews/doenet/PA-poly-rational-features-D.doenetml
@@ -0,0 +1,252 @@
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ (x^2-1)/(x^2-3x-4)
+
+
+ Consider the rational function r(x) = \frac{x^2 - 1}{x^2 - 3x - 4}, and let p(x) = x^2 - 1 and q(x) = x^2 - 3x - 4. Note that p(x) is the numerator of r(x) and q(x) is the denominator of r(x).
+
+
+
+ To enter a list of numbers, separate the numbers with commas.$mi1=1 -1
+
+
+ To enter a list of numbers, separate the numbers with commas.$mi2=4 -1
+
+
+ Type r(x) in the box below, and the following table values will automatically fill in.
+
+
+ $rEnt=$r
+
+
+
+
+
+ x
+
+
+ r(x)
+
+
+
+
+ 4.1
+
+
+ expression entered for r of x evaluated entered at 4.1$$rEnt(4.1)
+
+
+
+
+ 4.01
+
+
+ expression entered for r of x evaluated entered at 4.01$$rEnt(4.01)
+
+
+
+
+ 4.001
+
+
+ expression entered for r of x evaluated entered at 4.001$$rEnt(4.001)
+
+
+
+
+ 3.9
+
+
+ expression entered for r of x evaluated entered at 3.9$$rEnt(3.9)
+
+
+
+
+ 3.99
+
+
+ expression entered for r of x evaluated entered at 3.99$$rEnt(3.99)
+
+
+
+
+ 3.999
+
+
+ expression entered for r of x evaluated entered at 3.999$$rEnt(3.999)
+
+
+
+
+
+
+ x
+
+
+ r(x)
+
+
+
+
+ 1.1
+
+
+ expression entered for r of x evaluated entered at 1.1$$rEnt(1.1)
+
+
+
+
+ 1.01
+
+
+ expression entered for r of x evaluated entered at 1.01$$rEnt(1.01)
+
+
+
+
+ 1.001
+
+
+ expression entered for r of x evaluated entered at 1.001$$rEnt(1.001)
+
+
+
+
+ 0.9
+
+
+ expression entered for r of x evaluated entered at 0.9$$rEnt(0.9)
+
+
+
+
+ 0.99
+
+
+ expression entered for r of x evaluated entered at 0.99$$rEnt(0.99)
+
+
+
+
+ 0.999
+
+
+ expression entered for r of x evaluated entered at 0.999$$rEnt(0.999)
+
+
+
+
+
+
+ x
+
+
+ r(x)
+
+
+
+
+ -1.1
+
+
+ expression entered for r of x evaluated entered at -1.1$$rEnt(-1.1)
+
+
+
+
+ -1.01
+
+
+ expression entered for r of x evaluated entered at -1.01$$rEnt(-1.01)
+
+
+
+
+ -1.001
+
+
+ expression entered for r of x evaluated entered at -1.001$$rEnt(-1.001)
+
+
+
+
+ -0.9
+
+
+ expression entered for r of x evaluated entered at -0.9$$rEnt(-0.9)
+
+
+
+
+ -0.99
+
+
+ expression entered for r of x evaluated entered at -0.99$$rEnt(-0.99)
+
+
+
+
+ -0.999
+
+
+ expression entered for r of x evaluated entered at -0.999$$rEnt(-0.999)
+
+
+
+
+
+
+ Complete the sentences below to explain why r(x) behaves the way it does near x = 4, near x=1, and near x=-1.
+
+
+
+ As x gets closer and closer to 4, 15, while 0. approaches 0increases or decreases without boundapproaches 0.4approaches two different numbers near x = 4. zero
+ holejumpvertical asymptote at x=4.
+
+
+
+
+
+ As x gets closer and closer to 1, 0, while -6. approaches 0increases or decreases without boundapproaches 0.4approaches two different numbers near x = 1. zero
+ holejumpvertical asymptote at x=1.
+
+
+
+
+
+ As x gets closer and closer to -1, 0, and 0. Factoring both the numerator and denominator results in \frac{(x-1)(x+1)}{(x-4)(x+1)}. As long as x\neq -1, the two (x+1) terms cancel, resulting in \frac{x-1}{x-4}. approaches 0increases or decreases without boundapproaches \frac{-2}{-5}=0.4approaches two different numbers near x = -1, even though r(x) is undefined at x=-1. zero
+ holejumpvertical asymptote at x=-1.
+
+
+
+
+ Finally, the graph of r(x) appears below.
+
+
+
+ graph of the function r of x
+ $r
+
+
+
+
+ Can you see all the behavior we just discussed, or only some of it? Remember that you can zoom in and out using the "+" and "-" in the lower right corner of the graph, and that the "O" returns to the original view.
+ zeroholejumpvertical asymptote
+
\ No newline at end of file
diff --git a/source/previews/doenet/PA-trig-finding-angles-D.doenetml b/source/previews/doenet/PA-trig-finding-angles-D.doenetml
new file mode 100644
index 00000000..1073d3fe
--- /dev/null
+++ b/source/previews/doenet/PA-trig-finding-angles-D.doenetml
@@ -0,0 +1,99 @@
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ (0,0)
+ (3,0)
+ (3,sqrt(3))
+
+
+
+ Below is an image of a right triangle whose sides are labeled a, b, and c, and whose non-right angles are labeled x and y.
+
+
+ We will complete a sketch of a right triangle with one leg of length 3 and another leg of length \sqrt{3}. Let \theta be the angle that lies opposite the shorter leg.
+
+
+ a right triangle with sides labeled a, b, and c
+ The longest side of the right triangle is labeled c. The shortest side, which is vertical, is labeled a, and the remaining side of the triangle, which is horizontal, is labeled b. The angle across from the shortest side is labeled x, while the remaining non-right angle is labeled y.
+
+
+
+
+
+
+
+
+
+ (.35cos(t), .35sin(t))
+
+
+
+
+ (.35cos(t)+3, .35sin(t)+sqrt(3))
+
+
+
+
+
+
+
+
+ First, determine the exact values of the sides a, b and c.
+
+ Your answer is close to correct. Make sure you are entering an exact answer, not a decimal approximation.
+ Your answer is close to correct. Make sure you are entering an exact answer, not a decimal approximation.
+Your answer is close to correct. Make sure you are entering an exact answer, not a decimal approximation.
+ It's both more helpful and more accurate if we assume the lengths are drawn to scale, so make this answer the horizontal leg.
+ It's both more helpful and more accurate if we assume the lengths are drawn to scale, so make this answer the vertical leg.
+
+ xy
+
+
+ After answering all the questions in part a. correctly, you now have a correctly labeled right triangle meeting the specified conditions.
+
+
+
+
+ What is the exact value of \sin(\theta)?
+
+
+ 1/21/2
+
+ Your answer is correct but not in the right form or simplified enough, yet. It might help to rewrite \sqrt{12}=\sqrt{4\cdot 3}=2\sqrt{3}.
+
+
+
+ What special angle from the unit circle is \theta?
+
+
+ Which angle from the unit circle looks like the sketch of the triangle and has the value of \sin(\theta) you just calculated?pi/6
+
+
+
+
+ Use your answer to c. to rewrite your equation from b. using the arcsine function in the form \arcsin(\Box) = \Delta, where \Box and \Delta are numerical values.
+
+
+ 1/2pi/6
+
+
+
\ No newline at end of file
diff --git a/source/previews/doenet/PA-trig-inverse-D.doenetml b/source/previews/doenet/PA-trig-inverse-D.doenetml
new file mode 100644
index 00000000..9721770f
--- /dev/null
+++ b/source/previews/doenet/PA-trig-inverse-D.doenetml
@@ -0,0 +1,94 @@
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ cos(x)
+ cos(x)
+ cos(x)
+
+
+
+ Consider the plot of the standard cosine function in the following figure along with the emphasized portion of the graph on [0,\pi].
+
+
+
+ graph of cosine with the part of it between 0 and pi emphasized
+ t
+ y
+
+
+
+
+
+
+ $cosine1
+ $cosine2
+ $cosine0pi
+
+
+
+
+ Let g be the function whose domain is 0 \le t \le \pi and whose outputs are determined by the rule g(t) = \cos(t). Note well: g is defined in terms of the cosine function, but because it has a different domain, it is not the cosine function.
+
+
+
+
+
+ You can enter your answer using interval notation like [0,\pi], or using inequalities in the form 0\leq t \leq \pi.$mi1=[0,pi]
+
+
+
+
+ You can enter your answer using interval notation like [0,\pi], or using inequalities in the form 0\leq t \leq \pi.$mi2=[-1,1]
+
+
+
+
+ We can see from the graph that g(t)passes or doesn't pass the HLTpassesdoesn't pass the horizontal line test because 0 and \pi0 and 2\pi0 and 1-1 and 1[0,\pi][0,2\pi][0,1][-1,1].
+
+
+ This means that g(t)has or doesn't have an inversehasdoesn't have an inverse function, g^{-1}(t), [0,\pi][0,2\pi][0,1][-1,1] and [0,\pi][0,2\pi][0,1][-1,1].
+
+
+
+
+ We know that g(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}. That means
+ pi/4
+
+
+ 3pi/4
+
+
+
+
+ Determine the exact values of each of the quantities below.
+
+
+ 2pi/3
+
+
+ pi/6
+
+
+ pi/2
+
+
+ pi
+
+
g(t) comes from \cos(t), which takes in an angle and outputs the x-value of the point on the unit circle corresponding to that angle.
g^{-1}(t) swaps that; it takes in an x-value of a point on the unit circle and outputs the angle between 0 and \pi corresponding to the point.
+ Below is an image of a right triangle whose sides are labeled a, b, and c, and whose non-right angles are labeled x and y.
+
+
+ We will complete a sketch of a right triangle with hypotenuse of length 61 and one leg of length 11.
+ Let \alpha be the angle opposite the side of length 11.
+
+
+ a right triangle with sides labeled a, b, and c
+ The longest side of the right triangle is labeled c. The shortest side, which is horizontal, is labeled b, and the remaining side of the triangle, which is vertical, is labeled a. The angle across from the shortest side is labeled y, while the remaining non-right angle is labeled x.
+
+
+
+
+
+
+
+
+
+ (.3cos(t), .5sin(t))
+
+
+
+
+ (.75cos(t)+$f3.x, .9sin(t)+$f3.y)
+
+
+
+
+
+
+
+
+
+
+ First, determine the exact values of the sides a, b and c.
+
+
+ 606011
+ 111160
+ 6161
+
+ Your answer is close to correct. Make sure you are entering an exact answer, not a decimal approximation.
+ Your answer is close to correct. Make sure you are entering an exact answer, not a decimal approximation.
+Your answer is close to correct. Make sure you are entering an exact answer, not a decimal approximation.
+ It's both more helpful and more accurate if we assume the lengths are drawn to scale, so make this answer the horizontal leg.
+ It's both more helpful and more accurate if we assume the lengths are drawn to scale, so make this answer the vertical leg.
+
+ xy
+
+
+ After answering all the questions in part a. correctly, you now have a correctly labeled right triangle meeting the specified conditions.
+
+
+
+
+ Determine the exact value of each of the six trigonometric functions evaluated at \alpha.
+
+
+
+ Recall that \sin(\alpha)=\frac{\text{opp}}{\text{hyp}}. Which side is opposite to \alpha?11/6111/61
+ Recall that \csc(\alpha)=\frac{\text{hyp}}{\text{opp}}. Which side is opposite to \alpha?61/1161/11
+
+
+ Recall that \cos(\alpha)=\frac{\text{adj}}{\text{hyp}}. Which side is adjacent to \alpha?60/6160/61
+ Recall that \sec(\alpha)=\frac{\text{hyp}}{\text{adj}}. Which side is adjacent to \alpha?61/6061/60
+
+
+ Recall that \tan(\alpha)=\frac{\text{opp}}{\text{adj}}.11/6011/60
+ Recall that \cot(\alpha)=\frac{\text{adj}}{\text{opp}}. 60/1160/11
+
+
+ Your answer is close but not exact. You can leave your answer as a fraction.
+
+
+
+ What are the exact and approximate measures of the two non-right angles in the triangle?
+
+
+ Use an inverse trig function on one of your answers from part b.arccos(60/61)\approxdecimal approximation of alpha$alphaEnt
+
+
+ Use an inverse trig function.arcsin(60/61)\approxdecimal approximation of alpha$thetaEnt
+
+ What follows is 6 situations where some information about a right triangle is given, and some is missing.
+
+
+ In each of the situations, you will end up with an image of a right triangle that satisfies the given conditions, and you will determine the requested missing information in the triangle. Assume that all angles are being considered in radian measure.
+
+ Through the following questions, we work to understand the special values and overall behavior of the tangent function.
+
+
+
+
+ Use the unit circle below (not a computational device) to find the exact value of \tan(t) at the following values: t = \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \frac{2\pi}{3}, \frac{3\pi}{4}, \frac{5\pi}{6}.
+
+
+ Move the square point to each of the above angles and an answer box will appear. Enter a simplified exact expression, not a decimal approximation.
+
+ Recall that \tan(t)=\frac{\sin(t)}{\cos(t)}, and that for any point on a unit circle, the x-coordinate is \cos(\theta) and the y-coordinate is \sin(\theta).
+
+ unit circle with 24 equally spaced points
+
+ $points
+ $points(cos(pi/12),sin(pi/12))
+
+
+
+
+ (1/5cos(t), 1/5sin(t))
+
+
+ (1/5cos(t), 1/5sin(t))
+
+
+
+
+
+
+ (1,0)(0,1)(-1,0)(0,-1)
+ \cos\left( \frac{\pi}{2} \right)=0\sin\left( \frac{\pi}{2} \right)= 0 and so is 0is undefined because you can't divide by 0.
+
+
+ What are three other input values x for which \tan(x) is not defined? Enter your answers separated by commas. list of values of theta where tan theta is undefined$undefList.numValues=3 and cos($undefList[1])=0 and cos($undefList[2])=0 and cos($undefList[3])=0 and $undefList[1]!=$undefList[2] and $undefList[1]!=$undefList[3] and $undefList[3]!=$undefList[2] and $undefList[1]!=pi/2 and $undefList[2]!=pi/2 and $undefList[3]!=pi/2
+
+
+
+
+
+ unit circle with 24 equally spaced points
+
+ $points
+ $unitcir(cos(pi/12),sin(pi/12))
+
+
+
+ Drag the point to the approximate location of \theta=\frac{11\pi}{24}.
+
+ abs($P2.x-.125)< .05
+
+
+
+ 0\frac{\pi}{4}\frac{\pi}{2}\frac{3\pi}{4}\pi, and so (0,0)(1,0)(0,1)(-1,0).
+ As the angles get closer and closer to \frac{\pi}{2}, the function 01, and so closer to 0.closer to 1.larger and larger.
+
+
+
+
+
+ A table of values appears below, including the values from the angles asked about in the first graph above after you've answered them correctly.
+
+
+
+
+ x
+ y = \tan(x)
+
+
+
+ 0
+ \tan(0)
+ = \, 0
+
+
+ \frac{\pi}{12}
+ \tan(\frac{\pi}{12})
+ =\, tan(pi/12)
+
+
+ \frac{\pi}{6}
+ value of tangent of x1/sqrt(3)
+ =decimal value of tangent of xtan(pi/6)
+
+
+ \frac{\pi}{4}
+ value of tangent of x1
+ =decimal value of tangent of xtan(pi/4)
+
+
+ \frac{\pi}{3}
+ value of tangent of xsqrt(3)
+ =decimal value of tangent of xtan(pi/3)
+
+
+ \frac{5\pi}{12}
+ \tan(\frac{5\pi}{12})
+ = \, tan(5pi/12)
+
+
+ \frac{\pi}{2}
+ \tan(\frac{\pi}{2})
+ = \, undefined
+
+
+ \frac{7\pi}{12}
+ \tan(\frac{7\pi}{12})
+ = \, tan(7pi/12)
+
+
+ \frac{2\pi}{3}
+ value of tangent of x-sqrt(3)
+ =decimal value of tangent of xtan(2pi/3)
+
+
+ \frac{3\pi}{4}
+ value of tangent of x-1
+ =decimal value of tangent of xtan(3pi/4)
+
+
+ \frac{5\pi}{6}
+ value of tangent of x-1/sqrt(3)
+ =decimal value of tangent of xtan(5pi/6)
+
+
+ \pi
+ \tan(\pi)
+ = \, tan(pi)
+
+
+
+
+ Moreover, a graph of the values appears below, and you can click to show the function \tan(x).
+
+
+
+
+ unit circle with 24 equally spaced points and a movable point
+
+
+
+
+ $tanpoints
+ tan(x)
+
+
+
+
+
+
+
+ Use the graph and your work above to answer the following:
+
+
+
+ all real numbersall real numbers except x=0all real numbers except x=\pm\frac{\pi}{2},\pm\frac{3\pi}{2}, \pm\frac{5\pi}{2}, \cdotsall real numbers except x=0,\pm\pi,\pm 3\pi, \pm5\pi, \cdots
+
+
+ all real numbersall real numbers except x=0all real numbers except x=\pm\frac{\pi}{2},\pm\frac{3\pi}{2}, \pm\frac{5\pi}{2}, \cdotsall real numbers except x=0,\pm\pi,\pm 3\pi, \pm5\pi, \cdots
+
+
+ \frac{\pi}{2}\pi\frac{3\pi}{2}x=2\pi
+
+
\ No newline at end of file
diff --git a/source/sec-0-0.xml b/source/sec-0-0.xml
index fcf629b9..31bb928b 100755
--- a/source/sec-0-0.xml
+++ b/source/sec-0-0.xml
@@ -1,18 +1,18 @@
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/source/sec-changing-aroc-wb.xml b/source/sec-changing-aroc-wb.xml
index d2c16dd4..8709ec28 100644
--- a/source/sec-changing-aroc-wb.xml
+++ b/source/sec-changing-aroc-wb.xml
@@ -1,22 +1,22 @@
-
-
-
-
-
-
-
-
-
-
-
-
-
-
- The Average Rate of Change of a Function
-
-
-
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ The Average Rate of Change of a Function
+
+
+
+
+
+
+
diff --git a/source/sec-changing-aroc.xml b/source/sec-changing-aroc.xml
index 87e2d129..046677ac 100755
--- a/source/sec-changing-aroc.xml
+++ b/source/sec-changing-aroc.xml
@@ -1,198 +1,199 @@
-
-
-
-
-
-
-
-
-
-
-
-
-
-
- The Average Rate of Change of a Function
-
-
-
-
-
- What do we mean by the average rate of change of a function on an interval?
-
-
-
-
- What does the average rate of change of a function measure? How do we interpret its meaning in context?
-
-
-
-
- How is the average rate of change of a function connected to a line that passes through two points on the curve?
-
-
-
-
-
- Introduction
-
- Given a function that models a certain phenomenon,
- it's natural to ask such questions as
- how is the function changing on a given interval or
- on which interval is the function changing more rapidly?
- The concept of average rate of change
- enables us to make these questions more mathematically precise.
- Initially, we will focus on the average rate of change of an object moving along a straight-line path.
-
-
-
- For a function s that tells the location of a moving object along a straight path at time t, we define the average rate of change of s on the interval [a,b] to be the quantity
-
- AV_{[a,b]} = \frac{s(b)-s(a)}{b-a}
- .
- average rate of changeof position
- Note particularly that the average rate of change of s on [a,b] is measuring the
- change in position divided by the change in time.
-
-
-
-
-
-
-
- Defining and interpreting the average rate of change of a function
-
- In the context of a function that measures height or position of a moving object at a given time,
- the meaning of the average rate of change of the function on a given interval is the average velocityaverage velocity of the moving object
- because it is the ratio of change in position to change in time. For example, in Preview Activity, the units on AV_{[1.5,2.5]} = -32 are feet per second since the units on the numerator are feet and on the denominator seconds. Morever, -32 is numerically the same value as the slope of the line that connects the two corresponding points on the graph of the position function, as seen in Figure. The fact that the average rate of change is negative in this example indicates that the ball is falling.
-
-
-
-
-
The average rate of change of s on [1.5,2.5] for the function in Preview Activity.
-
The average rate of change of s on [1.5,2.5] for the function in Preview Activity.
-
-
-
-
The average rate of change of an abstract function f on the interval [a,b].
-
The average rate of change of an abstract function f on the interval [a,b].
-
-
-
-
-
- While the average rate of change of a position function tells us the moving object's average velocity, in other contexts,
- the average rate of change of a function can be similarly defined and has a related interpretation. We make the following formal definition.
-
-
-
- average rate of change
-
-
- For a function f defined on an interval [a,b], the average rate of change of f on [a,b] is the quantity
-
- AV_{[a,b]} = \frac{f(b) - f(a)}{b-a}
- .
-
-
-
-
-
- In every situation, the units on the average rate of change help us interpret its meaning,
- and those units are always units of output per unit of input.average rate of changeunits Moreover, the average rate of change of f on [a,b] always corresponds to the slope of the line between the points (a,f(a)) and (b,f(b)), as seen in Figure.
-
-
-
-
-
- The average rate of change of a function on an interval gives us an excellent way to describe how the function behaves, on average. For instance, if we compute AV_{[1970,2000]} for Kent County, we find that
-
- AV_{[1970,2000]} = \frac{574,336 - 411,044}{30} = 5443.07
- ,
- which tells us that in an average year from 1970 to 2000, the population of Kent County increased by about 5443 people. Said differently, we could also say that from 1970 to 2000, Kent County was growing at an average rate of 5443 people per year. These ideas also afford the opportunity to make comparisons over time. Since
-
- AV_{[1990,2000]} = \frac{574,336 - 500,631}{10} = 7370.5
- ,
- we can not only say that Kent county's population increased by about 7370 in an average year between 1990 and 2000, but also that the population was growing faster from 1990 to 2000 than it did from 1970 to 2000.
-
-
-
- Finally, we can even use the average rate of change of a function to predict future behavior. Since the population was changing on average by 7370.5 people per year from 1990 to 2000, we can estimate that the population in 2002 is
-
- K(2002) \approx K(2000) + 2 \cdot 7370.5 = 574,336 + 14,741 = 589,077
- .
-
-
-
-
- How average rate of change indicates function trends
-
-
- We have already seen that it is natural to use words such as increasing and decreasing to describe a function's behavior. For instance, for the tennis ball whose height is modeled by s(t) = 64 - 16(t-1)^2, we computed that AV_{[1.5,2.5]} = -32, which indicates that on the interval [1.5,2.5], the tennis ball's height is decreasing at an average rate of 32 feet per second. Similarly, for the population of Kent County, since AV_{[1990,2000]} = 7370.5, we know that on the interval [1990,2000] the population is increasing at an average rate of 7370.5 people per year.
-
-
-
- We make the following formal definitions to clarify what it means to say that a function is increasing or decreasing.
-
-
-
- function trendsincreasingfunction trendsdecreasing
-
-
- Let f be a function defined on an interval (a,b) (that is, on the set of all x for which a \lt x \lt b). We say that f is increasing on (a,b) provided that the function is always rising as we move from left to right. That is, for any x and y in (a,b), if x \lt y, then f(x) \lt f(y).
-
-
-
- Similarly, we say that f is decreasing on (a,b) provided that
- the function is always falling as we move from left to right. That is, for any x and y in (a,b), if x \lt y, then f(x) \gt f(y).
-
-
-
-
-
- If we compute the average rate of change of a function on an interval, we can decide if the function is increasing or decreasing on average on the interval, but it takes more workCalculus offers one way to justify that a function is always increasing or always decreasing on an interval. to decide if the function is increasing or decreasing always on the interval.
-
-
-
-
-
- It is helpful be able to connect information about a function's average rate of change and its graph. For instance, if we have determined that AV_{[-3,2]} = 1.75 for some function f, this tells us that, on average, the function rises between the points x = -3 and x = 2 and does so at an average rate of 1.75 vertical units for every horizontal unit. Moreover, we can even determine that the difference between f(2) and f(-3) is
-
- f(2)-f(-3) = 1.75 \cdot 5 = 8.75
-
- since \frac{f(2)-f(-3)}{2-(-3)} = 1.75.
-
-
-
-
-
-
-
- Summary
-
-
-
-
- For a function f defined on an interval [a,b], the average rate of change of f on [a,b] is the quantity
- AV_{[a,b]} = \frac{f(b) - f(a)}{b-a}.
-
-
-
-
- The value of AV_{[a,b]} = \frac{f(b) - f(a)}{b-a} tells us how much the function rises or falls, on average, for each additional unit we move to the right on the graph. For instance, if AV_{[3,7]} = 0.75, this means that for additional 1-unit increase in the value of x on the interval [3,7], the function increases, on average, by 0.75 units. In applied settings, the units of AV_{[a,b]} are units of output per unit of input.
-
-
-
-
- The value of AV_{[a,b]} = \frac{f(b) - f(a)}{b-a} is also the slope of the line that passes through the points (a,f(a)) and (b,f(b)) on the graph of f, as shown in Figure.
-
-
-
-
-
-
-
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ The Average Rate of Change of a Function
+
+
+
+
+
+ What do we mean by the average rate of change of a function on an interval?
+
+
+
+
+ What does the average rate of change of a function measure? How do we interpret its meaning in context?
+
+
+
+
+ How is the average rate of change of a function connected to a line that passes through two points on the curve?
+
+
+
+
+
+ Introduction
+
+ Given a function that models a certain phenomenon,
+ it's natural to ask such questions as
+ how is the function changing on a given interval or
+ on which interval is the function changing more rapidly?
+ The concept of average rate of change
+ enables us to make these questions more mathematically precise.
+ Initially, we will focus on the average rate of change of an object moving along a straight-line path.
+
+
+
+ For a function s that tells the location of a moving object along a straight path at time t, we define the average rate of change of s on the interval [a,b] to be the quantity
+
+ AV_{[a,b]} = \frac{s(b)-s(a)}{b-a}
+ .
+ average rate of changeof position
+ Note particularly that the average rate of change of s on [a,b] is measuring the
+ change in position divided by the change in time.
+
+
+
+
+
+
+
+
+ Defining and interpreting the average rate of change of a function
+
+ In the context of a function that measures height or position of a moving object at a given time,
+ the meaning of the average rate of change of the function on a given interval is the average velocityaverage velocity of the moving object
+ because it is the ratio of change in position to change in time. For example, in Preview Activity, the units on AV_{[1.5,2.5]} = -32 are feet per second since the units on the numerator are feet and on the denominator seconds. Morever, -32 is numerically the same value as the slope of the line that connects the two corresponding points on the graph of the position function, as seen in Figure. The fact that the average rate of change is negative in this example indicates that the ball is falling.
+
+
+
+
+
The average rate of change of s on [1.5,2.5] for the function in Preview Activity.
+
The average rate of change of s on [1.5,2.5] for the function in Preview Activity.
+
+
+
+
The average rate of change of an abstract function f on the interval [a,b].
+
The average rate of change of an abstract function f on the interval [a,b].
+
+
+
+
+
+ While the average rate of change of a position function tells us the moving object's average velocity, in other contexts,
+ the average rate of change of a function can be similarly defined and has a related interpretation. We make the following formal definition.
+
+
+
+ average rate of change
+
+
+ For a function f defined on an interval [a,b], the average rate of change of f on [a,b] is the quantity
+
+ AV_{[a,b]} = \frac{f(b) - f(a)}{b-a}
+ .
+
+
+
+
+
+ In every situation, the units on the average rate of change help us interpret its meaning,
+ and those units are always units of output per unit of input.average rate of changeunits Moreover, the average rate of change of f on [a,b] always corresponds to the slope of the line between the points (a,f(a)) and (b,f(b)), as seen in Figure.
+
+
+
+
+
+ The average rate of change of a function on an interval gives us an excellent way to describe how the function behaves, on average. For instance, if we compute AV_{[1970,2000]} for Kent County, we find that
+
+ AV_{[1970,2000]} = \frac{574,336 - 411,044}{30} = 5443.07
+ ,
+ which tells us that in an average year from 1970 to 2000, the population of Kent County increased by about 5443 people. Said differently, we could also say that from 1970 to 2000, Kent County was growing at an average rate of 5443 people per year. These ideas also afford the opportunity to make comparisons over time. Since
+
+ AV_{[1990,2000]} = \frac{574,336 - 500,631}{10} = 7370.5
+ ,
+ we can not only say that Kent county's population increased by about 7370 in an average year between 1990 and 2000, but also that the population was growing faster from 1990 to 2000 than it did from 1970 to 2000.
+
+
+
+ Finally, we can even use the average rate of change of a function to predict future behavior. Since the population was changing on average by 7370.5 people per year from 1990 to 2000, we can estimate that the population in 2002 is
+
+ K(2002) \approx K(2000) + 2 \cdot 7370.5 = 574,336 + 14,741 = 589,077
+ .
+
+
+
+
+ How average rate of change indicates function trends
+
+
+ We have already seen that it is natural to use words such as increasing and decreasing to describe a function's behavior. For instance, for the tennis ball whose height is modeled by s(t) = 64 - 16(t-1)^2, we computed that AV_{[1.5,2.5]} = -32, which indicates that on the interval [1.5,2.5], the tennis ball's height is decreasing at an average rate of 32 feet per second. Similarly, for the population of Kent County, since AV_{[1990,2000]} = 7370.5, we know that on the interval [1990,2000] the population is increasing at an average rate of 7370.5 people per year.
+
+
+
+ We make the following formal definitions to clarify what it means to say that a function is increasing or decreasing.
+
+
+
+ function trendsincreasingfunction trendsdecreasing
+
+
+ Let f be a function defined on an interval (a,b) (that is, on the set of all x for which a \lt x \lt b). We say that f is increasing on (a,b) provided that the function is always rising as we move from left to right. That is, for any x and y in (a,b), if x \lt y, then f(x) \lt f(y).
+
+
+
+ Similarly, we say that f is decreasing on (a,b) provided that
+ the function is always falling as we move from left to right. That is, for any x and y in (a,b), if x \lt y, then f(x) \gt f(y).
+
+
+
+
+
+ If we compute the average rate of change of a function on an interval, we can decide if the function is increasing or decreasing on average on the interval, but it takes more workCalculus offers one way to justify that a function is always increasing or always decreasing on an interval. to decide if the function is increasing or decreasing always on the interval.
+
+
+
+
+
+ It is helpful be able to connect information about a function's average rate of change and its graph. For instance, if we have determined that AV_{[-3,2]} = 1.75 for some function f, this tells us that, on average, the function rises between the points x = -3 and x = 2 and does so at an average rate of 1.75 vertical units for every horizontal unit. Moreover, we can even determine that the difference between f(2) and f(-3) is
+
+ f(2)-f(-3) = 1.75 \cdot 5 = 8.75
+
+ since \frac{f(2)-f(-3)}{2-(-3)} = 1.75.
+
+
+
+
+
+
+
+ Summary
+
+
+
+
+ For a function f defined on an interval [a,b], the average rate of change of f on [a,b] is the quantity
+ AV_{[a,b]} = \frac{f(b) - f(a)}{b-a}.
+
+
+
+
+ The value of AV_{[a,b]} = \frac{f(b) - f(a)}{b-a} tells us how much the function rises or falls, on average, for each additional unit we move to the right on the graph. For instance, if AV_{[3,7]} = 0.75, this means that for additional 1-unit increase in the value of x on the interval [3,7], the function increases, on average, by 0.75 units. In applied settings, the units of AV_{[a,b]} are units of output per unit of input.
+
+
+
+
+ The value of AV_{[a,b]} = \frac{f(b) - f(a)}{b-a} is also the slope of the line that passes through the points (a,f(a)) and (b,f(b)) on the graph of f, as shown in Figure.
+
- How can we create new functions by adding, subtracting, multiplying, or dividing given functions?
-
-
-
-
- What are piecewise functions and what are different ways we can represent them?
-
-
-
-
-
- Introduction
-
-
- In arithmetic, we execute processes where we take two numbers to generate a new number. For example, 2 + 3 = 5: the number 5 results from adding 2 and 3. Similarly, we can multiply two numbers to generate a new one: 2 \cdot 3 = 6.
-
-
-
- We can work similarly with functions. Indeed, we have already seen a sophisticated way to combine two functions to generate a new, related function through composition. If g : A \to B and f : B \to C, then we know there's a new, related function f \circ g : A \to C defined by the process (f \circ g)(x) = f(g(x)). Said differently, the new function f \circ g results from executing g first, followed by f.
-
-
-
- Just as we can add, subtract, multiply, and divide numbers, we can also add, subtract, multiply, and divide functions to create a new function from two or more given functions.
-
-
-
-
-
-
-
- Arithmetic with functions
-
- In most mathematics up until calculus,
- the main object we study is numbers.
- We ask questions such as
-
-
-
- what number(s) form solutions to the equation x^2 - 4x - 5 = 0?
-
-
-
-
- what number is the slope of the line 3x - 4y = 7?
-
-
-
-
- what number is generated as output by the function
- f(x) = \sqrt{x^2 + 1} by the input x = -2?
-
-
-
- Certainly we also study overall patterns as seen in functions and equations,
- but this usually occurs through an examination of numbers themselves,
- and we think of numbers as the main objects being acted upon.
-
-
-
- This changes in calculus. In calculus, the fundamental objects being studied are
- functions themselves. A function is a much more sophisticated mathematical object than a number,
- in part because a function can be thought of in terms of its graph, which is an infinite collection of ordered pairs of the form (x,f(x)).
-
-
-
- It is often helpful to look at a function's formula and observe algebraic structure. For instance, given the quadratic function
-
- q(x) = -3x^2 + 5x - 7
-
- we might benefit from thinking of this as the sum of three simpler functions: the constant function c(x) = -7, the linear function s(x) = 5x that passes through (0,0) with slope m = 5, and the concave down basic quadratic function w(x) = -3x^2. Indeed, each of the simpler functions c, s, and w contribute to making q be the function that it is. Likewise, if we were interested in the function p(x) = (3x^2 + 4)(9 - 2x^2), it might be natural to think about the two simpler functions f(x) = 3x^2 + 4 and g(x) = 9 - 2x^2 that are being multiplied to produce p.
-
-
-
- We thus naturally arrive at the ideas of adding, subtracting, multiplying, or dividing two or more functions, and hence introduce the following definitions and notation.
-
-
-
- function arithmeticsum of
- function arithmeticdifference of
- function arithmeticproduct of
- function arithmeticquotient of
-
-
- Let f and g be functions that share the same domain. Then,
-
-
-
- The sum of f and g is the function f + g defined by (f+g)(x) = f(x) + g(x).
-
-
-
-
- The difference of f and g is the function f - g defined by (f-g)(x) = f(x) - g(x).
-
-
-
-
- The product of f and g is the function f \cdot g defined by (f \cdot g)(x) = f(x) \cdot g(x).
-
-
-
-
- The quotient of f and g is the function \frac{f}{g} defined by \left( \frac{f}{g} \right)(x) = \frac{f(x)}{g(x)} for all x such that g(x) \ne 0.
-
- When we work in applied settings with functions that model phenomena in the world around us,
- it is often useful to think carefully about the units of various quantities.
- Analyzing units can help us both understand the algebraic structure of functions and the variables involved,
- as well as assist us in assigning meaning to quantities we compute.
- We have already seen this with the notion of average rate of change:
- if a function P(t) measures the population in a city in year t and we compute AV_{[5, 11]},
- then the units on AV_{[5, 11]} are
- people per year,
- and the value of AV_{[5, 11]} is telling us the average rate at which the population changes in people per year on the time interval from year 5 to year 11.
-
-
-
-
-
- Say that an investor is regularly purchasing stock in a particular company.This example is taken from Section 2.3 of Active Calculus.
- Let N(t) represent the number of shares owned on day t,
- where t = 0 represents the first day on which shares were purchased.
- Let S(t) give the value of one share of the stock on day t;
- note that the units on S(t) are dollars per share.
- How is the total value, V(t), of the held stock on day t determined?
-
-
-
- Solution. Observe that the units on N(t) are shares and the units on S(t) are dollars per share. Thus when we compute the product
-
- N(t) \, \text{shares} \cdot S(t) \, \text{dollars per share}
- ,
- it follows that the resulting units are dollars, which is the total value of held stock. Hence,
-
- V(t) = N(t) \cdot S(t)
- .
-
-
-
-
-
-
-
-
-
- Piecewise functions
-
-
- In both abstract and applied settings,
- we sometimes have to use different formulas on different intervals in order to define a function of interest.
-
-
-
-
-
- A familiar and important function that is defined piecewise
- is the absolute value function: A(x) = |x|.
- We know that if x \ge 0, |x| = x, whereas if x \lt 0, |x| = -x.
-
-
-
- absolute value function
-
-
- The absolute value of a real number, denoted by A(x) = |x|, is defined by the rule
-
- A(x) =
- \begin{cases}
- -x, \amp x \lt 0 \\
- x, \amp x \ge 0
- \end{cases}
-
-
-
-
-
-
-
-
A plot of the absolute value function, A(x) = |x|.
-
A plot of the absolute value function, A(x) = |x|.
-
-
-
-
-
- The absolute value function is one example of a piecewise-defined function. The bracket notation in Definition is how we express which piece of the function applies on which interval. As we can see in Figure, for x values less than 0, the function y = -x applies, whereas for x greater than or equal to 0, the rule is determined by y = x.
-
-
-
- As long as we are careful to make sure that each potential input has one and only one corresponding output, we can define a piecewise function using as many different functions on different intervals as we desire.
-
-
-
-
-
-
-
- Summary
-
-
-
-
- Just as we can generate a new number by adding, subtracting, multiplying, or dividing two given numbers, we can generate a new function by adding, subtracting, multiplying, or dividing two given functions. For instance, if we know formulas, graphs, or tables for functions f and g that share the same domain, we can create their product p according to the rule p(x) = (f \cdot g)(x) = f(x) \cdot g(x).
-
-
-
-
- A piecewise function is a function whose formula consists of at least two different formulas in such a way that which formula applies depends on where the input falls in the domain. For example, given two functions f and g each defined on all real numbers, we can define a new piecewise function P according to the rule
-
- P(x) =
- \begin{cases}
- f(x), \amp x \lt a \\
- g(x), \amp x \ge a
- \end{cases}
-
- This tells us that for any x to the left of a, we use the rule for f, whereas for any x to the right of or equal to a, we use the rule for g. We can use as many different functions as we want on different intervals, provided the intervals don't overlap.
-
+ How can we create new functions by adding, subtracting, multiplying, or dividing given functions?
+
+
+
+
+ What are piecewise functions and what are different ways we can represent them?
+
+
+
+
+
+ Introduction
+
+
+ In arithmetic, we execute processes where we take two numbers to generate a new number. For example, 2 + 3 = 5: the number 5 results from adding 2 and 3. Similarly, we can multiply two numbers to generate a new one: 2 \cdot 3 = 6.
+
+
+
+ We can work similarly with functions. Indeed, we have already seen a sophisticated way to combine two functions to generate a new, related function through composition. If g : A \to B and f : B \to C, then we know there's a new, related function f \circ g : A \to C defined by the process (f \circ g)(x) = f(g(x)). Said differently, the new function f \circ g results from executing g first, followed by f.
+
+
+
+ Just as we can add, subtract, multiply, and divide numbers, we can also add, subtract, multiply, and divide functions to create a new function from two or more given functions.
+
+
+
+
+
+
+
+
+ Arithmetic with functions
+
+ In most mathematics up until calculus,
+ the main object we study is numbers.
+ We ask questions such as
+
+
+
+ what number(s) form solutions to the equation x^2 - 4x - 5 = 0?
+
+
+
+
+ what number is the slope of the line 3x - 4y = 7?
+
+
+
+
+ what number is generated as output by the function
+ f(x) = \sqrt{x^2 + 1} by the input x = -2?
+
+
+
+ Certainly we also study overall patterns as seen in functions and equations,
+ but this usually occurs through an examination of numbers themselves,
+ and we think of numbers as the main objects being acted upon.
+
+
+
+ This changes in calculus. In calculus, the fundamental objects being studied are
+ functions themselves. A function is a much more sophisticated mathematical object than a number,
+ in part because a function can be thought of in terms of its graph, which is an infinite collection of ordered pairs of the form (x,f(x)).
+
+
+
+ It is often helpful to look at a function's formula and observe algebraic structure. For instance, given the quadratic function
+
+ q(x) = -3x^2 + 5x - 7
+
+ we might benefit from thinking of this as the sum of three simpler functions: the constant function c(x) = -7, the linear function s(x) = 5x that passes through (0,0) with slope m = 5, and the concave down basic quadratic function w(x) = -3x^2. Indeed, each of the simpler functions c, s, and w contribute to making q be the function that it is. Likewise, if we were interested in the function p(x) = (3x^2 + 4)(9 - 2x^2), it might be natural to think about the two simpler functions f(x) = 3x^2 + 4 and g(x) = 9 - 2x^2 that are being multiplied to produce p.
+
+
+
+ We thus naturally arrive at the ideas of adding, subtracting, multiplying, or dividing two or more functions, and hence introduce the following definitions and notation.
+
+
+
+ function arithmeticsum of
+ function arithmeticdifference of
+ function arithmeticproduct of
+ function arithmeticquotient of
+
+
+ Let f and g be functions that share the same domain. Then,
+
+
+
+ The sum of f and g is the function f + g defined by (f+g)(x) = f(x) + g(x).
+
+
+
+
+ The difference of f and g is the function f - g defined by (f-g)(x) = f(x) - g(x).
+
+
+
+
+ The product of f and g is the function f \cdot g defined by (f \cdot g)(x) = f(x) \cdot g(x).
+
+
+
+
+ The quotient of f and g is the function \frac{f}{g} defined by \left( \frac{f}{g} \right)(x) = \frac{f(x)}{g(x)} for all x such that g(x) \ne 0.
+
+ When we work in applied settings with functions that model phenomena in the world around us,
+ it is often useful to think carefully about the units of various quantities.
+ Analyzing units can help us both understand the algebraic structure of functions and the variables involved,
+ as well as assist us in assigning meaning to quantities we compute.
+ We have already seen this with the notion of average rate of change:
+ if a function P(t) measures the population in a city in year t and we compute AV_{[5, 11]},
+ then the units on AV_{[5, 11]} are
+ people per year,
+ and the value of AV_{[5, 11]} is telling us the average rate at which the population changes in people per year on the time interval from year 5 to year 11.
+
+
+
+
+
+ Say that an investor is regularly purchasing stock in a particular company.This example is taken from Section 2.3 of Active Calculus.
+ Let N(t) represent the number of shares owned on day t,
+ where t = 0 represents the first day on which shares were purchased.
+ Let S(t) give the value of one share of the stock on day t;
+ note that the units on S(t) are dollars per share.
+ How is the total value, V(t), of the held stock on day t determined?
+
+
+
+ Solution. Observe that the units on N(t) are shares and the units on S(t) are dollars per share. Thus when we compute the product
+
+ N(t) \, \text{shares} \cdot S(t) \, \text{dollars per share}
+ ,
+ it follows that the resulting units are dollars, which is the total value of held stock. Hence,
+
+ V(t) = N(t) \cdot S(t)
+ .
+
+
+
+
+
+
+
+
+
+ Piecewise functions
+
+
+ In both abstract and applied settings,
+ we sometimes have to use different formulas on different intervals in order to define a function of interest.
+
+
+
+
+
+ A familiar and important function that is defined piecewise
+ is the absolute value function: A(x) = |x|.
+ We know that if x \ge 0, |x| = x, whereas if x \lt 0, |x| = -x.
+
+
+
+ absolute value function
+
+
+ The absolute value of a real number, denoted by A(x) = |x|, is defined by the rule
+
+ A(x) =
+ \begin{cases}
+ -x, \amp x \lt 0 \\
+ x, \amp x \ge 0
+ \end{cases}
+
+
+
+
+
+
+
+
A plot of the absolute value function, A(x) = |x|.
+
A plot of the absolute value function, A(x) = |x|.
+
+
+
+
+
+ The absolute value function is one example of a piecewise-defined function. The bracket notation in Definition is how we express which piece of the function applies on which interval. As we can see in Figure, for x values less than 0, the function y = -x applies, whereas for x greater than or equal to 0, the rule is determined by y = x.
+
+
+
+ As long as we are careful to make sure that each potential input has one and only one corresponding output, we can define a piecewise function using as many different functions on different intervals as we desire.
+
+
+
+
+
+
+
+ Summary
+
+
+
+
+ Just as we can generate a new number by adding, subtracting, multiplying, or dividing two given numbers, we can generate a new function by adding, subtracting, multiplying, or dividing two given functions. For instance, if we know formulas, graphs, or tables for functions f and g that share the same domain, we can create their product p according to the rule p(x) = (f \cdot g)(x) = f(x) \cdot g(x).
+
+
+
+
+ A piecewise function is a function whose formula consists of at least two different formulas in such a way that which formula applies depends on where the input falls in the domain. For example, given two functions f and g each defined on all real numbers, we can define a new piecewise function P according to the rule
+
+ P(x) =
+ \begin{cases}
+ f(x), \amp x \lt a \\
+ g(x), \amp x \ge a
+ \end{cases}
+
+ This tells us that for any x to the left of a, we use the rule for f, whereas for any x to the right of or equal to a, we use the rule for g. We can use as many different functions as we want on different intervals, provided the intervals don't overlap.
+
- How does the process of function composition produce a new function from two other functions?
-
-
-
-
- In the composite function h(x) = f(g(x)), what do we mean by the inner and outer function? What role do the domain and codomain of f and g play in determining the domain and codomain of h?
-
-
-
-
- How does the expression for AV_{[a,a+h]} involve a composite function?
-
-
-
-
-
- Introduction
-
- Recall that a function, by definition, is a process that takes a collection of inputs and produces a corresponding collection of outputs in such a way that the process produces one and only one output value for any single input value. Because every function is a process, it makes sense to think that it may be possible to take two function processes and do one of the processes first, and then apply the second process to the result.
-
-
-
-
-
- Suppose we know that y is a function of x according to the process defined by y = f(x) = x^2 - 1 and, in turn, x is a function of t via x = g(t) = 3t-4. Is it possible to combine these processes to generate a new function so that y is a function of t?
-
-
-
-
- Since y depends on x and x depends on t, it follows that we can also think of y depending directly on t. We can use substitution and the notation of functions to determine this relationship.
-
-
-
- First, it's important to realize what the rule for f tells us. In words, f says to generate the output that corresponds to an input, take the input and square it, and then subtract 1. In symbols, we might express f more generally by writing f(\Box) = \Box^2 - 1.
-
-
-
- Now, observing that y = f(x) = x^2 - 1 and that x = g(t) = 3t - 4, we can substitute the expression g(t) for x in f. Doing so,
-
- y &= f(x)
- &= f(g(t))
- &= f(3t-4)
- .
- Applying the process defined by the function f to the input 3t-4, we see that
-
- y = (3t-4)^2 - 1
- ,
- which defines y as a function of t.
-
-
-
-
-
-
- When we have a situation such as in Example where we use the output of one function as the input of another, we often say that we have composed two functions. In addition, we use the notation h(t) = f(g(t)) to denote that a new function, h, results from composing the two functions f and g.
-
-
-
-
-
-
-
- Composing two functions
-
- Whenever we have two functions, say g : A \to B and f : B \to C, where the codomain of g matches the domain of f, it is possible to link the two processes together to create a new process that we call the composition of f and g.
-
-
- composite function
-
-
- If f and g are functions such that g : A \to B and f : B \to C, we define the composition of f and g to be the new function h: A \to C given by
-
- h(t) = f(g(t))
- .
- We also sometimes use the notation h = f \circ g, where f \circ g is the single function defined by (f \circ g)(t) = f(g(t)).
-
-
-
-
-
- We sometimes call g the inner function and f the outer function. It is important to note that the inner function is actually the first function that gets applied to a given input, and then outer function is applied to the output of the inner function. In addition, in order for a composite function to make sense, we need to ensure that the range of the inner function lies within the domain of the outer function so that the resulting composite function is defined at every possible input.
-
-
-
- In addition to the possibility that functions are given by formulas,
- functions can be given by tables or graphs.
- We can think about composite functions in these settings as well,
- and the following activities prompt us to consider functions given in this way.
-
-
-
-
-
-
- Composing functions in context
-
- Recall Dolbear's function, T = D(N) = 40 + 0.25N, that relates the number of chirps per minute from a snowy cricket to the Fahrenheit temperature, T. We earlier established that D has a domain of [40,160] and a corresponding range of [50,85]. In what follows, we replace T with F to emphasize that temperature is measured in Fahrenheit degrees.
-
-
-
- The Celsius and Fahrenheit temperature scales are connected by a linear function. Indeed, the function that converts Fahrenheit to Celsius is
-
- C = G(F) = \frac{5}{9}(F-32)
- .
- For instance, a Fahrenheit temperature of 32 degrees corresponds to C = G(32) = 0 degrees Celsius.
-
-
-
-
-
-
-
- Function composition and average rate of change
-
-
- Recall that the average rate of change of a function f on the interval [a,b] is given by
-
- AV_{[a,b]} = \frac{f(b) - f(a)}{b-a}
- .
- In Figure, we see the familiar representation of AV_{[a,b]} as the slope of the line joining the points (a,f(a)) and (b,f(b)) on the graph of f. In the study of calculus, we progress from the average rate of change on an interval to the instantaneous rate of change of a function at a single value; the core idea that allows us to move from an average rate to an instantaneous one is letting the interval [a,b] shrink in size.
-
-
-
-
-
AV_{[a,b]} is the slope of the line joining the points (a,f(a)) and (b,f(b)) on the graph of f.
-
ADD ALT TEXT TO THIS IMAGEAV_{[a,b]} is the slope of the line joining the points (a,f(a)) and (b,f(b)) on the graph of f.
-
-
-
-
AV_{[a,a+h]} is the slope of the line joining the points (a,f(a)) and (a,f(a+h)) on the graph of f.
-
ADD ALT TEXT TO THIS IMAGEAV_{[a,a+h]} is the slope of the line joining the points (a,f(a)) and (a,f(a+h)) on the graph of f.
-
-
-
-
-
- To think about the interval [a,b] shrinking while a stays fixed, we often change our perspective and think of b as b = a + h, where h measures the horizontal difference from b to a. This allows us to eventually think about h getting closer and closer to 0 (without every actually equalling 0), and in that context we consider the equivalent expression
-
- AV_{[a,a+h]} = \frac{f(a+h) - f(a)}{a+h-a} = \frac{f(a+h) - f(a)}{h}
-
- for the average rate of change of f on [a,a+h].
-
-
-
- In this most recent expression for AV_{[a,a+h]},
- we see the important role that the composite function
- f(a+h) plays. In particular, to understand the expression for AV_{[a,a+h]} we need to evaluate f at the quantity (a+h).
-
-
-
-
-
- Suppose that f(x) = x^2. Determine the simplest possible expression you can find for AV_{[3,3+h]}, the average rate of change of f on the interval [3,3+h].
-
-
-
-
- By definition, we know that
-
- AV_{[3,3+h]} = \frac{f(3+h)-f(3)}{h}.
-
- Using the formula for f, we see that
-
- AV_{[3,3+h]} = \frac{(3+h)^2-(3)^2}{h}.
-
- Expanding the numerator and combining like terms, it follows that
-
- AV_{[3,3+h]} &= \frac{(9+6h+h^2)-9}{h}
- &= \frac{6h + h^2}{h}
- .
- Removing a factor of h in the numerator and observing that h \ne 0, we can simplify and find that
-
- AV_{[3,3+h]} &= \frac{h(6 + h)}{h}
- &= 6+h
- .
- Hence, AV_{[3,3+h]} = 6+h, which is the average rate of change of f(x) = x^2 on the interval [3,3+h].Note that 6 + h is a linear function of h. This computation is connected to the observation we made in Table regarding how there's a linear aspect to how the average rate of change of a quadratic function changes as we modify the interval.
-
-
-
-
-
-
-
- In Activity, we see an important setting where algebraic simplification plays a crucial role in calculus. Because the expresssion
-
- AV_{[a,a+h]} = \frac{f(a+h) - f(a)}{h}
-
- always begins with an h in the denominator, in order to precisely understand how this quantity behaves when h gets close to 0, a simplified version of this expression is needed. For instance, as we found in part (b) of Activity, it's possible to show that for f(x) = 2x^2 - 3x + 1,
-
- AV_{[1,1+h]} = 2h + 1
- ,
- which is a much simpler expression to investigate.
-
-
-
-
-
- Summary
-
-
-
-
- When defined, the composition of two functions f and g produces a single new function f \circ g according to the rule (f \circ g)(x) = f(g(x)). We note that g is applied first to the input x, and then f is applied to the output g(x) that results from g.
-
-
-
-
- In the composite function h(x) = f(g(x)), the inner function is g and the outer function is f. Note that the inner function gets applied to x first, even though the outer function appears first when we read from left to right. The composite function is only defined provided that the codomain of g matches the domain of f: that is, we need any possible outputs of g to be among the allowed inputs for f. In particular, we can say that if g : A \to B and f : B \to C, then f \circ g : A \to C. Thus, the domain of the composite function is the domain of the inner function, and the codomain of the composite function is the codomain of the outer function.
-
-
-
-
- Because the expression AV_{[a,a+h]} is defined by
-
- AV_{[a,a+h]} = \frac{f(a+h) - f(a)}{h}
-
- and this includes the quantity f(a+h), the average rate of change of a function on the interval [a,a+h] always involves the evaluation of a composite function expression. This idea plays a crucial role in the study of calculus.
-
+ How does the process of function composition produce a new function from two other functions?
+
+
+
+
+ In the composite function h(x) = f(g(x)), what do we mean by the inner and outer function? What role do the domain and codomain of f and g play in determining the domain and codomain of h?
+
+
+
+
+ How does the expression for AV_{[a,a+h]} involve a composite function?
+
+
+
+
+
+ Introduction
+
+ Recall that a function, by definition, is a process that takes a collection of inputs and produces a corresponding collection of outputs in such a way that the process produces one and only one output value for any single input value. Because every function is a process, it makes sense to think that it may be possible to take two function processes and do one of the processes first, and then apply the second process to the result.
+
+
+
+
+
+ Suppose we know that y is a function of x according to the process defined by y = f(x) = x^2 - 1 and, in turn, x is a function of t via x = g(t) = 3t-4. Is it possible to combine these processes to generate a new function so that y is a function of t?
+
+
+
+
+ Since y depends on x and x depends on t, it follows that we can also think of y depending directly on t. We can use substitution and the notation of functions to determine this relationship.
+
+
+
+ First, it's important to realize what the rule for f tells us. In words, f says to generate the output that corresponds to an input, take the input and square it, and then subtract 1. In symbols, we might express f more generally by writing f(\Box) = \Box^2 - 1.
+
+
+
+ Now, observing that y = f(x) = x^2 - 1 and that x = g(t) = 3t - 4, we can substitute the expression g(t) for x in f. Doing so,
+
+ y &= f(x)
+ &= f(g(t))
+ &= f(3t-4)
+ .
+ Applying the process defined by the function f to the input 3t-4, we see that
+
+ y = (3t-4)^2 - 1
+ ,
+ which defines y as a function of t.
+
+
+
+
+
+
+ When we have a situation such as in Example where we use the output of one function as the input of another, we often say that we have composed two functions. In addition, we use the notation h(t) = f(g(t)) to denote that a new function, h, results from composing the two functions f and g.
+
+
+
+
+
+
+
+
+ Composing two functions
+
+ Whenever we have two functions, say g : A \to B and f : B \to C, where the codomain of g matches the domain of f, it is possible to link the two processes together to create a new process that we call the composition of f and g.
+
+
+ composite function
+
+
+ If f and g are functions such that g : A \to B and f : B \to C, we define the composition of f and g to be the new function h: A \to C given by
+
+ h(t) = f(g(t))
+ .
+ We also sometimes use the notation h = f \circ g, where f \circ g is the single function defined by (f \circ g)(t) = f(g(t)).
+
+
+
+
+
+ We sometimes call g the inner function and f the outer function. It is important to note that the inner function is actually the first function that gets applied to a given input, and then outer function is applied to the output of the inner function. In addition, in order for a composite function to make sense, we need to ensure that the range of the inner function lies within the domain of the outer function so that the resulting composite function is defined at every possible input.
+
+
+
+ In addition to the possibility that functions are given by formulas,
+ functions can be given by tables or graphs.
+ We can think about composite functions in these settings as well,
+ and the following activities prompt us to consider functions given in this way.
+
+
+
+
+
+
+ Composing functions in context
+
+ Recall Dolbear's function, T = D(N) = 40 + 0.25N, that relates the number of chirps per minute from a snowy cricket to the Fahrenheit temperature, T. We earlier established that D has a domain of [40,160] and a corresponding range of [50,85]. In what follows, we replace T with F to emphasize that temperature is measured in Fahrenheit degrees.
+
+
+
+ The Celsius and Fahrenheit temperature scales are connected by a linear function. Indeed, the function that converts Fahrenheit to Celsius is
+
+ C = G(F) = \frac{5}{9}(F-32)
+ .
+ For instance, a Fahrenheit temperature of 32 degrees corresponds to C = G(32) = 0 degrees Celsius.
+
+
+
+
+
+
+
+ Function composition and average rate of change
+
+
+ Recall that the average rate of change of a function f on the interval [a,b] is given by
+
+ AV_{[a,b]} = \frac{f(b) - f(a)}{b-a}
+ .
+ In Figure, we see the familiar representation of AV_{[a,b]} as the slope of the line joining the points (a,f(a)) and (b,f(b)) on the graph of f. In the study of calculus, we progress from the average rate of change on an interval to the instantaneous rate of change of a function at a single value; the core idea that allows us to move from an average rate to an instantaneous one is letting the interval [a,b] shrink in size.
+
+
+
+
+
AV_{[a,b]} is the slope of the line joining the points (a,f(a)) and (b,f(b)) on the graph of f.
+
ADD ALT TEXT TO THIS IMAGEAV_{[a,b]} is the slope of the line joining the points (a,f(a)) and (b,f(b)) on the graph of f.
+
+
+
+
AV_{[a,a+h]} is the slope of the line joining the points (a,f(a)) and (a,f(a+h)) on the graph of f.
+
ADD ALT TEXT TO THIS IMAGEAV_{[a,a+h]} is the slope of the line joining the points (a,f(a)) and (a,f(a+h)) on the graph of f.
+
+
+
+
+
+ To think about the interval [a,b] shrinking while a stays fixed, we often change our perspective and think of b as b = a + h, where h measures the horizontal difference from b to a. This allows us to eventually think about h getting closer and closer to 0 (without every actually equalling 0), and in that context we consider the equivalent expression
+
+ AV_{[a,a+h]} = \frac{f(a+h) - f(a)}{a+h-a} = \frac{f(a+h) - f(a)}{h}
+
+ for the average rate of change of f on [a,a+h].
+
+
+
+ In this most recent expression for AV_{[a,a+h]},
+ we see the important role that the composite function
+ f(a+h) plays. In particular, to understand the expression for AV_{[a,a+h]} we need to evaluate f at the quantity (a+h).
+
+
+
+
+
+ Suppose that f(x) = x^2. Determine the simplest possible expression you can find for AV_{[3,3+h]}, the average rate of change of f on the interval [3,3+h].
+
+
+
+
+ By definition, we know that
+
+ AV_{[3,3+h]} = \frac{f(3+h)-f(3)}{h}.
+
+ Using the formula for f, we see that
+
+ AV_{[3,3+h]} = \frac{(3+h)^2-(3)^2}{h}.
+
+ Expanding the numerator and combining like terms, it follows that
+
+ AV_{[3,3+h]} &= \frac{(9+6h+h^2)-9}{h}
+ &= \frac{6h + h^2}{h}
+ .
+ Removing a factor of h in the numerator and observing that h \ne 0, we can simplify and find that
+
+ AV_{[3,3+h]} &= \frac{h(6 + h)}{h}
+ &= 6+h
+ .
+ Hence, AV_{[3,3+h]} = 6+h, which is the average rate of change of f(x) = x^2 on the interval [3,3+h].Note that 6 + h is a linear function of h. This computation is connected to the observation we made in Table regarding how there's a linear aspect to how the average rate of change of a quadratic function changes as we modify the interval.
+
+
+
+
+
+
+
+ In Activity, we see an important setting where algebraic simplification plays a crucial role in calculus. Because the expresssion
+
+ AV_{[a,a+h]} = \frac{f(a+h) - f(a)}{h}
+
+ always begins with an h in the denominator, in order to precisely understand how this quantity behaves when h gets close to 0, a simplified version of this expression is needed. For instance, as we found in part (b) of Activity, it's possible to show that for f(x) = 2x^2 - 3x + 1,
+
+ AV_{[1,1+h]} = 2h + 1
+ ,
+ which is a much simpler expression to investigate.
+
+
+
+
+
+ Summary
+
+
+
+
+ When defined, the composition of two functions f and g produces a single new function f \circ g according to the rule (f \circ g)(x) = f(g(x)). We note that g is applied first to the input x, and then f is applied to the output g(x) that results from g.
+
+
+
+
+ In the composite function h(x) = f(g(x)), the inner function is g and the outer function is f. Note that the inner function gets applied to x first, even though the outer function appears first when we read from left to right. The composite function is only defined provided that the codomain of g matches the domain of f: that is, we need any possible outputs of g to be among the allowed inputs for f. In particular, we can say that if g : A \to B and f : B \to C, then f \circ g : A \to C. Thus, the domain of the composite function is the domain of the inner function, and the codomain of the composite function is the codomain of the outer function.
+
+
+
+
+ Because the expression AV_{[a,a+h]} is defined by
+
+ AV_{[a,a+h]} = \frac{f(a+h) - f(a)}{h}
+
+ and this includes the quantity f(a+h), the average rate of change of a function on the interval [a,a+h] always involves the evaluation of a composite function expression. This idea plays a crucial role in the study of calculus.
+
- How can we use the mathematical idea of a function to represent the relationship between two changing quantities?
-
-
-
-
- What are some formal characteristics of an abstract mathematical function? how do we think differently about these characteristics in the context of a physical model?
-
-
-
-
-
- Introduction
-
- A mathematical model is an abstract concept through which we use mathematical language and notation to describe a phenomenon in the world around us. One example of a mathematical model is found in Dolbear's LawYou can read more in the Wikipedia entry for Dolbear's Law, which has proven to be remarkably accurate for the behavior of snowy tree crickets. For even more of the story, including a reference to this phenomenon on the popular show The Big Bang Theory, see this article.. Dolbear's Law In the late 1800s, the physicist Amos Dolbear was listening to crickets chirp and noticed a pattern: how frequently the crickets chirped seemed to be connected to the outside temperature. If we let T represent the temperature in degrees Fahrenheit and N the number of chirps per minute, we can summarize Dolbear's observations in the following table.
-
-
-
- Data for Dolbear's observations.
-
-
- N (chirps per minute)
- 40
- 80
- 120
- 160
-
-
- T (^\circ Fahrenheit)
- 50^\circ
- 60^\circ
- 70^\circ
- 80^\circ
-
-
-
-
- For a mathematical model, we often seek an algebraic formula that captures observed behavior accurately and can be used to predict behavior not yet observed. For the data in Table, we observe that each of the ordered pairs in the table make the equation
-
- T = 40 + 0.25N
-
- true. For instance, 70 = 40 + 0.25(120). Indeed, scientists who made many additional cricket chirp observations following Dolbear's initial counts found that the formula in Equation holds with remarkable accuracy for the snowy tree cricket in temperatures ranging from about 50^\circ F to 85^\circ F.
-
-
-
-
-
-
-
- Functions
-
-
- The mathematical concept of a functionfunctionintroduction to is one of the most central ideas in all of mathematics, in part since functions provide an important tool for representing and explaining patterns. At its core, a function is a repeatable process that takes a collection of input values and generates a corresponding collection of output values with the property that if we use a particular single input, the process always produces exactly the same single output.
-
-
-
- For instance, Dolbear's Law in Equation provides a process that takes a given number of chirps between 40 and 180 per minute and reliably produces the corresponding temperature that corresponds to the number of chirps, and thus this equation generates a function. We often give functions shorthand names; using D for the Dolbear function, we can represent the process of taking inputs (observed chirp rates) to outputs (corresponding temperatures) using arrows:
-
-
- 80 &\xrightarrow{D} 60
-
-
- 120 &\xrightarrow{D} 70
-
-
- N &\xrightarrow{D} 40 + 0.25 N
-
-
- Alternatively, for the relationship 80 \xrightarrow{D} 60 we can also use the equivalent notation D(80) = 60 to indicate that Dolbear's Law takes an input of 80 chirps per minute and produces a corresponding output of 60 degrees Fahrenheit. More generally, we write T = D(N) = 40 + 0.25N to indicate that a certain temperature, T, is determined by a given number of chirps per minute, N, according to the process D(N) = 40 + 0.25N.
-
-
-
- Tables and graphs are particularly valuable ways to characterize and represent functions. For the current example, we summarize some of the data the Dolbear function generates in Table and plot that data along with the underlying curve in Figure.
-
-
-
-
- Data for the function <m>T = D(N) = 40 + 0.25N</m>.
-
-
- N
- T
-
-
- 40
- 50
-
-
- 80
- 60
-
-
- 120
- 70
-
-
- 160
- 80
-
-
- 180
- 85
-
-
-
-
-
Graph of data from the function T = D(N) = 40 + 0.25N and the underlying curve.
-
Graph of data from the function T = D(N) = 40 + 0.25N and the underlying curve.
-
-
-
-
-
- When a point such as (120,70) in Figure lies on a function's graph, this indicates the correspondence between input and output: when the value 120 chirps per minute is entered in the function D, the result is 70 degrees Fahrenheit. More concisely, D(120) = 70. Aloud, we read D of 120 is 70.
-
-
-
- For most important concepts in mathematics, the mathematical community decides on formal definitions to ensure that we have a shared language of understanding. In this text, we will use the following definition of the term function.
-
-
-
- functiondefinition
-
-
- A function is a process that may be applied to a collection of input values to produce a corresponding collection of output values in such a way that the process produces one and only one output value for any single input value.
-
-
-
-
-
- If we name a given function F and call the collection of possible inputs to F the set A and the corresponding collection of potential outputs B, we say F is a function from A to B, and sometimes write F : A \to B. When a particular input value to F, say t, produces a corresponding output z, we write F(t) = z and read this symbolic notation as F of t is z.functionnotation We often call t the independent variablefunctionindependent variable and z the dependent variablefunctiondependent variable, since z is a function of t.
-
-
-
- functiondomainfunctioncodomain
-
-
- Let F be a function from A to B. The set A of possible inputs to F is called the domain of F; the set B of potential outputs from F is called the codomain of F.
-
-
-
-
-
- For the Dolbear function D(N) = 40 + 0.25N in the context of modeling temperature as a function of the number of cricket chirps per minute, the domain of the function is A = [40,180]The notation [40,180] means the collection of all real numbers x that satisfy 40 \le x \le 80 and is sometimes called interval notation. and the codomain is all Fahrenheit temperatures. The codomain of a function is the collection of possible outputs, which we distinguish from the collection of actual ouputs.
-
-
-
- functionrange
-
-
- Let F be a function from A to B. The range of F is the collection of all actual outputs of the function. That is, the range is the collection of all elements y in B for which it is possible to find an element x in A such that F(x) = y.
-
-
-
-
-
- In many situations, the range of a function is much more challenging to determine than its codomain. For the Dolbear function, the range is straightforward to find by using the graph shown in Figure: since the actual outputs of D fall between T = 50 and T = 85 and include every value in that interval, the range of D is [50,85].
-
-
-
- The range of any function is always a subset of the codomain. It is possible for the range to equal the codomain.
-
- Again, a mathematical model is an abstract concept through which we use mathematical language and notation to describe a phenomenon in the world around us. So far, we have considered two different examples: the Dolbear function, T = D(N) = 40 + 0.25N, that models how Fahrenheit temperature is a function of the number of cricket chirps per minute and the function V = f(h) = \frac{\pi}{3}h^2(12-h) that models how the volume of water in a spherical tank of radius 4 m is a function of the depth of the water in the tank. While often we consider a function in the physical setting of some model, there are also many occasions where we consider an abstract function for its own sake in order to study and understand it.
-
-
-
- A parabola and a falling ball
-
-
- Calculus shows that for a tennis ball tossed vertically from a window 48 feet above the ground at an initial vertical velocity of 32 feet per second, the ball's height above the ground at time t (where t = 0 is the instant the ball is tossed) can be modeled by the function h = g(t) = -16t^2 + 32t + 48. Discuss the differences between the model g and the abstract function f determined by y = f(x) = -16x^2 + 32x + 48.
-
-
-
-
- We start with the abstract function y = f(x) = -16x^2 + 32x + 48. Absent a physical context, we can investigate the behavior of this function by computing function values, plotting points, and thinking about its overall behavior. We recognize the function f as quadraticWe will engage in a brief review of quadratic functions in Section, noting that it opens down because of the leading coefficient of -16, with vertex located at x = \frac{-32}{2(-16)} = 1, y-intercept at (0,48), and with x-intercepts at (-1,0) and (3,0) because
-
- -16x^2 + 32x + 48 = -16(x^2 - 2x - 3) = -16(x-3)(x+1)
- . Computing some additional points to gain more information, we see both the data in Table and the corresponding graph in Figure.
-
Graph of the function y = f(x) and some data from the table.
-
Graph of the function y = f(x) and some data from the table.
-
-
-
-
-
- For this abstract function, its domain is all real numbers since we may input any real number x we wish into the formula f(x) = -16x^2 + 32x + 48 and have the result be defined. Moreover, taking a real number x and processing it in the formula f(x) = -16x^2 + 32x + 48 will produce another real number. This tells us that the codomain of the abstract function f is also all real numbers. Finally, from the graph and the data, we observe that the largest possible output of the function f is y = 64. It is apparent that we can generate any y-value less than or equal to 64, and thus the range of the abstract function f is all real numbers less than or equal to 64. We denote this collection of real numbers using the shorthand interval notation (-\infty, 64].The notation (-\infty,64] stands for all the real numbers that lie to the left of and including 64. The -\infty indicates that there is no left-hand bound on the interval.
-
-
-
- Next, we turn our attention to the model h = g(t) = -16t^2 + 32t + 48 that represents the height of the ball, h, in feet t seconds after the ball in initially launched. Here, the big difference is the domain, codomain, and range associated with the model. Since the model takes effect once the ball is tossed, it only makes sense to consider the model for input values t \ge 0. Moreover, because the model ceases to apply once the ball lands, it is only valid for t \le 3. Thus, the domain of g is [0,3]. For the codomain, it only makes sense to consider values of h that are nonnegative. That is, as we think of potential outputs for the model, then can only be in the interval [0, \infty). Finally, we can consider the graph of the model on the given domain in Figure and see that the range of the model is [0,64], the collection of all heights between its lowest (ground level) and its largest (at the vertex).
-
-
-
-
- Data for the model <m>h = g(t) = -16t^2 + 32t + 48</m>.
-
-
- t
- g(t)
-
-
- 0
- 48
-
-
- 1
- 64
-
-
- 2
- 48
-
-
- 3
- 0
-
-
-
-
-
Graph of the model h = g(t) and some data from the table.
-
Graph of the model h = g(t) and some data from the table.
-
-
-
-
-
-
-
-
-
-
-
-
- Determining whether a relationship is a function or not
-
- To this point in our discussion of functions, we have mostly focused on what the function process may model and what the domain, codomain, and range of a model or abstract function are. It is also important to take note of another part of Definition: \ldots the process produces one and only one output value for any single input value. Said differently, if a relationship or process ever associates a single input with two or more different outputs, the process cannot be a function.
-
-
-
-
-
- Is the relationship between people and phone numbers a function?
-
-
-
- Solution. No, this relationship is not a function. A given individual person can be associated with more than one phone number, such as their cell phone and their work telephone. This means that we can't view phone numbers as a function of people: one input (a person) can lead to two different outputs (phone numbers). We also can't view people as a function of phone numbers, since more than one person can be associated with a phone number, such as when a family shares a single phone at home.
-
-
-
-
-
-
-
- The relationship between x and y that is given in the following table where we attempt to view y as depending on x.
-
-
-
- A table that relates <m>x</m> and <m>y</m> values.
-
-
- x
- 1
- 2
- 3
- 4
- 5
-
-
- y
- 13
- 11
- 10
- 11
- 13
-
-
-
-
-
- Solution. The relationship between y and x in Table allows us to think of y as a function of x since each particular input is associated with one and only one output. If we name the function f, we can say for instance that f(4) = 11. Moreover, the domain of f is the set of inputs \{1,2,3,4,5\}, and the codomain (which is also the range) is the set of outputs \{10,11,13\}.
-
-
-
-
-
-
-
- For a relationship or process to be a function, each individual input must be associated with one and only one output. Thus, the usual way that we demonstrate a relationship or process is not a function is to find a particular input that is associated with two or more outputs. When the relationship is given graphically, such as in the left graph in , we can use the vertical line test to determine whether or not the graph represents a function.
-
-
-
- Vertical Line Test
- vertical line test
-
- A graph in the plane represents a function if and only if every vertical line intersects the graph at most once. When the graph passes this test, the vertical coordinate of each point on the graph can be viewed as a function of the horizontal coordinate of the point.
-
-
-
-
- Since the vertical line x = -3 passes through the circle in the left graph in at both y = -\sqrt{7} and y = \sqrt{7}, the circle does not represent a relationship where y is a function of x. However, since any vertical line we draw in the right graph in intersects the blue curve at most one time, the graph indeed represents a function.
-
-
-
- We conclude with a formal definition of the graph of a function.
-
-
-
- functiongraph
-
-
- Let F : A \to B, where A and B are each collections of real numbers. The graph of F is the collection of all ordered pairs (x,y) that satisfy y = F(x).
-
-
-
-
-
- When we use a computing device such as Desmos to graph a function g, the program is generating a large collection of ordered pairs (x,g(x)), plotting them in the x-y plane, and connecting the points with short line segments.
-
-
-
-
-
- Summary
-
-
-
-
- A function is a process that generates a relationship between two collections of quantities. The function associates each member of a collection of input values with one and only one member of the collection of output values. A function can be described or defined by words, by a table of values, by a graph, or by a formula.
-
-
-
-
- Functions may be viewed as mathematical objects worthy of study for their own sake and also as models that represent physical phenomena in the world around us. Every function or model has a domain (the set of possible or allowable input values), a codomain (the set of possible output values), and a range (the set of all actual output values). Both the codomain and range depend on the domain. For an abstract function, the domain is usually viewed as the largest reasonable collection of input values; for a function that models a physical phenomenon, the domain is usually determined by the context of possibilities for the input in the phenomenon being considered.
-
+ How can we use the mathematical idea of a function to represent the relationship between two changing quantities?
+
+
+
+
+ What are some formal characteristics of an abstract mathematical function? how do we think differently about these characteristics in the context of a physical model?
+
+
+
+
+
+ Introduction
+
+ A mathematical model is an abstract concept through which we use mathematical language and notation to describe a phenomenon in the world around us. One example of a mathematical model is found in Dolbear's LawYou can read more in the Wikipedia entry for Dolbear's Law, which has proven to be remarkably accurate for the behavior of snowy tree crickets. For even more of the story, including a reference to this phenomenon on the popular show The Big Bang Theory, see this article.. Dolbear's Law In the late 1800s, the physicist Amos Dolbear was listening to crickets chirp and noticed a pattern: how frequently the crickets chirped seemed to be connected to the outside temperature. If we let T represent the temperature in degrees Fahrenheit and N the number of chirps per minute, we can summarize Dolbear's observations in the following table.
+
+
+
+ Data for Dolbear's observations.
+
+
+ N (chirps per minute)
+ 40
+ 80
+ 120
+ 160
+
+
+ T (^\circ Fahrenheit)
+ 50^\circ
+ 60^\circ
+ 70^\circ
+ 80^\circ
+
+
+
+
+ For a mathematical model, we often seek an algebraic formula that captures observed behavior accurately and can be used to predict behavior not yet observed. For the data in Table, we observe that each of the ordered pairs in the table make the equation
+
+ T = 40 + 0.25N
+
+ true. For instance, 70 = 40 + 0.25(120). Indeed, scientists who made many additional cricket chirp observations following Dolbear's initial counts found that the formula in Equation holds with remarkable accuracy for the snowy tree cricket in temperatures ranging from about 50^\circ F to 85^\circ F.
+
+
+
+
+
+
+
+
+ Functions
+
+
+ The mathematical concept of a functionfunctionintroduction to is one of the most central ideas in all of mathematics, in part since functions provide an important tool for representing and explaining patterns. At its core, a function is a repeatable process that takes a collection of input values and generates a corresponding collection of output values with the property that if we use a particular single input, the process always produces exactly the same single output.
+
+
+
+ For instance, Dolbear's Law in Equation provides a process that takes a given number of chirps between 40 and 180 per minute and reliably produces the corresponding temperature that corresponds to the number of chirps, and thus this equation generates a function. We often give functions shorthand names; using D for the Dolbear function, we can represent the process of taking inputs (observed chirp rates) to outputs (corresponding temperatures) using arrows:
+
+
+ 80 &\xrightarrow{D} 60
+
+
+ 120 &\xrightarrow{D} 70
+
+
+ N &\xrightarrow{D} 40 + 0.25 N
+
+
+ Alternatively, for the relationship 80 \xrightarrow{D} 60 we can also use the equivalent notation D(80) = 60 to indicate that Dolbear's Law takes an input of 80 chirps per minute and produces a corresponding output of 60 degrees Fahrenheit. More generally, we write T = D(N) = 40 + 0.25N to indicate that a certain temperature, T, is determined by a given number of chirps per minute, N, according to the process D(N) = 40 + 0.25N.
+
+
+
+ Tables and graphs are particularly valuable ways to characterize and represent functions. For the current example, we summarize some of the data the Dolbear function generates in Table and plot that data along with the underlying curve in Figure.
+
+
+
+
+ Data for the function <m>T = D(N) = 40 + 0.25N</m>.
+
+
+ N
+ T
+
+
+ 40
+ 50
+
+
+ 80
+ 60
+
+
+ 120
+ 70
+
+
+ 160
+ 80
+
+
+ 180
+ 85
+
+
+
+
+
Graph of data from the function T = D(N) = 40 + 0.25N and the underlying curve.
+
Graph of data from the function T = D(N) = 40 + 0.25N and the underlying curve.
+
+
+
+
+
+ When a point such as (120,70) in Figure lies on a function's graph, this indicates the correspondence between input and output: when the value 120 chirps per minute is entered in the function D, the result is 70 degrees Fahrenheit. More concisely, D(120) = 70. Aloud, we read D of 120 is 70.
+
+
+
+ For most important concepts in mathematics, the mathematical community decides on formal definitions to ensure that we have a shared language of understanding. In this text, we will use the following definition of the term function.
+
+
+
+ functiondefinition
+
+
+ A function is a process that may be applied to a collection of input values to produce a corresponding collection of output values in such a way that the process produces one and only one output value for any single input value.
+
+
+
+
+
+ If we name a given function F and call the collection of possible inputs to F the set A and the corresponding collection of potential outputs B, we say F is a function from A to B, and sometimes write F : A \to B. When a particular input value to F, say t, produces a corresponding output z, we write F(t) = z and read this symbolic notation as F of t is z.functionnotation We often call t the independent variablefunctionindependent variable and z the dependent variablefunctiondependent variable, since z is a function of t.
+
+
+
+ functiondomainfunctioncodomain
+
+
+ Let F be a function from A to B. The set A of possible inputs to F is called the domain of F; the set B of potential outputs from F is called the codomain of F.
+
+
+
+
+
+ For the Dolbear function D(N) = 40 + 0.25N in the context of modeling temperature as a function of the number of cricket chirps per minute, the domain of the function is A = [40,180]The notation [40,180] means the collection of all real numbers x that satisfy 40 \le x \le 80 and is sometimes called interval notation. and the codomain is all Fahrenheit temperatures. The codomain of a function is the collection of possible outputs, which we distinguish from the collection of actual ouputs.
+
+
+
+ functionrange
+
+
+ Let F be a function from A to B. The range of F is the collection of all actual outputs of the function. That is, the range is the collection of all elements y in B for which it is possible to find an element x in A such that F(x) = y.
+
+
+
+
+
+ In many situations, the range of a function is much more challenging to determine than its codomain. For the Dolbear function, the range is straightforward to find by using the graph shown in Figure: since the actual outputs of D fall between T = 50 and T = 85 and include every value in that interval, the range of D is [50,85].
+
+
+
+ The range of any function is always a subset of the codomain. It is possible for the range to equal the codomain.
+
+ Again, a mathematical model is an abstract concept through which we use mathematical language and notation to describe a phenomenon in the world around us. So far, we have considered two different examples: the Dolbear function, T = D(N) = 40 + 0.25N, that models how Fahrenheit temperature is a function of the number of cricket chirps per minute and the function V = f(h) = \frac{\pi}{3}h^2(12-h) that models how the volume of water in a spherical tank of radius 4 m is a function of the depth of the water in the tank. While often we consider a function in the physical setting of some model, there are also many occasions where we consider an abstract function for its own sake in order to study and understand it.
+
+
+
+ A parabola and a falling ball
+
+
+ Calculus shows that for a tennis ball tossed vertically from a window 48 feet above the ground at an initial vertical velocity of 32 feet per second, the ball's height above the ground at time t (where t = 0 is the instant the ball is tossed) can be modeled by the function h = g(t) = -16t^2 + 32t + 48. Discuss the differences between the model g and the abstract function f determined by y = f(x) = -16x^2 + 32x + 48.
+
+
+
+
+ We start with the abstract function y = f(x) = -16x^2 + 32x + 48. Absent a physical context, we can investigate the behavior of this function by computing function values, plotting points, and thinking about its overall behavior. We recognize the function f as quadraticWe will engage in a brief review of quadratic functions in Section, noting that it opens down because of the leading coefficient of -16, with vertex located at x = \frac{-32}{2(-16)} = 1, y-intercept at (0,48), and with x-intercepts at (-1,0) and (3,0) because
+
+ -16x^2 + 32x + 48 = -16(x^2 - 2x - 3) = -16(x-3)(x+1)
+ . Computing some additional points to gain more information, we see both the data in Table and the corresponding graph in Figure.
+
Graph of the function y = f(x) and some data from the table.
+
Graph of the function y = f(x) and some data from the table.
+
+
+
+
+
+ For this abstract function, its domain is all real numbers since we may input any real number x we wish into the formula f(x) = -16x^2 + 32x + 48 and have the result be defined. Moreover, taking a real number x and processing it in the formula f(x) = -16x^2 + 32x + 48 will produce another real number. This tells us that the codomain of the abstract function f is also all real numbers. Finally, from the graph and the data, we observe that the largest possible output of the function f is y = 64. It is apparent that we can generate any y-value less than or equal to 64, and thus the range of the abstract function f is all real numbers less than or equal to 64. We denote this collection of real numbers using the shorthand interval notation (-\infty, 64].The notation (-\infty,64] stands for all the real numbers that lie to the left of and including 64. The -\infty indicates that there is no left-hand bound on the interval.
+
+
+
+ Next, we turn our attention to the model h = g(t) = -16t^2 + 32t + 48 that represents the height of the ball, h, in feet t seconds after the ball in initially launched. Here, the big difference is the domain, codomain, and range associated with the model. Since the model takes effect once the ball is tossed, it only makes sense to consider the model for input values t \ge 0. Moreover, because the model ceases to apply once the ball lands, it is only valid for t \le 3. Thus, the domain of g is [0,3]. For the codomain, it only makes sense to consider values of h that are nonnegative. That is, as we think of potential outputs for the model, then can only be in the interval [0, \infty). Finally, we can consider the graph of the model on the given domain in Figure and see that the range of the model is [0,64], the collection of all heights between its lowest (ground level) and its largest (at the vertex).
+
+
+
+
+ Data for the model <m>h = g(t) = -16t^2 + 32t + 48</m>.
+
+
+ t
+ g(t)
+
+
+ 0
+ 48
+
+
+ 1
+ 64
+
+
+ 2
+ 48
+
+
+ 3
+ 0
+
+
+
+
+
Graph of the model h = g(t) and some data from the table.
+
Graph of the model h = g(t) and some data from the table.
+
+
+
+
+
+
+
+
+
+
+
+
+ Determining whether a relationship is a function or not
+
+ To this point in our discussion of functions, we have mostly focused on what the function process may model and what the domain, codomain, and range of a model or abstract function are. It is also important to take note of another part of Definition: \ldots the process produces one and only one output value for any single input value. Said differently, if a relationship or process ever associates a single input with two or more different outputs, the process cannot be a function.
+
+
+
+
+
+ Is the relationship between people and phone numbers a function?
+
+
+
+ Solution. No, this relationship is not a function. A given individual person can be associated with more than one phone number, such as their cell phone and their work telephone. This means that we can't view phone numbers as a function of people: one input (a person) can lead to two different outputs (phone numbers). We also can't view people as a function of phone numbers, since more than one person can be associated with a phone number, such as when a family shares a single phone at home.
+
+
+
+
+
+
+
+ The relationship between x and y that is given in the following table where we attempt to view y as depending on x.
+
+
+
+ A table that relates <m>x</m> and <m>y</m> values.
+
+
+ x
+ 1
+ 2
+ 3
+ 4
+ 5
+
+
+ y
+ 13
+ 11
+ 10
+ 11
+ 13
+
+
+
+
+
+ Solution. The relationship between y and x in Table allows us to think of y as a function of x since each particular input is associated with one and only one output. If we name the function f, we can say for instance that f(4) = 11. Moreover, the domain of f is the set of inputs \{1,2,3,4,5\}, and the codomain (which is also the range) is the set of outputs \{10,11,13\}.
+
+
+
+
+
+
+
+ For a relationship or process to be a function, each individual input must be associated with one and only one output. Thus, the usual way that we demonstrate a relationship or process is not a function is to find a particular input that is associated with two or more outputs. When the relationship is given graphically, such as in the left graph in , we can use the vertical line test to determine whether or not the graph represents a function.
+
+
+
+ Vertical Line Test
+ vertical line test
+
+ A graph in the plane represents a function if and only if every vertical line intersects the graph at most once. When the graph passes this test, the vertical coordinate of each point on the graph can be viewed as a function of the horizontal coordinate of the point.
+
+
+
+
+ Since the vertical line x = -3 passes through the circle in the left graph in at both y = -\sqrt{7} and y = \sqrt{7}, the circle does not represent a relationship where y is a function of x. However, since any vertical line we draw in the right graph in intersects the blue curve at most one time, the graph indeed represents a function.
+
+
+
+ We conclude with a formal definition of the graph of a function.
+
+
+
+ functiongraph
+
+
+ Let F : A \to B, where A and B are each collections of real numbers. The graph of F is the collection of all ordered pairs (x,y) that satisfy y = F(x).
+
+
+
+
+
+ When we use a computing device such as Desmos to graph a function g, the program is generating a large collection of ordered pairs (x,g(x)), plotting them in the x-y plane, and connecting the points with short line segments.
+
+
+
+
+
+ Summary
+
+
+
+
+ A function is a process that generates a relationship between two collections of quantities. The function associates each member of a collection of input values with one and only one member of the collection of output values. A function can be described or defined by words, by a table of values, by a graph, or by a formula.
+
+
+
+
+ Functions may be viewed as mathematical objects worthy of study for their own sake and also as models that represent physical phenomena in the world around us. Every function or model has a domain (the set of possible or allowable input values), a codomain (the set of possible output values), and a range (the set of all actual output values). Both the codomain and range depend on the domain. For an abstract function, the domain is usually viewed as the largest reasonable collection of input values; for a function that models a physical phenomenon, the domain is usually determined by the context of possibilities for the input in the phenomenon being considered.
+
- If we have two quantities that are changing in tandem, how can we connect the quantities and understand how change in one affects the other?
-
-
-
-
- When the amount of water in a tank is changing, what behaviors can we observe?
-
-
-
-
-
- Introduction
-
- Mathematics is the art of making sense of patterns. One way that patterns arise is when two quantities are changing in tandem. In this setting, we may make sense of the situation by expressing the relationship between the changing quantities through words, through images, through data, or through a formula.
-
-
-
-
-
-
-
- Using Graphs to Represent Relationships
-
-
- In Preview Activity, we saw how several changing quantities were related in the setting of an aquarium filling with water: time, the depth of the water, and the total amount of water in the tank are all changing, and any pair of these quantities changes in related ways. One way that we can make sense of the situation is to record some data in a table. For instance, observing that the tank is filling at a rate of 0.5 cubic feet per minute, this tells us that after 1 minute there will be 0.5 cubic feet of water in the tank, and after 2 minutes there will be 1 cubic foot of water in the tank, and so on. If we let t denote the time in minutes and V the amount of water in the tank at time t, we can represent the relationship between these quantities through Table.
-
-
-
-
-
- Data for how the volume of water in the tank changes with time.
-
-
- t
- V
-
-
- 0
- 0.0
-
-
- 1
- 0.5
-
-
- 2
- 1.0
-
-
- 3
- 1.5
-
-
- 4
- 2.0
-
-
- 5
- 2.5
-
-
-
-
-
A visual representation of the data in Table.
-
A visual representation of the data in Table.
-
-
-
-
-
- We can also represent this data in a graph by plotting ordered pairs (t,V) on a system of coordinate axes, where t represents the horizontal distance of the point from the origin, (0,0), and V represents the vertical distance from (0,0). The visual representation of the table of values from Table is seen in the graph in Figure.
-
-
-
- Sometimes it is possible to use variables and one or more equations to connect quantities that are changing in tandem. In the aquarium example from the preview activity, we can observe that the volume, V, of a rectangular box that has length l, width w, and height h is given by
-
- V = l \cdot w \cdot h
- ,
- and thus, since the water in the tank will always have length l = 4 feet and width w = 2 feet, the volume of water in the tank is directly related to the depth of water in the tank by the equation
-
- V = 4 \cdot 2 \cdot h = 8h
- .
- Depending on which variable we solve for, we can either see how V depends on h through the equation V = 8h, or how h depends on V via the equation h = \frac{1}{8}V. From either perspective, we observe that as depth or volume increases, so must volume or depth correspondingly increase.
-
-
-
-
-
-
-
- Using Algebra to Add Perspective
-
- One of the ways that we make sense of mathematical ideas is to view them from multiple perspectives. We may use different means to establish different points of view: words, numerical data, graphs, or symbols. In addition, sometimes by changing our perspective within a particular approach we gain deeper insight.
-
-
-
-
-
The empty conical tank.
-
The empty conical tank.
-
-
-
-
The conical tank, partially filled.
-
The conical tank, partially filled.
-
-
-
-
-
- If we consider the conical tank discussed in Activity, as seen in Figure and Figure, we can use algebra to better understand some of the relationships among changing quantities. The volume of a cone volumecone with radius r and height h is given by the formula
-
- V = \frac{1}{3}\pi r^2 h
- .
-
-
-
- Note that at any time while the tank is being filled, r (the radius of the surface of the water), h (the depth of the water), and V (the volume of the water) are all changing; moreover, all are connected to one another. Because of the constraints of the tank itself (with radius 2 feet and depth 4 feet), it follows that as the radius and height of the water change, they always do so in the proportion
-
- \frac{r}{h} = \frac{2}{4}
- .
- Solving this last equation for r, we see that r = \frac{1}{2}h; substituting this most recent result in the equation for volume, it follows that
-
- V = \frac{1}{3}\pi \left( \frac{1}{2}h \right)^2 h = \frac{\pi}{12} h^3
- .
-
-
-
- This most recent equation helps us understand how V and h change in tandem. We know from our earlier work that the volume of water in the tank increases at a constant rate of 0.75 cubic feet per minute. This leads to the data shown in Table.
-
-
-
- How time and volume change in tandem in a conical tank.
-
-
- t
- 0
- 1
- 2
- 3
- 4
- 5
-
-
- V
- 0.0
- 0.75
- 1.5
- 2.25
- 3.0
- 3.75
-
-
-
-
-
- With the equation V = \frac{\pi}{12} h^3, we can now also see how the height of the water changes in tandem with time. Solving the equation for h, note that h^3 = \frac{12}{\pi} V, and therefore
-
- h = \sqrt[3]{\frac{12}{\pi} V}
- . Thus, when V = 0.75, it follows that h = \sqrt[3]{\frac{12}{\pi} 0.75} \approx 1.42. Executing similar computations with the other values of V in Table, we get the following updated data that now includes h.
-
-
-
- How time, volume, and height change in concert in a conical tank.
-
-
- t
- 0
- 1
- 2
- 3
- 4
- 5
-
-
- V
- 0.0
- 0.75
- 1.5
- 2.25
- 3.0
- 3.75
-
-
- h
- 0.0
- 1.42
- 1.79
- 2.05
- 2.25
- 2.43
-
-
-
-
-
- Plotting this data on two different sets of axes, we see the different ways that V and h change with t. Whereas volume increases at a constant rate, as seen by the straight line appearance of the points in Figure, we observe that the water's height increases in a way that it rises more slowly as time goes on, as shown by the way the curve the points lie on in Figurebends down as time passes.
-
-
-
-
Plotting V versus t.
-
Plotting V versus t.
-
-
-
-
Plotting h versus t.
-
Plotting h versus t.
-
-
-
-
-
- These different behaviors make sense because of the shape of the tank. Since at first there is less volume relative to depth near the cone's point, as water flows in at a constant rate, the water's height will rise quickly. But as time goes on and more water is added at the same rate, there is more space for the water to fill in order to make the water level rise, and thus the water's height rises more and more slowly as time passes.
-
-
-
-
-
-
-
- Summary
-
-
-
-
- When two related quantities are changing in tandem, we can better understand how change in one affects the other by using data, graphs, words, or algebraic symbols to express the relationship between them. See, for instance, Table, Figure, , and Equation that together help explain how the height and volume of water in a conical tank change in tandem as time changes.
-
-
-
-
- When the amount of water in a tank is changing, we can observe other quantities that change, depending on the shape of the tank. For instance, if the tank is conical, we can consider both the changing height of the water and the changing radius of the surface of the water. In addition, whenever we think about a quantity that is changing as time passes, we note that time itself is changing.
-
+ If we have two quantities that are changing in tandem, how can we connect the quantities and understand how change in one affects the other?
+
+
+
+
+ When the amount of water in a tank is changing, what behaviors can we observe?
+
+
+
+
+
+ Introduction
+
+ Mathematics is the art of making sense of patterns. One way that patterns arise is when two quantities are changing in tandem. In this setting, we may make sense of the situation by expressing the relationship between the changing quantities through words, through images, through data, or through a formula.
+
+
+
+
+
+
+
+
+ Using Graphs to Represent Relationships
+
+
+ In Preview Activity, we saw how several changing quantities were related in the setting of an aquarium filling with water: time, the depth of the water, and the total amount of water in the tank are all changing, and any pair of these quantities changes in related ways. One way that we can make sense of the situation is to record some data in a table. For instance, observing that the tank is filling at a rate of 0.5 cubic feet per minute, this tells us that after 1 minute there will be 0.5 cubic feet of water in the tank, and after 2 minutes there will be 1 cubic foot of water in the tank, and so on. If we let t denote the time in minutes and V the amount of water in the tank at time t, we can represent the relationship between these quantities through Table.
+
+
+
+
+
+ Data for how the volume of water in the tank changes with time.
+
+
+ t
+ V
+
+
+ 0
+ 0.0
+
+
+ 1
+ 0.5
+
+
+ 2
+ 1.0
+
+
+ 3
+ 1.5
+
+
+ 4
+ 2.0
+
+
+ 5
+ 2.5
+
+
+
+
+
A visual representation of the data in Table.
+
A visual representation of the data in Table.
+
+
+
+
+
+ We can also represent this data in a graph by plotting ordered pairs (t,V) on a system of coordinate axes, where t represents the horizontal distance of the point from the origin, (0,0), and V represents the vertical distance from (0,0). The visual representation of the table of values from Table is seen in the graph in Figure.
+
+
+
+ Sometimes it is possible to use variables and one or more equations to connect quantities that are changing in tandem. In the aquarium example from the preview activity, we can observe that the volume, V, of a rectangular box that has length l, width w, and height h is given by
+
+ V = l \cdot w \cdot h
+ ,
+ and thus, since the water in the tank will always have length l = 4 feet and width w = 2 feet, the volume of water in the tank is directly related to the depth of water in the tank by the equation
+
+ V = 4 \cdot 2 \cdot h = 8h
+ .
+ Depending on which variable we solve for, we can either see how V depends on h through the equation V = 8h, or how h depends on V via the equation h = \frac{1}{8}V. From either perspective, we observe that as depth or volume increases, so must volume or depth correspondingly increase.
+
+
+
+
+
+
+
+ Using Algebra to Add Perspective
+
+ One of the ways that we make sense of mathematical ideas is to view them from multiple perspectives. We may use different means to establish different points of view: words, numerical data, graphs, or symbols. In addition, sometimes by changing our perspective within a particular approach we gain deeper insight.
+
+
+
+
+
The empty conical tank.
+
The empty conical tank.
+
+
+
+
The conical tank, partially filled.
+
The conical tank, partially filled.
+
+
+
+
+
+ If we consider the conical tank discussed in Activity, as seen in Figure and Figure, we can use algebra to better understand some of the relationships among changing quantities. The volume of a cone volumecone with radius r and height h is given by the formula
+
+ V = \frac{1}{3}\pi r^2 h
+ .
+
+
+
+ Note that at any time while the tank is being filled, r (the radius of the surface of the water), h (the depth of the water), and V (the volume of the water) are all changing; moreover, all are connected to one another. Because of the constraints of the tank itself (with radius 2 feet and depth 4 feet), it follows that as the radius and height of the water change, they always do so in the proportion
+
+ \frac{r}{h} = \frac{2}{4}
+ .
+ Solving this last equation for r, we see that r = \frac{1}{2}h; substituting this most recent result in the equation for volume, it follows that
+
+ V = \frac{1}{3}\pi \left( \frac{1}{2}h \right)^2 h = \frac{\pi}{12} h^3
+ .
+
+
+
+ This most recent equation helps us understand how V and h change in tandem. We know from our earlier work that the volume of water in the tank increases at a constant rate of 0.75 cubic feet per minute. This leads to the data shown in Table.
+
+
+
+ How time and volume change in tandem in a conical tank.
+
+
+ t
+ 0
+ 1
+ 2
+ 3
+ 4
+ 5
+
+
+ V
+ 0.0
+ 0.75
+ 1.5
+ 2.25
+ 3.0
+ 3.75
+
+
+
+
+
+ With the equation V = \frac{\pi}{12} h^3, we can now also see how the height of the water changes in tandem with time. Solving the equation for h, note that h^3 = \frac{12}{\pi} V, and therefore
+
+ h = \sqrt[3]{\frac{12}{\pi} V}
+ . Thus, when V = 0.75, it follows that h = \sqrt[3]{\frac{12}{\pi} 0.75} \approx 1.42. Executing similar computations with the other values of V in Table, we get the following updated data that now includes h.
+
+
+
+ How time, volume, and height change in concert in a conical tank.
+
+
+ t
+ 0
+ 1
+ 2
+ 3
+ 4
+ 5
+
+
+ V
+ 0.0
+ 0.75
+ 1.5
+ 2.25
+ 3.0
+ 3.75
+
+
+ h
+ 0.0
+ 1.42
+ 1.79
+ 2.05
+ 2.25
+ 2.43
+
+
+
+
+
+ Plotting this data on two different sets of axes, we see the different ways that V and h change with t. Whereas volume increases at a constant rate, as seen by the straight line appearance of the points in Figure, we observe that the water's height increases in a way that it rises more slowly as time goes on, as shown by the way the curve the points lie on in Figurebends down as time passes.
+
+
+
+
Plotting V versus t.
+
Plotting V versus t.
+
+
+
+
Plotting h versus t.
+
Plotting h versus t.
+
+
+
+
+
+ These different behaviors make sense because of the shape of the tank. Since at first there is less volume relative to depth near the cone's point, as water flows in at a constant rate, the water's height will rise quickly. But as time goes on and more water is added at the same rate, there is more space for the water to fill in order to make the water level rise, and thus the water's height rises more and more slowly as time passes.
+
+
+
+
+
+
+
+ Summary
+
+
+
+
+ When two related quantities are changing in tandem, we can better understand how change in one affects the other by using data, graphs, words, or algebraic symbols to express the relationship between them. See, for instance, Table, Figure, , and Equation that together help explain how the height and volume of water in a conical tank change in tandem as time changes.
+
+
+
+
+ When the amount of water in a tank is changing, we can observe other quantities that change, depending on the shape of the tank. For instance, if the tank is conical, we can consider both the changing height of the water and the changing radius of the surface of the water. In addition, whenever we think about a quantity that is changing as time passes, we note that time itself is changing.
+
- What does it mean to say that a given function has an inverse function?
-
-
-
-
- How can we determine whether or not a given function has a corresponding inverse function?
-
-
-
-
- When a function has an inverse function, what important properties does the inverse function have in comparison to the original function?
-
-
-
-
-
- Introduction
-
- Because every function is a process that converts a collection of inputs to a corresponding collection of outputs, a natural question is: for a particular function, can we change perspective and think of the original function's outputs as the inputs for a reverse process?
- If we phrase this question algebraically, it is analogous to asking: given an equation that defines y is a function of x, is it possible to find a corresponding equation where x is a function of y?
-
-
-
-
-
-
-
- When a function has an inverse function
-
- In Preview Activity, we found that for the function F = g(C) = \frac{9}{5}C + 32, it's also possible to solve for C in terms of F and write C = h(F) = \frac{5}{9}(F-32). The first function, g, converts Celsius temperatures to Fahrenheit ones; the second function, h, converts Fahrenheit temperatures to Celsius ones. Thus, the process h reverses the process of g, and likewise the process of g reverses the process of h. This is also why it makes sense that h(g(C)) = C and g(h(F)) = F. If, for instance, we take a Celsius temperature C, convert it to Fahrenheit, and convert the result back to Celsius, we arrive back at the Celsius temperature we started with: h(g(C)) = C.
-
-
-
- Similar work is sometimes possible with other functions. When we can find a new function that reverses the process of the original function, we say that the original function has an inverse function and make the following formal definition.
-
-
-
- inverse functiondefinition
-
-
- Let f : A \to B be a function. If there exists a function g : B \to A such that
-
- g(f(a)) = a \text{ and } f(g(b)) = b
-
- for each a in A and each b in B, then we say that f has an inverse function and that the function g is the inverse of f.
-
-
-
-
-
- Note particularly what the equation g(f(a)) = a says: for any input a in the domain of f, the function g will reverse the process of f (which converts a to f(a)) because g converts f(a) back to a.
-
-
-
- When a given function f has a corresponding inverse function g, we usually rename g as f^{-1}, which we read aloud as f-inverse. inverse functionnotation The equation g(f(a))=a now reads f^{-1}(f(a)) = a, which we interpret as saying f-inverse converts f(a) back to a. We similarly write that f(f^{-1}(b)) = b.
-
-
-
-
-
- When a given function has an inverse function, it allows us to express the same relationship from two different points of view. For instance, if y = f(t) = 2t+1, we can showObserve that g(f(t)) = g(2t+1) = \frac{(2t+1)-1}{2} = \frac{2t}{2} = t. Similarly, f(g(y)) = f\left(\frac{y-1}{2}\right) = 2\left(\frac{y-1}{2} \right) + 1 = y-1 + 1 = y. that the function t = g(y) = \frac{y-1}{2} reverses the effect of f (and vice versa), and thus g = f^{-1}. We observe that
-
- y = f(t) = 2t + 1 \text{ and } t = f^{-1}(y) = \frac{y-1}{2}
-
- are equivalent forms of the same equation, and thus they say the same thing from two different perspectives. The first version of the equation is solved for y in terms of t, while the second equation is solved for t in terms of y. This important principle holds in general whenever a function has an inverse function.
-
-
-
- Two perspectives from a function and its inverse function
- inverse functiontwo perspectives
-
- If y = f(t) has an inverse function, then the equations
-
- y = f(t) \text{ and } t = f^{-1}(y)
-
- say the exact same thing but from two different perspectives.
-
-
-
-
-
-
- Determining whether a function has an inverse function
-
- It's important to note in Definition that we say If there exists \ldots. That is, we don't guarantee that an inverse function exists for a given function. Thus, we might ask: how can we determine whether or not a given function has a corresponding inverse function? As with many questions about functions, there are often three different possible ways to explore such a question: through a table, through a graph, or through an algebraic formula.
-
-
-
-
-
-
- Do the functions f and g defined by Table and Table have corresponding inverse functions? Why or why not?
-
-
-
-
- The table that defines the function <m>f</m>.
-
-
- x
- 0
- 1
- 2
- 3
- 4
-
-
- f(x)
- 6
- 4
- 3
- 4
- 6
-
-
-
-
- The table that defines the function <m>g</m>.
-
-
- x
- 0
- 1
- 2
- 3
- 4
-
-
- g(x)
- 3
- 1
- 4
- 2
- 0
-
-
-
-
-
-
-
-
- For any function, the question of whether or not it has an inverse comes down to whether or not the process of the function can be reliably reversed. For functions given in table form such as f and g, we essentially ask if it's possible to switch the input and output rows and have the new resulting table also represent a function.
-
-
-
- The function f does not have an inverse function because there are two different inputs that lead to the same output: f(0) = 6 and f(4) = 6. If we attempt to reverse this process, we have a situation where the input 6 would correspond to two potential outputs, 0 and 4.
-
-
-
- However, the function g does have an inverse function because when we reverse the rows in Table, each input (in order, 3, 1, 4, 2, 0) indeed corresponds to one and only one output (in order, 0, 1, 2, 3, 4). We can thus make observations such as g^{-1}(4) = 2, which is the same as saying that g(2) = 4, just from a different perspective.
-
-
-
-
-
- In Example, we see that if we can identify one pair of distinct inputs that lead to the same output (such as f(0) = f(4) = 6 in Table), then the process of the function cannot be reversed and the function does not have an inverse.
-
-
-
-
-
-
- Do the functions p and q defined by Figure and Figure have corresponding inverse functions? Why or why not?
-
-
-
-
-
- The graph that defines function p.
-
-
- The graph that defines function p.
-
-
-
-
-
- The graph that defines function q.
-
-
- The graph that defines function q.
-
-
-
-
-
-
-
-
- Recall that when a point such as (a,c) lies on the graph of a function p, this means that the input x = a, which represents to a value on the horizontal axis, corresponds with the output y = c that is represented by a value on the vertical axis. In this situation, we write p(a) = c. We note explicitly that p is a function because its graph passes the : any vertical line intersects the graph of p exactly once, and thus each input from the domain corresponds to one and only one output.
-
-
-
- If we attempt to change perspective and use the graph of p to view x as a function of y, we see that this fails because the output value c is associated with two different inputs, a and b. Said differently, because the horizontal line y = c intersects the graph of p at both (a,c) and (b,c) (as shown in Figure), we cannot view y as the input to a function process that produces the corresonding x-value. Therefore, p does not have an inverse function.
-
-
-
- On the other hand, provided that the behavior seen in the figure continues, the function q does have an inverse because we can view x as a function of y via the graph given in Figure. This is because for any choice of y, there corresponds one and only one x that results from y. We can think of this visually by starting at a value such as y = c on the y-axis, moving horizontally to where the line intersects the graph of q, and then moving down to the corresonding location (here x = a) on the horizontal axis. From the behavior of the graph of q (a straight line that is always increasing), we see that this correspondence will hold for any choice of y, and thus indeed x is a function of y. From this, we can say that q indeed has an inverse function. We thus can write that q^{-1}(c) = a, which is a different way to express the equivalent fact that q(a) = c.
-
-
-
-
-
- The graphical observations that we made for the function q in Example provide a general test for whether or not a function given by a graph has a corresponding inverse function.
-
-
-
- Horizontal Line Test
- inverse functionhorizontal line test
-
- A function whose graph lies in the x-y plane has a corresponding inverse function if and only if every horizontal line intersects the graph at most once. When the graph passes this test, the horizontal coordinate of each point on the graph can be viewed as a function of the vertical coordinate of the point.
-
-
-
-
-
-
- Do the functions r and s defined by
-
- y = r(t) = 3 - \frac{1}{5}(t-1)^3 \text{ and } y = s(t) = 3 - \frac{1}{5}(t-1)^2
-
- have corresponding inverse functions? If not, use algebraic reasoning to explain why; if so, demonstrate by using algebra to find a formula for the inverse function.
-
-
-
-
- For any function of the form y = f(t), one way to determine if we can view the original input variable t as a function of the original output variable y is to attempt to solve the equation y = f(t) for t in terms of y.
-
-
-
- Taking y = 3 - \frac{1}{5}(t-1)^3, we try to solve for t by first subtracting 3 from both sides to get
-
- y - 3 = -\frac{1}{5}(t-1)^3
- .
- Next, multiplying both sides by -5, it follows that
-
- (t-1)^3 = -5(y-3)
- .
- Because the cube root function has the property that \sqrt[3]{z^3} = z for every real number z (since the cube root function is the inverse function for the cubing function, and each function has both a domain and range of all real numbers), we can take the cube root of both sides of the preceding equation to get
-
- t - 1 = \sqrt[3]{-5(y-3)}
- .
- Finally, adding 1 to both sides, we have determined that
-
- t = 1 + \sqrt[3]{-5(y-3)}
- .
- Because we have been able to express t as a single function of y for every possible value of y, this shows that r indeed has an inverse and that t = r^{-1}(y) = 1 + \sqrt[3]{-5(y-3)}.
-
-
-
- We attempt similar reasoning for the second function, y = 3 - \frac{1}{5}(t-1)^2. To solve for t, we first subtract 3 from both sides, so that
-
- y - 3 = -\frac{1}{5}(t-1)^2
- .
- After multiplying both sides by -5, we have
-
- (t-1)^2 = -5(y-3)
- .
- Next, it's necessary to take the square root of both sides in an effort to isolate t. Here, however, we encounter a crucial issue. Because the function g(x) = x^2 takes any nonzero number and its opposite to the same output (e.g. (-5)^2 = 25 = (5)^2), this means that we have to account for both possible inputs that result in the same output. Based on our last equation, this means that either
-
- t-1 = \sqrt{-5(y-3)} \ \text{ or } \ t-1 = -\sqrt{-5(y-3)}
- .
- As such, we find not a single equation that expresses t as a function of y, but rather two:
-
- t = 1 + \sqrt{-5(y-3)} \ \text{ or } \ t = 1 -\sqrt{-5(y-3)}
- .
- Since it appears that t can't be expressed as a single function of y, it seems to follow that y = s(t) = 3 - \frac{1}{5}(t-1)^2 does not have an inverse function.
-
-
-
-
-
- The graphs of y = r(t) = 3 - \frac{1}{5}(t-1)^3 and y = s(t) = 3 - \frac{1}{5}(t-1)^2 provide a different perspective to confirm the results of Example. Indeed, in Figure, we see that r appears to pass the horizontal line test because it is decreasingCalculus provides one way to fully justify that the graph of r is indeed always decreasing., and thus has an inverse function. On the other hand, the graph of s fails the horizontal line test (picture the line y = 2 in Figure) and therefore s does not have an inverse function.
-
-
-
-
-
A plot of y = r(t) = 3 - \frac{1}{5}(t-1)^3.
-
A plot of y = r(t) = 3 - \frac{1}{5}(t-1)^3.
-
-
-
-
A plot of y = s(t) = 3 - \frac{1}{5}(t-1)^2.
-
A plot of y = s(t) = 3 - \frac{1}{5}(t-1)^2.
-
-
-
-
-
-
-
-
-
- Properties of an inverse function
-
- When a function has an inverse function, we have observed several important relationships that hold between the original function and the corresponding inverse function.
-
-
-
- Properties of an inverse function
-
- Let f : A \to B be a function whose domain is A and whose range is B be such that f has an inverse function, f^{-1}. Then:
-
-
-
-
-
-
- f^{-1} : B \to A, so the domain of f^{-1} is B and its range is A.
-
-
-
-
- The functions f and f^{-1} reverse one anothers' processes. Symbolically,
- f^{-1}(f(a)) = a for every input a in the domain of f, and similarly,
- f(f^{-1}(b)) = b for every input b in the domain of f^{-1}.
-
-
-
-
- If y = f(t), then we can express the exact same relationship from a different perspective by writing t = f^{-1}(y).
-
-
-
-
- Consider the setting where A and B are collections of real numbers. If a point (x,y) lies on the graph of f, then it follows y = f(x). From this, we can equivalently say that x = f^{-1}(y). Hence, the point (y,x) lies on the graph of x = f^{-1}(y).
-
-
-
-
-
-
-
- The last item above leads to a special relationship between the graphs of f and f^{-1} when viewed on the same coordinate axes. In that setting, we need to view x as the input of each function (since it's the horizontal coordinate) and y as the output. If we know a particular input-output relationship for f, say f(-1) = \frac{1}{2}, then it follows that f^{-1} \left( \frac{1}{2} \right) = -1. We observe that the points \left(-1, \frac{1}{2} \right) and \left(\frac{1}{2}, -1 \right) are reflections of each other across the line y = x. Because such a relationship holds for every point (x,y) on the graph of f, this means that the graphs of f and f^{-1} are reflections of one another across the line y = x, as seen in Figure.
-
-
-
-
The graph of a function f along with its inverse, f^{-1}.
-
The graph of a function f along with its inverse, f^{-1}.
-
-
-
-
-
-
-
-
- Summary
-
-
-
-
- A given function f : A \to B has an inverse function whenever there exists a related function g : B \to A that reverses the process of f. Formally, this means that g must satisfy g(f(a)) = a for every a in the domain of f, and f(g(b)) = b for every b in the range of f.
-
-
-
-
- We determine whether or not a given function f has a corresponding inverse function by determining if the process that defines f can be reversed so that we can also think of the outputs as a function of the inputs. If we have a graph of the function f, we know f has an inverse function if the graph passes the . If we have a formula for the function f, say y = f(t), we know f has an inverse function if we can solve for t and write t = f^{-1}(y).
-
-
-
-
- A good summary of the properties of an inverse function is provided in the .
-
+ What does it mean to say that a given function has an inverse function?
+
+
+
+
+ How can we determine whether or not a given function has a corresponding inverse function?
+
+
+
+
+ When a function has an inverse function, what important properties does the inverse function have in comparison to the original function?
+
+
+
+
+
+ Introduction
+
+ Because every function is a process that converts a collection of inputs to a corresponding collection of outputs, a natural question is: for a particular function, can we change perspective and think of the original function's outputs as the inputs for a reverse process?
+ If we phrase this question algebraically, it is analogous to asking: given an equation that defines y is a function of x, is it possible to find a corresponding equation where x is a function of y?
+
+
+
+
+
+
+
+
+ When a function has an inverse function
+
+ In Preview Activity, we found that for the function F = g(C) = \frac{9}{5}C + 32, it's also possible to solve for C in terms of F and write C = h(F) = \frac{5}{9}(F-32). The first function, g, converts Celsius temperatures to Fahrenheit ones; the second function, h, converts Fahrenheit temperatures to Celsius ones. Thus, the process h reverses the process of g, and likewise the process of g reverses the process of h. This is also why it makes sense that h(g(C)) = C and g(h(F)) = F. If, for instance, we take a Celsius temperature C, convert it to Fahrenheit, and convert the result back to Celsius, we arrive back at the Celsius temperature we started with: h(g(C)) = C.
+
+
+
+ Similar work is sometimes possible with other functions. When we can find a new function that reverses the process of the original function, we say that the original function has an inverse function and make the following formal definition.
+
+
+
+ inverse functiondefinition
+
+
+ Let f : A \to B be a function. If there exists a function g : B \to A such that
+
+ g(f(a)) = a \text{ and } f(g(b)) = b
+
+ for each a in A and each b in B, then we say that f has an inverse function and that the function g is the inverse of f.
+
+
+
+
+
+ Note particularly what the equation g(f(a)) = a says: for any input a in the domain of f, the function g will reverse the process of f (which converts a to f(a)) because g converts f(a) back to a.
+
+
+
+ When a given function f has a corresponding inverse function g, we usually rename g as f^{-1}, which we read aloud as f-inverse. inverse functionnotation The equation g(f(a))=a now reads f^{-1}(f(a)) = a, which we interpret as saying f-inverse converts f(a) back to a. We similarly write that f(f^{-1}(b)) = b.
+
+
+
+
+
+ When a given function has an inverse function, it allows us to express the same relationship from two different points of view. For instance, if y = f(t) = 2t+1, we can showObserve that g(f(t)) = g(2t+1) = \frac{(2t+1)-1}{2} = \frac{2t}{2} = t. Similarly, f(g(y)) = f\left(\frac{y-1}{2}\right) = 2\left(\frac{y-1}{2} \right) + 1 = y-1 + 1 = y. that the function t = g(y) = \frac{y-1}{2} reverses the effect of f (and vice versa), and thus g = f^{-1}. We observe that
+
+ y = f(t) = 2t + 1 \text{ and } t = f^{-1}(y) = \frac{y-1}{2}
+
+ are equivalent forms of the same equation, and thus they say the same thing from two different perspectives. The first version of the equation is solved for y in terms of t, while the second equation is solved for t in terms of y. This important principle holds in general whenever a function has an inverse function.
+
+
+
+ Two perspectives from a function and its inverse function
+ inverse functiontwo perspectives
+
+ If y = f(t) has an inverse function, then the equations
+
+ y = f(t) \text{ and } t = f^{-1}(y)
+
+ say the exact same thing but from two different perspectives.
+
+
+
+
+
+
+ Determining whether a function has an inverse function
+
+ It's important to note in Definition that we say If there exists \ldots. That is, we don't guarantee that an inverse function exists for a given function. Thus, we might ask: how can we determine whether or not a given function has a corresponding inverse function? As with many questions about functions, there are often three different possible ways to explore such a question: through a table, through a graph, or through an algebraic formula.
+
+
+
+
+
+
+ Do the functions f and g defined by Table and Table have corresponding inverse functions? Why or why not?
+
+
+
+
+ The table that defines the function <m>f</m>.
+
+
+ x
+ 0
+ 1
+ 2
+ 3
+ 4
+
+
+ f(x)
+ 6
+ 4
+ 3
+ 4
+ 6
+
+
+
+
+ The table that defines the function <m>g</m>.
+
+
+ x
+ 0
+ 1
+ 2
+ 3
+ 4
+
+
+ g(x)
+ 3
+ 1
+ 4
+ 2
+ 0
+
+
+
+
+
+
+
+
+ For any function, the question of whether or not it has an inverse comes down to whether or not the process of the function can be reliably reversed. For functions given in table form such as f and g, we essentially ask if it's possible to switch the input and output rows and have the new resulting table also represent a function.
+
+
+
+ The function f does not have an inverse function because there are two different inputs that lead to the same output: f(0) = 6 and f(4) = 6. If we attempt to reverse this process, we have a situation where the input 6 would correspond to two potential outputs, 0 and 4.
+
+
+
+ However, the function g does have an inverse function because when we reverse the rows in Table, each input (in order, 3, 1, 4, 2, 0) indeed corresponds to one and only one output (in order, 0, 1, 2, 3, 4). We can thus make observations such as g^{-1}(4) = 2, which is the same as saying that g(2) = 4, just from a different perspective.
+
+
+
+
+
+ In Example, we see that if we can identify one pair of distinct inputs that lead to the same output (such as f(0) = f(4) = 6 in Table), then the process of the function cannot be reversed and the function does not have an inverse.
+
+
+
+
+
+
+ Do the functions p and q defined by Figure and Figure have corresponding inverse functions? Why or why not?
+
+
+
+
+
+ The graph that defines function p.
+
+
+ The graph that defines function p.
+
+
+
+
+
+ The graph that defines function q.
+
+
+ The graph that defines function q.
+
+
+
+
+
+
+
+
+ Recall that when a point such as (a,c) lies on the graph of a function p, this means that the input x = a, which represents to a value on the horizontal axis, corresponds with the output y = c that is represented by a value on the vertical axis. In this situation, we write p(a) = c. We note explicitly that p is a function because its graph passes the : any vertical line intersects the graph of p exactly once, and thus each input from the domain corresponds to one and only one output.
+
+
+
+ If we attempt to change perspective and use the graph of p to view x as a function of y, we see that this fails because the output value c is associated with two different inputs, a and b. Said differently, because the horizontal line y = c intersects the graph of p at both (a,c) and (b,c) (as shown in Figure), we cannot view y as the input to a function process that produces the corresonding x-value. Therefore, p does not have an inverse function.
+
+
+
+ On the other hand, provided that the behavior seen in the figure continues, the function q does have an inverse because we can view x as a function of y via the graph given in Figure. This is because for any choice of y, there corresponds one and only one x that results from y. We can think of this visually by starting at a value such as y = c on the y-axis, moving horizontally to where the line intersects the graph of q, and then moving down to the corresonding location (here x = a) on the horizontal axis. From the behavior of the graph of q (a straight line that is always increasing), we see that this correspondence will hold for any choice of y, and thus indeed x is a function of y. From this, we can say that q indeed has an inverse function. We thus can write that q^{-1}(c) = a, which is a different way to express the equivalent fact that q(a) = c.
+
+
+
+
+
+ The graphical observations that we made for the function q in Example provide a general test for whether or not a function given by a graph has a corresponding inverse function.
+
+
+
+ Horizontal Line Test
+ inverse functionhorizontal line test
+
+ A function whose graph lies in the x-y plane has a corresponding inverse function if and only if every horizontal line intersects the graph at most once. When the graph passes this test, the horizontal coordinate of each point on the graph can be viewed as a function of the vertical coordinate of the point.
+
+
+
+
+
+
+ Do the functions r and s defined by
+
+ y = r(t) = 3 - \frac{1}{5}(t-1)^3 \text{ and } y = s(t) = 3 - \frac{1}{5}(t-1)^2
+
+ have corresponding inverse functions? If not, use algebraic reasoning to explain why; if so, demonstrate by using algebra to find a formula for the inverse function.
+
+
+
+
+ For any function of the form y = f(t), one way to determine if we can view the original input variable t as a function of the original output variable y is to attempt to solve the equation y = f(t) for t in terms of y.
+
+
+
+ Taking y = 3 - \frac{1}{5}(t-1)^3, we try to solve for t by first subtracting 3 from both sides to get
+
+ y - 3 = -\frac{1}{5}(t-1)^3
+ .
+ Next, multiplying both sides by -5, it follows that
+
+ (t-1)^3 = -5(y-3)
+ .
+ Because the cube root function has the property that \sqrt[3]{z^3} = z for every real number z (since the cube root function is the inverse function for the cubing function, and each function has both a domain and range of all real numbers), we can take the cube root of both sides of the preceding equation to get
+
+ t - 1 = \sqrt[3]{-5(y-3)}
+ .
+ Finally, adding 1 to both sides, we have determined that
+
+ t = 1 + \sqrt[3]{-5(y-3)}
+ .
+ Because we have been able to express t as a single function of y for every possible value of y, this shows that r indeed has an inverse and that t = r^{-1}(y) = 1 + \sqrt[3]{-5(y-3)}.
+
+
+
+ We attempt similar reasoning for the second function, y = 3 - \frac{1}{5}(t-1)^2. To solve for t, we first subtract 3 from both sides, so that
+
+ y - 3 = -\frac{1}{5}(t-1)^2
+ .
+ After multiplying both sides by -5, we have
+
+ (t-1)^2 = -5(y-3)
+ .
+ Next, it's necessary to take the square root of both sides in an effort to isolate t. Here, however, we encounter a crucial issue. Because the function g(x) = x^2 takes any nonzero number and its opposite to the same output (e.g. (-5)^2 = 25 = (5)^2), this means that we have to account for both possible inputs that result in the same output. Based on our last equation, this means that either
+
+ t-1 = \sqrt{-5(y-3)} \ \text{ or } \ t-1 = -\sqrt{-5(y-3)}
+ .
+ As such, we find not a single equation that expresses t as a function of y, but rather two:
+
+ t = 1 + \sqrt{-5(y-3)} \ \text{ or } \ t = 1 -\sqrt{-5(y-3)}
+ .
+ Since it appears that t can't be expressed as a single function of y, it seems to follow that y = s(t) = 3 - \frac{1}{5}(t-1)^2 does not have an inverse function.
+
+
+
+
+
+ The graphs of y = r(t) = 3 - \frac{1}{5}(t-1)^3 and y = s(t) = 3 - \frac{1}{5}(t-1)^2 provide a different perspective to confirm the results of Example. Indeed, in Figure, we see that r appears to pass the horizontal line test because it is decreasingCalculus provides one way to fully justify that the graph of r is indeed always decreasing., and thus has an inverse function. On the other hand, the graph of s fails the horizontal line test (picture the line y = 2 in Figure) and therefore s does not have an inverse function.
+
+
+
+
+
A plot of y = r(t) = 3 - \frac{1}{5}(t-1)^3.
+
A plot of y = r(t) = 3 - \frac{1}{5}(t-1)^3.
+
+
+
+
A plot of y = s(t) = 3 - \frac{1}{5}(t-1)^2.
+
A plot of y = s(t) = 3 - \frac{1}{5}(t-1)^2.
+
+
+
+
+
+
+
+
+
+ Properties of an inverse function
+
+ When a function has an inverse function, we have observed several important relationships that hold between the original function and the corresponding inverse function.
+
+
+
+ Properties of an inverse function
+
+ Let f : A \to B be a function whose domain is A and whose range is B be such that f has an inverse function, f^{-1}. Then:
+
+
+
+
+
+
+ f^{-1} : B \to A, so the domain of f^{-1} is B and its range is A.
+
+
+
+
+ The functions f and f^{-1} reverse one anothers' processes. Symbolically,
+ f^{-1}(f(a)) = a for every input a in the domain of f, and similarly,
+ f(f^{-1}(b)) = b for every input b in the domain of f^{-1}.
+
+
+
+
+ If y = f(t), then we can express the exact same relationship from a different perspective by writing t = f^{-1}(y).
+
+
+
+
+ Consider the setting where A and B are collections of real numbers. If a point (x,y) lies on the graph of f, then it follows y = f(x). From this, we can equivalently say that x = f^{-1}(y). Hence, the point (y,x) lies on the graph of x = f^{-1}(y).
+
+
+
+
+
+
+
+ The last item above leads to a special relationship between the graphs of f and f^{-1} when viewed on the same coordinate axes. In that setting, we need to view x as the input of each function (since it's the horizontal coordinate) and y as the output. If we know a particular input-output relationship for f, say f(-1) = \frac{1}{2}, then it follows that f^{-1} \left( \frac{1}{2} \right) = -1. We observe that the points \left(-1, \frac{1}{2} \right) and \left(\frac{1}{2}, -1 \right) are reflections of each other across the line y = x. Because such a relationship holds for every point (x,y) on the graph of f, this means that the graphs of f and f^{-1} are reflections of one another across the line y = x, as seen in Figure.
+
+
+
+
The graph of a function f along with its inverse, f^{-1}.
+
The graph of a function f along with its inverse, f^{-1}.
+
+
+
+
+
+
+
+
+ Summary
+
+
+
+
+ A given function f : A \to B has an inverse function whenever there exists a related function g : B \to A that reverses the process of f. Formally, this means that g must satisfy g(f(a)) = a for every a in the domain of f, and f(g(b)) = b for every b in the range of f.
+
+
+
+
+ We determine whether or not a given function f has a corresponding inverse function by determining if the process that defines f can be reversed so that we can also think of the outputs as a function of the inputs. If we have a graph of the function f, we know f has an inverse function if the graph passes the . If we have a formula for the function f, say y = f(t), we know f has an inverse function if we can solve for t and write t = f^{-1}(y).
+
+
+
+
+ A good summary of the properties of an inverse function is provided in the .
+
- What behavior of a function makes its graph a straight line?
-
-
-
-
- For a function whose graph is a straight line, what structure does its formula have?
-
-
-
-
- How can we interpret the slope of a linear function in applied contexts?
-
-
-
-
-
- Introduction
-
- Functions whose graphs are straight lines are both the simplest and the most important functions in mathematics.
- Lines often model important phenomena,
- and even when they don't directly model phenomena,
- lines can often approximate other functions that do.
- Whether a function's graph is a straight line or not is connected directly to its average rate of change.
-
-
-
-
-
-
-
-
- Properties of linear functions
-
-
- In Preview Activity, we considered three different functions for which the average rate of change of each appeared to always be constant. For the first function in the preview activity, y = f(x) = 7 - 3x, we can compute its average rate of change on an arbitrary interval [a,b]. Doing so, we notice that
-
- AV_{[a,b]} &= \frac{f(b)-f(a)}{b-a}
- &= \frac{(7-3b) - (7-3a)}{b-a}
- &= \frac{7-3b - 7+3a}{b-a}
- &= \frac{-3b + 3a}{b-a}
- &= \frac{-3(b - a)}{b-a}
- &= -3
- .
- This result shows us that for the function y = f(x) = 7 - 3x, its average rate of change is always -3, regardless of the interval we choose. We will use the property of having constant rate of change as the defining property of a linear function.
-
-
-
- linear function
- linear functionslope
-
-
- A function f is linear provided that its average rate of change is constant on every choice of interval in its domainHere we are considering functions whose domain is the set of all real numbers.. That is, for any inputs a and b for which a \ne b, it follows that
-
- \frac{f(b) - f(a)}{b-a} = m
-
- for some fixed constant m. We call m the slope of the linear function f.
-
-
-
-
-
- From prior study, we already know a lot about linear functions. In this section, we work to understand some familiar properties in light of the new perspective of Definition.
-
-
-
- Let's suppose we know that a function f is linear with average rate of change AV_{[a,b]} = m and that we also know the function value is y_0 at some fixed input x_0. That is, we know that f(x_0) = y_0. From this information, we can find the formula for y = f(x) for any input x. Working with the known point (x_0, f(x_0)) and any other point (x,f(x)) on the function's graph, we know that the average rate of change between these two points must be the constant m. This tells us that
-
- \frac{f(x) - f(x_0)}{x-x_0} = m
- .
- Since we are interested in finding a formula for y = f(x), we solve this most recent equation for f(x). Multiplying both sides by (x-x_0), we see that
-
- f(x) - f(x_0) = m(x-x_0)
- .
- Adding f(x_0) to each side, it follows
-
- f(x) = f(x_0) + m(x-x_0)
- .
- This shows that to determine the formula for a linear function, all we need to know is its average rate of change (or slope) and a single point the function passes through.
-
-
-
-
-
- Find a formula for a linear function f whose average rate of change is m = -\frac{1}{4} and passes through the point (-7,-5).
-
-
-
- Solution. Using Equation and the facts that m = -\frac{1}{4} and f(-7) = -5 (that is, x_0 = -7 and f(x_0) = -5), we have
-
- f(x) = -5 -\frac{1}{4}(x - (-7)) = -5 - \frac{1}{4}(x+7)
- .
-
-
-
-
-
- Replacing f(x) with y and f(x_0) with y_0, we call Equation the point-slope formpoint-slope form of a line.
-
-
-
- Point-slope form of a line
- linepoint-slope form
-
- A line with slope m (equivalently, average rate of change m) that passes through the point (x_0,y_0) has equation
-
- y = y_0 + m(x-x_0)
- .
-
-
-
-
-
-
- Visualizing the various components of point-slope form is important. For a line through (x_0,y_0) with slope m, we know its equation is y = y_0 + m(x-x_0). In Figure, we see that the line passes through (x_0,y_0) along with an arbitary point (x,y), which makes the vertical change between the two points given by y - y_0 and the horizontal change between the points x - x_0. This is consistent with the fact that
-
- AV_{[x_0,x]} = m = \frac{y-y_0}{x-x_0}
- .
- Indeed, writing m = \frac{y-y_0}{x-x_0} is a rearrangement of the point-slope form of the line, y = y_0 + m(x-x_0).
-
-
-
- We naturally use the terms increasing and decreasing as from Definition to describe lines based on whether their slope is positive or negative. A line with positive slope, such as the one in Figure, is increasing because its constant rate of change is positive, while a line with negative slope, such as in Figure is decreasing because of its negative rate of change. We say that a horizontal line (one whose slope is m = 0) is neither increasing nor decreasing.
-
-
-
-
-
The point-slope form of a line's equation.
-
The point-slope form of a line's equation.
-
-
-
-
The slope-intercept form of a line's equation.
-
The slope-intercept form of a line's equation.
-
-
-
-
-
- A special case arises when the known point on a line satisfies x_0 = 0. In this situation, the known point lies on the y-axis, and thus we call the point the y-intercept of the line. liney-intercept The resulting form of the line's equation is called slope-intercept form, which is also demonstrated in Figure.
-
-
-
- Slope-intercept form
- lineslope-intercept form
-
- For the line with slope m and passing through (0,y_0), its equation is
-
- y = y_0 + mx
- .
-
-
-
-
- Slope-intercept form follows from point-slope form from the fact that replacing x_0 with 0 gives us y = y_0 + m(x-0) = y_0 + mx. In many textbooks, the slope-intercept form of a line (often written y = mx + b) is treated as if it is the most useful form of a line. Point-slope form is actually more important and valuable since we can easily write down the equation of a line as soon as we know its slope and any point that lies on it, as opposed to needing to find the y-intercept, which is needed for slope-intercept form. Moreover, point-slope form plays a prominent role in calculus.
-
-
-
- If a line is in slope-intercept or point-slope form, it is useful to be able to quickly interpret key information about the line from the form of its equation.
-
-
-
-
-
- For the line given by y = -3 - 2.5(x-5), determine its slope and a point that lies on the line.
-
-
-
- Solution. This line is in point-slope form. Its slope is m = -2.5 and a point on the line is (5,-3).
-
-
-
-
-
-
-
- For the line given by y = 6 + 0.25x, determine its slope and a point that lies on the line.
-
-
-
- Solution. This line is in slope-intercept form. Its slope is m = 0.25 and a point on the line is (0,6), which is also the line's y-intercept.
-
-
-
-
-
-
-
- Interpreting linear functions in context
-
-
- Since linear functions are defined by the property that their average rate of change is constant, linear functions perfectly model quantities that change at a constant rate. In context, we can often think of slope as a rate of change; analyzing units carefully often yields significant insight.
-
-
-
-
-
- The Dolbear function T = D(N) = 40 + 0.25N from Section is a linear function whose slope is m = 0.25. What is the meaning of the slope in this context?
-
-
-
-
- Recall that T is measured in degrees Fahrenheit and N in chirps per minute. We know that m = AV_{[a,b]} = 0.25 is the constant average rate of change of D. Its units are units of output per unit of input, and thus degrees Fahrenheit per chirp per minute. This tells us that the average rate of change of the temperature function is 0.25 degrees Fahrenheit per chirp per minute, which means that for each additional chirp per minute observed, we expect the temperature to rise by 0.25 degrees Fahrenheit.
-
-
-
-
- Indeed, we can observe this through function values. We note that T(60) = 55 and T(61) = 55.25: one additional observed chirp per minute corresponds to a 0.25 degree increase in temperature. We also see this in the graph of the line, as seen in Figure:
- the slope between the points (40,50) and (120,70) is
-
- m &= \frac{70-50}{120-40}
- &= \frac{20}{80}
- &= 0.25 \frac{\text{degrees F}}{\text{chirp per minute}}
- .
-
-
-
-
The linear Dolbear function with slope m = 0.25 degrees Fahrenheit per chirp per minute.
-
The linear Dolbear function with slope m = 0.25 degrees Fahrenheit per chirp per minute.
-
-
-
-
-
-
-
-
- Like with the Dolbear function, it is often useful to write a linear function (whose output is called y) that models a quantity changing at a constant rate (as a function of some input t) by writing the function relationship in the form
-
- y = b + mt
-
- where b and m are constants. We may think of the four quantities involved in the following way:
-
-
-
-
-
-
- The constant b is the starting value of the output that corresponds to an input of t = 0;
-
-
-
-
- The constant m is the rate at which the output changes with respect to changes in the input: for each additional 1-unit change in input, the output will change by m units.
-
-
-
-
- The variable t is the independent (input) variable. A nonzero value for t corresponds to how much the input variable has changed from an initial value of 0.
-
-
-
-
- The variable y is the dependent (output) variable. The value of y results from a particular choice of t, and can be thought of as the starting output value (b) plus the change in output that results from the corresponding change in input t.
-
-
-
-
-
-
-
-
-
-
-
-
- Summary
-
-
-
-
- Any function f with domain all real numbers that has a constant average rate of change on every interval [a,b] will have a straight line graph. We call such functions linear functions.
-
-
-
-
- A linear function y = f(x) can be written in the form y = f(x) = y_0 + m(x-x_0), where m is the slope of the line and (x_0,y_0) is a point that lies on the line. In particular, f(x_0) = y_0.
-
-
-
-
- In an applied context where we have a linear function that models a phenomenon in the world around us, the slope tells us the function's (constant) average rate of change. The units on the slope, m, are always units of output per unit of input and this enables us to articulate how the output changes in response to a 1-unit change in input.
-
+ What behavior of a function makes its graph a straight line?
+
+
+
+
+ For a function whose graph is a straight line, what structure does its formula have?
+
+
+
+
+ How can we interpret the slope of a linear function in applied contexts?
+
+
+
+
+
+ Introduction
+
+ Functions whose graphs are straight lines are both the simplest and the most important functions in mathematics.
+ Lines often model important phenomena,
+ and even when they don't directly model phenomena,
+ lines can often approximate other functions that do.
+ Whether a function's graph is a straight line or not is connected directly to its average rate of change.
+
+
+
+
+
+
+
+
+
+ Properties of linear functions
+
+
+ In Preview Activity, we considered three different functions for which the average rate of change of each appeared to always be constant. For the first function in the preview activity, y = f(x) = 7 - 3x, we can compute its average rate of change on an arbitrary interval [a,b]. Doing so, we notice that
+
+ AV_{[a,b]} &= \frac{f(b)-f(a)}{b-a}
+ &= \frac{(7-3b) - (7-3a)}{b-a}
+ &= \frac{7-3b - 7+3a}{b-a}
+ &= \frac{-3b + 3a}{b-a}
+ &= \frac{-3(b - a)}{b-a}
+ &= -3
+ .
+ This result shows us that for the function y = f(x) = 7 - 3x, its average rate of change is always -3, regardless of the interval we choose. We will use the property of having constant rate of change as the defining property of a linear function.
+
+
+
+ linear function
+ linear functionslope
+
+
+ A function f is linear provided that its average rate of change is constant on every choice of interval in its domainHere we are considering functions whose domain is the set of all real numbers.. That is, for any inputs a and b for which a \ne b, it follows that
+
+ \frac{f(b) - f(a)}{b-a} = m
+
+ for some fixed constant m. We call m the slope of the linear function f.
+
+
+
+
+
+ From prior study, we already know a lot about linear functions. In this section, we work to understand some familiar properties in light of the new perspective of Definition.
+
+
+
+ Let's suppose we know that a function f is linear with average rate of change AV_{[a,b]} = m and that we also know the function value is y_0 at some fixed input x_0. That is, we know that f(x_0) = y_0. From this information, we can find the formula for y = f(x) for any input x. Working with the known point (x_0, f(x_0)) and any other point (x,f(x)) on the function's graph, we know that the average rate of change between these two points must be the constant m. This tells us that
+
+ \frac{f(x) - f(x_0)}{x-x_0} = m
+ .
+ Since we are interested in finding a formula for y = f(x), we solve this most recent equation for f(x). Multiplying both sides by (x-x_0), we see that
+
+ f(x) - f(x_0) = m(x-x_0)
+ .
+ Adding f(x_0) to each side, it follows
+
+ f(x) = f(x_0) + m(x-x_0)
+ .
+ This shows that to determine the formula for a linear function, all we need to know is its average rate of change (or slope) and a single point the function passes through.
+
+
+
+
+
+ Find a formula for a linear function f whose average rate of change is m = -\frac{1}{4} and passes through the point (-7,-5).
+
+
+
+ Solution. Using Equation and the facts that m = -\frac{1}{4} and f(-7) = -5 (that is, x_0 = -7 and f(x_0) = -5), we have
+
+ f(x) = -5 -\frac{1}{4}(x - (-7)) = -5 - \frac{1}{4}(x+7)
+ .
+
+
+
+
+
+ Replacing f(x) with y and f(x_0) with y_0, we call Equation the point-slope formpoint-slope form of a line.
+
+
+
+ Point-slope form of a line
+ linepoint-slope form
+
+ A line with slope m (equivalently, average rate of change m) that passes through the point (x_0,y_0) has equation
+
+ y = y_0 + m(x-x_0)
+ .
+
+
+
+
+
+
+ Visualizing the various components of point-slope form is important. For a line through (x_0,y_0) with slope m, we know its equation is y = y_0 + m(x-x_0). In Figure, we see that the line passes through (x_0,y_0) along with an arbitary point (x,y), which makes the vertical change between the two points given by y - y_0 and the horizontal change between the points x - x_0. This is consistent with the fact that
+
+ AV_{[x_0,x]} = m = \frac{y-y_0}{x-x_0}
+ .
+ Indeed, writing m = \frac{y-y_0}{x-x_0} is a rearrangement of the point-slope form of the line, y = y_0 + m(x-x_0).
+
+
+
+ We naturally use the terms increasing and decreasing as from Definition to describe lines based on whether their slope is positive or negative. A line with positive slope, such as the one in Figure, is increasing because its constant rate of change is positive, while a line with negative slope, such as in Figure is decreasing because of its negative rate of change. We say that a horizontal line (one whose slope is m = 0) is neither increasing nor decreasing.
+
+
+
+
+
The point-slope form of a line's equation.
+
The point-slope form of a line's equation.
+
+
+
+
The slope-intercept form of a line's equation.
+
The slope-intercept form of a line's equation.
+
+
+
+
+
+ A special case arises when the known point on a line satisfies x_0 = 0. In this situation, the known point lies on the y-axis, and thus we call the point the y-intercept of the line. liney-intercept The resulting form of the line's equation is called slope-intercept form, which is also demonstrated in Figure.
+
+
+
+ Slope-intercept form
+ lineslope-intercept form
+
+ For the line with slope m and passing through (0,y_0), its equation is
+
+ y = y_0 + mx
+ .
+
+
+
+
+ Slope-intercept form follows from point-slope form from the fact that replacing x_0 with 0 gives us y = y_0 + m(x-0) = y_0 + mx. In many textbooks, the slope-intercept form of a line (often written y = mx + b) is treated as if it is the most useful form of a line. Point-slope form is actually more important and valuable since we can easily write down the equation of a line as soon as we know its slope and any point that lies on it, as opposed to needing to find the y-intercept, which is needed for slope-intercept form. Moreover, point-slope form plays a prominent role in calculus.
+
+
+
+ If a line is in slope-intercept or point-slope form, it is useful to be able to quickly interpret key information about the line from the form of its equation.
+
+
+
+
+
+ For the line given by y = -3 - 2.5(x-5), determine its slope and a point that lies on the line.
+
+
+
+ Solution. This line is in point-slope form. Its slope is m = -2.5 and a point on the line is (5,-3).
+
+
+
+
+
+
+
+ For the line given by y = 6 + 0.25x, determine its slope and a point that lies on the line.
+
+
+
+ Solution. This line is in slope-intercept form. Its slope is m = 0.25 and a point on the line is (0,6), which is also the line's y-intercept.
+
+
+
+
+
+
+
+ Interpreting linear functions in context
+
+
+ Since linear functions are defined by the property that their average rate of change is constant, linear functions perfectly model quantities that change at a constant rate. In context, we can often think of slope as a rate of change; analyzing units carefully often yields significant insight.
+
+
+
+
+
+ The Dolbear function T = D(N) = 40 + 0.25N from Section is a linear function whose slope is m = 0.25. What is the meaning of the slope in this context?
+
+
+
+
+ Recall that T is measured in degrees Fahrenheit and N in chirps per minute. We know that m = AV_{[a,b]} = 0.25 is the constant average rate of change of D. Its units are units of output per unit of input, and thus degrees Fahrenheit per chirp per minute. This tells us that the average rate of change of the temperature function is 0.25 degrees Fahrenheit per chirp per minute, which means that for each additional chirp per minute observed, we expect the temperature to rise by 0.25 degrees Fahrenheit.
+
+
+
+
+ Indeed, we can observe this through function values. We note that T(60) = 55 and T(61) = 55.25: one additional observed chirp per minute corresponds to a 0.25 degree increase in temperature. We also see this in the graph of the line, as seen in Figure:
+ the slope between the points (40,50) and (120,70) is
+
+ m &= \frac{70-50}{120-40}
+ &= \frac{20}{80}
+ &= 0.25 \frac{\text{degrees F}}{\text{chirp per minute}}
+ .
+
+
+
+
The linear Dolbear function with slope m = 0.25 degrees Fahrenheit per chirp per minute.
+
The linear Dolbear function with slope m = 0.25 degrees Fahrenheit per chirp per minute.
+
+
+
+
+
+
+
+
+ Like with the Dolbear function, it is often useful to write a linear function (whose output is called y) that models a quantity changing at a constant rate (as a function of some input t) by writing the function relationship in the form
+
+ y = b + mt
+
+ where b and m are constants. We may think of the four quantities involved in the following way:
+
+
+
+
+
+
+ The constant b is the starting value of the output that corresponds to an input of t = 0;
+
+
+
+
+ The constant m is the rate at which the output changes with respect to changes in the input: for each additional 1-unit change in input, the output will change by m units.
+
+
+
+
+ The variable t is the independent (input) variable. A nonzero value for t corresponds to how much the input variable has changed from an initial value of 0.
+
+
+
+
+ The variable y is the dependent (output) variable. The value of y results from a particular choice of t, and can be thought of as the starting output value (b) plus the change in output that results from the corresponding change in input t.
+
+
+
+
+
+
+
+
+
+
+
+
+ Summary
+
+
+
+
+ Any function f with domain all real numbers that has a constant average rate of change on every interval [a,b] will have a straight line graph. We call such functions linear functions.
+
+
+
+
+ A linear function y = f(x) can be written in the form y = f(x) = y_0 + m(x-x_0), where m is the slope of the line and (x_0,y_0) is a point that lies on the line. In particular, f(x_0) = y_0.
+
+
+
+
+ In an applied context where we have a linear function that models a phenomenon in the world around us, the slope tells us the function's (constant) average rate of change. The units on the slope, m, are always units of output per unit of input and this enables us to articulate how the output changes in response to a 1-unit change in input.
+
- What patterns can we observe in how a quadratic function changes?
-
-
-
-
- What are familiar and important properties of quadratic functions?
-
-
-
-
- How can quadratic functions be used to model objects falling under the influence of gravity?
-
-
-
-
-
- Introduction
-
-
- After linear functions,
- quadratic functions are arguably the next simplest functions in mathematics.
- A quadratic functionquadratic function is one that may be written in the form
-
- q(x) = ax^2 + bx + c
- ,
- where a,
- b, and c are real numbers with a \ne 0. One of the reasons that quadratic functions are especially important is that they model the height of an object falling under the force of gravity.
-
- Quadratic functions are likely familiar to you from experience in previous courses. Throughout, we let y = q(x) = ax^2 + bx + c where a, b, and c are real numbers with a \ne 0. From the outset, it is important to note that when we write q(x) = ax^2 + bx + c we are thinking of an infinite family of functions where each member depends on the three parameters a, b, and c.
-
-
-
-
-
- Because quadratic functions are familiar to us, we will quickly restate some of their important known properties.
-
-
-
- Solutions to <m>q(x) = 0</m>
- quadratic formula
-
- Let a, b, and c be real numbers with a \ne 0. The equation ax^2 + bx + c = 0 can have 0, 1, or 2 real solutions. These real solutions are given by the quadratic formula,
-
- x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}
- ,
- provided that b^2 - 4ac \ge 0.
-
-
-
-
- As we can see in Figure, by shifting the graph of a quadratic function vertically, we can make its graph cross the x-axis 0 times (as in the graph of p), exactly 1 time (q), or twice (r). These points are the x-intercepts of the graph.
-
-
-
-
-
Three examples of quadratic functions that open up.
-
Three examples of quadratic functions that open up.
-
-
-
-
One example of a quadratic function that opens down.
-
One example of a quadratic function that opens down.
-
-
-
-
-
- While the quadratic formula will always provide any real solutions to q(x) = 0, in practice it is often easier to attempt to factor before using the formula. For instance, given q(x) = x^2 - 5x + 6, we can find its x-intercepts quickly by factoring. Since
-
- x^2 - 5x + 6 = (x-2)(x-3)
- ,
- it follows that (2,0) and (3,0) are the x-intercepts of q. Note more generally that if we know the x-intercepts quadratic functionx-intercepts of a quadratic function are (r,0) and (s,0), it follows that we can write the quadratic function in the form q(x) = a(x-r)(x-s).
-
-
-
- Every quadratic function has a y-intercept; quadratic functiony-intercept for a function of form y = q(x) = ax^2 + bx + c, the y-intercept is the point (0,c), as demonstrated in Figure.
-
-
- In addition, every quadratic function has a symmetric graph that either always curves upward or always curves downward. The graph opens upward if and only if a \gt 0 and opens downward if and only if a \lt 0. We often call the graph of a quadratic function a parabola. parabola Every parabola is symmetric about a vertical line that runs through its lowest or highest point.
-
-
-
- The vertex of a parabolaparabolavertexquadratic functionvertex
-
- The quadratic function y = q(x) = ax^2 + bx + c has its vertex at the point \left( -\frac{b}{2a}, q\left( -\frac{b}{2a} \right) \right). When a \gt 0, the vertex is the lowest point on the graph of q, while if a \lt 0, the vertex is the highest point. Moreover, the graph of q is symmetric about the vertical line x = -\frac{b}{2a}.
-
-
-
-
-
-
The vertex of a quadratic function that opens up.
-
The vertex of a quadratic function that opens up.
-
-
-
-
The vertex of a quadratic function that opens down.
-
The vertex of a quadratic function that opens down.
-
-
-
-
-
- Note particularly that due to symmetry, the vertex of a quadratic function lies halfway between its x-intercepts (provided the function has x-intercepts). In both Figures and , we see how the parabola is symmetric about the vertical line that passes through the vertex. One way to understand this symmetry can be seen by writing a given quadratic function in a different algebraic form.
-
-
-
-
-
- Consider the quadratic function in standard form given by y = q(x) = 0.25x^2 - x + 3.5. Determine constants a, h, and k so that q(x) = a(x-h)^2 + k, and hence determine the vertex of q. How does this alternate form of q explain the symmetry in its graph?
-
-
-
-
- We first observe that we can write q(x) = 0.25x^2 - x + 3.5 in a form closer to q(x) = a(x-h)^2 + k by factoring 0.25 from the first two terms to get
-
- q(x) = 0.25(x^2 - 4x) + 3.5
- .
- Next, we want to add a constant inside the parentheses to form a perfect square. Noting that (x-2)^2 = x^2 - 4x + 4, we need to add 4. Since we are adding 4 inside the parentheses, the 4 is being multiplied by 0.25, which has the net effect of adding 1 to the function. To keep the function as given, we must also subtract 1, and thus we have
-
- q(x) = 0.25(x^2 - 4x + 4) + 3.5 - 1
- .
- It follows that
-
- q(x) = 0.25(x-2)^2 + 2.5
- .
-
-
- Next, observe that the vertex of q is (2,2.5). This holds because (x-2)^2 is always greater than or equal to 0, and thus its smallest possible value is 0 when x = 2. Moreover, when x = 2, q(2) = 2.5.We can also verify this point is the vertex using standard form. From q(x) = 0.25x^2 - x + 3.5, we see that a = 0.25 and b = -1, so x = -\frac{b}{2a} = \frac{1}{0.5} = 2. In addition, q(2) = 2.5.
-
-
-
- Finally, the form q(x) = 0.25(x-2)^2 + 2.5 explains the symmetry of q about the line x = 2. Consider the two points that lie equidistant from x = 2 on the x-axis, z units away: x = 2-z and x = 2 + z. Observe that for these values,
-
- q(2-z) &= 0.25(2-z-2)^2 + 2.5& q(2+z) &= 0.25(2+z-2)^2 + 2.5
- &= 0.25(-z)^2 + 2.5 & &= 0.25(z)^2 + 2.5
- &= 0.25z^2 + 2.5 & &= 0.25z^2 + 2.5
-
- Since q(2-z) = q(2+z) for any choice of z, this shows the parabola is symmetric about the vertical line through its vertex.
-
-
-
-
-
- In Example, we saw some of the advantages of writing a quadratic function in the form q(x) = a(x-h)^2 + k. We call this the vertex form of a quadratic function.
-
-
-
- Vertex form of a quadratic functionquadratic functionvertex form
-
- A quadratic function with vertex (h,k) may be written in the form y = a(x-h)^2 + k. The constant a may be determined from one other function value for an input x \ne h.
-
-
-
-
-
-
-
-
- Modeling falling objects
-
-
- One of the reasons that quadratic functions are so important is because of a physical fact of the universe we inhabit:
- for an object only being influenced by gravity, gravity
- acceleration due to gravity is constant.
- If we measure time in seconds and a rising or falling object's height in feet,
- the gravitational constant is g = 32 feet per second per second.
-
-
-
- One of the fantastic consequences of calculus which,
- like the realization that acceleration due to gravity is constant,
- is largely due to Sir Isaac Newton in the late 1600s is that the height of a falling object at time t is modeled by a quadratic function.
-
-
-
- Height of an object falling under the force of gravity
- gravityfalling object
-
- For an object tossed vertically from an initial height of s_0 feet with a velocity of v_0 feet per second,
- the object's height at time t (in seconds) is given by the formula
-
- h(t) = -16t^2 + v_0t + s_0
-
-
-
-
-
- If height is measured instead in meters and velocity in meters per second, the gravitational constant is g = 9.8 and the function h has form h(t) = -4.9t^2 + v_0t + s_0. gravitygravitational constant (When height is measured in feet, the gravitational constant is g = 32.)
-
- So far, we've seen that quadratic functions have many interesting properties. In Preview Activity, we discovered an additional pattern that is particularly noteworthy.
-
-
-
- Recall that we considered a water balloon tossed vertically from a fifth story window whose height, h, in meters, at time t, in seconds, is modeledHere we are using a = -5 rather than a = -4.9 for simplicity. by the function
-
- h = q(t) = -5t^2 + 20t + 25
- .
- We then completed Table and Table to investigate how both function values and averages rates of change varied as we changed the input to the function.
-
- Average rates of change for <m>h</m> on select intervals <m>[a,b]</m>.
-
-
- [a,b]
- AV_{[a,b]}
-
-
- [0,1]
- AV_{[0,1]} = 15 m/s
-
-
- [1,2]
- AV_{[1,2]} = 5 m/s
-
-
- [2,3]
- AV_{[2,3]} = -5 m/s
-
-
- [3,4]
- AV_{[3,4]} = -15 m/s
-
-
- [4,5]
- AV_{[4,5]} = -25 m/s
-
-
-
-
-
-
-
-
-
- In Table, we see an interesting pattern in the average velocities of the ball. Indeed, if we remove the AV notation and focus on the starting value of each interval, viewing the resulting average rate of change, r, as a function of the starting value, we may consider the related table seen in Table, where it is apparent that r is a linear function of a.
-
-
-
-
-
- Data from <xref ref="T-PA-changing-quadratic-AVs-complete">Table</xref>, slightly recast.
-
-
- a
- r(a)
-
-
- 0
- r(0) = 15 m/s
-
-
- 1
- r(1) = 5 m/s
-
-
- 2
- r(2) = -5 m/s
-
-
- 3
- r(3) = -15 m/s
-
-
- 4
- r(4) = -25 m/s
-
-
-
-
-
-
Plot of h(t) = -5t^2 + 20t + 25 along with line segments whose slopes correspond to average rates of change.
-
Plot of h(t) = -5t^2 + 20t + 25 along with line segments whose slopes correspond to average rates of change.
-
-
-
-
-
-
- Indeed, viewing this data graphically as in Figure, we observe that the average rate of change of h is itself changing in a way that seems to be represented by a linear function. While it takes key ideas from calculus to formalize this observation, for now we will simply note that for a quadratic function there seems to be a related linear function that tells us something about how the quadratic function changes. Moreover, we can also say that on the downward-opening quadratic function h that its average rate of change appears to be decreasing as we move from left to rightProvided that we consider the average rate of change on intervals of the same length. Again, it takes ideas from calculus to make this observation completely precise..
-
-
-
- A key closing observation here is that the fact the parabola bends down is apparently connected to the fact that its average rate of change decreases as we move left to right. By contrast, for a quadratic function that bends up, we can show that its average rate of change increases as we move left to right (see Exercise). Moreover, we also see that it's possible to view the average rate of change of a function on 1-unit intervals as itself being a function: a process that relates an input (the starting value of the interval) to a corresponding output (the average rate of change of the original function on the resulting 1-unit interval).
-
-
-
- For any function that consistently bends either exclusively upward or exclusively downward on a given interval (a,b), we use the following formal languageCalculus is needed to make Definition rigorous and precise. to describe it.
-
-
-
- function trendsconcave upfunction trendsconcave down
-
-
- If a function f always bends upward on an interval (a,b), we say that f is concave up on (a,b). Similarly, if f always bends downward on an interval (a,b), we say that f is concave down on (a,b).
-
-
-
-
-
- Thus, we now call a quadratic function q(x) = ax^2 + bx + c with a \gt 0concave up, while if a \lt 0 we say q is concave down.
-
-
-
-
-
- Summary
-
-
-
-
- Quadratic functions (of the form q(x) = ax^2 + bx + c with a \ne 0) are emphatically not linear: their average rate of change is not constant, but rather depends on the interval chosen. At the same time, quadratic functions appear to change in a very regimented way: if we compute the average rate of change on several consecutive 1-unit intervals, it appears that the average rate of change itself changes at a constant rate. Quadratic functions either bend upward (a \gt 0) or bend downward (a \lt 0) and these shapes are connected to whether the average rate of change on consecutive 1-unit intervals decreases or increases as we move left to right.
-
-
-
-
- For an object with height h measured in feet at time t in seconds, if the object was launched vertically at an initial velocity of v_0 feet per second and from an initial height of s_0 feet, the object's height is given by
-
- h = q(t) = -16t^2 + v_0t + s_0
- .
- That is, the object's height is completely determined by the initial height and initial velocity from which it was launched. The model is valid for the entire time until the object lands. If h is instead measured in meters and v_0 in meters per second, -16 is replaced with -4.9.
-
-
-
-
- A quadratic function q can be written in one of three familiar forms: standard, vertex, or factoredIt's not always possible to write a quadratic function in factored form involving only real numbers; this can only be done if it has 1 or 2x-intercepts.. Table shows how, depending on the algebraic form of the function, various properties may be (easily) read from the formula. In every case, the sign of a determines whether the function opens up or opens down.
-
-
- A summary of the information that can be read from the various algebraic forms of a quadratic function
-
-
-
- standard
- vertex
- factoredProvided q has 1 or 2x-intercepts. In the case of just one, we take r = s.
-
-
- form
- q(x) = ax^2 + bx + c
- q(x) = a(x-h)^2 + k
- q(x) = a(x-r)(x-s)
-
-
- y-int
- (0,c)
- (0,ah^2 + k)
- (0,ars)
-
-
- x-intProvided b^2 - 4ac \ge 0 for standard form; provided -\frac{k}{a} \ge 0 for vertex form.
- \left(\frac{-b \pm \sqrt{b^2 - 4ac}}{2a} , 0 \right)
- \left(h \pm \sqrt{-\frac{k}{a}} , 0 \right)
- (r,0), (s,0)
-
-
- vertex
- \left(-\frac{b}{2a}, q\left( -\frac{b}{2a} \right) \right)
- (h,k)
- \left( \frac{r+s}{2}, q\left( \frac{r+s}{2} \right) \right)
-
-
-
+ What patterns can we observe in how a quadratic function changes?
+
+
+
+
+ What are familiar and important properties of quadratic functions?
+
+
+
+
+ How can quadratic functions be used to model objects falling under the influence of gravity?
+
+
+
+
+
+ Introduction
+
+
+ After linear functions,
+ quadratic functions are arguably the next simplest functions in mathematics.
+ A quadratic functionquadratic function is one that may be written in the form
+
+ q(x) = ax^2 + bx + c
+ ,
+ where a,
+ b, and c are real numbers with a \ne 0. One of the reasons that quadratic functions are especially important is that they model the height of an object falling under the force of gravity.
+
+ Quadratic functions are likely familiar to you from experience in previous courses. Throughout, we let y = q(x) = ax^2 + bx + c where a, b, and c are real numbers with a \ne 0. From the outset, it is important to note that when we write q(x) = ax^2 + bx + c we are thinking of an infinite family of functions where each member depends on the three parameters a, b, and c.
+
+
+
+
+
+ Because quadratic functions are familiar to us, we will quickly restate some of their important known properties.
+
+
+
+ Solutions to <m>q(x) = 0</m>
+ quadratic formula
+
+ Let a, b, and c be real numbers with a \ne 0. The equation ax^2 + bx + c = 0 can have 0, 1, or 2 real solutions. These real solutions are given by the quadratic formula,
+
+ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}
+ ,
+ provided that b^2 - 4ac \ge 0.
+
+
+
+
+ As we can see in Figure, by shifting the graph of a quadratic function vertically, we can make its graph cross the x-axis 0 times (as in the graph of p), exactly 1 time (q), or twice (r). These points are the x-intercepts of the graph.
+
+
+
+
+
Three examples of quadratic functions that open up.
+
Three examples of quadratic functions that open up.
+
+
+
+
One example of a quadratic function that opens down.
+
One example of a quadratic function that opens down.
+
+
+
+
+
+ While the quadratic formula will always provide any real solutions to q(x) = 0, in practice it is often easier to attempt to factor before using the formula. For instance, given q(x) = x^2 - 5x + 6, we can find its x-intercepts quickly by factoring. Since
+
+ x^2 - 5x + 6 = (x-2)(x-3)
+ ,
+ it follows that (2,0) and (3,0) are the x-intercepts of q. Note more generally that if we know the x-intercepts quadratic functionx-intercepts of a quadratic function are (r,0) and (s,0), it follows that we can write the quadratic function in the form q(x) = a(x-r)(x-s).
+
+
+
+ Every quadratic function has a y-intercept; quadratic functiony-intercept for a function of form y = q(x) = ax^2 + bx + c, the y-intercept is the point (0,c), as demonstrated in Figure.
+
+
+ In addition, every quadratic function has a symmetric graph that either always curves upward or always curves downward. The graph opens upward if and only if a \gt 0 and opens downward if and only if a \lt 0. We often call the graph of a quadratic function a parabola. parabola Every parabola is symmetric about a vertical line that runs through its lowest or highest point.
+
+
+
+ The vertex of a parabolaparabolavertexquadratic functionvertex
+
+ The quadratic function y = q(x) = ax^2 + bx + c has its vertex at the point \left( -\frac{b}{2a}, q\left( -\frac{b}{2a} \right) \right). When a \gt 0, the vertex is the lowest point on the graph of q, while if a \lt 0, the vertex is the highest point. Moreover, the graph of q is symmetric about the vertical line x = -\frac{b}{2a}.
+
+
+
+
+
+
The vertex of a quadratic function that opens up.
+
The vertex of a quadratic function that opens up.
+
+
+
+
The vertex of a quadratic function that opens down.
+
The vertex of a quadratic function that opens down.
+
+
+
+
+
+ Note particularly that due to symmetry, the vertex of a quadratic function lies halfway between its x-intercepts (provided the function has x-intercepts). In both Figures and , we see how the parabola is symmetric about the vertical line that passes through the vertex. One way to understand this symmetry can be seen by writing a given quadratic function in a different algebraic form.
+
+
+
+
+
+ Consider the quadratic function in standard form given by y = q(x) = 0.25x^2 - x + 3.5. Determine constants a, h, and k so that q(x) = a(x-h)^2 + k, and hence determine the vertex of q. How does this alternate form of q explain the symmetry in its graph?
+
+
+
+
+ We first observe that we can write q(x) = 0.25x^2 - x + 3.5 in a form closer to q(x) = a(x-h)^2 + k by factoring 0.25 from the first two terms to get
+
+ q(x) = 0.25(x^2 - 4x) + 3.5
+ .
+ Next, we want to add a constant inside the parentheses to form a perfect square. Noting that (x-2)^2 = x^2 - 4x + 4, we need to add 4. Since we are adding 4 inside the parentheses, the 4 is being multiplied by 0.25, which has the net effect of adding 1 to the function. To keep the function as given, we must also subtract 1, and thus we have
+
+ q(x) = 0.25(x^2 - 4x + 4) + 3.5 - 1
+ .
+ It follows that
+
+ q(x) = 0.25(x-2)^2 + 2.5
+ .
+
+
+ Next, observe that the vertex of q is (2,2.5). This holds because (x-2)^2 is always greater than or equal to 0, and thus its smallest possible value is 0 when x = 2. Moreover, when x = 2, q(2) = 2.5.We can also verify this point is the vertex using standard form. From q(x) = 0.25x^2 - x + 3.5, we see that a = 0.25 and b = -1, so x = -\frac{b}{2a} = \frac{1}{0.5} = 2. In addition, q(2) = 2.5.
+
+
+
+ Finally, the form q(x) = 0.25(x-2)^2 + 2.5 explains the symmetry of q about the line x = 2. Consider the two points that lie equidistant from x = 2 on the x-axis, z units away: x = 2-z and x = 2 + z. Observe that for these values,
+
+ q(2-z) &= 0.25(2-z-2)^2 + 2.5& q(2+z) &= 0.25(2+z-2)^2 + 2.5
+ &= 0.25(-z)^2 + 2.5 & &= 0.25(z)^2 + 2.5
+ &= 0.25z^2 + 2.5 & &= 0.25z^2 + 2.5
+
+ Since q(2-z) = q(2+z) for any choice of z, this shows the parabola is symmetric about the vertical line through its vertex.
+
+
+
+
+
+ In Example, we saw some of the advantages of writing a quadratic function in the form q(x) = a(x-h)^2 + k. We call this the vertex form of a quadratic function.
+
+
+
+ Vertex form of a quadratic functionquadratic functionvertex form
+
+ A quadratic function with vertex (h,k) may be written in the form y = a(x-h)^2 + k. The constant a may be determined from one other function value for an input x \ne h.
+
+
+
+
+
+
+
+
+ Modeling falling objects
+
+
+ One of the reasons that quadratic functions are so important is because of a physical fact of the universe we inhabit:
+ for an object only being influenced by gravity, gravity
+ acceleration due to gravity is constant.
+ If we measure time in seconds and a rising or falling object's height in feet,
+ the gravitational constant is g = 32 feet per second per second.
+
+
+
+ One of the fantastic consequences of calculus which,
+ like the realization that acceleration due to gravity is constant,
+ is largely due to Sir Isaac Newton in the late 1600s is that the height of a falling object at time t is modeled by a quadratic function.
+
+
+
+ Height of an object falling under the force of gravity
+ gravityfalling object
+
+ For an object tossed vertically from an initial height of s_0 feet with a velocity of v_0 feet per second,
+ the object's height at time t (in seconds) is given by the formula
+
+ h(t) = -16t^2 + v_0t + s_0
+
+
+
+
+
+ If height is measured instead in meters and velocity in meters per second, the gravitational constant is g = 9.8 and the function h has form h(t) = -4.9t^2 + v_0t + s_0. gravitygravitational constant (When height is measured in feet, the gravitational constant is g = 32.)
+
+ So far, we've seen that quadratic functions have many interesting properties. In Preview Activity, we discovered an additional pattern that is particularly noteworthy.
+
+
+
+ Recall that we considered a water balloon tossed vertically from a fifth story window whose height, h, in meters, at time t, in seconds, is modeledHere we are using a = -5 rather than a = -4.9 for simplicity. by the function
+
+ h = q(t) = -5t^2 + 20t + 25
+ .
+ We then completed Table and Table to investigate how both function values and averages rates of change varied as we changed the input to the function.
+
+ Average rates of change for <m>h</m> on select intervals <m>[a,b]</m>.
+
+
+ [a,b]
+ AV_{[a,b]}
+
+
+ [0,1]
+ AV_{[0,1]} = 15 m/s
+
+
+ [1,2]
+ AV_{[1,2]} = 5 m/s
+
+
+ [2,3]
+ AV_{[2,3]} = -5 m/s
+
+
+ [3,4]
+ AV_{[3,4]} = -15 m/s
+
+
+ [4,5]
+ AV_{[4,5]} = -25 m/s
+
+
+
+
+
+
+
+
+
+ In Table, we see an interesting pattern in the average velocities of the ball. Indeed, if we remove the AV notation and focus on the starting value of each interval, viewing the resulting average rate of change, r, as a function of the starting value, we may consider the related table seen in Table, where it is apparent that r is a linear function of a.
+
+
+
+
+
+ Data from <xref ref="T-PA-changing-quadratic-AVs-complete">Table</xref>, slightly recast.
+
+
+ a
+ r(a)
+
+
+ 0
+ r(0) = 15 m/s
+
+
+ 1
+ r(1) = 5 m/s
+
+
+ 2
+ r(2) = -5 m/s
+
+
+ 3
+ r(3) = -15 m/s
+
+
+ 4
+ r(4) = -25 m/s
+
+
+
+
+
+
Plot of h(t) = -5t^2 + 20t + 25 along with line segments whose slopes correspond to average rates of change.
+
Plot of h(t) = -5t^2 + 20t + 25 along with line segments whose slopes correspond to average rates of change.
+
+
+
+
+
+
+ Indeed, viewing this data graphically as in Figure, we observe that the average rate of change of h is itself changing in a way that seems to be represented by a linear function. While it takes key ideas from calculus to formalize this observation, for now we will simply note that for a quadratic function there seems to be a related linear function that tells us something about how the quadratic function changes. Moreover, we can also say that on the downward-opening quadratic function h that its average rate of change appears to be decreasing as we move from left to rightProvided that we consider the average rate of change on intervals of the same length. Again, it takes ideas from calculus to make this observation completely precise..
+
+
+
+ A key closing observation here is that the fact the parabola bends down is apparently connected to the fact that its average rate of change decreases as we move left to right. By contrast, for a quadratic function that bends up, we can show that its average rate of change increases as we move left to right (see Exercise). Moreover, we also see that it's possible to view the average rate of change of a function on 1-unit intervals as itself being a function: a process that relates an input (the starting value of the interval) to a corresponding output (the average rate of change of the original function on the resulting 1-unit interval).
+
+
+
+ For any function that consistently bends either exclusively upward or exclusively downward on a given interval (a,b), we use the following formal languageCalculus is needed to make Definition rigorous and precise. to describe it.
+
+
+
+ function trendsconcave upfunction trendsconcave down
+
+
+ If a function f always bends upward on an interval (a,b), we say that f is concave up on (a,b). Similarly, if f always bends downward on an interval (a,b), we say that f is concave down on (a,b).
+
+
+
+
+
+ Thus, we now call a quadratic function q(x) = ax^2 + bx + c with a \gt 0concave up, while if a \lt 0 we say q is concave down.
+
+
+
+
+
+ Summary
+
+
+
+
+ Quadratic functions (of the form q(x) = ax^2 + bx + c with a \ne 0) are emphatically not linear: their average rate of change is not constant, but rather depends on the interval chosen. At the same time, quadratic functions appear to change in a very regimented way: if we compute the average rate of change on several consecutive 1-unit intervals, it appears that the average rate of change itself changes at a constant rate. Quadratic functions either bend upward (a \gt 0) or bend downward (a \lt 0) and these shapes are connected to whether the average rate of change on consecutive 1-unit intervals decreases or increases as we move left to right.
+
+
+
+
+ For an object with height h measured in feet at time t in seconds, if the object was launched vertically at an initial velocity of v_0 feet per second and from an initial height of s_0 feet, the object's height is given by
+
+ h = q(t) = -16t^2 + v_0t + s_0
+ .
+ That is, the object's height is completely determined by the initial height and initial velocity from which it was launched. The model is valid for the entire time until the object lands. If h is instead measured in meters and v_0 in meters per second, -16 is replaced with -4.9.
+
+
+
+
+ A quadratic function q can be written in one of three familiar forms: standard, vertex, or factoredIt's not always possible to write a quadratic function in factored form involving only real numbers; this can only be done if it has 1 or 2x-intercepts.. Table shows how, depending on the algebraic form of the function, various properties may be (easily) read from the formula. In every case, the sign of a determines whether the function opens up or opens down.
+
+
+ A summary of the information that can be read from the various algebraic forms of a quadratic function
+
+
+
+ standard
+ vertex
+ factoredProvided q has 1 or 2x-intercepts. In the case of just one, we take r = s.
+
+
+ form
+ q(x) = ax^2 + bx + c
+ q(x) = a(x-h)^2 + k
+ q(x) = a(x-r)(x-s)
+
+
+ y-int
+ (0,c)
+ (0,ah^2 + k)
+ (0,ars)
+
+
+ x-intProvided b^2 - 4ac \ge 0 for standard form; provided -\frac{k}{a} \ge 0 for vertex form.
+ \left(\frac{-b \pm \sqrt{b^2 - 4ac}}{2a} , 0 \right)
+ \left(h \pm \sqrt{-\frac{k}{a}} , 0 \right)
+ (r,0), (s,0)
+
+
+ vertex
+ \left(-\frac{b}{2a}, q\left( -\frac{b}{2a} \right) \right)
+ (h,k)
+ \left( \frac{r+s}{2}, q\left( \frac{r+s}{2} \right) \right)
+
+
+
- How is the graph of y = g(x) = af(x-b) + c related to the graph of y = f(x)?
-
-
-
-
- What do we mean by transformations of a given function f? How are translations and vertical stretches of a function examples of transformations?
-
-
-
-
-
- Introduction
-
- In our preparation for calculus, we aspire to understand functions from a wide range of perspectives and to become familiar with a library of basic functions. So far, two basic families of functions we have considered are linear functions and quadratic functions, the simplest of which are L(x) = x and Q(x) = x^2. As we progress further, we will endeavor to understand a parent function as the most fundamental member of a family of functions, as well as how other similar but more complicated functions are the result of transforming the parent function.
-
-
-
- Informally, a transformation transformation of a function of a given function is an algebraic process by which we change the function to a related function that has the same fundamental shape, but may be shifted, reflected, and/or stretched in a systematic way. For example, among all quadratic functions, the simplest is the parent function Q(x) = x^2, but any other quadratic function such as g(x) = -3(x-5)^2 + 4 can also be understood in relation to the parent function. We say that g is a transformation of Q.
-
-
-
- In Preview Activity, we investigate the effects of the constants a, b, and c in generating the function g(x) = af(x-b) + c in the context of already knowing the function f.
-
-
-
-
-
-
-
- Translations of Functions
-
- We begin by summarizing two of our findings in Preview Activity.
-
-
-
- Vertical Translation of a Function
- transformation of a functionvertical translation
-
- Given a function y = f(x) and a real number a, the transformed function y = g(x) = f(x) + a is a vertical translation of the graph of f. That is, every point (x,f(x)) on the graph of f gets shifted vertically to the corresponding point (x,f(x)+a) on the graph of g.
-
-
-
-
- As we found in our Desmos explorations in the preview activity, is especially helpful to see the effects of vertical translation dynamically.
-
-
-
-
Interactive vertical translations demonstration (in the HTML version only).
-
-
-
Interactive vertical translations demonstration (in the HTML version only).
-
-
-
-
-
- Move the sliderHuge thanks to the amazing David Austin for making these interactive javascript graphics for the text. by clicking and dragging on the red point to see how changing a affects the graph of y = f(x) + a, which appears in blue. The graph of y = f(x) will appear in grey and remain fixed.
-
-
-
-
-
-
- In a vertical translation, the graph of g lies above the graph of f whenever a \gt 0, while the graph of g lies below the graph of f whenever a \lt 0. In Figure, we see the original parent function f(x) = |x| along with the resulting transformation g(x) = f(x)-3, which is a downward vertical shift of 3 units. Note particularly that every point on the original graph of f is moved 3 units down; we often indicate this by an arrow and labeling at least one key point on each graph.
-
-
-
-
-
A vertical translation, g, of the function y = f(x) = |x|.
-
A vertical translation, g, of the function y = f(x) = |x|.
-
-
-
-
A horizontal translation, h, of a different function y = f(x).
-
A horizontal translation, h, of a different function y = f(x).
-
-
-
-
-
- In Figure, we see a horizontal translation of the original function f that shifts its graph 2 units to the right to form the function h. Observe that f is not a familiar basic function; transformations may be applied to any original function we desire.
-
-
-
- From an algebraic point of view, horizontal translations are slightly more complicated than vertical ones. Given y = f(x), if we define the transformed function y = h(x) = f(x-b), observe that
-
- h(x+b) = f( (x+b) - b ) = f(x)
- .
- This shows that for an input of x+b in h, the output of h is the same as the output of f that corresponds to an input of simply x. Hence, in Figure, the formula for h in terms of f is h(x) = f(x-2), since an input of x+2 in h will result in the same output as an input of x in f. For example, h(2) = f(0), which aligns with the graph of h being a shift of the graph of f to the right by 2 units.
-
-
-
- Again, it's instructive to see the effects of horizontal translation dynamically.
-
-
-
-
Interactive horizontal translations demonstration (in the HTML version only).
-
-
-
Interactive horizontal translations demonstration (in the HTML version only).
-
-
-
-
-
- Move the slider by clicking and dragging on the red point to see how changing b affects the graph of y = f(x-b), which appears in blue. The graph of y = f(x) will appear in grey and remain fixed.
-
-
-
-
-
-
- Overall, we have the following general principle.
-
-
-
- Horizontal Translation of a Function
- transformation of a functionhorizontal translation
-
- Given a function y = f(x) and a real number b, the transformed function y = h(x) = f(x-b) is a horizontal translation of the graph of f. That is, every point (x,f(x)) on the graph of f gets shifted horizontally to the corresponding point (x+b,f(x)) on the graph of g.
-
-
-
-
- We emphasize that in the horizontal translation h(x) = f(x-b), if b \gt 0 the graph of h lies b units to the right of f, while if b \lt 0, h lies b units to the left of f.
-
- So far, we have seen the possible effects of adding a constant value to function's output (that is, the new expression f(x)+a if given f(x)) and adding a constant value to function input (that is, the new expression f(x+b), given f(x)). Each of these actions results in a translation of the function's graph (either vertically or horizontally), but otherwise leaving the graph the same. Next, we investigate the effects of multiplication the function's output by a constant.
-
-
-
-
-
- Given the parent function y = f(x) pictured in Figure, what are the effects of the transformation y = v(x) = cf(x) for various values of c?
-
-
-
-
-
- We first investigate the effects of c = 2 and c = \frac{1}{2}. For v(x) = 2f(x), the algebraic impact of this transformation is that every output of f is multiplied by 2. This means that the only output that is unchanged is when f(x) = 0, while any other point on the graph of the original function f will be stretched vertically away from the x-axis by a factor of 2. We can see this in Figure where each point on the original dark blue graph is transformed to a corresponding point whose y-coordinate is twice as large, as partially indicated by the red arrows.
-
-
-
-
-
The parent function y = f(x) along with two different vertical stretches, v and u.
-
The parent function y = f(x) along with two different vertical stretches, v and u.
-
-
-
-
The parent function y = f(x) along with a vertical reflection, z, and a corresponding stretch, w.
-
The parent function y = f(x) along with a vertical reflection, z, and a corresponding stretch, w.
-
-
-
-
-
- In contrast, the transformation u(x) = \frac{1}{2}f(x) is stretched vertically by a factor of \frac{1}{2}, which has the effect of compressing the graph of f towards the x-axis, as all function outputs of f are multiplied by \frac{1}{2}. For instance, the point (0,-2) on the graph of f is transformed to the graph of (0,-1) on the graph of u, and others are transformed as indicated by the purple arrows.
-
-
-
- To consider the situation where c \lt 0, we first consider the simplest case where c = -1 in the transformation z(x) = -f(x). Here the impact of the transformation is to multiply every output of the parent function f by -1; this takes any point of form (x,y) and transforms it to (x,-y), which means we are reflecting each point on the original function's graph across the x-axis to generate the resulting function's graph. This is demonstrated in Figure where y = z(x) is the reflection of y = f(x) across the x-axis.
-
-
-
- Finally, we also investigate the case where c = -2, which generates y = w(x) = -2f(x). Here we can think of -2 as -2 = 2(-1): the effect of multiplying by -1 first reflects the graph of f across the x-axis (resulting in z), and then multiplying by 2 stretches the graph of z vertically to result in w, as shown in Figure.
-
-
-
-
-
-
- As with vertical and horizontal translation, it's particularly instructive to see the effects of vertical scaling in a dynamic way.
-
-
-
-
Interactive vertical scaling demonstration (in the HTML version only).
-
-
-
Interactive vertical scaling demonstration (in the HTML version only).
-
-
-
-
-
- Move the slider by clicking and dragging on the red point to see how changing c affects the graph of y = cf(x), which is shown in blue. The graph of y = f(x) will appear in grey and remain fixed.
-
-
-
-
-
-
- We summarize and generalize our observations from Example and Figure as follows.
-
-
-
- Vertical Scaling of a Function
- transformation of a functionvertical scaling
-
- Given a function y = f(x) and a real number c \gt 0, the transformed function y = v(x) = cf(x) is a vertical stretch of the graph of f. Every point (x,f(x)) on the graph of f gets stretched vertically to the corresponding point (x,cf(x)) on the graph of v. If 0 \lt c \lt 1, the graph of v is a compression of f toward the x-axis; if c \gt 1, the graph of v is a stretch of f away from the x-axis. Points where f(x) = 0 are unchanged by the transformation.
-
-
-
- Given a function y = f(x) and a real number c \lt 0, the transformed function y = v(x) = cf(x) is a reflection of the graph of f across the x-axis followed by a vertical stretch by a factor of |c|.
-
-
-
-
-
-
-
-
- Combining shifts and stretches: why order sometimes matters
-
-
- In the final question of Activity, we considered the transformation y = m(x) = 2r(x+1)-1 of the original function r. There are three different basic transformations involved: a vertical shift of 1 unit down, a horizontal shift of 1 unit left, and a vertical stretch by a factor of 2. To understand the order in which these transformations are applied, it's essential to remember that a function is a process that converts inputs to outputs.
-
-
-
- By the algebraic rule for m, m(x) = 2r(x+1)-1. In words, this means that given an input x for m, we do the following processes in this particular order:
-
-
-
-
- add 1 to x and then
- apply the function r to the quantity x+1;
-
-
-
-
- multiply the output of r(x+1) by 2;
-
-
-
-
- subtract 1 from the output of 2r(x+1).
-
-
-
-
- These three steps correspond to three basic transformations: (1) shift the graph of r to the left by 1 unit; (2) stretch the resulting graph vertically by a factor of 2; (3) shift the resulting graph vertically by -1 units. We can see the graphical impact of these algebraic steps by taking them one at a time. In Figure, we see the function p that results from a shift 1 unit left of the parent function in Figure. (Each time we take an additional step, we will de-emphasize the preceding function by having it appear in lighter color and dashed.)
-
-
-
-
-
The parent function y = r(x).
-
The parent function y = r(x).
-
-
-
-
The parent function y = r(x) along with the horizontal shift y = p(x) = r(x+1).
-
The parent function y = r(x) along with the horizontal shift y = p(x) = r(x+1).
-
-
-
-
-
- Continuing, we now consider the function q(x) = 2p(x) = 2r(x+1), which results in a vertical stretch of p away from the x-axis by a factor of 2, as seen in Figure.
-
-
-
-
-
The function y = q(x) = 2p(x) = 2r(x+1) along with graphs of p and r.
-
The function y = q(x) = 2p(x) = 2r(x+1) along with graphs of p and r.
-
-
-
-
The function y = m(x) = q(x)-1 = 2r(x+1) - 1 along with graphs of q, p and r.
-
The function y = m(x) = q(x)-1 = 2r(x+1) - 1 along with graphs of q, p and r.
-
-
-
-
-
- Finally, we arrive at y = m(x) = 2r(x+1) - 1 by subtracting 1 from q(x) = 2r(x+1); this of course is a vertical shift of -1 units, and produces the graph of m shown in red in Figure. We can also track the point (2,-1) on the original parent function: it first moves left 1 unit to (1,-1), then it is stretched vertically by a factor of 2 away from the x-axis to (1,-2), and lastly is shifted 1 unit down to the point (1,-3), which we see on the graph of m.
-
-
-
- While there are some transformations that can be executed in either order (such as a combination of a horizontal translation and a vertical translation, as seen in part (b) of Activity), in other situations order matters. For instance, in our preceding discussion, we have to apply the vertical stretch before applying the vertical shift. Algebraically, this is because
-
- 2r(x+1) - 1 \ne 2[r(x+1)-1]
- .
- The quantity 2r(x+1) - 1 multiplies the function r(x+1) by 2 first (the stretch) and then the vertical shift follows; the quantity 2[r(x+1) - 1] shifts the function r(x+1) down 1 unit first, and then executes a vertical stretch by a factor of 2. In the latter scenario, the point (1,-1) that lies on r(x+1) gets transformed first to (1,-2) and then to (1,-4), which is not the same as the point (1,-3) that lies on m(x) = 2r(x+1) - 1.
-
-
-
-
-
-
-
- Summary
-
-
-
-
- The graph of y = g(x) = af(x-b) + c is related to the graph of y = f(x) by a sequence of transformations. First, there is horizontal shift of |b| units to the right (b \gt 0) or left (b \lt 0). Next, there is a vertical stretch by a factor of |a| (along with a reflection across y = 0 in the case where a \lt 0). Finally, there's a vertical shift of c units.
-
-
-
-
- A transformation of a given function f is a process by which the graph may be shifted or stretched to generate a new, related function with fundamentally the same shape. In this section we considered four different ways this can occur: through a horizontal translation (shift), through a reflection across the line y = 0 (the x-axis), through a vertical scaling (stretch) that multiplies every output of a function by the same constant, and through a vertical translation (shift). Each of these individual processes is itself a transformation, and they may be combined in various ways to create more complicated transformations.
-
+ How is the graph of y = g(x) = af(x-b) + c related to the graph of y = f(x)?
+
+
+
+
+ What do we mean by transformations of a given function f? How are translations and vertical stretches of a function examples of transformations?
+
+
+
+
+
+ Introduction
+
+ In our preparation for calculus, we aspire to understand functions from a wide range of perspectives and to become familiar with a library of basic functions. So far, two basic families of functions we have considered are linear functions and quadratic functions, the simplest of which are L(x) = x and Q(x) = x^2. As we progress further, we will endeavor to understand a parent function as the most fundamental member of a family of functions, as well as how other similar but more complicated functions are the result of transforming the parent function.
+
+
+
+ Informally, a transformation transformation of a function of a given function is an algebraic process by which we change the function to a related function that has the same fundamental shape, but may be shifted, reflected, and/or stretched in a systematic way. For example, among all quadratic functions, the simplest is the parent function Q(x) = x^2, but any other quadratic function such as g(x) = -3(x-5)^2 + 4 can also be understood in relation to the parent function. We say that g is a transformation of Q.
+
+
+
+ In Preview Activity, we investigate the effects of the constants a, b, and c in generating the function g(x) = af(x-b) + c in the context of already knowing the function f.
+
+
+
+
+
+
+
+
+
+ Translations of Functions
+
+ We begin by summarizing two of our findings in Preview Activity.
+
+
+
+ Vertical Translation of a Function
+ transformation of a functionvertical translation
+
+ Given a function y = f(x) and a real number a, the transformed function y = g(x) = f(x) + a is a vertical translation of the graph of f. That is, every point (x,f(x)) on the graph of f gets shifted vertically to the corresponding point (x,f(x)+a) on the graph of g.
+
+
+
+
+ As we found in our Desmos explorations in the preview activity, is especially helpful to see the effects of vertical translation dynamically.
+
+
+
+
Interactive vertical translations demonstration (in the HTML version only).
+
+
+
Interactive vertical translations demonstration (in the HTML version only).
+
+
+
+
+
+ Move the sliderHuge thanks to the amazing David Austin for making these interactive javascript graphics for the text. by clicking and dragging on the red point to see how changing a affects the graph of y = f(x) + a, which appears in blue. The graph of y = f(x) will appear in grey and remain fixed.
+
+
+
+
+
+
+ In a vertical translation, the graph of g lies above the graph of f whenever a \gt 0, while the graph of g lies below the graph of f whenever a \lt 0. In Figure, we see the original parent function f(x) = |x| along with the resulting transformation g(x) = f(x)-3, which is a downward vertical shift of 3 units. Note particularly that every point on the original graph of f is moved 3 units down; we often indicate this by an arrow and labeling at least one key point on each graph.
+
+
+
+
+
A vertical translation, g, of the function y = f(x) = |x|.
+
+
+
+
+
+
A vertical translation, g, of the function y = f(x) = |x|.
+
+
+
+
A horizontal translation, h, of a different function y = f(x).
+
+
+
+
+
+
A horizontal translation, h, of a different function y = f(x).
+
+
+
+
+
+
+ In Figure, we see a horizontal translation of the original function f that shifts its graph 2 units to the right to form the function h. Observe that f is not a familiar basic function; transformations may be applied to any original function we desire.
+
+
+
+ From an algebraic point of view, horizontal translations are slightly more complicated than vertical ones. Given y = f(x), if we define the transformed function y = h(x) = f(x-b), observe that
+
+ h(x+b) = f( (x+b) - b ) = f(x)
+ .
+ This shows that for an input of x+b in h, the output of h is the same as the output of f that corresponds to an input of simply x. Hence, in Figure, the formula for h in terms of f is h(x) = f(x-2), since an input of x+2 in h will result in the same output as an input of x in f. For example, h(2) = f(0), which aligns with the graph of h being a shift of the graph of f to the right by 2 units.
+
+
+
+ Again, it's instructive to see the effects of horizontal translation dynamically.
+
+
+
+
Interactive horizontal translations demonstration (in the HTML version only).
+
+
+
Interactive horizontal translations demonstration (in the HTML version only).
+
+
+
+
+
+ Move the slider by clicking and dragging on the red point to see how changing b affects the graph of y = f(x-b), which appears in blue. The graph of y = f(x) will appear in grey and remain fixed.
+
+
+
+
+
+
+ Overall, we have the following general principle.
+
+
+
+ Horizontal Translation of a Function
+ transformation of a functionhorizontal translation
+
+ Given a function y = f(x) and a real number b, the transformed function y = h(x) = f(x-b) is a horizontal translation of the graph of f. That is, every point (x,f(x)) on the graph of f gets shifted horizontally to the corresponding point (x+b,f(x)) on the graph of g.
+
+
+
+
+ We emphasize that in the horizontal translation h(x) = f(x-b), if b \gt 0 the graph of h lies b units to the right of f, while if b \lt 0, h lies b units to the left of f.
+
+ So far, we have seen the possible effects of adding a constant value to function's output (that is, the new expression f(x)+a if given f(x)) and adding a constant value to function input (that is, the new expression f(x+b), given f(x)). Each of these actions results in a translation of the function's graph (either vertically or horizontally), but otherwise leaving the graph the same. Next, we investigate the effects of multiplication the function's output by a constant.
+
+
+
+
+
+ Given the parent function y = f(x) pictured in Figure, what are the effects of the transformation y = v(x) = cf(x) for various values of c?
+
+
+
+
+
+ We first investigate the effects of c = 2 and c = \frac{1}{2}. For v(x) = 2f(x), the algebraic impact of this transformation is that every output of f is multiplied by 2. This means that the only output that is unchanged is when f(x) = 0, while any other point on the graph of the original function f will be stretched vertically away from the x-axis by a factor of 2. We can see this in Figure where each point on the original dark blue graph is transformed to a corresponding point whose y-coordinate is twice as large, as partially indicated by the red arrows.
+
+
+
+
+
The parent function y = f(x) along with two different vertical stretches, v and u.
+
The parent function y = f(x) along with two different vertical stretches, v and u.
+
+
+
+
The parent function y = f(x) along with a vertical reflection, z, and a corresponding stretch, w.
+
The parent function y = f(x) along with a vertical reflection, z, and a corresponding stretch, w.
+
+
+
+
+
+ In contrast, the transformation u(x) = \frac{1}{2}f(x) is stretched vertically by a factor of \frac{1}{2}, which has the effect of compressing the graph of f towards the x-axis, as all function outputs of f are multiplied by \frac{1}{2}. For instance, the point (0,-2) on the graph of f is transformed to the graph of (0,-1) on the graph of u, and others are transformed as indicated by the purple arrows.
+
+
+
+ To consider the situation where c \lt 0, we first consider the simplest case where c = -1 in the transformation z(x) = -f(x). Here the impact of the transformation is to multiply every output of the parent function f by -1; this takes any point of form (x,y) and transforms it to (x,-y), which means we are reflecting each point on the original function's graph across the x-axis to generate the resulting function's graph. This is demonstrated in Figure where y = z(x) is the reflection of y = f(x) across the x-axis.
+
+
+
+ Finally, we also investigate the case where c = -2, which generates y = w(x) = -2f(x). Here we can think of -2 as -2 = 2(-1): the effect of multiplying by -1 first reflects the graph of f across the x-axis (resulting in z), and then multiplying by 2 stretches the graph of z vertically to result in w, as shown in Figure.
+
+
+
+
+
+
+ As with vertical and horizontal translation, it's particularly instructive to see the effects of vertical scaling in a dynamic way.
+
+
+
+
Interactive vertical scaling demonstration (in the HTML version only).
+
+
+
Interactive vertical scaling demonstration (in the HTML version only).
+
+
+
+
+
+ Move the slider by clicking and dragging on the red point to see how changing c affects the graph of y = cf(x), which is shown in blue. The graph of y = f(x) will appear in grey and remain fixed.
+
+
+
+
+
+
+ We summarize and generalize our observations from Example and Figure as follows.
+
+
+
+ Vertical Scaling of a Function
+ transformation of a functionvertical scaling
+
+ Given a function y = f(x) and a real number c \gt 0, the transformed function y = v(x) = cf(x) is a vertical stretch of the graph of f. Every point (x,f(x)) on the graph of f gets stretched vertically to the corresponding point (x,cf(x)) on the graph of v. If 0 \lt c \lt 1, the graph of v is a compression of f toward the x-axis; if c \gt 1, the graph of v is a stretch of f away from the x-axis. Points where f(x) = 0 are unchanged by the transformation.
+
+
+
+ Given a function y = f(x) and a real number c \lt 0, the transformed function y = v(x) = cf(x) is a reflection of the graph of f across the x-axis followed by a vertical stretch by a factor of |c|.
+
+
+
+
+
+
+
+
+ Combining shifts and stretches: why order sometimes matters
+
+
+ In the final question of Activity, we considered the transformation y = m(x) = 2r(x+1)-1 of the original function r. There are three different basic transformations involved: a vertical shift of 1 unit down, a horizontal shift of 1 unit left, and a vertical stretch by a factor of 2. To understand the order in which these transformations are applied, it's essential to remember that a function is a process that converts inputs to outputs.
+
+
+
+ By the algebraic rule for m, m(x) = 2r(x+1)-1. In words, this means that given an input x for m, we do the following processes in this particular order:
+
+
+
+
+ add 1 to x and then
+ apply the function r to the quantity x+1;
+
+
+
+
+ multiply the output of r(x+1) by 2;
+
+
+
+
+ subtract 1 from the output of 2r(x+1).
+
+
+
+
+ These three steps correspond to three basic transformations: (1) shift the graph of r to the left by 1 unit; (2) stretch the resulting graph vertically by a factor of 2; (3) shift the resulting graph vertically by -1 units. We can see the graphical impact of these algebraic steps by taking them one at a time. In Figure, we see the function p that results from a shift 1 unit left of the parent function in Figure. (Each time we take an additional step, we will de-emphasize the preceding function by having it appear in lighter color and dashed.)
+
+
+
+
+
The parent function y = r(x).
+
+
+
+
+
The parent function y = r(x).
+
+
+
+
The parent function y = r(x) along with the horizontal shift y = s(x) = r(x+1).
+
+
+
+
+
The parent function y = r(x) along with the horizontal shift y = s(x) = r(x+1).
+
+
+
+
+
+ Continuing, we now consider the function q(x) = 2s(x) = 2r(x+1), which results in a vertical stretch of s away from the x-axis by a factor of 2, as seen in Figure.
+
+
+
+
+
The function y = q(x) = 2s(x) = 2r(x+1) along with the graph of s.
+
+
+
+
+
The function y = q(x) = 2s(x) = 2r(x+1) along with the graph of s.
+
+
+
+
The function y = m(x) = q(x)-1 = 2r(x+1) - 1 along with the graph of q.
+
+
+
+
+
The function y = m(x) = q(x)-1 = 2r(x+1) - 1 along with the graph of q.
+
+
+
+
+
+ Finally, we arrive at y = m(x) = 2r(x+1) - 1 by subtracting 1 from q(x) = 2r(x+1); this of course is a vertical shift of -1 units, and produces the graph of m shown in Figure. We can also track the point (2,-1) on the original parent function: it first moves left 1 unit to (1,-1), then it is stretched vertically by a factor of 2 away from the x-axis to (1,-2), and lastly is shifted 1 unit down to the point (1,-3), which we see on the graph of m.
+
+
+
+ While there are some transformations that can be executed in either order (such as a combination of a horizontal translation and a vertical translation, as seen in part (b) of Activity), in other situations order matters. For instance, in our preceding discussion, we have to apply the vertical stretch before applying the vertical shift. Algebraically, this is because
+
+ 2r(x+1) - 1 \ne 2[r(x+1)-1]
+ .
+ The quantity 2r(x+1) - 1 multiplies the function r(x+1) by 2 first (the stretch) and then the vertical shift follows; the quantity 2[r(x+1) - 1] shifts the function r(x+1) down 1 unit first, and then executes a vertical stretch by a factor of 2. In the latter scenario, the point (1,-1) that lies on r(x+1) gets transformed first to (1,-2) and then to (1,-4), which is not the same as the point (1,-3) that lies on m(x) = 2r(x+1) - 1.
+
+
+
+
+
+
+
+ Summary
+
+
+
+
+ The graph of y = g(x) = af(x-b) + c is related to the graph of y = f(x) by a sequence of transformations. First, there is horizontal shift of |b| units to the right (b \gt 0) or left (b \lt 0). Next, there is a vertical stretch by a factor of |a| (along with a reflection across y = 0 in the case where a \lt 0). Finally, there's a vertical shift of c units.
+
+
+
+
+ A transformation of a given function f is a process by which the graph may be shifted or stretched to generate a new, related function with fundamentally the same shape. In this section we considered four different ways this can occur: through a horizontal translation (shift), through a reflection across the line y = 0 (the x-axis), through a vertical scaling (stretch) that multiplies every output of a function by the same constant, and through a vertical translation (shift). Each of these individual processes is itself a transformation, and they may be combined in various ways to create more complicated transformations.
+
Given a central angle in the unit circle that measures t radians and that intersects the circle at both (1,0) and (a,b), as shown in Figure, we define the sine of t, denoted \sin(t), by the rule
-
+
\sin(t) = b
- .
+ .
@@ -172,9 +173,9 @@
Given a central angle in the unit circle that measures t radians and that intersects the circle at both (1,0) and (a,b), as shown in Figure, we define the cosine of t, denoted \cos(t), by the rule
-
+
\cos(t) = a
- .
+ .
@@ -256,13 +257,13 @@
In particular, since the sine graph can be viewed as the cosine graph shifted \frac{\pi}{2} units to the right, it follows that for any value of t,
-
+
\sin(t) = \cos(t-\frac{\pi}{2})
- .
+ .
Similarly, since the cosine graph can be viewed as the sine graph shifted left,
-
+
\cos(t) = \sin(t + \frac{\pi}{2})
- .
+ .
Because each of the two preceding equations hold for every value of t, they are often referred to as identities.
@@ -274,9 +275,9 @@
The Fundamental Trigonometric Identity
For any real number t,
-
+
\cos^2(t) + \sin^2(t) = 1
- .
+ .
- How do the three standard transformations (vertical translation, horizontal translation, and vertical scaling) affect the midline, amplitude, range, and period of sine and cosine curves?
-
-
-
-
- What algebraic transformation results in horizontal stretching or scaling of a function?
-
-
-
-
- How can we determine a formula involving sine or cosine that models any circular periodic function for which the midline, amplitude, period, and an anchor point are known?
-
-
-
-
- Introduction
-
- Recall our work in Section, where we studied how the graph of the function g defined by g(x) = af(x-b) + c is related to the graph of f, where a, b, and c are real numbers with a \ne 0. Because such transformations can shift and stretch a function, we are interested in understanding how we can use transformations of the sine and cosine functions to fit formulas to circular functions.
-
-
-
-
-
-
- Shifts and vertical stretches of the sine and cosine functions
-
- We know that the standard functions f(t) = \sin(t) and g(t) = \cos(t) are circular functions that each have midline y = 0, amplitude a = 1, period p = 2\pi, and range [-1,1]. Our work in Preview Activity suggests the following general principles.
-
-
-
- Transformations of sine and cosine
-
- Given real numbers a, b, and c with a \ne 0, the functions
-
- k(t) = a\cos(t-b)+c \ \text{ and } \ h(t) = a\sin(t-b) + c
-
- each represent a horizontal shift by b units to the right, followed by a vertical stretch by |a| units (if a \lt 0, there is also a reflection across the x-axis), followed by a vertical shift of c units, applied to the parent function (\cos(t) or \sin(t), respectively). The resulting circular functions have midline y = c, amplitude |a|, range [c-|a|,c+|a|], and period p = 2\pi. In addition, the point (b,a+c) lies on the graph of k and the point (b,c) lies on the graph of h.
-
-
-
-
- In Figure, we see how the overall transformation k(t) = a\cos(t-b)+c comes from executing a sequence of simpler ones. The original parent function y = \cos(t) (in dark gray) is first shifted b units right to generate the light red graph of y = \cos(t - b). In turn, that graph is then scaled vertically by a to generate the purple graph of y = a\cos(t-b). Finally, the purple graph is shifted c units vertically to result in the final graph of y = a\cos(t-b) + c in blue.
-
-
-
-
A sequence of transformations of y = \cos(t).
-
A sequence of transformations of y = \cos(t).
-
-
-
-
- It is often useful to follow one particular point through a sequence of transformations. In Figure, we see the red point that is located at (0,1) on the original function y = \cos(t), as well as the point (b, a+c) that is the corresponding point on k(t) = a\cos(t-b) + c under the overall transformation. Note that the point (b,a+c) results from the input, t = b, that makes the argument of the cosine function zero: k(b) = a\cos(b-b) + c = a\cos(0) + c.
-
-
-
- While the sine and cosine functions extend infinitely in either direction, it's natural to think of the point (0,1) as the starting point of the cosine function, and similarly the point (0,0) as the starting point of the sine function. We will refer to the corresponding point (b,a+c) on k(t) = a\cos(t-b) + c, and (b,c) on h(t) = a\sin(t-b) + c, respectively, as the anchor point. sinusoidal functionanchor point The anchor point, along with other information about a circular function's amplitude, midline, and period help us to determine a formula for a function that fits a given situation.
-
-
-
- For example, in Figure, the anchor point (b,a+c) on y = a\cos(t-b)+c corresponds to the starting point(0,1) on y = \cos(t).
-
-
-
-
-
-
-
- Horizontal scaling
-
-
- There is one more very important transformation of a function that we've not yet explored. Given a function y = f(x), we want to understand the related function g(x) = f(kx), where k is a positive real number. The sine and cosine functions are ideal functions with which to explore these effects; moreover, this transformation is crucial for being able to use the sine and cosine functions to model phenomena that oscillate at different frequencies.
-
-
-
- In the interactive Figure, we can explore the effect of the transformation g(t) = f(kt), where f(t) = \sin(t).
-
-
-
-
Interactive horizontal scaling demonstration (in the HTML version only).
-
-
-
Interactive horizontal scaling demonstration (in the HTML version only).
-
-
-
-
-
- Move the slider by clicking and dragging on the red point to see how changing k affects the graph of y = f(kt). The graph of y = f(t) will appear in grey and remain fixed.
-
-
-
-
-
-
- By experimenting with the slider, we gain an intuitive sense for how the value of k affects the graph of h(t) = f(kt) in comparision to the graph of f(t). When k = 2, we see that the graph of h is oscillating twice as fast as the graph of f since h(t) = f(2t) completes two full cycles over an interval in which f completes one full cycle. In contrast, when k = \frac{1}{2}, the graph of h oscillates half as fast as the graph of f, as h(t) = f(\frac{1}{2}t) completes only half of one cycle over an interval where f(t) completes a full one.
-
-
-
- We can also understand this from the perspective of function composition. To evaluate h(t) = f(2t), at a given value of t, we first multiply the input t by a factor of 2, and then evaluate the function f at the result. An important observation is that
-
- h\left( \frac{1}{2}t \right) = f\left( 2 \cdot \frac{1}{2}t \right) = f(t)
- .
- This tells us that the point (\frac{1}{2}t, f(t)) lies on the graph of h since an input of \frac{1}{2}t in h results in the value f(t). At the same time, the point (t,f(t)) lies on the graph of f. Thus we see that the correlation between points on the graphs of f and h (where h(t) = f(2t)) is
-
- (t, f(t)) \rightarrow \left( \frac{1}{2}t, f(t) \right)
- .
- We can therefore think of the transformation h(t) = f(2t) as achieving the output values of f twice as fast as the original function f(t) does. Analogously, the transformation h(t) = f(\frac{1}{2}t) will achieve the output values of f only half as quickly as the original function.
-
-
-
- Horizontal scaling
- horizontal scaling
-
- Given a function y = f(t) and a real number k \gt 0, the transformed function y = h(t) = f(kt) is a horizontal stretch of the graph of f. Every point (t,f(t)) on the graph of f gets stretched horizontally to the corresponding point (\frac{1}{k}t,f(t)) on the graph of h. If 0 \lt k \lt 1, the graph of h is a stretch of f away from the y-axis by a factor of \frac{1}{k}; if k \gt 1, the graph of h is a compression of f toward the y-axis by a factor of \frac{1}{k}. The only point on the graph of f that is unchanged by the transformation is (0,f(0)).
-
-
-
-
- While we will soon focus on horizontal stretches of the sine and cosine functions for the remainder of this section, it's important to note that horizontal scaling follows the same principles for any function we choose.
-
-
-
-
-
-
-
- Circular functions with different periods
-
- Because the circumference of the unit circle is 2\pi, the sine and cosine functions each have period 2\pi. Of course, as we think about using transformations of the sine and cosine functions to model different phenomena, it is apparent that we will need to generate functions with different periods than 2\pi. For instance, if a ferris wheel makes one revolution every 5 minutes, we'd want the period of the function that models the height of one car as a function of time to be P = 5. Horizontal scaling of functions enables us to generate circular functions with any period we desire.
-
-
-
- We begin by considering two basic examples. First, let f(t) = \sin(t) and g(t) = f(2t) = \sin(2t). We know from our most recent work that this transformation results in a horizontal compression of the graph of \sin(t) by a factor of \frac{1}{2} toward the y-axis. If we plot the two functions on the same axes as seen in Figure, it becomes apparent how this transformation affects the period of f.
-
-
-
-
A plot of the parent function, f(t) = \sin(t) (dashed, in gray), and the transformed function g(t) = f(2t) = \sin(2t) (in blue).
-
A plot of the parent function, f(t) = \sin(t) (dashed, in gray), and the transformed function g(t) = f(2t) = \sin(2t) (in blue).
-
-
-
-
- From the graph, we see that g(t) = \sin(2t) oscillates twice as frequently as f(t) = \sin(t), and that g completes a full cycle on the interval [0,\pi], which is half the length of the period of f. Thus, the 2 in f(2t) causes the period of f to be \frac{1}{2} as long; specifially, the period of g is P = \frac{1}{2} (2\pi) = \pi.
-
-
-
-
A plot of the parent function, f(t) = \sin(t) (dashed, in gray), and the transformed function h(t) = f(\frac{1}{2}t) = \sin(\frac{1}{2}t) (in blue).
-
A plot of the parent function, f(t) = \sin(t) (dashed, in gray), and the transformed function h(t) = f(\frac{1}{2}t) = \sin(\frac{1}{2}t) (in blue).
-
-
-
-
- On the other hand, if we let h(t) = f(\frac{1}{2}t) = \sin(\frac{1}{2}t), the transformed graph h is stretched away from the y-axis by a factor of 2. This has the effect of doubling the period of f, so that the period of h is P = 2 \cdot 2\pi = 4\pi, as seen in Figure.
-
-
-
- Our observations generalize for any positive constant k \gt 0. In the case where k = 2, we saw that the period of g(t) = \sin(2t) is P = \frac{1}{2} \cdot 2\pi, whereas in the case where k = \frac{1}{2}, the period of h(t) = \sin(\frac{1}{2}t) is P = 2 \cdot 2\pi = \frac{1}{\frac{1}{2}} \cdot 2\pi. Identical reasoning holds if we are instead working with the cosine function. In general, we can say the following.
-
-
-
- The period of a circular function
-
- For any constant k \gt 0, the period of the functions \sin(kt) and \cos(kt) is
-
- P = \frac{2\pi}{k}
- .
-
-
-
-
- Thus, if we know the k-value from the given function, we can deduce the period. If instead we know the desired period, we can determine k by the rule k = \frac{2\pi}{P}.
-
-
-
-
-
-
-
-
-
- Summary
-
-
-
-
- Given real numbers a, b, and c with a \ne 0, the functions
-
- k(t) = a\cos(t-b)+c \text{ and } h(t) = a\sin(t-b) + c
-
- each represent a horizontal shift by b units to the right, followed by a vertical stretch by |a| units (with a reflection across the x-axis if a \lt 0), followed by a vertical shift of c units, applied to the parent function (\cos(t) or \sin(t), respectively). The resulting circular functions have midline y = c, amplitude |a|, range [c-|a|,c+|a|], and period p = 2\pi. In addition, the anchor point (b,a+c) lies on the graph of k and the anchor point (b,c) lies on the graph of h.
-
-
-
-
- Given a function f and a constant k \gt 0, the algebraic transformation h(t) = f(kt) results in horizontal scaling of f by a factor of \frac{1}{k}. In particular, when k \gt 1, the graph of f is compressed toward the y-axis by a factor of \frac{1}{k} to create the graph of h, while when 0 \lt k \lt 1, the graph of f is stretched away from the y-axis by a factor of \frac{1}{k} to create the graph of h.
-
-
-
-
- Given any circular periodic function for which the midline, amplitude, period, and an anchor point are known, we can find a corresponding formula for the function of the form
-
- k(t) = a\cos(k(t-b))+c \text{ or } h(t) = a\sin(k(t-b)) + c
- .
- Each of these functions has midline y = c, amplitude |a|, and period P = \frac{2\pi}{k}. The point (b,a+c) lies on k and the point (b,c) lies on h.
-
+ How do the three standard transformations (vertical translation, horizontal translation, and vertical scaling) affect the midline, amplitude, range, and period of sine and cosine curves?
+
+
+
+
+ What algebraic transformation results in horizontal stretching or scaling of a function?
+
+
+
+
+ How can we determine a formula involving sine or cosine that models any circular periodic function for which the midline, amplitude, period, and an anchor point are known?
+
+
+
+
+ Introduction
+
+ Recall our work in Section, where we studied how the graph of the function g defined by g(x) = af(x-b) + c is related to the graph of f, where a, b, and c are real numbers with a \ne 0. Because such transformations can shift and stretch a function, we are interested in understanding how we can use transformations of the sine and cosine functions to fit formulas to circular functions.
+
+
+
+
+
+
+
+ Shifts and vertical stretches of the sine and cosine functions
+
+ We know that the standard functions f(t) = \sin(t) and g(t) = \cos(t) are circular functions that each have midline y = 0, amplitude a = 1, period p = 2\pi, and range [-1,1]. Our work in Preview Activity suggests the following general principles.
+
+
+
+ Transformations of sine and cosine
+
+ Given real numbers a, b, and c with a \ne 0, the functions
+
+ k(t) = a\cos(t-b)+c \ \text{ and } \ h(t) = a\sin(t-b) + c
+
+ each represent a horizontal shift by b units to the right, followed by a vertical stretch by |a| units (if a \lt 0, there is also a reflection across the x-axis), followed by a vertical shift of c units, applied to the parent function (\cos(t) or \sin(t), respectively). The resulting circular functions have midline y = c, amplitude |a|, range [c-|a|,c+|a|], and period p = 2\pi. In addition, the point (b,a+c) lies on the graph of k and the point (b,c) lies on the graph of h.
+
+
+
+
+ In Figure, we see how the overall transformation k(t) = a\cos(t-b)+c comes from executing a sequence of simpler ones. The original parent function y = \cos(t) (in dark gray) is first shifted b units right to generate the light red graph of y = \cos(t - b). In turn, that graph is then scaled vertically by a to generate the purple graph of y = a\cos(t-b). Finally, the purple graph is shifted c units vertically to result in the final graph of y = a\cos(t-b) + c in blue.
+
+
+
+
A sequence of transformations of y = \cos(t).
+
A sequence of transformations of y = \cos(t).
+
+
+
+
+ It is often useful to follow one particular point through a sequence of transformations. In Figure, we see the red point that is located at (0,1) on the original function y = \cos(t), as well as the point (b, a+c) that is the corresponding point on k(t) = a\cos(t-b) + c under the overall transformation. Note that the point (b,a+c) results from the input, t = b, that makes the argument of the cosine function zero: k(b) = a\cos(b-b) + c = a\cos(0) + c.
+
+
+
+ While the sine and cosine functions extend infinitely in either direction, it's natural to think of the point (0,1) as the starting point of the cosine function, and similarly the point (0,0) as the starting point of the sine function. We will refer to the corresponding point (b,a+c) on k(t) = a\cos(t-b) + c, and (b,c) on h(t) = a\sin(t-b) + c, respectively, as the anchor point. sinusoidal functionanchor point The anchor point, along with other information about a circular function's amplitude, midline, and period help us to determine a formula for a function that fits a given situation.
+
+
+
+ For example, in Figure, the anchor point (b,a+c) on y = a\cos(t-b)+c corresponds to the starting point(0,1) on y = \cos(t).
+
+
+
+
+
+
+
+ Horizontal scaling
+
+
+ There is one more very important transformation of a function that we've not yet explored. Given a function y = f(x), we want to understand the related function g(x) = f(kx), where k is a positive real number. The sine and cosine functions are ideal functions with which to explore these effects; moreover, this transformation is crucial for being able to use the sine and cosine functions to model phenomena that oscillate at different frequencies.
+
+
+
+ In the interactive Figure, we can explore the effect of the transformation g(t) = f(kt), where f(t) = \sin(t).
+
+
+
+
Interactive horizontal scaling demonstration (in the HTML version only).
+
+
+
Interactive horizontal scaling demonstration (in the HTML version only).
+
+
+
+
+
+ Move the slider by clicking and dragging on the red point to see how changing k affects the graph of y = f(kt). The graph of y = f(t) will appear in grey and remain fixed.
+
+
+
+
+
+
+ By experimenting with the slider, we gain an intuitive sense for how the value of k affects the graph of h(t) = f(kt) in comparision to the graph of f(t). When k = 2, we see that the graph of h is oscillating twice as fast as the graph of f since h(t) = f(2t) completes two full cycles over an interval in which f completes one full cycle. In contrast, when k = \frac{1}{2}, the graph of h oscillates half as fast as the graph of f, as h(t) = f(\frac{1}{2}t) completes only half of one cycle over an interval where f(t) completes a full one.
+
+
+
+ We can also understand this from the perspective of function composition. To evaluate h(t) = f(2t), at a given value of t, we first multiply the input t by a factor of 2, and then evaluate the function f at the result. An important observation is that
+
+ h\left( \frac{1}{2}t \right) = f\left( 2 \cdot \frac{1}{2}t \right) = f(t)
+ .
+ This tells us that the point (\frac{1}{2}t, f(t)) lies on the graph of h since an input of \frac{1}{2}t in h results in the value f(t). At the same time, the point (t,f(t)) lies on the graph of f. Thus we see that the correlation between points on the graphs of f and h (where h(t) = f(2t)) is
+
+ (t, f(t)) \rightarrow \left( \frac{1}{2}t, f(t) \right)
+ .
+ We can therefore think of the transformation h(t) = f(2t) as achieving the output values of f twice as fast as the original function f(t) does. Analogously, the transformation h(t) = f(\frac{1}{2}t) will achieve the output values of f only half as quickly as the original function.
+
+
+
+ Horizontal scaling
+ horizontal scaling
+
+ Given a function y = f(t) and a real number k \gt 0, the transformed function y = h(t) = f(kt) is a horizontal stretch of the graph of f. Every point (t,f(t)) on the graph of f gets stretched horizontally to the corresponding point (\frac{1}{k}t,f(t)) on the graph of h. If 0 \lt k \lt 1, the graph of h is a stretch of f away from the y-axis by a factor of \frac{1}{k}; if k \gt 1, the graph of h is a compression of f toward the y-axis by a factor of \frac{1}{k}. The only point on the graph of f that is unchanged by the transformation is (0,f(0)).
+
+
+
+
+ While we will soon focus on horizontal stretches of the sine and cosine functions for the remainder of this section, it's important to note that horizontal scaling follows the same principles for any function we choose.
+
+
+
+
+
+
+
+ Circular functions with different periods
+
+ Because the circumference of the unit circle is 2\pi, the sine and cosine functions each have period 2\pi. Of course, as we think about using transformations of the sine and cosine functions to model different phenomena, it is apparent that we will need to generate functions with different periods than 2\pi. For instance, if a ferris wheel makes one revolution every 5 minutes, we'd want the period of the function that models the height of one car as a function of time to be P = 5. Horizontal scaling of functions enables us to generate circular functions with any period we desire.
+
+
+
+ We begin by considering two basic examples. First, let f(t) = \sin(t) and g(t) = f(2t) = \sin(2t). We know from our most recent work that this transformation results in a horizontal compression of the graph of \sin(t) by a factor of \frac{1}{2} toward the y-axis. If we plot the two functions on the same axes as seen in Figure, it becomes apparent how this transformation affects the period of f.
+
+
+
+
A plot of the parent function, f(t) = \sin(t) (dashed, in gray), and the transformed function g(t) = f(2t) = \sin(2t) (in blue).
+
A plot of the parent function, f(t) = \sin(t) (dashed, in gray), and the transformed function g(t) = f(2t) = \sin(2t) (in blue).
+
+
+
+
+ From the graph, we see that g(t) = \sin(2t) oscillates twice as frequently as f(t) = \sin(t), and that g completes a full cycle on the interval [0,\pi], which is half the length of the period of f. Thus, the 2 in f(2t) causes the period of f to be \frac{1}{2} as long; specifially, the period of g is P = \frac{1}{2} (2\pi) = \pi.
+
+
+
+
A plot of the parent function, f(t) = \sin(t) (dashed, in gray), and the transformed function h(t) = f(\frac{1}{2}t) = \sin(\frac{1}{2}t) (in blue).
+
A plot of the parent function, f(t) = \sin(t) (dashed, in gray), and the transformed function h(t) = f(\frac{1}{2}t) = \sin(\frac{1}{2}t) (in blue).
+
+
+
+
+ On the other hand, if we let h(t) = f(\frac{1}{2}t) = \sin(\frac{1}{2}t), the transformed graph h is stretched away from the y-axis by a factor of 2. This has the effect of doubling the period of f, so that the period of h is P = 2 \cdot 2\pi = 4\pi, as seen in Figure.
+
+
+
+ Our observations generalize for any positive constant k \gt 0. In the case where k = 2, we saw that the period of g(t) = \sin(2t) is P = \frac{1}{2} \cdot 2\pi, whereas in the case where k = \frac{1}{2}, the period of h(t) = \sin(\frac{1}{2}t) is P = 2 \cdot 2\pi = \frac{1}{\frac{1}{2}} \cdot 2\pi. Identical reasoning holds if we are instead working with the cosine function. In general, we can say the following.
+
+
+
+ The period of a circular function
+
+ For any constant k \gt 0, the period of the functions \sin(kt) and \cos(kt) is
+
+ P = \frac{2\pi}{k}
+ .
+
+
+
+
+ Thus, if we know the k-value from the given function, we can deduce the period. If instead we know the desired period, we can determine k by the rule k = \frac{2\pi}{P}.
+
+
+
+
+
+
+
+
+
+ Summary
+
+
+
+
+ Given real numbers a, b, and c with a \ne 0, the functions
+
+ k(t) = a\cos(t-b)+c \text{ and } h(t) = a\sin(t-b) + c
+
+ each represent a horizontal shift by b units to the right, followed by a vertical stretch by |a| units (with a reflection across the x-axis if a \lt 0), followed by a vertical shift of c units, applied to the parent function (\cos(t) or \sin(t), respectively). The resulting circular functions have midline y = c, amplitude |a|, range [c-|a|,c+|a|], and period p = 2\pi. In addition, the anchor point (b,a+c) lies on the graph of k and the anchor point (b,c) lies on the graph of h.
+
+
+
+
+ Given a function f and a constant k \gt 0, the algebraic transformation h(t) = f(kt) results in horizontal scaling of f by a factor of \frac{1}{k}. In particular, when k \gt 1, the graph of f is compressed toward the y-axis by a factor of \frac{1}{k} to create the graph of h, while when 0 \lt k \lt 1, the graph of f is stretched away from the y-axis by a factor of \frac{1}{k} to create the graph of h.
+
+
+
+
+ Given any circular periodic function for which the midline, amplitude, period, and an anchor point are known, we can find a corresponding formula for the function of the form
+
+ k(t) = a\cos(k(t-b))+c \text{ or } h(t) = a\sin(k(t-b)) + c
+ .
+ Each of these functions has midline y = c, amplitude |a|, and period P = \frac{2\pi}{k}. The point (b,a+c) lies on k and the point (b,c) lies on h.
+
Remembering that h is a function of distance traversed along the circle, it follows that the average rate of change of h on any interval of distance between two points P and Q on the circle is given by
-
+
AV_{[P,Q]} = \frac{\text{change in height}}{\text{distance along the circle}}
- ,
+ ,
where both quantities are measured from point P to point Q.
First, in Figure, we consider points P, Q, and R where Q results from traversing 1/8 of the circumference from P, and R1/8 of the circumference from Q. In particular, we note that the distance d_1 along the circle from P to Q is the same as the distance d_2 along the circle from Q to R, and thus d_1 = d_2. At the same time, it is apparent from the geometry of the circle that the change in height h_1 from P to Q is greater than the change in height h_2 from Q to R, so h_1 \gt h_2. Thus, we can say that
-
+
AV_{[P,Q]} = \frac{h_1}{d_1} \gt \frac{h_2}{d_2} = AV_{[Q,R]}
- .
+ .
@@ -281,9 +282,9 @@
The differences in certain average rates of change appear to become more extreme if we consider shorter arcs along the circle. Next we consider traveling 1/20 of the circumference along the circle. In Figure, points P and Q lie 1/20 of the circumference apart, as do R and S, so here d_1 = d_5. In this situation, it is the case that h_1 \gt h_5 for the same reasons as above, but we can say even more. From the green triangle in Figure, we see that h_1 \approx d_1 (while h_1 \lt d_1), so that AV_{[P,Q]} = \frac{h_1}{d_1} \approx 1. At the same time, in the magenta triangle in the figure we see that h_5 is very small, especially in comparison to d_5, and thus AV_{[R,S]} = \frac{h_5}{d_5} \approx 0. Hence, in Figure,
-
+
AV_{[P,Q]} \approx 1 \text{ and } AV_{[R,S]} \approx 0
- .
+ .
This information tells us that a circular function appears to change most rapidly for points near its midline and to change least rapidly for points near its highest and lowest values.
diff --git a/source/sec-circular-unit-circle-wb.xml b/source/sec-circular-unit-circle-wb.xml
index a0d0295f..70079ff1 100755
--- a/source/sec-circular-unit-circle-wb.xml
+++ b/source/sec-circular-unit-circle-wb.xml
@@ -1,22 +1,22 @@
-
-
-
-
-
-
-
-
-
-
-
-
-
-
- The Unit Circle
-
-
-
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ The Unit Circle
+
+
+
+
+
+
+
diff --git a/source/sec-circular-unit-circle.xml b/source/sec-circular-unit-circle.xml
index c85c0dac..ecebd9c1 100755
--- a/source/sec-circular-unit-circle.xml
+++ b/source/sec-circular-unit-circle.xml
@@ -1,200 +1,201 @@
-
-
-
-
-
-
-
-
-
-
-
-
-
-
- The Unit Circle
-
-
-
-
- What is the radian measure of an angle?
-
-
-
-
- Are there natural special points on the unit circle whose coordinates we can identify exactly?
-
-
-
-
- How can we determine arc length and the location of special points in circles other than the unit circle?
-
-
-
-
- Introduction
-
- As demonstrated by several different examples in Section, certain periodic phenomena are closely linked to circles and circular motion. Rather than regularly work with circles of different center and radius, it turns out to be ideal to work with one standard circle and build all circular functions from it. The unit circleunit circleis the circle of radius 1 that is centered at the origin, (0,0).
-
-
-
- If we pick any point (x,y) that lies on the unit circle, the point is associated with a right triangle whose horizontal leg has length |x| and whose vertical leg has length |y|, as seen in Figure. By the Pythagorean Theorem, it follows that
-
- x^2 + y^2 = 1
- ,
- and this is the equation of the unit circle: a point (x,y) lies on the unit circle if and only if x^2 + y^2 = 1.
-
-
-
-
-
Coordinates of a point on the unit circle.
-
Coordinates of a point on the unit circle.
-
-
-
-
A point traversing the unit circle.
-
A point traversing the unit circle.
-
-
-
-
-
- To study the circular functions generated by the unit circle, we will also animate a point and let it traverse the circle. Starting at (1,0) indicated by t_0 in Figure, we see a sequence of points that result from traveling a distance along the circle that is 1/24 the circumference of the unit circle. Since the unit circle's circumference is C = 2\pi r = 2\pi, it follows that the distance from t_0 to t_1 is
-
- d = \frac{1}{24} \cdot 2\pi = \frac{\pi}{12}
- .
- As we work to better understand the unit circle, we will commonly use fractional multiples of \pi as these result in natural distances traveled along the unit circle.
-
-
-
-
-
-
-
- Radians and degrees
-
- In Preview Activity, we introduced the idea of radian measure of an angle. Here we state the formal definition of this term.
-
-
-
- radian measuredefinition of
-
-
- An angle whose vertex is at the center of a circleWe often call such an angle a central angle. measures 1 radian provided that the arc the angle intercepts on the circle equals the radius of the circle.
-
-
-
-
-
- As seen in , in the unit circle this means that a central angle has measure 1 radian whenever it intercepts an arc of length 1 unit along the circumference. Because of this important correspondence between the unit circle and radian measure (one unit of arc length on the unit circle corresponds to 1 radian), we focus our discussion of radian measure within the unit circle.
-
-
-
- Since there are 2\pi units of length along the unit circle's circumference it follows there are \frac{1}{4} \cdot 2\pi = \frac{\pi}{2} units of length in \frac{1}{4} of a revolution. We also know that \frac{1}{4} of a revolution corresponds to a central angle that is a right angle, whose familiar degree measure is 90^\circ. If we extend to a central angle that intercepts half the circle, we see similarly that \pi radians corresponds to 180^\circ; this relationship enables us to convert angle measures from radians to degrees and vice versa.
-
-
-
- Converting between radians and degrees
-
- An angle whose radian measure is 1 radian has degree measure \frac{180}{\pi} ^\circ. An angle whose degree measure is 1^\circ has radian measure \frac{\pi}{180}.
-
-
-
-
-
-
- Note that in , we labeled 24 equally spaced points with their respective distances around the unit circle counterclockwise from (1,0). Because these distances are on the unit circle, they also correspond to the radian measure of the central angles that intercept them. In particular, each central angle with one of its sides on the positive x-axis generates a unique point on the unit circle, and with it, an associated length intercepted along the circumference of the circle. A good exercise at this point is to return to and label each of the noted points with the degree measure that is intercepted by a central angle with one side on the positive x-axis, in addition to the arc lengths (radian measures) already identified.
-
-
-
-
-
- Special points on the unit circle
-
-
- Our in-depth study of the unit circle is motivated by our desire to better understand the behavior of circular functions. Recall that as we traverse a circle, the height of the point moving along the circle generates a function that depends on distance traveled along the circle. Wherever possible, we'd like to be able to identify the exact height of a given point on the unit circle. Two special right triangles enable us to locate exactly an important collection of points on the unit circle.
-
-
-
-
-
- Our work in Activity enables us to identify exactly the location of 12 special points on the unit circle. In part (d) of the activity, we located the three noted points in Figure along with their respective radian measures. By symmetry across the coordinate axes and thinking about the signs of coordinates in the other three quadrants, we can now identify all of the coordinates of the remaining 9 points.
-
-
-
-
The unit circle with 16 special points whose location we can determine exactly.
-
The unit circle with 16 special points whose location we can determine exactly.
-
-
-
-
- In addition, we note that there are four additional points on the circle that we can locate exactly: the four points that correspond to angle measures of 0, \frac{\pi}{2}, \pi, and \frac{3\pi}{2} radians, which lie where the coordinate axes intersect the circle. Each such point has 0 for one coordinate and \pm 1 for the other. Labeling all of the remaining points in Figure is an important exercise that you should do on your own.
-
-
-
- Finally, we note that we can identify any point on the unit circle exactly simply by choosing one of its coordinates. Since every point (x,y) on the unit circle satisfies the equation x^2 + y^2 = 1, if we know the value of x or y and the quadrant in which the point lies, we can determine the other coordinate exactly.
-
-
-
-
-
- Special points and arc length in non-unit circles
-
-
- All of our work with the unit circle can be extended to circles centered at the origin with different radii, since a circle with a larger or smaller radius is a scaled version of the unit circle. For instance, if we instead consider a circle of radius 7, the coordinates of every point on the unit circle are magnified by a factor of 7, so the point that corresponds to an angle such as \theta = \frac{2\pi}{3} has coordinates \left( -\frac{7}{2}, \frac{7\sqrt{3}}{2} \right). Distance along the circle is magnified by the same factor: the arc length along the unit circle from (0,0) to \left( -\frac{7}{2}, \frac{7\sqrt{3}}{2} \right) is 7 \cdot \frac{2\pi}{3}, since the arc length along the unit circle for this angle is \frac{2\pi}{3}.
-
-
-
- If we think more generally about a circle of radius r with a central angle \theta that intercepts an arc of length s, we see how the magnification factor r (in comparison to the unit circle) connects arc length and the central angle according to the following principle.
-
-
-
- Connecting arc length and angles in non-unit circles
-
- If a central angle measuring \theta radians intercepts an arc of length s in a circle of radius r, then
-
- s = r \theta
- .
-
-
-
- In the unit circle, where r = 1, the equation s = r\theta demonstrates the familiar fact that arc length matches the radian measure of the central angle. Moreover, we also see how this formula aligns with the definition of radian measure: if the arc length and radius are equal, then the angle measures 1 radian.
-
-
-
-
-
-
-
-
- Summary
-
-
-
-
- The radian measure of an angle connects the measure of a central angle in a circle to the radius of the circle. A central angle has radian measure 1 provided that it intercepts an arc of length equal to the circle's radius. In the unit circle, a central angle's radian measure is precisely the same numerical value as the length of the arc it intercepts along the circle.
-
-
-
-
- If we begin at the point (1,0) and move counterclockwise along the unit circle, there are natural special points on the unit circle that correspond to angles of measure 30^\circ, 45^\circ, 60^\circ, and their multiples. We can count in 30^\circ increments and identify special points that correspond to angles of measure 30^\circ, 60^\circ, 90^\circ, 120^\circ, and so on; doing likewise with 45^\circ, these correspond to angles of 45^\circ, 90^\circ, 135^\circ, etc. In radian measure, these sequences together give us the important angles \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \frac{\pi}{2}, \frac{2\pi}{3}, \frac{3\pi}{4}, and so on. Together with our work involving 45^\circ-45^\circ-90^\circ and 30^\circ-60^\circ-90^\circ triangles in Activity, we are able to identify the exact locations of all of the points in Figure.
-
-
-
-
- In any circle of radius r, if a central angle of measure \theta radians intercepts an arc of length s, then it follows that
-
- s = r\theta
- .
- This shows that arc length, s, is magnified along with the size of the radius, r, of the circle.
-
-
-
-
-
-
-
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ The Unit Circle
+
+
+
+
+ What is the radian measure of an angle?
+
+
+
+
+ Are there natural special points on the unit circle whose coordinates we can identify exactly?
+
+
+
+
+ How can we determine arc length and the location of special points in circles other than the unit circle?
+
+
+
+
+ Introduction
+
+ As demonstrated by several different examples in Section, certain periodic phenomena are closely linked to circles and circular motion. Rather than regularly work with circles of different center and radius, it turns out to be ideal to work with one standard circle and build all circular functions from it. The unit circleunit circleis the circle of radius 1 that is centered at the origin, (0,0).
+
+
+
+ If we pick any point (x,y) that lies on the unit circle, the point is associated with a right triangle whose horizontal leg has length |x| and whose vertical leg has length |y|, as seen in Figure. By the Pythagorean Theorem, it follows that
+
+ x^2 + y^2 = 1
+ ,
+ and this is the equation of the unit circle: a point (x,y) lies on the unit circle if and only if x^2 + y^2 = 1.
+
+
+
+
+
Coordinates of a point on the unit circle.
+
Coordinates of a point on the unit circle.
+
+
+
+
A point traversing the unit circle.
+
A point traversing the unit circle.
+
+
+
+
+
+ To study the circular functions generated by the unit circle, we will also animate a point and let it traverse the circle. Starting at (1,0) indicated by t_0 in Figure, we see a sequence of points that result from traveling a distance along the circle that is 1/24 the circumference of the unit circle. Since the unit circle's circumference is C = 2\pi r = 2\pi, it follows that the distance from t_0 to t_1 is
+
+ d = \frac{1}{24} \cdot 2\pi = \frac{\pi}{12}
+ .
+ As we work to better understand the unit circle, we will commonly use fractional multiples of \pi as these result in natural distances traveled along the unit circle.
+
+
+
+
+
+
+
+
+ Radians and degrees
+
+ In Preview Activity, we introduced the idea of radian measure of an angle. Here we state the formal definition of this term.
+
+
+
+ radian measuredefinition of
+
+
+ An angle whose vertex is at the center of a circleWe often call such an angle a central angle. measures 1 radian provided that the arc the angle intercepts on the circle equals the radius of the circle.
+
+
+
+
+
+ As seen in , in the unit circle this means that a central angle has measure 1 radian whenever it intercepts an arc of length 1 unit along the circumference. Because of this important correspondence between the unit circle and radian measure (one unit of arc length on the unit circle corresponds to 1 radian), we focus our discussion of radian measure within the unit circle.
+
+
+
+ Since there are 2\pi units of length along the unit circle's circumference it follows there are \frac{1}{4} \cdot 2\pi = \frac{\pi}{2} units of length in \frac{1}{4} of a revolution. We also know that \frac{1}{4} of a revolution corresponds to a central angle that is a right angle, whose familiar degree measure is 90^\circ. If we extend to a central angle that intercepts half the circle, we see similarly that \pi radians corresponds to 180^\circ; this relationship enables us to convert angle measures from radians to degrees and vice versa.
+
+
+
+ Converting between radians and degrees
+
+ An angle whose radian measure is 1 radian has degree measure \frac{180}{\pi} ^\circ. An angle whose degree measure is 1^\circ has radian measure \frac{\pi}{180}.
+
+
+
+
+
+
+ Note that in , we labeled 24 equally spaced points with their respective distances around the unit circle counterclockwise from (1,0). Because these distances are on the unit circle, they also correspond to the radian measure of the central angles that intercept them. In particular, each central angle with one of its sides on the positive x-axis generates a unique point on the unit circle, and with it, an associated length intercepted along the circumference of the circle. A good exercise at this point is to return to and label each of the noted points with the degree measure that is intercepted by a central angle with one side on the positive x-axis, in addition to the arc lengths (radian measures) already identified.
+
+
+
+
+
+ Special points on the unit circle
+
+
+ Our in-depth study of the unit circle is motivated by our desire to better understand the behavior of circular functions. Recall that as we traverse a circle, the height of the point moving along the circle generates a function that depends on distance traveled along the circle. Wherever possible, we'd like to be able to identify the exact height of a given point on the unit circle. Two special right triangles enable us to locate exactly an important collection of points on the unit circle.
+
+
+
+
+
+ Our work in Activity enables us to identify exactly the location of 12 special points on the unit circle. In part (d) of the activity, we located the three noted points in Figure along with their respective radian measures. By symmetry across the coordinate axes and thinking about the signs of coordinates in the other three quadrants, we can now identify all of the coordinates of the remaining 9 points.
+
+
+
+
The unit circle with 16 special points whose location we can determine exactly.
+
The unit circle with 16 special points whose location we can determine exactly.
+
+
+
+
+ In addition, we note that there are four additional points on the circle that we can locate exactly: the four points that correspond to angle measures of 0, \frac{\pi}{2}, \pi, and \frac{3\pi}{2} radians, which lie where the coordinate axes intersect the circle. Each such point has 0 for one coordinate and \pm 1 for the other. Labeling all of the remaining points in Figure is an important exercise that you should do on your own.
+
+
+
+ Finally, we note that we can identify any point on the unit circle exactly simply by choosing one of its coordinates. Since every point (x,y) on the unit circle satisfies the equation x^2 + y^2 = 1, if we know the value of x or y and the quadrant in which the point lies, we can determine the other coordinate exactly.
+
+
+
+
+
+ Special points and arc length in non-unit circles
+
+
+ All of our work with the unit circle can be extended to circles centered at the origin with different radii, since a circle with a larger or smaller radius is a scaled version of the unit circle. For instance, if we instead consider a circle of radius 7, the coordinates of every point on the unit circle are magnified by a factor of 7, so the point that corresponds to an angle such as \theta = \frac{2\pi}{3} has coordinates \left( -\frac{7}{2}, \frac{7\sqrt{3}}{2} \right). Distance along the circle is magnified by the same factor: the arc length along the unit circle from (0,0) to \left( -\frac{7}{2}, \frac{7\sqrt{3}}{2} \right) is 7 \cdot \frac{2\pi}{3}, since the arc length along the unit circle for this angle is \frac{2\pi}{3}.
+
+
+
+ If we think more generally about a circle of radius r with a central angle \theta that intercepts an arc of length s, we see how the magnification factor r (in comparison to the unit circle) connects arc length and the central angle according to the following principle.
+
+
+
+ Connecting arc length and angles in non-unit circles
+
+ If a central angle measuring \theta radians intercepts an arc of length s in a circle of radius r, then
+
+ s = r \theta
+ .
+
+
+
+ In the unit circle, where r = 1, the equation s = r\theta demonstrates the familiar fact that arc length matches the radian measure of the central angle. Moreover, we also see how this formula aligns with the definition of radian measure: if the arc length and radius are equal, then the angle measures 1 radian.
+
+
+
+
+
+
+
+
+ Summary
+
+
+
+
+ The radian measure of an angle connects the measure of a central angle in a circle to the radius of the circle. A central angle has radian measure 1 provided that it intercepts an arc of length equal to the circle's radius. In the unit circle, a central angle's radian measure is precisely the same numerical value as the length of the arc it intercepts along the circle.
+
+
+
+
+ If we begin at the point (1,0) and move counterclockwise along the unit circle, there are natural special points on the unit circle that correspond to angles of measure 30^\circ, 45^\circ, 60^\circ, and their multiples. We can count in 30^\circ increments and identify special points that correspond to angles of measure 30^\circ, 60^\circ, 90^\circ, 120^\circ, and so on; doing likewise with 45^\circ, these correspond to angles of 45^\circ, 90^\circ, 135^\circ, etc. In radian measure, these sequences together give us the important angles \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \frac{\pi}{2}, \frac{2\pi}{3}, \frac{3\pi}{4}, and so on. Together with our work involving 45^\circ-45^\circ-90^\circ and 30^\circ-60^\circ-90^\circ triangles in Activity, we are able to identify the exact locations of all of the points in Figure.
+
+
+
+
+ In any circle of radius r, if a central angle of measure \theta radians intercepts an arc of length s, then it follows that
+
+ s = r\theta
+ .
+ This shows that arc length, s, is magnified along with the size of the radius, r, of the circle.
+
+
+
+
+
+
+
+
+
+
+
diff --git a/source/sec-exp-e-wb.xml b/source/sec-exp-e-wb.xml
index 590ca36e..75d1b6bd 100755
--- a/source/sec-exp-e-wb.xml
+++ b/source/sec-exp-e-wb.xml
@@ -1,21 +1,21 @@
-
-
-
-
-
-
-
-
-
-
-
-
-
-
- The special number <m>e</m>
-
-
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ The special number <m>e</m>
+
+
+
+
+
+
diff --git a/source/sec-exp-e.xml b/source/sec-exp-e.xml
index 2a73a8bc..55f0ca3a 100755
--- a/source/sec-exp-e.xml
+++ b/source/sec-exp-e.xml
@@ -1,246 +1,247 @@
-
-
-
-
-
-
-
-
-
-
-
-
-
-
- The special number <m>e</m>
-
-
-
-
- Why can every exponential function of form f(t) = b^t (where b \gt 0 and b \ne 1) be thought of as a horizontal scaling of a single special exponential function?
-
-
-
-
- What is the natural base e and what makes this number special?
-
-
-
-
-
- Introduction
-
- We have observed that the behavior of functions of the form f(t) = b^t is very consistent, where the only major differences depend on whether b \lt 1 or b \gt 1. Indeed, if we stipulate that b \gt 1, the graphs of functions with different bases b look nearly identical, as seen in the plots of p, q, r, and s in Figure.
-
-
-
-
Plots of four different exponential functions of form b^t with b \gt 1.
-
Plots of four different exponential functions of form b^t with b \gt 1.
-
-
-
-
- Because the point (0,1) lies on the graph of each of the four functions in Figure, the functions cannot be vertical scalings of one another. However, it is possible that the functions are horizontal scalings of one another. This leads us to a natural question: might it be possible to find a single exponential function with a special base, say e, for which every other exponential function f(t) = b^t can be expressed as a horizontal scaling of E(t) = e^t?
-
-
-
-
-
-
-
- The natural base <m>e</m>
-
- In Preview Activity, we found that it appears possible to find a value of k so that given any base b, we can write the function b^t as the horizontal scaling of 2^t given by
-
- b^t = 2^{kt}
- .
- It's also apparent that there's nothing particularly special about 2: we could similarly write any function b^t as a horizontal scaling of 3^t or 4^t, albeit with a different scaling factor k for each. Thus, we might also ask: is there a best possible single base to use?
-
-
-
- Through the central topic of the rate of change of a function, calculus helps us decide which base is best to use to represent all exponential functions. While we study average rate of change extensively in this course, in calculus there is more emphasis on the instantaneous rate of change. In that context, a natural question arises: is there a nonzero function that grows in such a way that its height is exactly how fast its height is increasing?
-
-
-
- Amazingly, it turns out that the answer to this questions is yes, and the function with this property is
- the exponential function with the natural base, denoted e^t. The number e (named in homage to the great Swiss mathematician Leonard Euler (1707-1783)) is complicated to define. Like \pi, e is an irrational number that cannot be represented exactly by a ratio of integers and whose decimal expansion never repeats. Advanced mathematics is needed in order to make the following formal definition of e.
-
-
-
- The natural base, <m>e</m>
- exponential functionwith the natural base, e
- natural base, e
- e
-
-
- The number e is the infinite sumInfinite sums are usually studied in second semester calculus.
-
- e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \cdots
-
- From this, e \approx 2.718281828.
-
-
-
-
-
- For instance, 1 + \frac{1}{1} + \frac{1}{2} + \frac{1}{6} + \frac{1}{24} + \frac{1}{120} = \frac{163}{60} \approx 2.7167 is an approximation of e generated by taking the first 6 terms in the infinite sum that defines it. Every computational device knows the number e and we will normally work with this number by using technology appropriately.
-
-
-
- Initially, it's important to note that 2 \lt e \lt 3, and thus we expect the function e^t to lie between 2^t and 3^t.
-
- If we compare the graphs and some selected outputs of each function, as in Table and Figure, we see that the function e^t satisfies the inequality
-
- 2^t \lt e^t \lt 3^t
-
- for all positive values of t. When t is negative, we can view the values of each function as being reciprocals of powers of 2, e, and 3. For instance, since 2^2 \lt e^2 \lt 3^2, it follows \frac{1}{3^2} \lt \frac{1}{e^2} \lt \frac{1}{2^2}, or
-
- 3^{-2} \lt e^{-2} \lt 2^{-2}
- . Thus, for any t \lt 0,
-
- 3^t \lt e^t \lt 2^t
-
- Like 2^t and 3^t, the function e^t passes through (0,1) is always increasing and always concave up, and its range is the set of all positive real numbers.
-
-
-
-
-
-
-
- Why any exponential function can be written in terms of <m>e</m>
-
-
- In Preview Activity, we saw graphical evidence that any exponential function f(t) = b^t can be written as a horizontal scaling of the function g(t) = 2^t, plus we observed that there wasn't anything particularly special about 2^t. Because of the importance of e^t in calculus, we will choose instead to use the natural exponential function, E(t) = e^t as the function we scale to generate any other exponential function f(t) = b^t. We claim that for any choice of b \gt 0 (with b \ne 1), there exists a horizontal scaling factor k such that b^t = f(t) = E(kt) = e^{kt}.
-
-
-
- By the rules of exponents, we can rewrite this last equation equivalently as
-
- b^t = (e^k)^t
- .
- Since this equation has to hold for every value of t, it follows that b = e^k. Thus, our claim that we can scale E(t) to get f(t) requires us to show that regardless of the choice of the positive number b, there exists a single corresponding value of k such that b = e^k.
-
-
-
- Given b \gt 0, we can always find a corresponding value of k such that e^k = b because the function f(t) = e^t passes the , as seen in Figure.
-
-
-
-
A plot of f(t) = e^t along with several choices of positive constants b viewed on the vertical axis.
-
A plot of f(t) = e^t along with several choices of positive constants b viewed on the vertical axis.
-
-
-
-
- In Figure, we can think of b as a point on the positive vertical axis. From there, we draw a horizontal line over to the graph of f(t) = e^t, and then from the (unique) point of intersection we drop a vertical line to the x-axis. At that corresponding point on the x-axis we have found the input value k that corresponds to b. We see that there is always exactly one such k value that corresponds to each chosen b because f(t) = e^t is always increasing, and any always increasing function passes the Horizontal Line Test.
-
-
-
- It follows that the function f(t) = e^t has an inverse function, and hence there must be some other function g such that writing y = f(t) is the same as writing t = g(y). This important function g will be developed in Section and will enable us to find the value of k exactly for a given b. For now, we are content to work with these observations graphically and to hence find estimates for the value of k.
-
-
-
-
-
-
-
- Summary
-
-
-
- Any exponential function f(t) = b^t can be viewed as a horizontal scaling of E(t) = e^t because there exists a unique constant k such that E(kt) = e^{kt} = b^t = f(t) is true for every value of t. This holds since the exponential function e^t is always increasing, so given an output b there exists a unique input k such that e^k = b, from which it follows that e^{kt} = b^t.
-
-
-
-
- The natural base e is the special number that defines an increasing exponential function whose rate of change at any point is the same as its height at that point, a fact that is established using calculus. The number e turns out to be given exactly by an infinite sum and approximately by e \approx 2.7182818.
-
-
-
-
-
-
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ The special number <m>e</m>
+
+
+
+
+ Why can every exponential function of form f(t) = b^t (where b \gt 0 and b \ne 1) be thought of as a horizontal scaling of a single special exponential function?
+
+
+
+
+ What is the natural base e and what makes this number special?
+
+
+
+
+
+ Introduction
+
+ We have observed that the behavior of functions of the form f(t) = b^t is very consistent, where the only major differences depend on whether b \lt 1 or b \gt 1. Indeed, if we stipulate that b \gt 1, the graphs of functions with different bases b look nearly identical, as seen in the plots of p, q, r, and s in Figure.
+
+
+
+
Plots of four different exponential functions of form b^t with b \gt 1.
+
Plots of four different exponential functions of form b^t with b \gt 1.
+
+
+
+
+ Because the point (0,1) lies on the graph of each of the four functions in Figure, the functions cannot be vertical scalings of one another. However, it is possible that the functions are horizontal scalings of one another. This leads us to a natural question: might it be possible to find a single exponential function with a special base, say e, for which every other exponential function f(t) = b^t can be expressed as a horizontal scaling of E(t) = e^t?
+
+
+
+
+
+
+
+
+ The natural base <m>e</m>
+
+ In Preview Activity, we found that it appears possible to find a value of k so that given any base b, we can write the function b^t as the horizontal scaling of 2^t given by
+
+ b^t = 2^{kt}
+ .
+ It's also apparent that there's nothing particularly special about 2: we could similarly write any function b^t as a horizontal scaling of 3^t or 4^t, albeit with a different scaling factor k for each. Thus, we might also ask: is there a best possible single base to use?
+
+
+
+ Through the central topic of the rate of change of a function, calculus helps us decide which base is best to use to represent all exponential functions. While we study average rate of change extensively in this course, in calculus there is more emphasis on the instantaneous rate of change. In that context, a natural question arises: is there a nonzero function that grows in such a way that its height is exactly how fast its height is increasing?
+
+
+
+ Amazingly, it turns out that the answer to this questions is yes, and the function with this property is
+ the exponential function with the natural base, denoted e^t. The number e (named in homage to the great Swiss mathematician Leonard Euler (1707-1783)) is complicated to define. Like \pi, e is an irrational number that cannot be represented exactly by a ratio of integers and whose decimal expansion never repeats. Advanced mathematics is needed in order to make the following formal definition of e.
+
+
+
+ The natural base, <m>e</m>
+ exponential functionwith the natural base, e
+ natural base, e
+ e
+
+
+ The number e is the infinite sumInfinite sums are usually studied in second semester calculus.
+
+ e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \cdots
+
+ From this, e \approx 2.718281828.
+
+
+
+
+
+ For instance, 1 + \frac{1}{1} + \frac{1}{2} + \frac{1}{6} + \frac{1}{24} + \frac{1}{120} = \frac{163}{60} \approx 2.7167 is an approximation of e generated by taking the first 6 terms in the infinite sum that defines it. Every computational device knows the number e and we will normally work with this number by using technology appropriately.
+
+
+
+ Initially, it's important to note that 2 \lt e \lt 3, and thus we expect the function e^t to lie between 2^t and 3^t.
+
+ If we compare the graphs and some selected outputs of each function, as in Table and Figure, we see that the function e^t satisfies the inequality
+
+ 2^t \lt e^t \lt 3^t
+
+ for all positive values of t. When t is negative, we can view the values of each function as being reciprocals of powers of 2, e, and 3. For instance, since 2^2 \lt e^2 \lt 3^2, it follows \frac{1}{3^2} \lt \frac{1}{e^2} \lt \frac{1}{2^2}, or
+
+ 3^{-2} \lt e^{-2} \lt 2^{-2}
+ . Thus, for any t \lt 0,
+
+ 3^t \lt e^t \lt 2^t
+
+ Like 2^t and 3^t, the function e^t passes through (0,1) is always increasing and always concave up, and its range is the set of all positive real numbers.
+
+
+
+
+
+
+
+ Why any exponential function can be written in terms of <m>e</m>
+
+
+ In Preview Activity, we saw graphical evidence that any exponential function f(t) = b^t can be written as a horizontal scaling of the function g(t) = 2^t, plus we observed that there wasn't anything particularly special about 2^t. Because of the importance of e^t in calculus, we will choose instead to use the natural exponential function, E(t) = e^t as the function we scale to generate any other exponential function f(t) = b^t. We claim that for any choice of b \gt 0 (with b \ne 1), there exists a horizontal scaling factor k such that b^t = f(t) = E(kt) = e^{kt}.
+
+
+
+ By the rules of exponents, we can rewrite this last equation equivalently as
+
+ b^t = (e^k)^t
+ .
+ Since this equation has to hold for every value of t, it follows that b = e^k. Thus, our claim that we can scale E(t) to get f(t) requires us to show that regardless of the choice of the positive number b, there exists a single corresponding value of k such that b = e^k.
+
+
+
+ Given b \gt 0, we can always find a corresponding value of k such that e^k = b because the function f(t) = e^t passes the , as seen in Figure.
+
+
+
+
A plot of f(t) = e^t along with several choices of positive constants b viewed on the vertical axis.
+
A plot of f(t) = e^t along with several choices of positive constants b viewed on the vertical axis.
+
+
+
+
+ In Figure, we can think of b as a point on the positive vertical axis. From there, we draw a horizontal line over to the graph of f(t) = e^t, and then from the (unique) point of intersection we drop a vertical line to the x-axis. At that corresponding point on the x-axis we have found the input value k that corresponds to b. We see that there is always exactly one such k value that corresponds to each chosen b because f(t) = e^t is always increasing, and any always increasing function passes the Horizontal Line Test.
+
+
+
+ It follows that the function f(t) = e^t has an inverse function, and hence there must be some other function g such that writing y = f(t) is the same as writing t = g(y). This important function g will be developed in Section and will enable us to find the value of k exactly for a given b. For now, we are content to work with these observations graphically and to hence find estimates for the value of k.
+
+
+
+
+
+
+
+ Summary
+
+
+
+ Any exponential function f(t) = b^t can be viewed as a horizontal scaling of E(t) = e^t because there exists a unique constant k such that E(kt) = e^{kt} = b^t = f(t) is true for every value of t. This holds since the exponential function e^t is always increasing, so given an output b there exists a unique input k such that e^k = b, from which it follows that e^{kt} = b^t.
+
+
+
+
+ The natural base e is the special number that defines an increasing exponential function whose rate of change at any point is the same as its height at that point, a fact that is established using calculus. The number e turns out to be given exactly by an infinite sum and approximately by e \approx 2.7182818.
+
- What does it mean to say that a function is exponential?
-
-
-
-
- How much data do we need to know in order to determine the formula for an exponential function?
-
-
-
-
- Are there important trends that all exponential functions exhibit?
-
-
-
-
-
- Introduction
-
- Linear functions have constant average rate of change and model many important phenomena. In other settings, it is natural for a quantity to change at a rate that is proportional to the amount of the quantity present. For instance, whether you put $100 or $100000 or any other amount in a mutual fund, the investment's value changes at a rate proportional the amount present. We often measure that rate in terms of the annual percentage rate of return.
-
-
-
- Suppose that a certain mutual fund has a 10% annual return. If we invest $100, after 1 year we still have the original $100, plus we gain 10% of $100, so
-
- 100 \overset{\text{year } 1}{\longrightarrow} 100 + 0.1(100) = 1.1(100)
- .
- If we instead invested $100000, after 1 year we again have the original $100000, but now we gain 10% of $100000, and thus
-
- 100000 \overset{\text{year } 1}{\longrightarrow} 100000 + 0.1(100000) = 1.1(100000)
- .
- We therefore see that regardless of the amount of money originally invested, say P, the amount of money we have after 1 year is 1.1P.
-
-
-
- If we repeat our computations for the second year, we observe that
-
- 1.1(100) \overset{\text{year } 2}{\longrightarrow} 1.1(100) + 0.1(1.1(100)) = 1.1(1.1(100)) = 1.1^2 (100)
- .
- The ideas are identical with the larger dollar value, so
-
- 1.1(100000) \overset{\text{year } 2}{\longrightarrow} 1.1(100000) + 0.1(1.1(100000)) = 1.1(1.1(100000)) = 1.1^2 (100000)
- ,
- and we see that if we invest P dollars, in 2 years our investment will grow to 1.1^2 P.
-
-
-
- Of course, in 3 years at 10%, the original investment P will have grown to 1.1^3 P. Here we see a new kind of pattern developing: annual growth of 10% is leading to powers of the base 1.1, where the power to which we raise 1.1 corresponds to the number of years the investment has grown. We often call this phenomenon exponential growth. exponential growthintroduction
-
-
-
-
-
-
- Exponential functions of form <m>f(t) = ab^t</m>
-
- In Preview Activity, we encountered the functions I(t) and V(t) that had the same basic structure. Each can be written in the form g(t) = ab^t where a and b are positive constants and b \ne 1. Based on our earlier work with transformations, we know that the constant a is a vertical scaling factor, and thus the main behavior of the function comes from b^t, which we call an exponential function.
-
- Let b be a real number such that b \gt 0 and b \ne 1. We call the function defined by
- f(t) = b^t
- an exponential function with baseb.
-
-
-
-
-
- For an exponential function f(t) = b^t, we note that f(0) = b^0 = 1, so an exponential function of this form always passes through (0,1). In addition, because a positive number raised to any power is always positive (for instance, 2^{10} = 1024 and 2^{-10} = \frac{1}{2^{10}} = \frac{1}{1024}), the output of an exponential function is also always positive. In particular, f(t) = b^t is never zero and thus has no x-intercepts.
-
-
-
- Because we will be frequently interested in functions such as I(t) and V(t) with the form ab^t, we will also refer to functions of this form as exponential, understanding that technically these are vertical stretches of exponential functions according to Definition. In Preview Activity, we found that I(t) = 20000(1.08)^t and V(t) = 20000(0.88)^t. It is natural to call 1.08 the growth factor of I and similarly 0.88 the growth factor of V. In addition, we note that these values stem from the actual growth rates: 0.08 for I and -0.12 for V, the latter being negative because value is depreciating. In general, for a function of form f(t) = ab^t, we call b the growth factor. exponential functiongrowth factor Moreover, if b = 1+r, we call r the growth rate. exponential functiongrowth rate Whenever b \gt 1, we often say that the function f is exhibiting exponential growth, wherease if 0 \lt b \lt 1, we say f exhibits exponential decay. exponential functionexponential decay
-
-
-
- We explore the properties of functions of form f(t) = ab^t further in Activity.
-
- To better understand the roles that a and b play in an exponential function, let's compare exponential and linear functions. In Table and Table, we see output for two different functions r and s that correspond to equally spaced input.
-
-
-
-
- Data for the function <m>r</m>.
-
-
- t
- 0
- 3
- 6
- 9
-
-
- r(t)
- 12
- 10
- 8
- 6
-
-
-
-
- Data for the function <m>s</m>.
-
-
- t
- 0
- 3
- 6
- 9
-
-
- s(t)
- 12
- 9
- 6.75
- 5.0625
-
-
-
-
-
-
- In Table, we see a function that exhibits constant average rate of change since the change in output is always \triangle r = -2 for any change in input of \triangle t = 3. Said differently, r is a linear function with slope m = -\frac{2}{3}. Since its y-intercept is (0,12), the function's formula is y = r(t) = 12 - \frac{2}{3}t.
-
-
-
- In contrast, the function s given by Table does not exhibit constant average rate of change. Instead, another pattern is present. Observe that if we consider the ratios of consecutive outputs in the table, we see that
-
- \frac{9} {12}= \frac{3}{4}, \frac{6.75}{9} = 0.75 = \frac{3}{4}, \text{ and } \frac{5.0625}{6.75} = 0.75 = \frac{3}{4}
- .
- So, where the differences in the outputs in Table are constant, the ratios in the outputs in Table are constant. The latter is a hallmark of exponential functions and may be used to help us determine the formula of a function for which we have certain information.
-
-
-
- If we know that a certain function is linear, it suffices to know two points that lie on the line to determine the function's formula. It turns out that exponential functions are similar: knowing two points on the graph of a function known to be exponential is enough information to determine the function's formula. In the following example, we show how knowing two values of an exponential function enables us to find both a and b exactly.
-
-
-
-
-
- Suppose that p is an exponential function and we know that p(2) = 11 and p(5) = 18. Determine the exact values of a and b for which p(t) = ab^t.
-
-
-
-
-
- Since we know that p(t) = ab^t, the two data points give us two equations in the unknowns a and b. First, using t = 2,
-
- ab^2 = 11
- ,
- and using t = 5 we also have
-
- ab^5 = 18
- .
- Because we know that the quotient of outputs of an exponential function corresponding to equally-spaced inputs must be constant, we thus naturally consider the quotient \frac{18}{11}. Using Equation and Equation, it follows that
-
- \frac{18}{11} = \frac{ab^5}{ab^2}
- .
- Simplifying the fraction on the right, we see that
-
- \frac{18}{11} = b^3
- .
- Solving for b, we find that b = \sqrt[3]{\frac{18}{11}} is the exact value of b. Substituting this value for b in Equation, it then follows that a \left( \sqrt[3]{\frac{18}{11}} \right)^2 = 11, so
-
- a = \frac{11}{\left( \frac{18}{11} \right)^{2/3}}
- . Therefore,
-
- p(t) = \frac{11}{\left( \frac{18}{11} \right)^{2/3}} \left( \sqrt[3]{\frac{18}{11}} \right)^t \approx 7.9215 \cdot 1.1784^t
- , and a plot of y = p(t) confirms that the function indeed passes through (2,11) and (5,18) as shown in Figure.
-
-
-
-
Plot of p(t) = ab^t that passes through (2,11) and (5,18).
-
Plot of p(t) = ab^t that passes through (2,11) and (5,18).
-
-
-
-
-
-
-
-
-
-
- Trends in the behavior of exponential functions
-
-
- Recall that a function is increasing on an interval if its value always increases as we move from left to right. Similarly, a function is decreasing on an interval provided that its value always decreases as we move from left to right.
-
-
-
-
-
The exponential function f.
-
The exponential function f.
-
-
-
-
The exponential function g.
-
The exponential function g.
-
-
-
-
-
- If we consider an exponential function f with a growth factor b > 1, such as the function pictured in Figure, then the function is always increasing because higher powers of b are greater than lesser powers (for example, (1.2)^3 \gt (1.2)^2). On the other hand, if 0 \lt b \lt 1, then the exponential function will be decreasing because higher powers of positive numbers less than 1 get smaller (e.g., (0.9)^3 \lt (0.9)^2), as seen for the exponential function in Figure.
-
-
-
- An additional trend is apparent in the graphs in Figure and Figure. Each graph bends upward and is therefore concave up. We can better understand why this is so by considering the average rate of change of both f and g on consecutive intervals of the same width. We choose adjacent intervals of length 1 and note particularly that as we compute the average rate of change of each function on such intervals,
-
- AV_{[t,t+1]} = \frac{f(t+1) - f(t)}{t+1-t} = f(t+1) - f(t)
- . Thus, these average rates of change are also measuring the total change in the function across an interval that is 1-unit wide. We now assume that f(t) = 2 (1.25)^t and g(t) = 8(0.75)^t and compute the rate of change of each function on several consecutive intervals.
-
- From the data in Table, we see that the average rate of change is increasing as we increase the value of t. We naturally say that f appears to be increasing at an increasing rate. For the function g, we first notice that its average rate of change is always negative, but also that the average rate of change gets less negative as we increase the value of t. Said differently, the average rate of change of g is also increasing as we increase the value of t. Since g is always decreasing but its average rate of change is increasing, we say that g appears to be decreasing at an increasing rate. These trends hold for exponential functions generallyIt takes calculus to justify this claim fully and rigorously. according to the following conditions.
-
-
-
- Trends in exponential function behavior
-
- For an exponential function of the form f(t) = ab^t where a and b are both positive with b \ne 1,
-
-
-
- if b \gt 1, then f is always increasing and always increases at an increasing rate;
-
-
-
-
- if 0 \lt b \lt 1, then f is always decreasing and always decreases at an increasing rate.
-
-
-
-
-
-
-
- Observe how a function's average rate of change helps us classify the function's behavior on an interval: whether the average rate of change is always positive or always negative on the interval enables us to say if the function is always increasing or always decreasing, and then how the average rate of change itself changes enables us to potentially say how the function is increasing or decreasing through phrases such as decreasing at an increasing rate.
-
-
-
-
-
-
-
- Summary
-
-
-
-
- We say that a function is exponential whenever its algebraic form is f(t) = ab^t for some positive constants a and b where b \ne 1. (Technically, the formal definition of an exponential function is one of form f(t) = b^t, but in our everyday usage of the term exponential we include vertical stretches of these functions and thus allow a to be any positive constant, not just a = 1.)
-
-
-
-
- To determine the formula for an exponential function of form f(t) = ab^t, we need to know two pieces of information. Typically this information is presented in one of two ways.
-
-
-
- If we know the amount, a, of a quantity at time t = 0 and the rate, r, at which the quantity grows or decays per unit time, then it follows f(t) = a(1+r)^t. In this setting, r is often given as a percentage that we convert to a decimal (e.g., if the quantity grows at a rate of 7% per year, we set r = 0.07, so b = 1.07).
-
-
-
-
- If we know any two points on the exponential function's graph, then we can set up a system of two equations in two unknowns and solve for both a and b exactly. In this situation, it is useful to consider the quotient of the two known outputs, as demonstrated in Example.
-
-
-
-
-
-
-
- Exponential functions of the form f(t) = ab^t (where a and b are both positive and b \ne 1) exhibit the following important characteristics:
-
-
-
-
-
- The domain of any exponential function is the set of all real numbers and the range of any exponential function is the set of all positive real numbers.
-
-
-
-
- The y-intercept of the exponential function f(t) = ab^t is (0,a) and the function has no x-intercepts.
-
-
-
-
- If b \gt 1, then the exponential function is always increasing and always increases at an increasing rate. If 0 \lt b \lt 1, then the exponential function is always decreasing and always decreases at an increasing rate.
-
+ What does it mean to say that a function is exponential?
+
+
+
+
+ How much data do we need to know in order to determine the formula for an exponential function?
+
+
+
+
+ Are there important trends that all exponential functions exhibit?
+
+
+
+
+
+ Introduction
+
+ Linear functions have constant average rate of change and model many important phenomena. In other settings, it is natural for a quantity to change at a rate that is proportional to the amount of the quantity present. For instance, whether you put $100 or $100000 or any other amount in a mutual fund, the investment's value changes at a rate proportional the amount present. We often measure that rate in terms of the annual percentage rate of return.
+
+
+
+ Suppose that a certain mutual fund has a 10% annual return. If we invest $100, after 1 year we still have the original $100, plus we gain 10% of $100, so
+
+ 100 \overset{\text{year } 1}{\longrightarrow} 100 + 0.1(100) = 1.1(100)
+ .
+ If we instead invested $100000, after 1 year we again have the original $100000, but now we gain 10% of $100000, and thus
+
+ 100000 \overset{\text{year } 1}{\longrightarrow} 100000 + 0.1(100000) = 1.1(100000)
+ .
+ We therefore see that regardless of the amount of money originally invested, say P, the amount of money we have after 1 year is 1.1P.
+
+
+
+ If we repeat our computations for the second year, we observe that
+
+ 1.1(100) \overset{\text{year } 2}{\longrightarrow} 1.1(100) + 0.1(1.1(100)) = 1.1(1.1(100)) = 1.1^2 (100)
+ .
+ The ideas are identical with the larger dollar value, so
+
+ 1.1(100000) \overset{\text{year } 2}{\longrightarrow} 1.1(100000) + 0.1(1.1(100000)) = 1.1(1.1(100000)) = 1.1^2 (100000)
+ ,
+ and we see that if we invest P dollars, in 2 years our investment will grow to 1.1^2 P.
+
+
+
+ Of course, in 3 years at 10%, the original investment P will have grown to 1.1^3 P. Here we see a new kind of pattern developing: annual growth of 10% is leading to powers of the base 1.1, where the power to which we raise 1.1 corresponds to the number of years the investment has grown. We often call this phenomenon exponential growth. exponential growthintroduction
+
+
+
+
+
+
+
+
+ Exponential functions of form <m>f(t) = ab^t</m>
+
+ In Preview Activity, we encountered the functions I(t) and V(t) that had the same basic structure. Each can be written in the form g(t) = ab^t where a and b are positive constants and b \ne 1. Based on our earlier work with transformations, we know that the constant a is a vertical scaling factor, and thus the main behavior of the function comes from b^t, which we call an exponential function.
+
+ Let b be a real number such that b \gt 0 and b \ne 1. We call the function defined by
+ f(t) = b^t
+ an exponential function with baseb.
+
+
+
+
+
+ For an exponential function f(t) = b^t, we note that f(0) = b^0 = 1, so an exponential function of this form always passes through (0,1). In addition, because a positive number raised to any power is always positive (for instance, 2^{10} = 1024 and 2^{-10} = \frac{1}{2^{10}} = \frac{1}{1024}), the output of an exponential function is also always positive. In particular, f(t) = b^t is never zero and thus has no x-intercepts.
+
+
+
+ Because we will be frequently interested in functions such as I(t) and V(t) with the form ab^t, we will also refer to functions of this form as exponential, understanding that technically these are vertical stretches of exponential functions according to Definition. In Preview Activity, we found that I(t) = 20000(1.08)^t and V(t) = 20000(0.88)^t. It is natural to call 1.08 the growth factor of I and similarly 0.88 the growth factor of V. In addition, we note that these values stem from the actual growth rates: 0.08 for I and -0.12 for V, the latter being negative because value is depreciating. In general, for a function of form f(t) = ab^t, we call b the growth factor. exponential functiongrowth factor Moreover, if b = 1+r, we call r the growth rate. exponential functiongrowth rate Whenever b \gt 1, we often say that the function f is exhibiting exponential growth, wherease if 0 \lt b \lt 1, we say f exhibits exponential decay. exponential functionexponential decay
+
+
+
+ We explore the properties of functions of form f(t) = ab^t further in Activity.
+
+ To better understand the roles that a and b play in an exponential function, let's compare exponential and linear functions. In Table and Table, we see output for two different functions r and s that correspond to equally spaced input.
+
+
+
+
+ Data for the function <m>r</m>.
+
+
+ t
+ 0
+ 3
+ 6
+ 9
+
+
+ r(t)
+ 12
+ 10
+ 8
+ 6
+
+
+
+
+ Data for the function <m>s</m>.
+
+
+ t
+ 0
+ 3
+ 6
+ 9
+
+
+ s(t)
+ 12
+ 9
+ 6.75
+ 5.0625
+
+
+
+
+
+
+ In Table, we see a function that exhibits constant average rate of change since the change in output is always \triangle r = -2 for any change in input of \triangle t = 3. Said differently, r is a linear function with slope m = -\frac{2}{3}. Since its y-intercept is (0,12), the function's formula is y = r(t) = 12 - \frac{2}{3}t.
+
+
+
+ In contrast, the function s given by Table does not exhibit constant average rate of change. Instead, another pattern is present. Observe that if we consider the ratios of consecutive outputs in the table, we see that
+
+ \frac{9} {12}= \frac{3}{4}, \frac{6.75}{9} = 0.75 = \frac{3}{4}, \text{ and } \frac{5.0625}{6.75} = 0.75 = \frac{3}{4}
+ .
+ So, where the differences in the outputs in Table are constant, the ratios in the outputs in Table are constant. The latter is a hallmark of exponential functions and may be used to help us determine the formula of a function for which we have certain information.
+
+
+
+ If we know that a certain function is linear, it suffices to know two points that lie on the line to determine the function's formula. It turns out that exponential functions are similar: knowing two points on the graph of a function known to be exponential is enough information to determine the function's formula. In the following example, we show how knowing two values of an exponential function enables us to find both a and b exactly.
+
+
+
+
+
+ Suppose that p is an exponential function and we know that p(2) = 11 and p(5) = 18. Determine the exact values of a and b for which p(t) = ab^t.
+
+
+
+
+
+ Since we know that p(t) = ab^t, the two data points give us two equations in the unknowns a and b. First, using t = 2,
+
+ ab^2 = 11
+ ,
+ and using t = 5 we also have
+
+ ab^5 = 18
+ .
+ Because we know that the quotient of outputs of an exponential function corresponding to equally-spaced inputs must be constant, we thus naturally consider the quotient \frac{18}{11}. Using Equation and Equation, it follows that
+
+ \frac{18}{11} = \frac{ab^5}{ab^2}
+ .
+ Simplifying the fraction on the right, we see that
+
+ \frac{18}{11} = b^3
+ .
+ Solving for b, we find that b = \sqrt[3]{\frac{18}{11}} is the exact value of b. Substituting this value for b in Equation, it then follows that a \left( \sqrt[3]{\frac{18}{11}} \right)^2 = 11, so
+
+ a = \frac{11}{\left( \frac{18}{11} \right)^{2/3}}
+ . Therefore,
+
+ p(t) = \frac{11}{\left( \frac{18}{11} \right)^{2/3}} \left( \sqrt[3]{\frac{18}{11}} \right)^t \approx 7.9215 \cdot 1.1784^t
+ , and a plot of y = p(t) confirms that the function indeed passes through (2,11) and (5,18) as shown in Figure.
+
+
+
+
Plot of p(t) = ab^t that passes through (2,11) and (5,18).
+
Plot of p(t) = ab^t that passes through (2,11) and (5,18).
+
+
+
+
+
+
+
+
+
+
+ Trends in the behavior of exponential functions
+
+
+ Recall that a function is increasing on an interval if its value always increases as we move from left to right. Similarly, a function is decreasing on an interval provided that its value always decreases as we move from left to right.
+
+
+
+
+
The exponential function f.
+
The exponential function f.
+
+
+
+
The exponential function g.
+
The exponential function g.
+
+
+
+
+
+ If we consider an exponential function f with a growth factor b > 1, such as the function pictured in Figure, then the function is always increasing because higher powers of b are greater than lesser powers (for example, (1.2)^3 \gt (1.2)^2). On the other hand, if 0 \lt b \lt 1, then the exponential function will be decreasing because higher powers of positive numbers less than 1 get smaller (e.g., (0.9)^3 \lt (0.9)^2), as seen for the exponential function in Figure.
+
+
+
+ An additional trend is apparent in the graphs in Figure and Figure. Each graph bends upward and is therefore concave up. We can better understand why this is so by considering the average rate of change of both f and g on consecutive intervals of the same width. We choose adjacent intervals of length 1 and note particularly that as we compute the average rate of change of each function on such intervals,
+
+ AV_{[t,t+1]} = \frac{f(t+1) - f(t)}{t+1-t} = f(t+1) - f(t)
+ . Thus, these average rates of change are also measuring the total change in the function across an interval that is 1-unit wide. We now assume that f(t) = 2 (1.25)^t and g(t) = 8(0.75)^t and compute the rate of change of each function on several consecutive intervals.
+
+ From the data in Table, we see that the average rate of change is increasing as we increase the value of t. We naturally say that f appears to be increasing at an increasing rate. For the function g, we first notice that its average rate of change is always negative, but also that the average rate of change gets less negative as we increase the value of t. Said differently, the average rate of change of g is also increasing as we increase the value of t. Since g is always decreasing but its average rate of change is increasing, we say that g appears to be decreasing at an increasing rate. These trends hold for exponential functions generallyIt takes calculus to justify this claim fully and rigorously. according to the following conditions.
+
+
+
+ Trends in exponential function behavior
+
+ For an exponential function of the form f(t) = ab^t where a and b are both positive with b \ne 1,
+
+
+
+ if b \gt 1, then f is always increasing and always increases at an increasing rate;
+
+
+
+
+ if 0 \lt b \lt 1, then f is always decreasing and always decreases at an increasing rate.
+
+
+
+
+
+
+
+ Observe how a function's average rate of change helps us classify the function's behavior on an interval: whether the average rate of change is always positive or always negative on the interval enables us to say if the function is always increasing or always decreasing, and then how the average rate of change itself changes enables us to potentially say how the function is increasing or decreasing through phrases such as decreasing at an increasing rate.
+
+
+
+
+
+
+
+ Summary
+
+
+
+
+ We say that a function is exponential whenever its algebraic form is f(t) = ab^t for some positive constants a and b where b \ne 1. (Technically, the formal definition of an exponential function is one of form f(t) = b^t, but in our everyday usage of the term exponential we include vertical stretches of these functions and thus allow a to be any positive constant, not just a = 1.)
+
+
+
+
+ To determine the formula for an exponential function of form f(t) = ab^t, we need to know two pieces of information. Typically this information is presented in one of two ways.
+
+
+
+ If we know the amount, a, of a quantity at time t = 0 and the rate, r, at which the quantity grows or decays per unit time, then it follows f(t) = a(1+r)^t. In this setting, r is often given as a percentage that we convert to a decimal (e.g., if the quantity grows at a rate of 7% per year, we set r = 0.07, so b = 1.07).
+
+
+
+
+ If we know any two points on the exponential function's graph, then we can set up a system of two equations in two unknowns and solve for both a and b exactly. In this situation, it is useful to consider the quotient of the two known outputs, as demonstrated in Example.
+
+
+
+
+
+
+
+ Exponential functions of the form f(t) = ab^t (where a and b are both positive and b \ne 1) exhibit the following important characteristics:
+
+
+
+
+
+ The domain of any exponential function is the set of all real numbers and the range of any exponential function is the set of all positive real numbers.
+
+
+
+
+ The y-intercept of the exponential function f(t) = ab^t is (0,a) and the function has no x-intercepts.
+
+
+
+
+ If b \gt 1, then the exponential function is always increasing and always increases at an increasing rate. If 0 \lt b \lt 1, then the exponential function is always decreasing and always decreases at an increasing rate.
+
- What structural rules do logarithms obey that are similar to rules for exponents?
-
-
-
-
- What are the key properties of the graph of the natural logarithm function?
-
-
-
-
- How do logarithms enable us to solve exponential equations?
-
-
-
-
-
- Introduction
-
- Logarithms arise as inverses of exponential functions. In addition, we have motivated their development by our desire to solve exponential equations such as e^k = 3 for k. Because of the inverse relationship between exponential and logarithmic functions, there are several important properties logarithms have that are analogous to ones held by exponential functions. We will work to develop these properties and then show how they are useful in applied settings.
-
-
-
-
-
-
-
- Key properties of logarithms
-
- In Preview Activity, we considered an argument for why \log_{10}(ab) = \log_{10}(a) + \log_{10}(b) for any choice of positive numbers a and b. In what follows, we develop this and other properties of the natural logarithm function; similar reasoning shows the same properties hold for logarithms of any base.
-
-
-
- Let a and b be any positive real numbers so that x = \ln(a) and y = \ln(b) are both defined. Observe that we can rewrite these two equations using the definition of the natural logarithm so that
-
- a = e^x \ \text{ and } \ b = e^y
- .
- Using substitution, we can now say that
-
- \ln(a \cdot b) = \ln(e^x \cdot e^y)
- .
- By exponent rules, we know that \ln(e^x \cdot e^y) = \ln(e^{x+y}), and because the natural logarithm and natural exponential function are inverses, \ln(e^{x+y}) = x+y. Combining the three most recent equations,
-
- \ln(a \cdot b) = x + y
- .
- Finally, recalling that x = \ln(a) and y = \ln(b), we have shown that
-
- \ln(a \cdot b) = \ln(a) + \ln(b)
-
- for any choice of positive real numbers a and b.
-
-
-
- A similar property holds for \ln(\frac{a}{b}). By nearly the same argument, we can say that
-
- \ln\left( \frac{a}{b} \right) &= \ln\left( \frac{e^x}{e^y} \right)
- &= \ln \left( e^{x-y} \right)
- &= x-y
- &= \ln(a) - \ln(b)
- .
-
-
-
- We have thus shown the following general principles.
-
-
-
- Logarithms of products and quotients
- logarithmof a product
- logarithmof a quotient
-
- Because positive integer exponents are a shorthand way to express repeated multiplication, we can use the multiplication rule for logarithms to think about exponents as well. For example,
-
- \ln(a^3) = \ln(a \cdot a \cdot a)
- ,
- and by repeated application of the rule for the natural logarithm of a product, we see
-
- \ln(a^3) = \ln(a) + \ln(a) + \ln(a) = 3\ln(a)
- .
- A similar argument works to show that for every natural number n,
-
- \ln(a^n) = n\ln(a)
- .
- More sophisticated mathematics can be used to prove that the following property holds for every real number exponent t.
-
-
-
- Logarithms of exponential expressions
-
- For any positive real number a and any real number t,
-
- \ln(a^t) = t\ln(a)
- .
-
-
-
-
- The rule that \ln(a^t) = t\ln(a) is extremely powerful: by working with logarithms appropriately, it enables us to move from having a variable in an exponential expression to the variable being part of a linear expression. Moreover, it enables us to solve exponential equations exactly, regardless of the base involved.
-
-
-
-
-
- Solve the equation 7 \cdot 3^t - 1 = 5 exactly for t.
-
-
-
-
- To solve for t, we first solve for 3^t. Adding 1 to both sides and dividing by 7, we find that
-
- 3^t = \frac{6}{7}
- .
- Next, we take the natural logarithm of both sides of the equation. Doing so, we have
-
- \ln \left( 3^t \right) = \ln \left( \frac{6}{7} \right)
- .
- Applying the rule for the logarithm of an exponential expression on the left, we see that
-
- t \ln(3) = \ln \left( \frac{6}{7} \right)
- .
- Both \ln(3) and \ln \left( \frac{6}{7} \right) are simply numbers, and thus we conclude that
-
- t = \frac{\ln \left( \frac{6}{7} \right)}{\ln(3)}
- .
-
-
-
-
-
- The approach used in Example works in a wide range of settings: any time we have an exponential equation of the form p \cdot q^t + r = s, we can solve for t by first isolating the exponential expression q^t and then by taking the natural logarithm of both sides of the equation.
-
-
-
-
-
-
-
- The graph of the natural logarithm
-
- As the inverse of the natural exponential function E(x) = e^x, we have already established that the natural logarithm N(x) = \ln(x) has the set of all positive real numbers as its domain and the set of all real numbers as its range. In addition, being the inverse of E(x) = e^x, we know that when we plot the natural logarithm and natural exponential functions on the same coordinate axes, their graphs are reflections of one another across the line y = x, as seen in Figure and Figure.
-
-
-
-
-
The natural exponential and natural logarithm functions on the interval [-3,3].
-
The natural exponential and natural logarithm functions on the interval [-3,3].
-
-
-
-
The natural exponential and natural logarithm functions on the interval [-15,15].
-
The natural exponential and natural logarithm functions on the interval [-15,15].
-
-
-
-
-
- Indeed, for any point (a,b) that lies on the graph of E(x) = e^x, it follows that the point (b,a) lies on the graph of the inverse N(x) = \ln(x). From this, we see several important properties of the graph of the logarithm function.
-
-
-
- The graph of <m>y = \ln(x)</m>
-
- The graph of y = \ln(x)
-
-
-
- passes through the point (1,0);
-
-
-
-
- is always increasing;
-
-
-
-
- is always concave down; and
-
-
-
-
- increases without bound.
-
-
-
-
-
-
-
- Because the graph of E(x) = e^x increases more and more rapidly as x increases, the graph of N(x) = \ln(x) increases more and more slowly as x increases. Even though the natural logarithm function grows very slowly, it does grow without bound because we can make \ln(x) as large as we want by making x sufficiently large. For instance, if we want x such that \ln(x) = 100, we choose x = e^{100}, since \ln(e^{100}) = 100.
-
-
-
- While the natural exponential function and the natural logarithm (and transformations of these functions) are connected and have certain similar properties, it's also important to be able to distinguish between behavior that is fundamentally exponential and fundamentally logarithmic.
-
-
-
-
-
-
-
- Putting logarithms to work
-
- We've seen in several different settings that the function e^{kt} plays a key role in modeling phenomena in the world around us. We also understand that the value of k controls whether e^{kt} is increasing (k \gt 0) or decreasing (k \lt 0) and how fast the function is increasing or decreasing. As such, we often need to determine the value of k from data that is presented to us; doing so almost always requires the use of logarithms.
-
-
-
-
-
- A population of bacteria cells is growing at a rate proportionate to the number of cells present at a given time t (in hours). Suppose that the number of cells, P, in the population is measured in millions of cells and we know that P(0) = 2.475 and P(10) = 4.298. Find a model of the form P(t) = Ae^{kt} that fits this data and use it to determine the value of k and how long it will take for the population to reach 1 billion cells.
-
-
-
-
- Since the model has form P(t) = Ae^{kt}, we know that P(0) = A. Because we are given that P(0) = 2.475, this shows that A = 2.475. To find k, we use the fact that P(10) = 4.298. Applying this information, A = 2.475, and the form of the model, P(t) = Ae^{kt}, we see that
-
- 4.298 = 2.475 e^{k \cdot 10}
- .
- To solve for k, we first isolate e^{10k} by dividing both sides by 2.475 to get
-
- e^{10k} = \frac{4.298}{2.475}
- .
- Taking the natural logarithm of each side, we find
-
- 10k = \ln \left( \frac{4.298}{2.475} \right)
- ,
- and thus k = \frac{1}{10}\ln \left( \frac{4.298}{2.475} \right) \approx 0.05519.
-
-
-
- To determine how long it takes for the population to reach 1 billion cells, we need to solve the equation P(t) = 1000. Using our preceding work to find A and k, we know that we need to solve the equation
-
- 1000 = 2.475 e^{\frac{1}{10}\ln \left( \frac{4.298}{2.475} \right)t}
- .
- We divide both sides by 2.475 to get e^{\frac{1}{10}\ln \left( \frac{4.298}{2.475} \right)t} = \frac{1000}{2.475}, and after taking the natural logarithm of each side, we see
-
- \frac{1}{10}\ln \left( \frac{4.298}{2.475} \right)t = \ln \left( \frac{1000}{2.475} \right)
- ,
- so that
-
- t = \frac{10 \ln \left( \frac{1000}{2.475} \right)}{\ln \left( \frac{4.298}{2.475} \right)} \approx 108.741
- .
-
-
-
-
-
-
-
-
-
- Summary
-
-
-
-
- There are three fundamental rules for exponents given nonzero base a and exponents m and n:
-
- a^m \cdot a^n = a^{m+n}, \frac{a^m}{a^n} = a^{m-n}, \text{ and } (a^m)^n = a^{mn}
- .
- For logarithmsWe state these rules for the natural logarithm, but they hold for any logarithm of any base., we have the following analogous structural rules for positive real numbers a and b and any real number t:
-
- \ln(a \cdot b) = \ln(a) + \ln(b), \ln \left( \frac{a}{b} \right) = \ln(a) - \ln(b), \text{ and } \ln(a^t) = t \ln(a)
- .
-
-
-
-
- The natural logarithm's domain is the set of all positive real numbers and its range is the set of all real numbers. Its graph passes through (1,0), is always increasing, is always concave down, and increases without bound.
-
-
-
-
- Logarithms are very important in determining values that arise in equations of the form
-
- a^b = c
- ,
- where a and c are known, but b is not. In this context, we can take the natural logarithm of both sides of the equation to find that
-
- \ln(a^b) = \ln(c)
-
- and thus b\ln(a) = \ln(c), so that b = \frac{\ln(c)}{\ln(a)}.
-
+ What structural rules do logarithms obey that are similar to rules for exponents?
+
+
+
+
+ What are the key properties of the graph of the natural logarithm function?
+
+
+
+
+ How do logarithms enable us to solve exponential equations?
+
+
+
+
+
+ Introduction
+
+ Logarithms arise as inverses of exponential functions. In addition, we have motivated their development by our desire to solve exponential equations such as e^k = 3 for k. Because of the inverse relationship between exponential and logarithmic functions, there are several important properties logarithms have that are analogous to ones held by exponential functions. We will work to develop these properties and then show how they are useful in applied settings.
+
+
+
+
+
+
+
+
+ Key properties of logarithms
+
+ In Preview Activity, we considered an argument for why \log_{10}(ab) = \log_{10}(a) + \log_{10}(b) for any choice of positive numbers a and b. In what follows, we develop this and other properties of the natural logarithm function; similar reasoning shows the same properties hold for logarithms of any base.
+
+
+
+ Let a and b be any positive real numbers so that x = \ln(a) and y = \ln(b) are both defined. Observe that we can rewrite these two equations using the definition of the natural logarithm so that
+
+ a = e^x \ \text{ and } \ b = e^y
+ .
+ Using substitution, we can now say that
+
+ \ln(a \cdot b) = \ln(e^x \cdot e^y)
+ .
+ By exponent rules, we know that \ln(e^x \cdot e^y) = \ln(e^{x+y}), and because the natural logarithm and natural exponential function are inverses, \ln(e^{x+y}) = x+y. Combining the three most recent equations,
+
+ \ln(a \cdot b) = x + y
+ .
+ Finally, recalling that x = \ln(a) and y = \ln(b), we have shown that
+
+ \ln(a \cdot b) = \ln(a) + \ln(b)
+
+ for any choice of positive real numbers a and b.
+
+
+
+ A similar property holds for \ln(\frac{a}{b}). By nearly the same argument, we can say that
+
+ \ln\left( \frac{a}{b} \right) &= \ln\left( \frac{e^x}{e^y} \right)
+ &= \ln \left( e^{x-y} \right)
+ &= x-y
+ &= \ln(a) - \ln(b)
+ .
+
+
+
+ We have thus shown the following general principles.
+
+
+
+ Logarithms of products and quotients
+ logarithmof a product
+ logarithmof a quotient
+
+ Because positive integer exponents are a shorthand way to express repeated multiplication, we can use the multiplication rule for logarithms to think about exponents as well. For example,
+
+ \ln(a^3) = \ln(a \cdot a \cdot a)
+ ,
+ and by repeated application of the rule for the natural logarithm of a product, we see
+
+ \ln(a^3) = \ln(a) + \ln(a) + \ln(a) = 3\ln(a)
+ .
+ A similar argument works to show that for every natural number n,
+
+ \ln(a^n) = n\ln(a)
+ .
+ More sophisticated mathematics can be used to prove that the following property holds for every real number exponent t.
+
+
+
+ Logarithms of exponential expressions
+
+ For any positive real number a and any real number t,
+
+ \ln(a^t) = t\ln(a)
+ .
+
+
+
+
+ The rule that \ln(a^t) = t\ln(a) is extremely powerful: by working with logarithms appropriately, it enables us to move from having a variable in an exponential expression to the variable being part of a linear expression. Moreover, it enables us to solve exponential equations exactly, regardless of the base involved.
+
+
+
+
+
+ Solve the equation 7 \cdot 3^t - 1 = 5 exactly for t.
+
+
+
+
+ To solve for t, we first solve for 3^t. Adding 1 to both sides and dividing by 7, we find that
+
+ 3^t = \frac{6}{7}
+ .
+ Next, we take the natural logarithm of both sides of the equation. Doing so, we have
+
+ \ln \left( 3^t \right) = \ln \left( \frac{6}{7} \right)
+ .
+ Applying the rule for the logarithm of an exponential expression on the left, we see that
+
+ t \ln(3) = \ln \left( \frac{6}{7} \right)
+ .
+ Both \ln(3) and \ln \left( \frac{6}{7} \right) are simply numbers, and thus we conclude that
+
+ t = \frac{\ln \left( \frac{6}{7} \right)}{\ln(3)}
+ .
+
+
+
+
+
+ The approach used in Example works in a wide range of settings: any time we have an exponential equation of the form p \cdot q^t + r = s, we can solve for t by first isolating the exponential expression q^t and then by taking the natural logarithm of both sides of the equation.
+
+
+
+
+
+
+
+ The graph of the natural logarithm
+
+ As the inverse of the natural exponential function E(x) = e^x, we have already established that the natural logarithm N(x) = \ln(x) has the set of all positive real numbers as its domain and the set of all real numbers as its range. In addition, being the inverse of E(x) = e^x, we know that when we plot the natural logarithm and natural exponential functions on the same coordinate axes, their graphs are reflections of one another across the line y = x, as seen in Figure and Figure.
+
+
+
+
+
The natural exponential and natural logarithm functions on the interval [-3,3].
+
The natural exponential and natural logarithm functions on the interval [-3,3].
+
+
+
+
The natural exponential and natural logarithm functions on the interval [-15,15].
+
The natural exponential and natural logarithm functions on the interval [-15,15].
+
+
+
+
+
+ Indeed, for any point (a,b) that lies on the graph of E(x) = e^x, it follows that the point (b,a) lies on the graph of the inverse N(x) = \ln(x). From this, we see several important properties of the graph of the logarithm function.
+
+
+
+ The graph of <m>y = \ln(x)</m>
+
+ The graph of y = \ln(x)
+
+
+
+ passes through the point (1,0);
+
+
+
+
+ is always increasing;
+
+
+
+
+ is always concave down; and
+
+
+
+
+ increases without bound.
+
+
+
+
+
+
+
+ Because the graph of E(x) = e^x increases more and more rapidly as x increases, the graph of N(x) = \ln(x) increases more and more slowly as x increases. Even though the natural logarithm function grows very slowly, it does grow without bound because we can make \ln(x) as large as we want by making x sufficiently large. For instance, if we want x such that \ln(x) = 100, we choose x = e^{100}, since \ln(e^{100}) = 100.
+
+
+
+ While the natural exponential function and the natural logarithm (and transformations of these functions) are connected and have certain similar properties, it's also important to be able to distinguish between behavior that is fundamentally exponential and fundamentally logarithmic.
+
+
+
+
+
+
+
+ Putting logarithms to work
+
+ We've seen in several different settings that the function e^{kt} plays a key role in modeling phenomena in the world around us. We also understand that the value of k controls whether e^{kt} is increasing (k \gt 0) or decreasing (k \lt 0) and how fast the function is increasing or decreasing. As such, we often need to determine the value of k from data that is presented to us; doing so almost always requires the use of logarithms.
+
+
+
+
+
+ A population of bacteria cells is growing at a rate proportionate to the number of cells present at a given time t (in hours). Suppose that the number of cells, P, in the population is measured in millions of cells and we know that P(0) = 2.475 and P(10) = 4.298. Find a model of the form P(t) = Ae^{kt} that fits this data and use it to determine the value of k and how long it will take for the population to reach 1 billion cells.
+
+
+
+
+ Since the model has form P(t) = Ae^{kt}, we know that P(0) = A. Because we are given that P(0) = 2.475, this shows that A = 2.475. To find k, we use the fact that P(10) = 4.298. Applying this information, A = 2.475, and the form of the model, P(t) = Ae^{kt}, we see that
+
+ 4.298 = 2.475 e^{k \cdot 10}
+ .
+ To solve for k, we first isolate e^{10k} by dividing both sides by 2.475 to get
+
+ e^{10k} = \frac{4.298}{2.475}
+ .
+ Taking the natural logarithm of each side, we find
+
+ 10k = \ln \left( \frac{4.298}{2.475} \right)
+ ,
+ and thus k = \frac{1}{10}\ln \left( \frac{4.298}{2.475} \right) \approx 0.05519.
+
+
+
+ To determine how long it takes for the population to reach 1 billion cells, we need to solve the equation P(t) = 1000. Using our preceding work to find A and k, we know that we need to solve the equation
+
+ 1000 = 2.475 e^{\frac{1}{10}\ln \left( \frac{4.298}{2.475} \right)t}
+ .
+ We divide both sides by 2.475 to get e^{\frac{1}{10}\ln \left( \frac{4.298}{2.475} \right)t} = \frac{1000}{2.475}, and after taking the natural logarithm of each side, we see
+
+ \frac{1}{10}\ln \left( \frac{4.298}{2.475} \right)t = \ln \left( \frac{1000}{2.475} \right)
+ ,
+ so that
+
+ t = \frac{10 \ln \left( \frac{1000}{2.475} \right)}{\ln \left( \frac{4.298}{2.475} \right)} \approx 108.741
+ .
+
+
+
+
+
+
+
+
+
+ Summary
+
+
+
+
+ There are three fundamental rules for exponents given nonzero base a and exponents m and n:
+
+ a^m \cdot a^n = a^{m+n}, \frac{a^m}{a^n} = a^{m-n}, \text{ and } (a^m)^n = a^{mn}
+ .
+ For logarithmsWe state these rules for the natural logarithm, but they hold for any logarithm of any base., we have the following analogous structural rules for positive real numbers a and b and any real number t:
+
+ \ln(a \cdot b) = \ln(a) + \ln(b), \ln \left( \frac{a}{b} \right) = \ln(a) - \ln(b), \text{ and } \ln(a^t) = t \ln(a)
+ .
+
+
+
+
+ The natural logarithm's domain is the set of all positive real numbers and its range is the set of all real numbers. Its graph passes through (1,0), is always increasing, is always concave down, and increases without bound.
+
+
+
+
+ Logarithms are very important in determining values that arise in equations of the form
+
+ a^b = c
+ ,
+ where a and c are known, but b is not. In this context, we can take the natural logarithm of both sides of the equation to find that
+
+ \ln(a^b) = \ln(c)
+
+ and thus b\ln(a) = \ln(c), so that b = \frac{\ln(c)}{\ln(a)}.
+
- What is the natural logarithm and how is it different from the base-10 logarithm?
-
-
-
-
- How can we solve an equation that involves e to some unknown quantity?
-
-
-
-
-
- Introduction
-
- In Section, we introduced the idea of an inverse function. The fundamental idea is that f has an inverse function if and only if there exists another function g such that f and gundo one another's respective processes. In other words, the process of the function f is reversible, and reversing f generates a related function g.
-
-
-
- More formally, recall that a function y = f(x) (where f : A \to B) has an inverse function if and only if there exists another function g : B \to A such that g(f(x)) = x for every x in A, and f(g(y)) = y for every y in B. We know that given a function f, we can use the to determine whether or not f has an inverse function. Finally, whenever a function f has an inverse function, we call its inverse function f^{-1} and know that the two equations y = f(x) and x = f^{-1}(y) say the same thing from different perspectives.
-
-
-
-
-
-
-
- The base-<m>10</m> logarithm
-
- The powers-of-10 function P(t) = 10^t is an exponential function with base b \gt 1. As such, P is always increasing, and thus its graph passes the , so P has an inverse function. We therefore know there exists some other function, L, such that writing y = P(t) is equivalent to writing t = L(y). For instance, we know that P(2)=100 and P(-3)=\frac{1}{1000}, so it's equivalent to say that L(100) = 2 and L(\frac{1}{1000}) = -3. This new function L we call the base 10 logarithm, which is formally defined as follows.
-
-
-
- logarithmbase 10definition
-
-
- Given a positive real number y, the base-10 logarithm of y is the power to which we raise 10 to get y. We use the notation \log_{10}(y) to denote the base-10 logarithm of y.
-
-
-
-
-
- The base-10 logarithm is therefore the inverse of the powers of 10 function. Whereas P(t) = 10^t takes an input whose value is an exponent and produces the result of taking 10 to that power, the base-10 logarithm takes an input number we view as a power of 10 and produces the corresponding exponent such that 10 to that exponent is the input number.
-
-
-
- In the notation of logarithms, we can now update our earlier observations with the functions P and L and see how exponential equations can be written in two equivalent ways. For instance,
-
- 10^2 = 100 \text{ and } \log_{10}(100) = 2
-
- each say the same thing from two different perspectives. The first says 100 is 10 to the power 2, while the second says 2 is the power to which we raise 10 to get 100. Similarly,
-
- 10^{-3} = \frac{1}{1000} \text{ and } \log_{10} \left( \frac{1}{1000} \right) = -3
- .
-
-
-
- If we rearrange the statements of the facts in Equation, we can see yet another important relationship between the powers of 10 and base-10 logarithm function. Noting that \log_{10}(100) = 2 and 100 = 10^2 are equivalent statements, and substituting the latter equation into the former, we see that
-
- \log_{10}(10^2) = 2
- .
- In words, Equation says that the power to which we raise 10 to get 10^2, is 2. That is, the base-10 logarithm function undoes the work of the powers of 10 function.
-
-
-
- In a similar way, if we rearrange the statements in Equation, we can observe that by replacing -3 with \log_{10}(\frac{1}{1000}) we have
-
- 10^{\log_{10}(\frac{1}{1000})} = \frac{1}{1000}
- .
- In words, Equation says that when 10 is raised to the power to which we raise 10 in order to get \frac{1}{1000}, we get \frac{1}{1000}.
-
-
-
- We summarize the key relationships between the powers-of-10 function and its inverse, the base-10 logarithm function, more generally as follows.
-
- The domain of P is the set of all real numbers and the range of P is the set of all positive real numbers.
-
-
-
-
- The domain of L is the set of all positive real numbers and the range of L is the set of all real numbers.
-
-
-
-
- For any real number t, \log_{10}(10^t) = t. That is, L(P(t)) = t.
-
-
-
-
- For any positive real number y, 10^{\log_{10}(y)} = y. That is, P(L(y)) = y.
-
-
-
-
- 10^0 = 1 and \log_{10}(1) = 0.
-
-
-
-
-
-
-
- The base-10 logarithm function is like the sine or cosine function in this way: for certain special values, it's easy to know the value of the logarithm function. While for sine and cosine the familiar points come from specially placed points on the unit circle, for the base-10 logarithm function, the familiar points come from powers of 10.
- In addition, like sine and cosine, for all other input values, (a) calculus ultimately determines the value of the base-10 logarithm function at other values, and (b) we use computational technology in order to compute these values. For most computational devices, the command log(y) produces the result of the base-10 logarithm of y.
-
-
-
- It's important to note that the logarithm function produces exact values. For instance, if we want to solve the equation 10^t = 5, then it follows that t = \log_{10}(5) is the exact solution to the equation. Like \sqrt{2} or \cos(1), \log_{10}(5) is a number that is an exact value. A computational device can give us a decimal approximation, and we normally want to distinguish between the exact value and the approximate one. For the three different numbers here, \sqrt{2} \approx 1.414, \cos(1) \approx 0.540, and \log_{10}(5) \approx 0.699.
-
-
-
-
-
-
-
- The natural logarithm
-
- The base-10 logarithm is a good starting point for understanding how logarithmic functions work because powers of 10 are easy to mentally compute. We could similarly consider the powers of 2 or powers of 3 function and develop a corresponding logarithm of base 2 or 3. But rather than have a whole collection of different logarithm functions, in the same way that we now use the function e^t and appropriate scaling to represent any exponential function, we develop a single logarithm function that we can use to represent any other logarithmic function through scaling. In correspondence with the natural exponential function, e^t, we now develop its inverse function, and call this inverse function the natural logarithm.
-
-
-
- logarithmnaturaldefinition
-
-
- Given a positive real number y, the natural logarithm of y is the power to which we raise e to get y. We use the notation \ln(y) to denote the natural logarithm of y.
-
-
-
-
-
- We can think of the natural logarithm, \ln(y), as the base-e logarithm. For instance,
-
- \ln(e^{-1}) = -1
-
- and
-
- e^{\ln(2)} = 2
- .
- The former equation is true since the power to which we raise e to get e^{-1} is -1; the latter equation is true since when we raise e to the power to which we raise e to get 2, we get 2. The key relationships between the natural exponential and the natural logarithm function are investigated in Activity.
-
-
-
-
-
-
-
- <m>f(t) = b^t</m> revisited
-
- In Section and Section, we saw that that function f(t) = b^t plays a key role in modeling exponential growth and decay, and that the value of b not only determines whether the function models growth (b \gt 1) or decay (0 \lt b \lt 1), but also how fast the growth or decay occurs. Furthermore, once we introduced the natural base e in Section, we realized that we could write every exponential function of form f(t) = b^t as a horizontal scaling of the function E(t) = e^t by writing
-
- b^t = f(t) = E(kt) = e^{kt}
-
- for some value k. Our development of the natural logarithm function in the current section enables us to now determine k exactly.
-
-
-
-
-
- Determine the exact value of k for which f(t) = 3^t = e^{kt}.
-
-
-
- Solution. Since we want 3^t = e^{kt} to hold for every value of t and e^{kt} = (e^k)^t, we need to have 3^t = (e^k)^t, and thus 3 = e^k. Therefore, k is the power to which we raise e to get 3, which by definition means that k = \ln(3).
-
-
-
-
-
- In modeling important phenomena using exponential functions, we will frequently encounter equations where the variable is in the exponent, like in Example where we had to solve e^k = 3. It is in this context where logarithms find one of their most powerful applications. Activity provides some opportunities to practice solving equations involving the natural base, e, and the natural logarithm.
-
-
-
-
-
-
-
- Summary
-
-
-
-
- The base-10 logarithm of y, denoted \log_{10}(y), is defined to be the power to which we raise 10 to get y. For instance, \log_{10}(1000) = 3, since 10^3 = 1000. The function L(y) = \log_{10}(y) is thus the inverse of the powers-of-10 function, P(t) = 10^t.
-
-
-
-
- The natural logarithm N(y) = \ln(y) differs from the base-10 logarithm in that it is the logarithm with base e instead of 10, and thus \ln(y) is the power to which we raise e to get y. The function N(y) = \ln(y) is the inverse of the natural exponential function E(t) = e^t.
-
-
-
-
- The natural logarithm often enables us to solve an equation that involves e to some unknown quantity. For instance, to solve 2e^{3t-4} + 5 = 13, we can first solve for e^{3t-4} by subtracting 5 from each side and dividing by 2 to get
-
- e^{3t-4} = 4
- .
- This last equation says e to some power is 4. We know that it is equivalent to say
-
- \ln(4) = 3t-4
- .
- Since \ln(4) is a number, we can solve this most recent linear equation for t. In particular, 3t = 4 + \ln(4), so
-
- t = \frac{1}{3}(4 + \ln(4))
- .
-
-
-
-
-
-
-
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ What a logarithm is
+
+
+
+
+
+ How is the base-10 logarithm defined?
+
+
+
+
+ What is the natural logarithm and how is it different from the base-10 logarithm?
+
+
+
+
+ How can we solve an equation that involves e to some unknown quantity?
+
+
+
+
+
+ Introduction
+
+ In Section, we introduced the idea of an inverse function. The fundamental idea is that f has an inverse function if and only if there exists another function g such that f and gundo one another's respective processes. In other words, the process of the function f is reversible, and reversing f generates a related function g.
+
+
+
+ More formally, recall that a function y = f(x) (where f : A \to B) has an inverse function if and only if there exists another function g : B \to A such that g(f(x)) = x for every x in A, and f(g(y)) = y for every y in B. We know that given a function f, we can use the to determine whether or not f has an inverse function. Finally, whenever a function f has an inverse function, we call its inverse function f^{-1} and know that the two equations y = f(x) and x = f^{-1}(y) say the same thing from different perspectives.
+
+
+
+
+
+
+
+
+ The base-<m>10</m> logarithm
+
+ The powers-of-10 function P(t) = 10^t is an exponential function with base b \gt 1. As such, P is always increasing, and thus its graph passes the , so P has an inverse function. We therefore know there exists some other function, L, such that writing y = P(t) is equivalent to writing t = L(y). For instance, we know that P(2)=100 and P(-3)=\frac{1}{1000}, so it's equivalent to say that L(100) = 2 and L(\frac{1}{1000}) = -3. This new function L we call the base 10 logarithm, which is formally defined as follows.
+
+
+
+ logarithmbase 10definition
+
+
+ Given a positive real number y, the base-10 logarithm of y is the power to which we raise 10 to get y. We use the notation \log_{10}(y) to denote the base-10 logarithm of y.
+
+
+
+
+
+ The base-10 logarithm is therefore the inverse of the powers of 10 function. Whereas P(t) = 10^t takes an input whose value is an exponent and produces the result of taking 10 to that power, the base-10 logarithm takes an input number we view as a power of 10 and produces the corresponding exponent such that 10 to that exponent is the input number.
+
+
+
+ In the notation of logarithms, we can now update our earlier observations with the functions P and L and see how exponential equations can be written in two equivalent ways. For instance,
+
+ 10^2 = 100 \text{ and } \log_{10}(100) = 2
+
+ each say the same thing from two different perspectives. The first says 100 is 10 to the power 2, while the second says 2 is the power to which we raise 10 to get 100. Similarly,
+
+ 10^{-3} = \frac{1}{1000} \text{ and } \log_{10} \left( \frac{1}{1000} \right) = -3
+ .
+
+
+
+ If we rearrange the statements of the facts in Equation, we can see yet another important relationship between the powers of 10 and base-10 logarithm function. Noting that \log_{10}(100) = 2 and 100 = 10^2 are equivalent statements, and substituting the latter equation into the former, we see that
+
+ \log_{10}(10^2) = 2
+ .
+ In words, Equation says that the power to which we raise 10 to get 10^2, is 2. That is, the base-10 logarithm function undoes the work of the powers of 10 function.
+
+
+
+ In a similar way, if we rearrange the statements in Equation, we can observe that by replacing -3 with \log_{10}(\frac{1}{1000}) we have
+
+ 10^{\log_{10}(\frac{1}{1000})} = \frac{1}{1000}
+ .
+ In words, Equation says that when 10 is raised to the power to which we raise 10 in order to get \frac{1}{1000}, we get \frac{1}{1000}.
+
+
+
+ We summarize the key relationships between the powers-of-10 function and its inverse, the base-10 logarithm function, more generally as follows.
+
+ The domain of P is the set of all real numbers and the range of P is the set of all positive real numbers.
+
+
+
+
+ The domain of L is the set of all positive real numbers and the range of L is the set of all real numbers.
+
+
+
+
+ For any real number t, \log_{10}(10^t) = t. That is, L(P(t)) = t.
+
+
+
+
+ For any positive real number y, 10^{\log_{10}(y)} = y. That is, P(L(y)) = y.
+
+
+
+
+ 10^0 = 1 and \log_{10}(1) = 0.
+
+
+
+
+
+
+
+ The base-10 logarithm function is like the sine or cosine function in this way: for certain special values, it's easy to know the value of the logarithm function. While for sine and cosine the familiar points come from specially placed points on the unit circle, for the base-10 logarithm function, the familiar points come from powers of 10.
+ In addition, like sine and cosine, for all other input values, (a) calculus ultimately determines the value of the base-10 logarithm function at other values, and (b) we use computational technology in order to compute these values. For most computational devices, the command log(y) produces the result of the base-10 logarithm of y.
+
+
+
+ It's important to note that the logarithm function produces exact values. For instance, if we want to solve the equation 10^t = 5, then it follows that t = \log_{10}(5) is the exact solution to the equation. Like \sqrt{2} or \cos(1), \log_{10}(5) is a number that is an exact value. A computational device can give us a decimal approximation, and we normally want to distinguish between the exact value and the approximate one. For the three different numbers here, \sqrt{2} \approx 1.414, \cos(1) \approx 0.540, and \log_{10}(5) \approx 0.699.
+
+
+
+
+
+
+
+ The natural logarithm
+
+ The base-10 logarithm is a good starting point for understanding how logarithmic functions work because powers of 10 are easy to mentally compute. We could similarly consider the powers of 2 or powers of 3 function and develop a corresponding logarithm of base 2 or 3. But rather than have a whole collection of different logarithm functions, in the same way that we now use the function e^t and appropriate scaling to represent any exponential function, we develop a single logarithm function that we can use to represent any other logarithmic function through scaling. In correspondence with the natural exponential function, e^t, we now develop its inverse function, and call this inverse function the natural logarithm.
+
+
+
+ logarithmnaturaldefinition
+
+
+ Given a positive real number y, the natural logarithm of y is the power to which we raise e to get y. We use the notation \ln(y) to denote the natural logarithm of y.
+
+
+
+
+
+ We can think of the natural logarithm, \ln(y), as the base-e logarithm. For instance,
+
+ \ln(e^{-1}) = -1
+
+ and
+
+ e^{\ln(2)} = 2
+ .
+ The former equation is true since the power to which we raise e to get e^{-1} is -1; the latter equation is true since when we raise e to the power to which we raise e to get 2, we get 2. The key relationships between the natural exponential and the natural logarithm function are investigated in Activity.
+
+
+
+
+
+
+
+ <m>f(t) = b^t</m> revisited
+
+ In Section and Section, we saw that that function f(t) = b^t plays a key role in modeling exponential growth and decay, and that the value of b not only determines whether the function models growth (b \gt 1) or decay (0 \lt b \lt 1), but also how fast the growth or decay occurs. Furthermore, once we introduced the natural base e in Section, we realized that we could write every exponential function of form f(t) = b^t as a horizontal scaling of the function E(t) = e^t by writing
+
+ b^t = f(t) = E(kt) = e^{kt}
+
+ for some value k. Our development of the natural logarithm function in the current section enables us to now determine k exactly.
+
+
+
+
+
+ Determine the exact value of k for which f(t) = 3^t = e^{kt}.
+
+
+
+ Solution. Since we want 3^t = e^{kt} to hold for every value of t and e^{kt} = (e^k)^t, we need to have 3^t = (e^k)^t, and thus 3 = e^k. Therefore, k is the power to which we raise e to get 3, which by definition means that k = \ln(3).
+
+
+
+
+
+ In modeling important phenomena using exponential functions, we will frequently encounter equations where the variable is in the exponent, like in Example where we had to solve e^k = 3. It is in this context where logarithms find one of their most powerful applications. Activity provides some opportunities to practice solving equations involving the natural base, e, and the natural logarithm.
+
+
+
+
+
+
+
+ Summary
+
+
+
+
+ The base-10 logarithm of y, denoted \log_{10}(y), is defined to be the power to which we raise 10 to get y. For instance, \log_{10}(1000) = 3, since 10^3 = 1000. The function L(y) = \log_{10}(y) is thus the inverse of the powers-of-10 function, P(t) = 10^t.
+
+
+
+
+ The natural logarithm N(y) = \ln(y) differs from the base-10 logarithm in that it is the logarithm with base e instead of 10, and thus \ln(y) is the power to which we raise e to get y. The function N(y) = \ln(y) is the inverse of the natural exponential function E(t) = e^t.
+
+
+
+
+ The natural logarithm often enables us to solve an equation that involves e to some unknown quantity. For instance, to solve 2e^{3t-4} + 5 = 13, we can first solve for e^{3t-4} by subtracting 5 from each side and dividing by 2 to get
+
+ e^{3t-4} = 4
+ .
+ This last equation says e to some power is 4. We know that it is equivalent to say
+
+ \ln(4) = 3t-4
+ .
+ Since \ln(4) is a number, we can solve this most recent linear equation for t. In particular, 3t = 4 + \ln(4), so
+
+ t = \frac{1}{3}(4 + \ln(4))
+ .
+
- What can we say about the behavior of an exponential function as the input gets larger and larger?
-
-
-
-
- How do vertical stretches and shifts of an exponential function affect its behavior?
-
-
-
-
- Why is the temperature of a cooling or warming object modeled by a function of the form F(t) = ab^t + c?
-
-
-
-
-
- Introduction
-
- If a quantity changes so that its growth or decay occurs at a constant percentage rate with respect to time, the function is exponential. This is because if the growth or decay rate is r, the total amount of the quantity at time t is given by
-
- A(t) = a(1+r)^t
- ,
- where a is the amount present at time t = 0. Many different natural quantities change according to exponential models: money growth through compounding interest, the growth of a population of cells, and the decay of radioactive elements.
-
-
-
- A related situation arises when an object's temperature changes in response to its surroundings. For instance, if we have a cup of coffee at an initial temperature of 186^\circ Fahrenheit and the cup is placed in a room where the surrounding temperature is 71^\circ, our intuition and experience tell us that over time the coffee will cool and eventually tend to the 71^\circ temperature of the surroundings. From an experiment See http://gvsu.edu/s/0SB for this data. with an actual temperature probe, we have the data in Table that is plotted in Figure.
-
- In one sense, the data looks exponential: the points appear to lie on a curve that is always decreasing and decreasing at an increasing rate. However, we know that the function can't have the form f(t) = ab^t because such a function's range is the set of all positive real numbers, and it's impossible for the coffee's temperature to fall below room temperature (71^\circ). It is natural to wonder if a function of the form g(t) = ab^t + c will work. Thus, in order to find a function that fits the data in a situation such as Figure, we begin by investigating and understanding the roles of a, b, and c in the behavior of g(t) = ab^t + c.
-
- We have already established that any exponential function of the form f(t) = ab^t where a and b are positive real numbers with b \ne 1 is always concave up and is either always increasing or always decreasing. We next introduce precise language to describe the behavior of an exponential function's value as t gets bigger and bigger. To start, let's consider the two basic exponential functions p(t) = 2^t and q(t) = (\frac{1}{2})^t and their respective values at t = 10, t = 20, and t = 30, as displayed in Table and Table.
-
- For the increasing function p(t) = 2^t, we see that the output of the function gets very large very quickly. In addition, there is no upper bound to how large the function can be. Indeed, we can make the value of p(t) as large as we'd like by taking t sufficiently big. We thus say that as t increases, p(t)increases without bound. increasingwithout bound
-
-
-
- For the decreasing function q(t) = (\frac{1}{2})^t, we see that the output q(t) is always positive but getting closer and closer to 0. Indeed, because we can make 2^t as large as we like, it follows that we can make its reciprocal \frac{1}{2^t} = (\frac{1}{2})^t as small as we'd like. We thus say that as t increases, q(t)approaches 0. approaching 0
-
-
-
- To represent these two common phenomena with exponential functionsthe value increasing without bound or the value approaching 0we will use shorthand notation. First, it is natural to write q(t) \to 0 as t increases without bound. Moreover, since we have the notion of the infinite to represent quantities without bound, we use the symbol for infinity and arrow notation infinity (\infty) and write p(t) \to \infty as t increases without bound in order to indicate that p(t) increases without bound.
-
-
-
- In Preview Activity, we saw how the value of b affects the steepness of the graph of f(t) = ab^t, as well as how all graphs with b \gt 1 have the similar increasing behavior, and all graphs with 0 \lt b \lt 1 have similar decreasing behavior. For instance, by taking t sufficiently large, we can make (1.01)^t as large as we want; it just takes much larger t to make (1.01)^t big in comparison to 2^t. In the same way, we can make (0.99)^t as close to 0 as we wish by taking t sufficiently big, even though it takes longer for (0.99)^t to get close to 0 in comparison to (\frac{1}{2})^t. For an arbitrary choice of b, we can say the following.
-
-
-
- Long-term behavior of exponential functions
-
- Let f(t) = b^t with b \gt 0 and b \ne 1.
-
-
-
-
- If 0 \lt b \lt 1, then b^t \to 0 as t \to \infty. We read this notation as b^t tends to 0 as t increases without bound.
-
-
-
-
- If b \gt 1, then b^t \to \infty as t \to \infty. We read this notation as b^t increases without bound as t increases without bound.
-
-
-
-
-
-
-
- In addition, we make a key observation about the use of exponents. For the function q(t) = (\frac{1}{2})^t, there are three equivalent ways we may write the function:
-
- \left( \frac{1}{2} \right)^t = \frac{1}{2^t} = 2^{-t}
- .
- In our work with transformations involving horizontal scaling in Exercise, we saw that the graph of y = h(-t) is the reflection of the graph of y = h(t) across the y-axis. Therefore, we can say that the graphs of p(t) = 2^t and q(t) = (\frac{1}{2})^t = 2^{-t} are reflections of one another in the y-axis since p(-t) = 2^{-t} = q(t). We see this fact verified in Figure.
-
-
-
-
Plots of p(t) = 2^t and q(t) = 2^{-t}.
-
Plots of p(t) = 2^t and q(t) = 2^{-t}.
-
-
-
-
- Similar observations hold for the relationship between the graphs of b^{t} and \frac{1}{b^t} = b^{-t} for any positive b \ne 1.
-
-
-
-
-
- The role of <m>c</m> in <m>g(t) = ab^t + c</m>
-
- The function g(t) = ab^t + c is a vertical translation of the function f(t) = ab^t. We now have extensive understanding of the behavior of f(t) and how that behavior depends on a and b. Since a vertical translation by c does not change the shape of any graph, we expect that g will exhibit very similar behavior to f. Indeed, we can compare the two functions' graphs as shown in Figure and Figure and then make the following general observations.
-
- Let g(t) = ab^t + c with a \gt 0, b \gt 0 and b \ne 1, and c any real number.
-
-
-
-
- If 0 \lt b \lt 1, then g(t) = ab^t + c \to c as t \to \infty. The function g is always decreasing, always concave up, and has y-intercept (0,a+c). The range of the function is all real numbers greater than c.
-
-
-
-
- If b \gt 1, then g(t) = ab^t + c \to \infty as t \to \infty. The function g is always increasing, always concave up, and has y-intercept (0,a+c). The range of the function is all real numbers greater than c.
-
-
-
-
-
-
-
- It is also possible to have a \lt 0. In this situation, because g(t) = ab^t is both a reflection of f(t) = b^t across the x-axis and a vertical stretch by |a|, the function g is always concave down. If 0 \lt b \lt 1 so that f is always decreasing, then g is always increasing; if instead b \gt 1 so f is increasing, then g is decreasing. Moreover, instead of the range of the function g having a lower bound as when a \gt 0, in this setting the range of g has an upper bound. These ideas are explored further in Activity.
-
-
-
- It's an important skill to be able to look at an exponential function of the form g(t) = ab^t + c and form an accurate mental picture of the graph's main features in light of the values of a, b, and c.
-
-
-
-
-
-
-
- Modeling temperature data
-
-
- Newton's Law of Cooling Newton's Law of Cooling states that the rate that an object warms or cools occurs in direct proportion to the difference between its own temperature and the temperature of its surroundings. If we return to the coffee temperature data in Table and recall that the room temperature in that experiment was 71^\circ, we can see how to use a transformed exponential function to model the data. In Table, we add a row of information to the table where we compute F(t)-71 to subtract the room temperature from each reading.
-
- The data in the bottom row of Table appears exponential, and if we test the data by computing the quotients of output values that correspond to equally-spaced input, we see a nearly constant ratio. In particular,
-
- \frac{73}{85} \approx 0.86, \ \frac{64}{73} \approx 0.88, \ \frac{56}{64} \approx 0.88, \ \frac{49}{56} \approx 0.88, \ \frac{45}{49} \approx 0.92, \text{and} \frac{40}{45} \approx 0.89
- .
- Of course, there is some measurement error in the data (plus it is only recorded to accuracy of whole degrees), so these computations provide convincing evidence that the underlying function is exponential. In addition, we expect that if the data continued in the bottom row of Table, the values would approach 0 because F(t) will approach 71.
-
-
-
-
-
Plot of f(t) = 103.503 (0.974)^t.
-
Plot of f(t) = 103.503 (0.974)^t.
-
-
-
-
Plot of F(t) = 103.503 (0.974)^t + 71.
-
Plot of F(t) = 103.503 (0.974)^t + 71.
-
-
-
-
-
- If we choose two of the data points, say (18,64) and (23,56), and assume that f(t) = ab^t, we can determine the values of a and b. Doing so, it turns out that a \approx 103.503 and b \approx 0.974, so f(t) = 103.503 ( 0.974)^t. Since f(t) = F(t) - 71, we see that F(t) = f(t) + 71, so F(t) = 103.503 (0.974)^t + 71. Plotting f against the shifted data and F along with the original data in Figure and Figure, we see that the curves go exactly through the points where t = 18 and t = 23 as expected, but also that the function provides a reasonable model for the observed behavior at any time t. If our data was even more accurate, we would expect that the curve's fit would be even better.
-
-
-
- Our preceding work with the coffee data can be done similarly with data for any cooling or warming object whose temperature initially differs from its surroundings. Indeed, it is possible to show that Newton's Law of Cooling implies that the object's temperature is given by a function of the form F(t) = ab^t + c.
-
-
-
-
-
-
-
-
-
- Summary
-
-
-
-
- For an exponential function of the form f(t) = b^t, the function either approaches zero or grows without bound as the input gets larger and larger. In particular, if 0 \lt b \lt 1, then f(t) = b^t \to 0 as t \to \infty, while if b \gt 1, then f(t) = b^t \to \infty as t \to \infty. Scaling f by a positive value a (that is, the transformed function ab^t) does not affect the long-range behavior: whether the function tends to 0 or increases without bound depends solely on whether b is less than or greater than 1.
-
-
-
-
- The function f(t) = b^t passes through (0,1), is always concave up, is either always increasing or always decreasing, and its range is the set of all positive real numbers. Among these properties, a vertical stretch by a positive value a only affects the y-intercept, which is instead (0,a). If we include a vertical shift and write g(t) = ab^t + c, the biggest change is that the range of g is the set of all real numbers greater than c. In addition, the y-intercept of g is (0,a+c).
-
-
-
- In the situation where a \lt 0, several other changes are induced. Here, because g(t) = ab^t is both a reflection of f(t) = b^t across the x-axis and a vertical stretch by |a|, the function g is now always concave down. If 0 \lt b \lt 1 so that f is always decreasing, then g (the reflected function) is now always increasing; if instead b \gt 1 so f is increasing, then g is decreasing. Finally, if a \lt 0, then the range of g(t) = ab^t + c is the set of all real numbers less than c.
-
-
-
-
- An exponential function can be thought of as a function that changes at a rate proportional to itself, like how money grows with compound interest or the amount of a radioactive quantity decays. Newton's Law of Cooling says that the rate of change of an object's temperature is proportional to the difference between its own temperature and the temperature of its surroundings. This leads to the function that measures the difference between the object's temperature and room temperature being exponential, and hence the object's temperature itself is a vertically-shifted exponential function of the form F(t) = ab^t + c.
-
+ What can we say about the behavior of an exponential function as the input gets larger and larger?
+
+
+
+
+ How do vertical stretches and shifts of an exponential function affect its behavior?
+
+
+
+
+ Why is the temperature of a cooling or warming object modeled by a function of the form F(t) = ab^t + c?
+
+
+
+
+
+ Introduction
+
+ If a quantity changes so that its growth or decay occurs at a constant percentage rate with respect to time, the function is exponential. This is because if the growth or decay rate is r, the total amount of the quantity at time t is given by
+
+ A(t) = a(1+r)^t
+ ,
+ where a is the amount present at time t = 0. Many different natural quantities change according to exponential models: money growth through compounding interest, the growth of a population of cells, and the decay of radioactive elements.
+
+
+
+ A related situation arises when an object's temperature changes in response to its surroundings. For instance, if we have a cup of coffee at an initial temperature of 186^\circ Fahrenheit and the cup is placed in a room where the surrounding temperature is 71^\circ, our intuition and experience tell us that over time the coffee will cool and eventually tend to the 71^\circ temperature of the surroundings. From an experiment See http://gvsu.edu/s/0SB for this data. with an actual temperature probe, we have the data in Table that is plotted in Figure.
+
+ In one sense, the data looks exponential: the points appear to lie on a curve that is always decreasing and decreasing at an increasing rate. However, we know that the function can't have the form f(t) = ab^t because such a function's range is the set of all positive real numbers, and it's impossible for the coffee's temperature to fall below room temperature (71^\circ). It is natural to wonder if a function of the form g(t) = ab^t + c will work. Thus, in order to find a function that fits the data in a situation such as Figure, we begin by investigating and understanding the roles of a, b, and c in the behavior of g(t) = ab^t + c.
+
+ We have already established that any exponential function of the form f(t) = ab^t where a and b are positive real numbers with b \ne 1 is always concave up and is either always increasing or always decreasing. We next introduce precise language to describe the behavior of an exponential function's value as t gets bigger and bigger. To start, let's consider the two basic exponential functions p(t) = 2^t and q(t) = (\frac{1}{2})^t and their respective values at t = 10, t = 20, and t = 30, as displayed in Table and Table.
+
+ For the increasing function p(t) = 2^t, we see that the output of the function gets very large very quickly. In addition, there is no upper bound to how large the function can be. Indeed, we can make the value of p(t) as large as we'd like by taking t sufficiently big. We thus say that as t increases, p(t)increases without bound. increasingwithout bound
+
+
+
+ For the decreasing function q(t) = (\frac{1}{2})^t, we see that the output q(t) is always positive but getting closer and closer to 0. Indeed, because we can make 2^t as large as we like, it follows that we can make its reciprocal \frac{1}{2^t} = (\frac{1}{2})^t as small as we'd like. We thus say that as t increases, q(t)approaches 0. approaching 0
+
+
+
+ To represent these two common phenomena with exponential functionsthe value increasing without bound or the value approaching 0we will use shorthand notation. First, it is natural to write q(t) \to 0 as t increases without bound. Moreover, since we have the notion of the infinite to represent quantities without bound, we use the symbol for infinity and arrow notation infinity (\infty) and write p(t) \to \infty as t increases without bound in order to indicate that p(t) increases without bound.
+
+
+
+ In Preview Activity, we saw how the value of b affects the steepness of the graph of f(t) = ab^t, as well as how all graphs with b \gt 1 have the similar increasing behavior, and all graphs with 0 \lt b \lt 1 have similar decreasing behavior. For instance, by taking t sufficiently large, we can make (1.01)^t as large as we want; it just takes much larger t to make (1.01)^t big in comparison to 2^t. In the same way, we can make (0.99)^t as close to 0 as we wish by taking t sufficiently big, even though it takes longer for (0.99)^t to get close to 0 in comparison to (\frac{1}{2})^t. For an arbitrary choice of b, we can say the following.
+
+
+
+ Long-term behavior of exponential functions
+
+ Let f(t) = b^t with b \gt 0 and b \ne 1.
+
+
+
+
+ If 0 \lt b \lt 1, then b^t \to 0 as t \to \infty. We read this notation as b^t tends to 0 as t increases without bound.
+
+
+
+
+ If b \gt 1, then b^t \to \infty as t \to \infty. We read this notation as b^t increases without bound as t increases without bound.
+
+
+
+
+
+
+
+ In addition, we make a key observation about the use of exponents. For the function q(t) = (\frac{1}{2})^t, there are three equivalent ways we may write the function:
+
+ \left( \frac{1}{2} \right)^t = \frac{1}{2^t} = 2^{-t}
+ .
+ In our work with transformations involving horizontal scaling in Exercise, we saw that the graph of y = h(-t) is the reflection of the graph of y = h(t) across the y-axis. Therefore, we can say that the graphs of p(t) = 2^t and q(t) = (\frac{1}{2})^t = 2^{-t} are reflections of one another in the y-axis since p(-t) = 2^{-t} = q(t). We see this fact verified in Figure.
+
+
+
+
Plots of p(t) = 2^t and q(t) = 2^{-t}.
+
Plots of p(t) = 2^t and q(t) = 2^{-t}.
+
+
+
+
+ Similar observations hold for the relationship between the graphs of b^{t} and \frac{1}{b^t} = b^{-t} for any positive b \ne 1.
+
+
+
+
+
+ The role of <m>c</m> in <m>g(t) = ab^t + c</m>
+
+ The function g(t) = ab^t + c is a vertical translation of the function f(t) = ab^t. We now have extensive understanding of the behavior of f(t) and how that behavior depends on a and b. Since a vertical translation by c does not change the shape of any graph, we expect that g will exhibit very similar behavior to f. Indeed, we can compare the two functions' graphs as shown in Figure and Figure and then make the following general observations.
+
+ Let g(t) = ab^t + c with a \gt 0, b \gt 0 and b \ne 1, and c any real number.
+
+
+
+
+ If 0 \lt b \lt 1, then g(t) = ab^t + c \to c as t \to \infty. The function g is always decreasing, always concave up, and has y-intercept (0,a+c). The range of the function is all real numbers greater than c.
+
+
+
+
+ If b \gt 1, then g(t) = ab^t + c \to \infty as t \to \infty. The function g is always increasing, always concave up, and has y-intercept (0,a+c). The range of the function is all real numbers greater than c.
+
+
+
+
+
+
+
+ It is also possible to have a \lt 0. In this situation, because g(t) = ab^t is both a reflection of f(t) = b^t across the x-axis and a vertical stretch by |a|, the function g is always concave down. If 0 \lt b \lt 1 so that f is always decreasing, then g is always increasing; if instead b \gt 1 so f is increasing, then g is decreasing. Moreover, instead of the range of the function g having a lower bound as when a \gt 0, in this setting the range of g has an upper bound. These ideas are explored further in Activity.
+
+
+
+ It's an important skill to be able to look at an exponential function of the form g(t) = ab^t + c and form an accurate mental picture of the graph's main features in light of the values of a, b, and c.
+
+
+
+
+
+
+
+ Modeling temperature data
+
+
+ Newton's Law of Cooling Newton's Law of Cooling states that the rate that an object warms or cools occurs in direct proportion to the difference between its own temperature and the temperature of its surroundings. If we return to the coffee temperature data in Table and recall that the room temperature in that experiment was 71^\circ, we can see how to use a transformed exponential function to model the data. In Table, we add a row of information to the table where we compute F(t)-71 to subtract the room temperature from each reading.
+
+ The data in the bottom row of Table appears exponential, and if we test the data by computing the quotients of output values that correspond to equally-spaced input, we see a nearly constant ratio. In particular,
+
+ \frac{73}{85} \approx 0.86, \ \frac{64}{73} \approx 0.88, \ \frac{56}{64} \approx 0.88, \ \frac{49}{56} \approx 0.88, \ \frac{45}{49} \approx 0.92, \text{and} \frac{40}{45} \approx 0.89
+ .
+ Of course, there is some measurement error in the data (plus it is only recorded to accuracy of whole degrees), so these computations provide convincing evidence that the underlying function is exponential. In addition, we expect that if the data continued in the bottom row of Table, the values would approach 0 because F(t) will approach 71.
+
+
+
+
+
Plot of f(t) = 103.503 (0.974)^t.
+
Plot of f(t) = 103.503 (0.974)^t.
+
+
+
+
Plot of F(t) = 103.503 (0.974)^t + 71.
+
Plot of F(t) = 103.503 (0.974)^t + 71.
+
+
+
+
+
+ If we choose two of the data points, say (18,64) and (23,56), and assume that f(t) = ab^t, we can determine the values of a and b. Doing so, it turns out that a \approx 103.503 and b \approx 0.974, so f(t) = 103.503 ( 0.974)^t. Since f(t) = F(t) - 71, we see that F(t) = f(t) + 71, so F(t) = 103.503 (0.974)^t + 71. Plotting f against the shifted data and F along with the original data in Figure and Figure, we see that the curves go exactly through the points where t = 18 and t = 23 as expected, but also that the function provides a reasonable model for the observed behavior at any time t. If our data was even more accurate, we would expect that the curve's fit would be even better.
+
+
+
+ Our preceding work with the coffee data can be done similarly with data for any cooling or warming object whose temperature initially differs from its surroundings. Indeed, it is possible to show that Newton's Law of Cooling implies that the object's temperature is given by a function of the form F(t) = ab^t + c.
+
+
+
+
+
+
+
+
+
+ Summary
+
+
+
+
+ For an exponential function of the form f(t) = b^t, the function either approaches zero or grows without bound as the input gets larger and larger. In particular, if 0 \lt b \lt 1, then f(t) = b^t \to 0 as t \to \infty, while if b \gt 1, then f(t) = b^t \to \infty as t \to \infty. Scaling f by a positive value a (that is, the transformed function ab^t) does not affect the long-range behavior: whether the function tends to 0 or increases without bound depends solely on whether b is less than or greater than 1.
+
+
+
+
+ The function f(t) = b^t passes through (0,1), is always concave up, is either always increasing or always decreasing, and its range is the set of all positive real numbers. Among these properties, a vertical stretch by a positive value a only affects the y-intercept, which is instead (0,a). If we include a vertical shift and write g(t) = ab^t + c, the biggest change is that the range of g is the set of all real numbers greater than c. In addition, the y-intercept of g is (0,a+c).
+
+
+
+ In the situation where a \lt 0, several other changes are induced. Here, because g(t) = ab^t is both a reflection of f(t) = b^t across the x-axis and a vertical stretch by |a|, the function g is now always concave down. If 0 \lt b \lt 1 so that f is always decreasing, then g (the reflected function) is now always increasing; if instead b \gt 1 so f is increasing, then g is decreasing. Finally, if a \lt 0, then the range of g(t) = ab^t + c is the set of all real numbers less than c.
+
+
+
+
+ An exponential function can be thought of as a function that changes at a rate proportional to itself, like how money grows with compound interest or the amount of a radioactive quantity decays. Newton's Law of Cooling says that the rate of change of an object's temperature is proportional to the difference between its own temperature and the temperature of its surroundings. This leads to the function that measures the difference between the object's temperature and room temperature being exponential, and hence the object's temperature itself is a vertically-shifted exponential function of the form F(t) = ab^t + c.
+
- What roles do the parameters a, k, and c play in how the function F(t) = c + ae^{-kt} models the temperature of an object that is cooling or warming in its surroundings?
-
-
-
-
- How can we use an exponential function to more realistically model a population whose growth levels off?
-
-
-
-
-
- Introduction
-
-
- We've seen that exponential functions can be used to model several different important phenomena, such as the growth of money due to continuously compounded interest, the decay of radioactive quanitities, and the temperature of an object that is cooling or warming due to its surroundings. From initial work with functions of the form f(t) = ab^t where b \gt 0 and b \ne 1, we found that shifted exponential functions of form g(t) = ab^t + c are also important. Moreover, the special base e allows us to represent all of these functions through horizontal scaling by writing
-
- g(t) = ae^{kt} + c
-
- where k is the constant such that e^k = b.
- Functions of the form of Equation are either always increasing or always decreasing, always have the same concavity, are defined on the set of all real numbers, and have their range as the set of all real numbers greater than c or all real numbers less than c. In whatever setting we are using a model of this form, the crucial task is to identify the values of a, k, and c; that endeavor is the focus of this section.
-
-
-
- We have also begun to see the important role that logarithms play in work with exponential models. The natural logarithm is the inverse of the natural exponential function and satisfies the important rule that \ln(b^k) = k\ln (b). This rule enables us to solve equations with the structure a^k = b for k in the context where a and b are known but k is not. Indeed, we can first take the natural log of both sides of the equation to get
-
- \ln(a^k) = \ln(b)
- ,
- from which it follows that k \ln(a) = \ln(b), and therefore
-
- k = \frac{\ln(b)}{\ln(a)}
- .
- Finding k is often central to determining an exponential model, and logarithms make finding the exact value of k possible.
-
-
-
- In Preview Activity, we revisit some key algebraic ideas with exponential and logarithmic equations in preparation for using these concepts in models for temperature and population.
-
-
-
-
-
-
-
- Newton's Law of Cooling revisited
-
- In Section, we learned that Newton's Law of Cooling, which states that an object's temperature changes at a rate proportional to the difference between its own temperature and the surrounding temperature, results in the object's temperature being modeled by functions of the form F(t) = ab^t + c. In light of our subsequent work in Section with the natural base e, as well as the fact that 0 \lt b \lt 1 in this model, we know that Newton's Law of Cooling implies that the object's temperature is modeled by a function of the form
-
- F(t) = ae^{-kt} + c
-
- for some constants a, c, and k, where k \gt 0.
-
-
-
- From Equation, we can determine several different characteristics of how the constants a, b, and k are connected to the behavior of F by thinking about what happens at t = 0, at one additional value of t, and as t increases without bound. In particular, note that e^{-kt} will tend to 0 as t increases without bound.
-
-
-
- Modeling temperature with Newton's Law of Cooling
-
- For the function F(t) = ae^{-kt} + c that models the temperature of a cooling or warming object, the constants a, c, and k play the following roles. Note that k \gt 0.
-
-
-
- Since e^{-kt} tends to 0 as t increases without bound, F(t) tends to c as t increases without bound, and thus c represents the temperature of the object's surroundings.
-
-
-
-
- Since e^0 = 1, F(0) = a + c, and thus the object's initial temperature is a + c. Said differently, a is the difference between the object's initial temperature and the temperature of the surroundings.
-
-
-
-
- Once we know the values of a and c, the value of k is determined by knowing the value of the temperature function F(t) at one nonzero value of t.
-
-
-
-
-
-
-
-
-
-
-
- A more realistic model for population growth
-
- If we assume that a population grows at a rate that is proportionate to the size of the population, it follows that the population grows exponentially according to the model
-
- P(t) = Ae^{kt}
-
- where A is the initial population and k is tied to the rate at which the population grows. Since k \gt 0, we know that e^{kt} is an always increasing, always concave up function that grows without bound. While P(t) = Ae^{kt} may be a reasonable model for how a population grows when it is relatively small, because the function grows without bound as time increases, it can't be a realistic long-term representation of what happens in reality. Indeed, whether it is the number of fish who can survive in a lake, the number of cells in a petri dish, or the number of human beings on earth, the size of the surroundings and the limitations of resources will keep the population from being able to grow without bound.
-
-
-
- In light of these observations, a different model is needed for population, one that grows exponentially at first, but that levels off later. Calculus can be used to develop such a model, and the resulting function is usually called the
- logistic function, logistic function which has form
-
- P(t) = \frac{A}{1 + Me^{-kt}}
- ,
- where A, M, and k are positive constants.
- Since k \gt 0, it follows that e^{-kt} \to 0 as t increases without bound, and thus the denominator of P approaches 1 as time goes on. Thus, we observe that P(t) tends to A as t increases without bound. We sometimes refer to A as the carrying capacity of the population.
-
-
-
-
-
-
-
-
-
- Summary
-
-
-
- When a function of form F(t) = c + ae^{-kt} models the temperature of an object that is cooling or warming in its surroundings, the temperature of the surroundings is c because e^{-kt} \to 0 as time goes on, the object's initial temperature is a+c, and the constant k is connected to how rapidly the object's temperature changes. Once a and c are known, the constant k can be determined by knowing the temperature at one additional time, t.
-
-
-
-
- Because the exponential function P(t) = Ae^{kt} grows without bound as t increases, such a function is not a realistic model of a population that we expect to level off as time goes on. The logistic function
-
- P(t) = \frac{A}{1 + Me^{-kt}}
-
- more appropriately models a population that grows roughly exponentially when P is small but whose size levels off as it approaches the carrying capacity of the surrounding environment, which is the value of the constant A.
-
-
-
-
-
-
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ Modeling temperature and population
+
+
+
+
+
+ What roles do the parameters a, k, and c play in how the function F(t) = c + ae^{-kt} models the temperature of an object that is cooling or warming in its surroundings?
+
+
+
+
+ How can we use an exponential function to more realistically model a population whose growth levels off?
+
+
+
+
+
+ Introduction
+
+
+ We've seen that exponential functions can be used to model several different important phenomena, such as the growth of money due to continuously compounded interest, the decay of radioactive quanitities, and the temperature of an object that is cooling or warming due to its surroundings. From initial work with functions of the form f(t) = ab^t where b \gt 0 and b \ne 1, we found that shifted exponential functions of form g(t) = ab^t + c are also important. Moreover, the special base e allows us to represent all of these functions through horizontal scaling by writing
+
+ g(t) = ae^{kt} + c
+
+ where k is the constant such that e^k = b.
+ Functions of the form of Equation are either always increasing or always decreasing, always have the same concavity, are defined on the set of all real numbers, and have their range as the set of all real numbers greater than c or all real numbers less than c. In whatever setting we are using a model of this form, the crucial task is to identify the values of a, k, and c; that endeavor is the focus of this section.
+
+
+
+ We have also begun to see the important role that logarithms play in work with exponential models. The natural logarithm is the inverse of the natural exponential function and satisfies the important rule that \ln(b^k) = k\ln (b). This rule enables us to solve equations with the structure a^k = b for k in the context where a and b are known but k is not. Indeed, we can first take the natural log of both sides of the equation to get
+
+ \ln(a^k) = \ln(b)
+ ,
+ from which it follows that k \ln(a) = \ln(b), and therefore
+
+ k = \frac{\ln(b)}{\ln(a)}
+ .
+ Finding k is often central to determining an exponential model, and logarithms make finding the exact value of k possible.
+
+
+
+ In Preview Activity, we revisit some key algebraic ideas with exponential and logarithmic equations in preparation for using these concepts in models for temperature and population.
+
+
+
+
+
+
+
+
+ Newton's Law of Cooling revisited
+
+ In Section, we learned that Newton's Law of Cooling, which states that an object's temperature changes at a rate proportional to the difference between its own temperature and the surrounding temperature, results in the object's temperature being modeled by functions of the form F(t) = ab^t + c. In light of our subsequent work in Section with the natural base e, as well as the fact that 0 \lt b \lt 1 in this model, we know that Newton's Law of Cooling implies that the object's temperature is modeled by a function of the form
+
+ F(t) = ae^{-kt} + c
+
+ for some constants a, c, and k, where k \gt 0.
+
+
+
+ From Equation, we can determine several different characteristics of how the constants a, b, and k are connected to the behavior of F by thinking about what happens at t = 0, at one additional value of t, and as t increases without bound. In particular, note that e^{-kt} will tend to 0 as t increases without bound.
+
+
+
+ Modeling temperature with Newton's Law of Cooling
+
+ For the function F(t) = ae^{-kt} + c that models the temperature of a cooling or warming object, the constants a, c, and k play the following roles. Note that k \gt 0.
+
+
+
+ Since e^{-kt} tends to 0 as t increases without bound, F(t) tends to c as t increases without bound, and thus c represents the temperature of the object's surroundings.
+
+
+
+
+ Since e^0 = 1, F(0) = a + c, and thus the object's initial temperature is a + c. Said differently, a is the difference between the object's initial temperature and the temperature of the surroundings.
+
+
+
+
+ Once we know the values of a and c, the value of k is determined by knowing the value of the temperature function F(t) at one nonzero value of t.
+
+
+
+
+
+
+
+
+
+
+
+ A more realistic model for population growth
+
+ If we assume that a population grows at a rate that is proportionate to the size of the population, it follows that the population grows exponentially according to the model
+
+ P(t) = Ae^{kt}
+
+ where A is the initial population and k is tied to the rate at which the population grows. Since k \gt 0, we know that e^{kt} is an always increasing, always concave up function that grows without bound. While P(t) = Ae^{kt} may be a reasonable model for how a population grows when it is relatively small, because the function grows without bound as time increases, it can't be a realistic long-term representation of what happens in reality. Indeed, whether it is the number of fish who can survive in a lake, the number of cells in a petri dish, or the number of human beings on earth, the size of the surroundings and the limitations of resources will keep the population from being able to grow without bound.
+
+
+
+ In light of these observations, a different model is needed for population, one that grows exponentially at first, but that levels off later. Calculus can be used to develop such a model, and the resulting function is usually called the
+ logistic function, logistic function which has form
+
+ P(t) = \frac{A}{1 + Me^{-kt}}
+ ,
+ where A, M, and k are positive constants.
+ Since k \gt 0, it follows that e^{-kt} \to 0 as t increases without bound, and thus the denominator of P approaches 1 as time goes on. Thus, we observe that P(t) tends to A as t increases without bound. We sometimes refer to A as the carrying capacity of the population.
+
+
+
+
+
+
+
+
+
+ Summary
+
+
+
+ When a function of form F(t) = c + ae^{-kt} models the temperature of an object that is cooling or warming in its surroundings, the temperature of the surroundings is c because e^{-kt} \to 0 as time goes on, the object's initial temperature is a+c, and the constant k is connected to how rapidly the object's temperature changes. Once a and c are known, the constant k can be determined by knowing the temperature at one additional time, t.
+
+
+
+
+ Because the exponential function P(t) = Ae^{kt} grows without bound as t increases, such a function is not a realistic model of a population that we expect to level off as time goes on. The logistic function
+
+ P(t) = \frac{A}{1 + Me^{-kt}}
+
+ more appropriately models a population that grows roughly exponentially when P is small but whose size levels off as it approaches the carrying capacity of the surrounding environment, which is the value of the constant A.
+
- How can we use limit notation to succinctly express a function's behavior as the input increases without bound or as the function's value increases without bound?
-
-
-
-
- What are some important limits and trends involving \infty that we can observe for familiar functions such as e^x, \ln(x), x^2, and \frac{1}{x}?
-
-
-
-
- What is a power function and how does the value of the power determine the function's overall behavior?
-
-
-
-
-
- Introduction
-
- In Section, we compared the behavior of the exponential functions p(t) = 2^t and q(t) = (\frac{1}{2})^t, and observed in Figure that as t increases without bound, p(t) also increases without bound, while q(t) approaches 0 (while having its value be always positive). We also introduced shorthand notation for describing these phenomena, writing
-
- p(t) \to \infty \text{ as } t \to \infty
-
- and
-
- q(t) \to 0 \text{ as } t \to \infty
- .
- It's important to remember that infinity is not itself a number. We use the \infty symbol to represent a quantity that gets larger and larger with no bound on its growth.
-
-
-
- We also know that the concept of infinity plays a key role in understanding the graphical behavior of functions. For instance, we've seen that for a function such as F(t) = 72 - 45e^{-0.05t}, F(t) \to 72 as t \to \infty, since e^{-0.05t} \to 0 as t increases without bound. The function F can be viewed as modeling the temperature of an object that is initially F(0) = 72-45 = 27 degrees that eventually warms to 72 degrees. The line y = 72 is thus a horizontal asymptote of the function F.
-
-
-
- In Preview, we review some familiar functions and portions of their behavior that involve \infty.
-
-
-
-
-
-
-
- Limit notation
-
- When observing a pattern in the values of a function that correspond to letting the inputs get closer and closer to a fixed value or letting the inputs increase or decrease without bound, we are often interested in the behavior of the function in the limit. In either case, we are considering an infinite collection of inputs that are themselves following a pattern, and we ask the question how can we expect the function's output to behave if we continue?
-
-
-
- For instance, we have regularly observed that as t \to \infty, e^{-t} \to 0, which means that by allowing t to get bigger and bigger without bound, we can make e^{-t} get as close to 0 as we'd like (without e^{-t} ever equalling 0, since e^{-t} is always positive).
-
-
-
-
-
Plots of y = e^t and y = e^{-t}.
-
Plots of y = e^t and y = e^{-t}.
-
-
-
-
Plots of y = e^t and y = \ln(t).
-
Plots of y = e^t and y = \ln(t).
-
-
-
-
-
- Similarly, as seen in Figure and Figure, we can make such observations as e^t \to \infty as t \to \infty, \ln(t) \to \infty as t \to \infty, and \ln(t) \to -\infty as t \to 0^+. We introduce formal limit notation in order to be able to express these patterns even more succinctly.
-
- Let L be a real number and f be a function. If we can make the value of f(t) as close to L as we want by letting t increase without bound, we write
-
- \lim_{t \to \infty} f(t) = L
-
- and say that the limit of f as t increases without bound is L.
-
-
-
- If the value of f(t) increases without bound as t increases without bound, we instead write
-
- \lim_{t \to \infty} f(t) = \infty
- .
-
-
-
- Finally, if f doesn't increase without bound, doesn't decrease without bound, and doesn't approach a single value L as t \to \infty, we say that f does not have a limit as t \to \infty.
-
-
-
-
-
- We use limit notation in related, natural ways to express patterns we see in function behavior. For instance, we write t \to -\infty when we let t decrease without bound, and f(t) \to -\infty if f decreases without bound. We can also think about an input value t approaching a value a at which the function f is not defined. As one example, we write
-
- \lim_{t \to 0^+} \ln(t) = -\infty
-
- because the natural logarithm function decreases without bound as input values get closer and closer to 0 (while always being positive), as seen in Figure.
-
-
-
- In the situation where \lim_{t \to \infty} f(t) = L, this tells us that f has a horizontal asymptote horizontal asymptote at y = L since the function's value approaches this fixed number as t increases without bound. Similarly, if we can say that \lim_{t \to a} f(t) = \infty, this shows that f has a vertical asymptote vertical asymptote at x = a since the function's value increases without bound as inputs approach the fixed number a.
-
-
-
- For now, we are going to focus on the long-range behavior of certain basic, familiar functions and work to understand how they behave as the input increases or decreases without bound. Above we've used the input variable t in most of our previous work; going forward, we'll regularly use x as well.
-
-
-
-
-
-
-
- Power functions
-
- To date, we have worked with several families of functions:
- linear functions of form y = mx + b,
- quadratic functions in standard form, y = ax^2 + bx + c,
- the sinusoidal (trigonometric) functions y = a\sin(k(x-b))+c or y = a\cos(k(x-b))+c,
- transformed exponential functions such as y = ae^{kx} + c,
- and transformed logarithmic functions of form y = a\ln(x) + c. For trigonometric, exponential, and logarithmic functions, it was essential that we first understood the behavior of the basic parent functions \sin(x), \cos(x), e^x, and \ln(x). In order to build on our prior work with linear and quadratic functions, we now consider basic functions such as x, x^2, and additional powers of x.
-
-
-
- power function
-
-
A function of the form f(x) = x^p where p is any real number is called a power function.
-
-
-
-
- We first focus on the case where p is a natural number (that is, a positive whole number).
-
-
-
-
-
- In the situation where the power p is a negative integer (i.e., a negative whole number), power functions behave very differently. This is because of the property of exponents that states
-
- x^{-n} = \frac{1}{x^n}
-
- so for a power function such as p(x) = x^{-2}, we can equivalently consider p(x) = \frac{1}{x^2}. Note well that for these functions, their domain is the set of all real numbers except x = 0. Like with power functions with positive whole number powers, we want to know how power functions with negative whole number powers behave as x increases without bound, as well as how the functions behave near x = 0.
-
-
-
-
-
-
-
- Summary
-
-
-
- The notation
-
- \lim_{x \to \infty} f(x) = L
-
- means that we can make the value of f(x) as close to L as we'd like by letting x be sufficiently large. This indicates that the value of f eventually stops changing much and tends to a single value, and thus y = L is a horizontal asymptote of the function f.
-
-
-
- Similarly, the notation
-
- \lim_{x \to a} f(x) = \infty
-
- means that we can make the value of f(x) as large as we'd like by letting x be sufficiently close, but not equal, to a. This unbounded behavior of f near a finite value a indicates that f has a vertical asymptote at x = a.
-
-
-
-
- We summarize some key behavior of familiar basic functions with limits as x increases without bound in Table.
-
-
-
- Some familiar functions and their limits as <m>x \to \infty</m> or <m>x \to -\infty</m>.
-
-
- f(x)
- \lim_{x \to \infty} f(x)
- \lim_{x \to -\infty} f(x)
-
-
- e^x
- \infty
- 0
-
-
- e^{-x}
- 0
- \infty
-
-
- \ln(x)
- \infty
- NABecause the domain of the natural logarithm function is only positive real numbers, it doesn't make sense to even consider this limit.
-
-
- x
- \infty
- -\infty
-
-
- x^2
- \infty
- \infty
-
-
- x^3
- \infty
- -\infty
-
-
- x^4
- \infty
- \infty
-
-
- \frac{1}{x}
- 0
- 0
-
-
- \frac{1}{x^2}
- 0
- 0
-
-
- \sin(x)
- no limitBecause the sine function neither increases without bound nor approaches a single value, but rather keeps oscillating through every value between -1 and 1 repeatedly, the sine function does not have a limit as x \to \infty.
- no limit
-
-
-
-
-
- Additionally, Table summarizes some key familiar function behavior where the function's output increases or decreases without bound as x approaches a fixed number not in the function's domain.
-
- A power function is a function of the form f(x) = x^p where p is any real number. For the two cases where p is a positive whole number or a negative whole number, it is straightforward to summarize key trends in power functions' behavior.
-
-
-
-
-
-
- If p = 1, 2, 3, \ldots, then the domain of f(x) = x^p is the set of all real numbers, and as x \to \infty, f(x) \to \infty. For the limit as x \to -\infty, it matters whether p is even or odd: if p is even, f(x) \to \infty as x \to -\infty; if p is odd, f(x) \to -\infty as x \to -\infty. Informally, all power functions of form f(x) = x^p where p is a positive even number are U-shaped, while all power functions of form f(x) = x^p where p is a positive odd number are chair-shaped.
-
-
-
-
- If p = -1, -2, -3, \ldots, then the domain of f(x) = x^p is the set of all real numbers exceptx=0, and as x \to \pm \infty, f(x) \to 0. This means that each such power function with a negative whole number exponent has a horizontal asymptote of y = 0. Regardless of the value of p (p = -1, -2, -3, \ldots), \lim_{x \to 0^+} f(x) = \infty. But when we approach 0 from the negative side, it matters whether p is even or odd: if p is even, f(x) \to \infty as x \to 0^-; if p is odd, f(x) \to -\infty as x \to 0^-. Informally, all power functions of form f(x) = x^p where p is a negative odd number look similar to \frac{1}{x}, while all power functions of form f(x) = x^p where p is a negative even number look similar to \frac{1}{x^2}.
-
+ How can we use limit notation to succinctly express a function's behavior as the input increases without bound or as the function's value increases without bound?
+
+
+
+
+ What are some important limits and trends involving \infty that we can observe for familiar functions such as e^x, \ln(x), x^2, and \frac{1}{x}?
+
+
+
+
+ What is a power function and how does the value of the power determine the function's overall behavior?
+
+
+
+
+
+ Introduction
+
+ In Section, we compared the behavior of the exponential functions p(t) = 2^t and q(t) = (\frac{1}{2})^t, and observed in Figure that as t increases without bound, p(t) also increases without bound, while q(t) approaches 0 (while having its value be always positive). We also introduced shorthand notation for describing these phenomena, writing
+
+ p(t) \to \infty \text{ as } t \to \infty
+
+ and
+
+ q(t) \to 0 \text{ as } t \to \infty
+ .
+ It's important to remember that infinity is not itself a number. We use the \infty symbol to represent a quantity that gets larger and larger with no bound on its growth.
+
+
+
+ We also know that the concept of infinity plays a key role in understanding the graphical behavior of functions. For instance, we've seen that for a function such as F(t) = 72 - 45e^{-0.05t}, F(t) \to 72 as t \to \infty, since e^{-0.05t} \to 0 as t increases without bound. The function F can be viewed as modeling the temperature of an object that is initially F(0) = 72-45 = 27 degrees that eventually warms to 72 degrees. The line y = 72 is thus a horizontal asymptote of the function F.
+
+
+
+ In Preview, we review some familiar functions and portions of their behavior that involve \infty.
+
+
+
+
+
+
+
+
+ Limit notation
+
+ When observing a pattern in the values of a function that correspond to letting the inputs get closer and closer to a fixed value or letting the inputs increase or decrease without bound, we are often interested in the behavior of the function in the limit. In either case, we are considering an infinite collection of inputs that are themselves following a pattern, and we ask the question how can we expect the function's output to behave if we continue?
+
+
+
+ For instance, we have regularly observed that as t \to \infty, e^{-t} \to 0, which means that by allowing t to get bigger and bigger without bound, we can make e^{-t} get as close to 0 as we'd like (without e^{-t} ever equalling 0, since e^{-t} is always positive).
+
+
+
+
+
Plots of y = e^t and y = e^{-t}.
+
Plots of y = e^t and y = e^{-t}.
+
+
+
+
Plots of y = e^t and y = \ln(t).
+
Plots of y = e^t and y = \ln(t).
+
+
+
+
+
+ Similarly, as seen in Figure and Figure, we can make such observations as e^t \to \infty as t \to \infty, \ln(t) \to \infty as t \to \infty, and \ln(t) \to -\infty as t \to 0^+. We introduce formal limit notation in order to be able to express these patterns even more succinctly.
+
+ Let L be a real number and f be a function. If we can make the value of f(t) as close to L as we want by letting t increase without bound, we write
+
+ \lim_{t \to \infty} f(t) = L
+
+ and say that the limit of f as t increases without bound is L.
+
+
+
+ If the value of f(t) increases without bound as t increases without bound, we instead write
+
+ \lim_{t \to \infty} f(t) = \infty
+ .
+
+
+
+ Finally, if f doesn't increase without bound, doesn't decrease without bound, and doesn't approach a single value L as t \to \infty, we say that f does not have a limit as t \to \infty.
+
+
+
+
+
+ We use limit notation in related, natural ways to express patterns we see in function behavior. For instance, we write t \to -\infty when we let t decrease without bound, and f(t) \to -\infty if f decreases without bound. We can also think about an input value t approaching a value a at which the function f is not defined. As one example, we write
+
+ \lim_{t \to 0^+} \ln(t) = -\infty
+
+ because the natural logarithm function decreases without bound as input values get closer and closer to 0 (while always being positive), as seen in Figure.
+
+
+
+ In the situation where \lim_{t \to \infty} f(t) = L, this tells us that f has a horizontal asymptote horizontal asymptote at y = L since the function's value approaches this fixed number as t increases without bound. Similarly, if we can say that \lim_{t \to a} f(t) = \infty, this shows that f has a vertical asymptote vertical asymptote at x = a since the function's value increases without bound as inputs approach the fixed number a.
+
+
+
+ For now, we are going to focus on the long-range behavior of certain basic, familiar functions and work to understand how they behave as the input increases or decreases without bound. Above we've used the input variable t in most of our previous work; going forward, we'll regularly use x as well.
+
+
+
+
+
+
+
+ Power functions
+
+ To date, we have worked with several families of functions:
+ linear functions of form y = mx + b,
+ quadratic functions in standard form, y = ax^2 + bx + c,
+ the sinusoidal (trigonometric) functions y = a\sin(k(x-b))+c or y = a\cos(k(x-b))+c,
+ transformed exponential functions such as y = ae^{kx} + c,
+ and transformed logarithmic functions of form y = a\ln(x) + c. For trigonometric, exponential, and logarithmic functions, it was essential that we first understood the behavior of the basic parent functions \sin(x), \cos(x), e^x, and \ln(x). In order to build on our prior work with linear and quadratic functions, we now consider basic functions such as x, x^2, and additional powers of x.
+
+
+
+ power function
+
+
A function of the form f(x) = x^p where p is any real number is called a power function.
+
+
+
+
+ We first focus on the case where p is a natural number (that is, a positive whole number).
+
+
+
+
+
+ In the situation where the power p is a negative integer (i.e., a negative whole number), power functions behave very differently. This is because of the property of exponents that states
+
+ x^{-n} = \frac{1}{x^n}
+
+ so for a power function such as p(x) = x^{-2}, we can equivalently consider p(x) = \frac{1}{x^2}. Note well that for these functions, their domain is the set of all real numbers except x = 0. Like with power functions with positive whole number powers, we want to know how power functions with negative whole number powers behave as x increases without bound, as well as how the functions behave near x = 0.
+
+
+
+
+
+
+
+ Summary
+
+
+
+ The notation
+
+ \lim_{x \to \infty} f(x) = L
+
+ means that we can make the value of f(x) as close to L as we'd like by letting x be sufficiently large. This indicates that the value of f eventually stops changing much and tends to a single value, and thus y = L is a horizontal asymptote of the function f.
+
+
+
+ Similarly, the notation
+
+ \lim_{x \to a} f(x) = \infty
+
+ means that we can make the value of f(x) as large as we'd like by letting x be sufficiently close, but not equal, to a. This unbounded behavior of f near a finite value a indicates that f has a vertical asymptote at x = a.
+
+
+
+
+ We summarize some key behavior of familiar basic functions with limits as x increases without bound in Table.
+
+
+
+ Some familiar functions and their limits as <m>x \to \infty</m> or <m>x \to -\infty</m>.
+
+
+ f(x)
+ \lim_{x \to \infty} f(x)
+ \lim_{x \to -\infty} f(x)
+
+
+ e^x
+ \infty
+ 0
+
+
+ e^{-x}
+ 0
+ \infty
+
+
+ \ln(x)
+ \infty
+ NABecause the domain of the natural logarithm function is only positive real numbers, it doesn't make sense to even consider this limit.
+
+
+ x
+ \infty
+ -\infty
+
+
+ x^2
+ \infty
+ \infty
+
+
+ x^3
+ \infty
+ -\infty
+
+
+ x^4
+ \infty
+ \infty
+
+
+ \frac{1}{x}
+ 0
+ 0
+
+
+ \frac{1}{x^2}
+ 0
+ 0
+
+
+ \sin(x)
+ no limitBecause the sine function neither increases without bound nor approaches a single value, but rather keeps oscillating through every value between -1 and 1 repeatedly, the sine function does not have a limit as x \to \infty.
+ no limit
+
+
+
+
+
+ Additionally, Table summarizes some key familiar function behavior where the function's output increases or decreases without bound as x approaches a fixed number not in the function's domain.
+
+ A power function is a function of the form f(x) = x^p where p is any real number. For the two cases where p is a positive whole number or a negative whole number, it is straightforward to summarize key trends in power functions' behavior.
+
+
+
+
+
+
+ If p = 1, 2, 3, \ldots, then the domain of f(x) = x^p is the set of all real numbers, and as x \to \infty, f(x) \to \infty. For the limit as x \to -\infty, it matters whether p is even or odd: if p is even, f(x) \to \infty as x \to -\infty; if p is odd, f(x) \to -\infty as x \to -\infty. Informally, all power functions of form f(x) = x^p where p is a positive even number are U-shaped, while all power functions of form f(x) = x^p where p is a positive odd number are chair-shaped.
+
+
+
+
+ If p = -1, -2, -3, \ldots, then the domain of f(x) = x^p is the set of all real numbers exceptx=0, and as x \to \pm \infty, f(x) \to 0. This means that each such power function with a negative whole number exponent has a horizontal asymptote of y = 0. Regardless of the value of p (p = -1, -2, -3, \ldots), \lim_{x \to 0^+} f(x) = \infty. But when we approach 0 from the negative side, it matters whether p is even or odd: if p is even, f(x) \to \infty as x \to 0^-; if p is odd, f(x) \to -\infty as x \to 0^-. Informally, all power functions of form f(x) = x^p where p is a negative odd number look similar to \frac{1}{x}, while all power functions of form f(x) = x^p where p is a negative even number look similar to \frac{1}{x^2}.
+
- Why do polynomials arise naturally in the study of problems involving the volume and surface area of three-dimensional containers such as boxes and cylinders?
-
-
-
-
- How can polynomial functions be used to approximate non-polynomial curves and functions?
-
-
-
-
-
- Introduction
-
- Polynomial functions are the simplest of all functions in mathematics in part because they only involve multiplication and addition. In any applied setting where we can formulate key ideas using only those arithmetic operations, it's natural that polynomial functions model the corresponding phenomena. For example, in Activity, we saw that for a spherical tank of radius 4 m filling with water, the volume of water in the tank at a given instant, V, is a function of the depth, h, of the water in the tank at the same moment according to the formula
-
- V = f(h) = \frac{\pi}{3} h^2(12-h)
- .
- The function f is a polynomial of degree 3 with a repeated zero at h = 0 and an additional zero at h = 12. Because the tank has a radius of 4, its total height is 8, and thus the model V = f(h) = \frac{\pi}{3} h^2(12-h) is only valid on the domain 0 \le h \le 8. This polynomial function tells us how the volume of water in the tank changes as h changes.
-
-
-
- In other similar situations where we consider the volume of a box, tank, or other three-dimensional container, polynomial functions frequently arise. To develop a model function that represents a physical situation, we almost always begin by drawing one or more diagrams of the situation and then introduce one or more variables to represent quantities that are changing. From there, we explore relationships that are present and work to express one of the quantities in terms of the other(s).
-
- In Preview Activity, we worked with a rectangular box being built by folding cardboard. One of the key principles we needed to use was the fact that the volume of a rectangular box of length l, width w, and height h is
-
- V = lwh
- . rectangular boxvolumevolumebox
-
-
-
-
-
A rectangular box.
-
A rectangular box.
-
-
-
-
A circular cylinder.
-
A circular cylinder.
-
-
-
-
-
- One way to remember the formula for the volume of a rectangular box is area of the base times the height. This principle extends to other three-dimensional shapes that have constant cross-sectional area. For instance, the volume of a circular cylinder with radius r and height h is
-
- V = \pi r^2 h
- circular cylindervolumevolumecylinder
- since the area of the base is \pi r^2.
-
-
-
- We'll also often consider the surface area of a three-dimensional container. For a rectangular box with side lengths of l, w, and h, its surface area consists of 3 pairs of rectangles: the top and bottom, each of area lw, the two sides that are the front and back when we look right at the box, each of area lh, and the remaining two sides of area wh. Thus the total surface area of the box is
- rectangular boxsurface areasurface areabox
- SA = 2lw + 2lh + 2wh
- .
- For a circular cylinder, its surface area is the sum of the areas of the top and bottom (\pi r^2 each), plus the area of the sides. If we think of cutting the cylinder vertically and unfurling it, the resulting figure is a rectangle whose dimensions are the height of the cylinder, h, by the circumference of the base, 2\pi r. The rectangle's area is therefore 2\pi r \cdot h, and hence the total surface area of a cylinder is
-
- SA = 2\pi r^2 + 2\pi r h
- . circular cylindersurface areasurface areacylinder
-
-
-
- Each of the volume and surface area equations (Equation, Equation, Equation, and Equation) involve only multiplication and addition, and thus have the potential to result in polynomial functions. At present, however, each of these equations involves at least two variables. The inclusion of additional constraints can enable us to use these formulas to generate polynomial functions of a single variable.
-
-
-
-
-
-
-
-
-
- Other applications of polynomial functions
-
-
- A different use of polynomial functions arises with Bezier curves. Bezier curves The most common type of Bezier curve used in applications is the cubic Bezier curve, which is a curve given parametrically by a formula of the form (x(t), y(t)), where
-
- x(t) = (1-t)^3 x_0 + 3(1-t)^2 t x_1 + 3(1-t) t^2 x_2 + t^3 x_3
-
- and
-
- y(t) = (1-t)^3 y_0 + 3(1-t)^2 t y_1 + 3(1-t) t^2 y_2 + t^3 y_3
- .
- The curve passes through the points A = (x_0,y_0) and B = (x_3, y_3) and the points C = (x_1, y_1) and D = (x_2, y_2) are called control points. At http://gvsu.edu/s/0zC, you can explore the effects of moving the control points (in gray) and the points on the curve (in black) to generate different curves in the plane, similar to the one shown in Figure.
-
-
-
-
-
A cubic Bezier curve with control points in gray.
-
A cubic Bezier curve with control points in gray.
-
-
-
-
The letter S in Palatino font, generated by Bezier curves.
-
The letter S in Palatino font, generated by Bezier curves.
-
-
-
-
-
- The main issue to realize is that the form of the curve depends on a special family of cubic polynomials:
-
- (1-t)^3, 3(1-t)^2 t, 3(1-t) t^2, \ \text{ and } \ t^3
- .
- These four cubic polynomials play a key role in graphic design and are used in all sorts of important ways, including in font design, as seen in Figure.
-
-
-
- Another important application of polynomial functions is found in how they can be used to approximate the sine and cosine functions.
-
-
-
-
-
-
-
- Summary
-
-
-
- Polynomials arise naturally in the study of problems involving the volume and surface area of three-dimensional containers such as boxes and cylinders because these formulas fundamentally involve sums and products of variables. For instance, the volume of a cylinder is V = \pi r^2 h. In the presence of a surface area constraint that tells us that h = \frac{100-2\pi r^2}{2\pi r}, it follows that
-
- V = \pi r^2 \frac{100-2\pi r^2}{2\pi r} = r(50-\pi r^2)
- ,
- which is a cubic polynomial.
-
-
-
-
- Polynomial functions can be used to approximate non-polynomial curves and functions in many different ways. One example is found in cubic Bezier curves which use a collection of control points to enable the user to manipulate curves to pass through select points in such a way that the curve first travels in a certain direction. Another example is in the remarkable approximation of non-polynomial functions like the sine function, as given by
-
- \sin(x) \approx x - \frac{1}{3!}x^3 + \frac{1}{5!}x^5 - \frac{1}{7!}x^7
- ,
- where the approximation is good for x-values near x = 0.
-
+ Why do polynomials arise naturally in the study of problems involving the volume and surface area of three-dimensional containers such as boxes and cylinders?
+
+
+
+
+ How can polynomial functions be used to approximate non-polynomial curves and functions?
+
+
+
+
+
+ Introduction
+
+ Polynomial functions are the simplest of all functions in mathematics in part because they only involve multiplication and addition. In any applied setting where we can formulate key ideas using only those arithmetic operations, it's natural that polynomial functions model the corresponding phenomena. For example, in Activity, we saw that for a spherical tank of radius 4 m filling with water, the volume of water in the tank at a given instant, V, is a function of the depth, h, of the water in the tank at the same moment according to the formula
+
+ V = f(h) = \frac{\pi}{3} h^2(12-h)
+ .
+ The function f is a polynomial of degree 3 with a repeated zero at h = 0 and an additional zero at h = 12. Because the tank has a radius of 4, its total height is 8, and thus the model V = f(h) = \frac{\pi}{3} h^2(12-h) is only valid on the domain 0 \le h \le 8. This polynomial function tells us how the volume of water in the tank changes as h changes.
+
+
+
+ In other similar situations where we consider the volume of a box, tank, or other three-dimensional container, polynomial functions frequently arise. To develop a model function that represents a physical situation, we almost always begin by drawing one or more diagrams of the situation and then introduce one or more variables to represent quantities that are changing. From there, we explore relationships that are present and work to express one of the quantities in terms of the other(s).
+
+ In Preview Activity, we worked with a rectangular box being built by folding cardboard. One of the key principles we needed to use was the fact that the volume of a rectangular box of length l, width w, and height h is
+
+ V = lwh
+ . rectangular boxvolumevolumebox
+
+
+
+
+
A rectangular box.
+
A rectangular box.
+
+
+
+
A circular cylinder.
+
A circular cylinder.
+
+
+
+
+
+ One way to remember the formula for the volume of a rectangular box is area of the base times the height. This principle extends to other three-dimensional shapes that have constant cross-sectional area. For instance, the volume of a circular cylinder with radius r and height h is
+
+ V = \pi r^2 h
+ circular cylindervolumevolumecylinder
+ since the area of the base is \pi r^2.
+
+
+
+ We'll also often consider the surface area of a three-dimensional container. For a rectangular box with side lengths of l, w, and h, its surface area consists of 3 pairs of rectangles: the top and bottom, each of area lw, the two sides that are the front and back when we look right at the box, each of area lh, and the remaining two sides of area wh. Thus the total surface area of the box is
+ rectangular boxsurface areasurface areabox
+ SA = 2lw + 2lh + 2wh
+ .
+ For a circular cylinder, its surface area is the sum of the areas of the top and bottom (\pi r^2 each), plus the area of the sides. If we think of cutting the cylinder vertically and unfurling it, the resulting figure is a rectangle whose dimensions are the height of the cylinder, h, by the circumference of the base, 2\pi r. The rectangle's area is therefore 2\pi r \cdot h, and hence the total surface area of a cylinder is
+
+ SA = 2\pi r^2 + 2\pi r h
+ . circular cylindersurface areasurface areacylinder
+
+
+
+ Each of the volume and surface area equations (Equation, Equation, Equation, and Equation) involve only multiplication and addition, and thus have the potential to result in polynomial functions. At present, however, each of these equations involves at least two variables. The inclusion of additional constraints can enable us to use these formulas to generate polynomial functions of a single variable.
+
+
+
+
+
+
+
+
+
+ Other applications of polynomial functions
+
+
+ A different use of polynomial functions arises with Bezier curves. Bezier curves The most common type of Bezier curve used in applications is the cubic Bezier curve, which is a curve given parametrically by a formula of the form (x(t), y(t)), where
+
+ x(t) = (1-t)^3 x_0 + 3(1-t)^2 t x_1 + 3(1-t) t^2 x_2 + t^3 x_3
+
+ and
+
+ y(t) = (1-t)^3 y_0 + 3(1-t)^2 t y_1 + 3(1-t) t^2 y_2 + t^3 y_3
+ .
+ The curve passes through the points A = (x_0,y_0) and B = (x_3, y_3) and the points C = (x_1, y_1) and D = (x_2, y_2) are called control points. At http://gvsu.edu/s/0zC, you can explore the effects of moving the control points (in gray) and the points on the curve (in black) to generate different curves in the plane, similar to the one shown in Figure.
+
+
+
+
+
A cubic Bezier curve with control points in gray.
+
A cubic Bezier curve with control points in gray.
+
+
+
+
The letter S in Palatino font, generated by Bezier curves.
+
The letter S in Palatino font, generated by Bezier curves.
+
+
+
+
+
+ The main issue to realize is that the form of the curve depends on a special family of cubic polynomials:
+
+ (1-t)^3, 3(1-t)^2 t, 3(1-t) t^2, \ \text{ and } \ t^3
+ .
+ These four cubic polynomials play a key role in graphic design and are used in all sorts of important ways, including in font design, as seen in Figure.
+
+
+
+ Another important application of polynomial functions is found in how they can be used to approximate the sine and cosine functions.
+
+
+
+
+
+
+
+ Summary
+
+
+
+ Polynomials arise naturally in the study of problems involving the volume and surface area of three-dimensional containers such as boxes and cylinders because these formulas fundamentally involve sums and products of variables. For instance, the volume of a cylinder is V = \pi r^2 h. In the presence of a surface area constraint that tells us that h = \frac{100-2\pi r^2}{2\pi r}, it follows that
+
+ V = \pi r^2 \frac{100-2\pi r^2}{2\pi r} = r(50-\pi r^2)
+ ,
+ which is a cubic polynomial.
+
+
+
+
+ Polynomial functions can be used to approximate non-polynomial curves and functions in many different ways. One example is found in cubic Bezier curves which use a collection of control points to enable the user to manipulate curves to pass through select points in such a way that the curve first travels in a certain direction. Another example is in the remarkable approximation of non-polynomial functions like the sine function, as given by
+
+ \sin(x) \approx x - \frac{1}{3!}x^3 + \frac{1}{5!}x^5 - \frac{1}{7!}x^7
+ ,
+ where the approximation is good for x-values near x = 0.
+
- What properties of a polynomial function can we deduce from its algebraic structure?
-
-
-
-
- What is a sign chart and how does it help us understand a polynomial function's behavior?
-
-
-
-
- How do zeros of multiplicity other than 1 impact the graph of a polynomial function?
-
-
-
-
-
- Introduction
-
- We know that linear functions are the simplest of all functions we can consider: their graphs have the simplest shape, their average rate of change is always constant (regardless of the interval chosen), and their formula is elementary. Moreover, computing the value of a linear function only requires multiplication and addition.
-
-
-
- If we think of a linear function as having formula L(x) = b + mx, and the next-simplest functions, quadratic functions, as having form Q(x) = c + bx + ax^2, we can see immediate parallels between their respective forms and realize that it's natural to consider slightly more complicated functions by adding additional power functions.
-
-
-
- Indeed, if we instead view linear functions as having form
-
- L(x) = a_0 + a_1 x
-
- (for some constants a_0 and a_1) and quadratic functions as having form
-
- Q(x) = a_0 + a_1 x + a_2 x^2
-
- (for some constants a_0, a_1, and a_2),
- then it's natural to think about more general functions of this same form,
- but with additional power functions included.
-
-
-
- polynomial function
-
-
- Given real numbers a_0, a_1, \ldots, a_n where a_n \ne 0, we say that the function
-
- P(x) = a_0 + a_1 x + a_2 x^2 + \cdots + a_{n-1}x^{n-1} + a_n x^n
-
- is a polynomial of degree n. polynomialdegree In addition, we say that the values of a_i are the coefficients of the polynomial and the individual power functions a_i x^i are the terms of the polynomial. polynomialcoefficientspolynomialterms Any value of x for which P(x) = 0 is called a zero of the polynomial. polynomialzero
-
-
-
-
-
-
-
- The polyomial function P(x) = 3 - 7x + 4x^2 - 2x^3 + 9x^5 has degree 5, its constant term is 3, and its linear term is -7x.
-
-
-
-
-
- Since a polynomial is simply a sum of constant multiples of various power functions with positive integer powers, we often refer to those individual terms by referring to their individual degrees: the linear term, the quadratic term, and so on. In addition, since the domain of any power function of the form p(x) = x^n where n is a positive whole number is the set of all real numbers, it's also true the the domain of any polynomial function is the set of all real numbers.
-
- Our observations in Preview Activity generalize to polynomials of any degree. In particular, it is possible to prove the following general conclusions regarding the number of zeros, the long-range behavior, and the number of turning points for any polynomial of degree n.
-
-
-
- The number of real zeros of a polynomial
-
- For any degree n polynomial p(x) = a_0 + a_1 x + \cdots + a_{n-1}x^{n-1} + a_n x^n, has at most n real zeros.We can actually say even more: if we allow the zeros to be complex numbers, then every degree n polynomial has exactlyn zeros, provided we count zeros according to their multiplicity. For example, the polynomial p(x) = (x-1)^2 = x^2 - 2x + 1 has a zero of multiplicity two at x = 1.
-
-
-
-
- We know that each of the power functions x, x^2, \ldots, x^n grow without bound as x \to \infty. Intuitively, we sense that x^5 grows faster than x^4 (and likewise for any comparison of a higher power to a lower one). This means that for large values of x, the most important term in any polynomial is its highest order term, as we saw in Preview Activity when we compared p(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + a_4 x^4 and y = a_4 x^4.
-
-
-
- polynomiallong-range behavior
- The long-range behavior of a polynomial
-
- For any degree n polynomial p(x) = a_0 + a_1 x + \cdots + a_{n-1}x^{n-1} + a_n x^n, its long-range behavior is the same as its highest-order term q(x) = a_n x^n. Thus, any polynomial of even degree appears U-shaped (\cup or \cap, like x^2 or -x^2) when we zoom way out, and any polynomial of odd degree appears chair-shaped (like x^3 or -x^3) when we zoom way out.
-
-
-
-
- In Figure, we see how the degree 7 polynomial pictured there (and in Figure as well) appears to look like q(x) = -x^7 as we zoom out.
-
-
-
-
-
Plot of a degree 7 polynomial function p.
-
Plot of a degree 7 polynomial function p.
-
-
-
-
Plot of the same degree 7 polynomial function p, but zoomed out.
-
Plot of the same degree 7 polynomial function p, but zoomed out.
-
-
-
-
-
- Finally, a key idea from calculus justifies the fact that the maximum number of turning points of a degree n polynomial is n-1, as we conjectured in the degree 4 case in Preview Activity. Moreover, the only possible numbers of turning points must have the same parity as n-1; that is, if n-1 is even, then the number of turning points must be even, and if instead n-1 is odd, the number of turning points must also be odd. For instance, for the degree 7 polynomial in Figure, we know that it is chair-shaped, with one end up and one end down. There could be zero turning points and the function could always decrease. But if there is at least one, then there must be a second, since if there were only one the function would decrease and then increase without turning back, which would force the graph to appear U-shaped.
-
-
-
- polynomialnumber of turning points
- The turning points of a polynomial
-
- For any degree n polynomial p(x) = a_0 + a_1 x + \cdots + a_{n-1}x^{n-1} + a_n x^n, if n is even, its number of turning points is exactly one of n-1, n-3, \ldots, 1, and if n is odd, its number of turning points is exactly one of n-1, n-3, \ldots, 0.
-
-
-
-
-
-
-
-
- Using zeros and signs to understand polynomial behavior
-
-
- Just like a quadratic function can be written in different forms (standard: q(x) = ax^2 + bx + c, vertex: q(x) = a(x-h)^2 + k, and factored: q(x) = a(x-r_1)(x-r_2)), it's possible to write a polynomial function in different forms and to gain information about its behavior from those different forms. In particular, if we know all of the zeros of a polynomial function, we can write its formula in factored form, which gives us a deeper understanding of its graph.
-
-
-
- The Zero Product Property Zero Product Property states that if two or more numbers are multiplied together and the result is 0, then at least one of the numbers must be 0. We use the Zero Product Property regularly with polynomial functions. If we can determine all n zeros of a degree n polynomial, and we call those zeros r_1, r_2, \ldots, r_n, we can write
-
- p(x) = a_n(x-r_1)(x-r_2) \cdots (x-r_n)
- .
- Moreover, if we are given a polynomial in this factored form, we can quickly determine its zeros. For instance, if p(x) = 2(x+7)(x+1)(x-2)(x-5), we know that the only way p(x) = 0 is if at least one of the factors (x+7), (x+1), (x-2), or (x-5) equals 0, which implies that x = -7, x = -1, x = 2, or x = 5. Hence, from the factored form of a polynomial, it is straightforward to identify the polynomial's zeros, the x-values at which its graph crosses the x-axis. We can also use the factored form of a polynomial to develop what we call a sign chart, which we demonstrate in Example. sign chart
-
-
-
-
-
- Consider the polynomial function p(x) = k(x-1)(x-a)(x-b). Suppose we know that 1 \lt a \lt b and that k \lt 0. Fully describe the graph of p without the aid of a graphing utility.
-
-
-
-
- Since p(x) = k(x-1)(x-a)(x-b), we immediately know that p is a degree 3 polynomial with 3 real zeros: x = 1, a, b. We are given that 1 \lt a \lt b and in addition that k \lt 0. If we expand the factored form of p(x), it has form p(x) = kx^3 + \cdots, and since we know that when we zoom out, p(x) behaves like kx^3, we know that with k \lt 0 it follows \lim_{x \to -\infty} p(x) = +\infty and \lim_{x \to \infty} p(x) = -\infty.
-
-
-
- Since p is degree 3 and we know it has zeros at x = 1, a, b, we know there are no other locations where p(x) = 0. Thus, on any interval between two zeros (or to the left of the least or the right of the greatest), the polynomial cannot change sign. We now investigate, interval by interval, the sign of the function.
-
-
-
- When x \lt 1, it follows that x - 1 \lt 0. In addition, since 1 \lt a \lt b, when x \lt 1, x lies to the left of 1, a, and b, which also makes x-a and x-b negative. Moreover, we know that the constant k \lt 0. Hence, on the interval x \lt 1, all four terms in p(x) = k(x-1)(x-a)(x-b) are negative, which we indicate by writing ---- in that location on the sign chart pictured in Figure.
-
-
-
- In addition, since there are an even number of negative terms in the product, the overall product's sign is positive, which we indicate by the single + beneath ----, and by writing POS below the coordinate axis.
-
-
-
-
A sign chart for the polynomial function p(x) = k(x-1)(x-a)(x-b).
-
A sign chart for the polynomial function p(x) = k(x-1)(x-a)(x-b).
-
-
-
-
- We now proceed to the other intervals created by the zeros. On 1 \lt x \lt a, the term (x-1) has become positive, since x \gt 1. But both x-a and x-b are negative, as is the constant k, and thus we write -+-- for this interval, which has overall sign -, as noted in the figure. Similar reasoning completes the diagram.
-
-
-
- From all of the information we have deduced about p, we conclude that regardless of the locations of a and b, the graph of p must look like the curve shown in Figure.
-
-
-
-
The graph of the polynomial function p(x) = k(x-1)(x-a)(x-b).
-
The graph of the polynomial function p(x) = k(x-1)(x-a)(x-b).
- In Activity, we found that one of the zeros of the polynomial p(x) = 4692(x + 1520)(x^2 + 10000)(x - 3471)^2 (x - 9738) leads to different behavior of the function near that zero than we've seen in other situations. We now consider the more general situation where a polynomial has a repeated factor of the form (x-r)^n. When (x-r)^n is a factor of a polynomial p, we say that p has a zero of multiplicity n at x = r. polynomialzero of multiplicity n
-
-
-
- To see the impact of repeated factors, we examine a collection of degree 4 polynomials that each have 4 real zeros. We start with the simplest of all, the function f(x) = x^4, whose zeros are x = 0, 0, 0, 0. Because the factor x-0 is repeated 4 times, the zero x = 0 has multiplicity 4.
-
-
-
- Next we consider the degree 4 polynomial g(x) = x^3 (x-1), which has a zero of multiplicity 3 at x = 0 and a zero of multiplicity 1 at x = 1.
-
-
-
-
A plot of g(x) = x^3(x-1) with zero x = 0 of multiplicity 3 and x = 1 of multiplicity 1.
-
A plot of g(x) = x^3(x-1) with zero x = 0 of multiplicity 3 and x = 1 of multiplicity 1.
-
-
-
-
-
- Observe that in Figure, the up-close plot near the zero x = 0 of multiplicity 3, the polynomial function g looks similar to the basic cubic polynomial -x^3. In addition, in Figure, we observe that if we zoom in even futher on the zero of multiplicity 1, the function g looks roughly linear, like a degree 1 polynomial. This type of behavior near repeated zeros turns out to hold in other cases as well.
-
-
-
-
-
A plot of g(x) = x^3(x-1) zoomed in on the zero x = 0 of multiplicity 3.
-
A plot of g(x) = x^3(x-1) zoomed in on the zero x = 0 of multiplicity 3.
-
-
-
-
A plot of g(x) = x^3(x-1) zoomed in on the zero x = 1 of multiplicity 1.
-
A plot of g(x) = x^3(x-1) zoomed in on the zero x = 1 of multiplicity 1.
-
-
-
-
-
- If we next let h(x) = x^2 (x-1)^2, we see that h has two distinct real zeros, each of multiplicity 2. The graph of h in Figure shows that h behaves similar to a basic quadratic function near each of those zeros and thus shows U-shaped behavior nearby. If instead we let k(x) = x^2(x-1)(x+1), we see approximately linear behavior near x = -1 and x = 1 (the zeros of multiplicity 1), and quadratic (U-shaped) behavior near x = 0 (the zero of multiplicity 2), as seen in Figure.
-
-
-
-
-
Plot of h(x) = x^2 (x-1)^2 with zeros x = 0 and x = 1 of multiplicity 2.
-
Plot of h(x) = x^2 (x-1)^2 with zeros x = 0 and x = 1 of multiplicity 2.
-
-
-
-
Plot of k(x) = x^2(x-1)(x+1) with zeros x = 0 of multiplicity 2 and x = -1 and x = 1 of multiplicity 1.
-
Plot of k(x) = x^2(x-1)(x+1) with zeros x = 0 of multiplicity 2 and x = -1 and x = 1 of multiplicity 1.
-
-
-
-
-
- Finally, if we consider m(x) = (x+1)x(x-1)(x-2), which has 4 distinct real zeros each of multiplicity 1, we observe in Figure that zooming in on each zero individually, the function demonstrates approximately linear behavior as it passes through the x-axis.
-
-
-
-
Plot of m(x) = (x+1)x(x-1)(x-2) with 4 distinct zeros of multiplicity 1.
-
Plot of m(x) = (x+1)x(x-1)(x-2) with 4 distinct zeros of multiplicity 1.
-
-
-
-
- Our observations with polynomials of degree 4 in the various figures above generalize to polynomials of any degree.
-
- If (x-r)^n is a factor of a polynomial p, then x = r is a zero of p of multiplicity n, and near x = r the graph of p looks like either -x^n or x^n does near x = 0. That is, the shape of the graph near the zero is determined by the multiplicity of the zero.
-
-
-
-
-
-
-
-
- Summary
-
-
-
-
- From a polynomial function's algebraic structure, we can deduce several key traits of the function.
-
-
-
-
-
- If the function is in standard form, say p(x) = a_0 + a_1 x + a_2 x^2 + \cdots + a_{n-1}x^{n-1} + a_n x^n, we know that its degree is n and that when we zoom out, p looks like a_n x^n and thus has the same long-range behavior as a_n x^n. Thus, p is chair-shaped if n is odd and U-shaped if n is even. Whether \lim_{n \to \infty} p(x) is +\infty or -\infty depends on the sign of a_n.
-
-
-
-
-
- If the function is in factored form, say p(x) = a_n(x-r_1)(x-r_2) \cdots (x-r_n) (where the r_i's are possibly not distinct and possibly complex), we can quickly determine both the degree of the polynomial (n) and the locations of its zeros, as well as their multiplicities.
-
-
-
-
-
-
-
- A sign chart is a visual way to identify all of the locations where a function is zero along with the sign of the function on the various intervals the zeros create. A sign chart gives us an overall sense of the graph of the function, but without concerning ourselves with any specific values of the function besides the zeros. For a sample sign chart, see Figure.
-
-
-
-
- When a polynomial p has a repeated factor such as p(x) = (x-5)(x-5)(x-5) = (x-5)^3, we say that x = 5 is a zero of multiplicity 3. At the point x = 5 where p will cross the x-axis, up close it will look like a cubic polynomial and thus be chair-shaped. In general, if (x-r)^n is a factor of a polynomial p so that x = r is a zero of multiplicity n, the polynomial will behave near x = r like the polynomial x^n behaves near x = 0.
-
+ What properties of a polynomial function can we deduce from its algebraic structure?
+
+
+
+
+ What is a sign chart and how does it help us understand a polynomial function's behavior?
+
+
+
+
+ How do zeros of multiplicity other than 1 impact the graph of a polynomial function?
+
+
+
+
+
+ Introduction
+
+ We know that linear functions are the simplest of all functions we can consider: their graphs have the simplest shape, their average rate of change is always constant (regardless of the interval chosen), and their formula is elementary. Moreover, computing the value of a linear function only requires multiplication and addition.
+
+
+
+ If we think of a linear function as having formula L(x) = b + mx, and the next-simplest functions, quadratic functions, as having form Q(x) = c + bx + ax^2, we can see immediate parallels between their respective forms and realize that it's natural to consider slightly more complicated functions by adding additional power functions.
+
+
+
+ Indeed, if we instead view linear functions as having form
+
+ L(x) = a_0 + a_1 x
+
+ (for some constants a_0 and a_1) and quadratic functions as having form
+
+ Q(x) = a_0 + a_1 x + a_2 x^2
+
+ (for some constants a_0, a_1, and a_2),
+ then it's natural to think about more general functions of this same form,
+ but with additional power functions included.
+
+
+
+ polynomial function
+
+
+ Given real numbers a_0, a_1, \ldots, a_n where a_n \ne 0, we say that the function
+
+ P(x) = a_0 + a_1 x + a_2 x^2 + \cdots + a_{n-1}x^{n-1} + a_n x^n
+
+ is a polynomial of degree n. polynomialdegree In addition, we say that the values of a_i are the coefficients of the polynomial and the individual power functions a_i x^i are the terms of the polynomial. polynomialcoefficientspolynomialterms Any value of x for which P(x) = 0 is called a zero of the polynomial. polynomialzero
+
+
+
+
+
+
+
+ The polyomial function P(x) = 3 - 7x + 4x^2 - 2x^3 + 9x^5 has degree 5, its constant term is 3, and its linear term is -7x.
+
+
+
+
+
+ Since a polynomial is simply a sum of constant multiples of various power functions with positive integer powers, we often refer to those individual terms by referring to their individual degrees: the linear term, the quadratic term, and so on. In addition, since the domain of any power function of the form p(x) = x^n where n is a positive whole number is the set of all real numbers, it's also true the the domain of any polynomial function is the set of all real numbers.
+
+ Our observations in Preview Activity generalize to polynomials of any degree. In particular, it is possible to prove the following general conclusions regarding the number of zeros, the long-range behavior, and the number of turning points for any polynomial of degree n.
+
+
+
+ The number of real zeros of a polynomial
+
+ For any degree n polynomial p(x) = a_0 + a_1 x + \cdots + a_{n-1}x^{n-1} + a_n x^n, has at most n real zeros.We can actually say even more: if we allow the zeros to be complex numbers, then every degree n polynomial has exactlyn zeros, provided we count zeros according to their multiplicity. For example, the polynomial p(x) = (x-1)^2 = x^2 - 2x + 1 has a zero of multiplicity two at x = 1.
+
+
+
+
+ We know that each of the power functions x, x^2, \ldots, x^n grow without bound as x \to \infty. Intuitively, we sense that x^5 grows faster than x^4 (and likewise for any comparison of a higher power to a lower one). This means that for large values of x, the most important term in any polynomial is its highest order term, as we saw in Preview Activity when we compared p(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + a_4 x^4 and y = a_4 x^4.
+
+
+
+ polynomiallong-range behavior
+ The long-range behavior of a polynomial
+
+ For any degree n polynomial p(x) = a_0 + a_1 x + \cdots + a_{n-1}x^{n-1} + a_n x^n, its long-range behavior is the same as its highest-order term q(x) = a_n x^n. Thus, any polynomial of even degree appears U-shaped (\cup or \cap, like x^2 or -x^2) when we zoom way out, and any polynomial of odd degree appears chair-shaped (like x^3 or -x^3) when we zoom way out.
+
+
+
+
+ In Figure, we see how the degree 7 polynomial pictured there (and in Figure as well) appears to look like q(x) = -x^7 as we zoom out.
+
+
+
+
+
Plot of a degree 7 polynomial function p.
+
Plot of a degree 7 polynomial function p.
+
+
+
+
Plot of the same degree 7 polynomial function p, but zoomed out.
+
Plot of the same degree 7 polynomial function p, but zoomed out.
+
+
+
+
+
+ Finally, a key idea from calculus justifies the fact that the maximum number of turning points of a degree n polynomial is n-1, as we conjectured in the degree 4 case in Preview Activity. Moreover, the only possible numbers of turning points must have the same parity as n-1; that is, if n-1 is even, then the number of turning points must be even, and if instead n-1 is odd, the number of turning points must also be odd. For instance, for the degree 7 polynomial in Figure, we know that it is chair-shaped, with one end up and one end down. There could be zero turning points and the function could always decrease. But if there is at least one, then there must be a second, since if there were only one the function would decrease and then increase without turning back, which would force the graph to appear U-shaped.
+
+
+
+ polynomialnumber of turning points
+ The turning points of a polynomial
+
+ For any degree n polynomial p(x) = a_0 + a_1 x + \cdots + a_{n-1}x^{n-1} + a_n x^n, if n is even, its number of turning points is exactly one of n-1, n-3, \ldots, 1, and if n is odd, its number of turning points is exactly one of n-1, n-3, \ldots, 0.
+
+
+
+
+
+
+
+
+ Using zeros and signs to understand polynomial behavior
+
+
+ Just like a quadratic function can be written in different forms (standard: q(x) = ax^2 + bx + c, vertex: q(x) = a(x-h)^2 + k, and factored: q(x) = a(x-r_1)(x-r_2)), it's possible to write a polynomial function in different forms and to gain information about its behavior from those different forms. In particular, if we know all of the zeros of a polynomial function, we can write its formula in factored form, which gives us a deeper understanding of its graph.
+
+
+
+ The Zero Product Property Zero Product Property states that if two or more numbers are multiplied together and the result is 0, then at least one of the numbers must be 0. We use the Zero Product Property regularly with polynomial functions. If we can determine all n zeros of a degree n polynomial, and we call those zeros r_1, r_2, \ldots, r_n, we can write
+
+ p(x) = a_n(x-r_1)(x-r_2) \cdots (x-r_n)
+ .
+ Moreover, if we are given a polynomial in this factored form, we can quickly determine its zeros. For instance, if p(x) = 2(x+7)(x+1)(x-2)(x-5), we know that the only way p(x) = 0 is if at least one of the factors (x+7), (x+1), (x-2), or (x-5) equals 0, which implies that x = -7, x = -1, x = 2, or x = 5. Hence, from the factored form of a polynomial, it is straightforward to identify the polynomial's zeros, the x-values at which its graph crosses the x-axis. We can also use the factored form of a polynomial to develop what we call a sign chart, which we demonstrate in Example. sign chart
+
+
+
+
+
+ Consider the polynomial function p(x) = k(x-1)(x-a)(x-b). Suppose we know that 1 \lt a \lt b and that k \lt 0. Fully describe the graph of p without the aid of a graphing utility.
+
+
+
+
+ Since p(x) = k(x-1)(x-a)(x-b), we immediately know that p is a degree 3 polynomial with 3 real zeros: x = 1, a, b. We are given that 1 \lt a \lt b and in addition that k \lt 0. If we expand the factored form of p(x), it has form p(x) = kx^3 + \cdots, and since we know that when we zoom out, p(x) behaves like kx^3, we know that with k \lt 0 it follows \lim_{x \to -\infty} p(x) = +\infty and \lim_{x \to \infty} p(x) = -\infty.
+
+
+
+ Since p is degree 3 and we know it has zeros at x = 1, a, b, we know there are no other locations where p(x) = 0. Thus, on any interval between two zeros (or to the left of the least or the right of the greatest), the polynomial cannot change sign. We now investigate, interval by interval, the sign of the function.
+
+
+
+ When x \lt 1, it follows that x - 1 \lt 0. In addition, since 1 \lt a \lt b, when x \lt 1, x lies to the left of 1, a, and b, which also makes x-a and x-b negative. Moreover, we know that the constant k \lt 0. Hence, on the interval x \lt 1, all four terms in p(x) = k(x-1)(x-a)(x-b) are negative, which we indicate by writing ---- in that location on the sign chart pictured in Figure.
+
+
+
+ In addition, since there are an even number of negative terms in the product, the overall product's sign is positive, which we indicate by the single + beneath ----, and by writing POS below the coordinate axis.
+
+
+
+
A sign chart for the polynomial function p(x) = k(x-1)(x-a)(x-b).
+
A sign chart for the polynomial function p(x) = k(x-1)(x-a)(x-b).
+
+
+
+
+ We now proceed to the other intervals created by the zeros. On 1 \lt x \lt a, the term (x-1) has become positive, since x \gt 1. But both x-a and x-b are negative, as is the constant k, and thus we write -+-- for this interval, which has overall sign -, as noted in the figure. Similar reasoning completes the diagram.
+
+
+
+ From all of the information we have deduced about p, we conclude that regardless of the locations of a and b, the graph of p must look like the curve shown in Figure.
+
+
+
+
The graph of the polynomial function p(x) = k(x-1)(x-a)(x-b).
+
The graph of the polynomial function p(x) = k(x-1)(x-a)(x-b).
+ In Activity, we found that one of the zeros of the polynomial p(x) = 4692(x + 1520)(x^2 + 10000)(x - 3471)^2 (x - 9738) leads to different behavior of the function near that zero than we've seen in other situations. We now consider the more general situation where a polynomial has a repeated factor of the form (x-r)^n. When (x-r)^n is a factor of a polynomial p, we say that p has a zero of multiplicity n at x = r. polynomialzero of multiplicity n
+
+
+
+ To see the impact of repeated factors, we examine a collection of degree 4 polynomials that each have 4 real zeros. We start with the simplest of all, the function f(x) = x^4, whose zeros are x = 0, 0, 0, 0. Because the factor x-0 is repeated 4 times, the zero x = 0 has multiplicity 4.
+
+
+
+ Next we consider the degree 4 polynomial g(x) = x^3 (x-1), which has a zero of multiplicity 3 at x = 0 and a zero of multiplicity 1 at x = 1.
+
+
+
+
A plot of g(x) = x^3(x-1) with zero x = 0 of multiplicity 3 and x = 1 of multiplicity 1.
+
A plot of g(x) = x^3(x-1) with zero x = 0 of multiplicity 3 and x = 1 of multiplicity 1.
+
+
+
+
+
+ Observe that in Figure, the up-close plot near the zero x = 0 of multiplicity 3, the polynomial function g looks similar to the basic cubic polynomial -x^3. In addition, in Figure, we observe that if we zoom in even futher on the zero of multiplicity 1, the function g looks roughly linear, like a degree 1 polynomial. This type of behavior near repeated zeros turns out to hold in other cases as well.
+
+
+
+
+
A plot of g(x) = x^3(x-1) zoomed in on the zero x = 0 of multiplicity 3.
+
A plot of g(x) = x^3(x-1) zoomed in on the zero x = 0 of multiplicity 3.
+
+
+
+
A plot of g(x) = x^3(x-1) zoomed in on the zero x = 1 of multiplicity 1.
+
A plot of g(x) = x^3(x-1) zoomed in on the zero x = 1 of multiplicity 1.
+
+
+
+
+
+ If we next let h(x) = x^2 (x-1)^2, we see that h has two distinct real zeros, each of multiplicity 2. The graph of h in Figure shows that h behaves similar to a basic quadratic function near each of those zeros and thus shows U-shaped behavior nearby. If instead we let k(x) = x^2(x-1)(x+1), we see approximately linear behavior near x = -1 and x = 1 (the zeros of multiplicity 1), and quadratic (U-shaped) behavior near x = 0 (the zero of multiplicity 2), as seen in Figure.
+
+
+
+
+
Plot of h(x) = x^2 (x-1)^2 with zeros x = 0 and x = 1 of multiplicity 2.
+
Plot of h(x) = x^2 (x-1)^2 with zeros x = 0 and x = 1 of multiplicity 2.
+
+
+
+
Plot of k(x) = x^2(x-1)(x+1) with zeros x = 0 of multiplicity 2 and x = -1 and x = 1 of multiplicity 1.
+
Plot of k(x) = x^2(x-1)(x+1) with zeros x = 0 of multiplicity 2 and x = -1 and x = 1 of multiplicity 1.
+
+
+
+
+
+ Finally, if we consider m(x) = (x+1)x(x-1)(x-2), which has 4 distinct real zeros each of multiplicity 1, we observe in Figure that zooming in on each zero individually, the function demonstrates approximately linear behavior as it passes through the x-axis.
+
+
+
+
Plot of m(x) = (x+1)x(x-1)(x-2) with 4 distinct zeros of multiplicity 1.
+
Plot of m(x) = (x+1)x(x-1)(x-2) with 4 distinct zeros of multiplicity 1.
+
+
+
+
+ Our observations with polynomials of degree 4 in the various figures above generalize to polynomials of any degree.
+
+ If (x-r)^n is a factor of a polynomial p, then x = r is a zero of p of multiplicity n, and near x = r the graph of p looks like either -x^n or x^n does near x = 0. That is, the shape of the graph near the zero is determined by the multiplicity of the zero.
+
+
+
+
+
+
+
+
+ Summary
+
+
+
+
+ From a polynomial function's algebraic structure, we can deduce several key traits of the function.
+
+
+
+
+
+ If the function is in standard form, say p(x) = a_0 + a_1 x + a_2 x^2 + \cdots + a_{n-1}x^{n-1} + a_n x^n, we know that its degree is n and that when we zoom out, p looks like a_n x^n and thus has the same long-range behavior as a_n x^n. Thus, p is chair-shaped if n is odd and U-shaped if n is even. Whether \lim_{n \to \infty} p(x) is +\infty or -\infty depends on the sign of a_n.
+
+
+
+
+
+ If the function is in factored form, say p(x) = a_n(x-r_1)(x-r_2) \cdots (x-r_n) (where the r_i's are possibly not distinct and possibly complex), we can quickly determine both the degree of the polynomial (n) and the locations of its zeros, as well as their multiplicities.
+
+
+
+
+
+
+
+ A sign chart is a visual way to identify all of the locations where a function is zero along with the sign of the function on the various intervals the zeros create. A sign chart gives us an overall sense of the graph of the function, but without concerning ourselves with any specific values of the function besides the zeros. For a sample sign chart, see Figure.
+
+
+
+
+ When a polynomial p has a repeated factor such as p(x) = (x-5)(x-5)(x-5) = (x-5)^3, we say that x = 5 is a zero of multiplicity 3. At the point x = 5 where p will cross the x-axis, up close it will look like a cubic polynomial and thus be chair-shaped. In general, if (x-r)^n is a factor of a polynomial p so that x = r is a zero of multiplicity n, the polynomial will behave near x = r like the polynomial x^n behaves near x = 0.
+
- What does it mean to say that a rational function has a hole at a certain point, and what algebraic structure leads to such behavior?
-
-
-
-
- How do we determine where a rational function has zeros and where it has vertical asymptotes?
-
-
-
-
- What does a sign chart reveal about the behavior of a rational function and how do we develop a sign chart from a given formula?
-
-
-
-
-
- Introduction
-
- Because any rational function is the ratio of two polynomial functions, it's natural to ask questions about rational functions similar to those we ask about polynomials. With polynomials, it is often helpful to know where the function's value is zero. In a rational function r(x) = \frac{p(x)}{q(x)}, we are curious to know where both p(x) = 0 and where q(x) = 0.
-
-
-
- Connected to these questions, we want to understand both where a rational function's output value is zero, as well as where the function is undefined. In addition, from the behavior of simple rational power functions such as \frac{1}{x}, we expect that rational functions may not only have horizontal asymptotes (as investigated in Section), but also vertical asymptotes. At first glance, these questions about zeros and vertical asymptotes of rational functions may appear to be elementary ones whose answers simply depend on where the numerator and denominator of the rational function are zero. But in fact, rational functions often admit very subtle behavior that can escape the human eye and the graph generated by a computer.
-
-
-
-
-
-
-
- When a rational function has a <q>hole</q>
-
-
- Two important features of any rational function r(x) = \frac{p(x)}{q(x)} are any zeros and vertical asymptotes the function may have. These aspects of a rational function are closely connected to where the numerator and denominator, respectively, are zero. At the same time, a subtle related issue can lead to radically different behavior. To understand why, we first remind ourselves of a few key facts about fractions that involve 0. Because we are working with a function, we'll think about fractions whose numerator and denominator are approaching particular values.
-
-
-
- If the numerator of a fraction approaches 0 while the denominator approaches a nonzero value, then the overall fraction values will approach zero. For instance, consider the sequence of values
-
- \frac{0.1}{0.9} = 0.111111\cdots, \frac{0.01}{0.99} = 0.010101\cdots, \frac{0.001}{0.999} = 0.001001\cdots
- .
- Because the numerator gets closer and closer to 0 and the denominator stays away from 0, the quotients tend to 0.
-
-
-
- Similarly, if the denominator of a fraction approaches 0 while the numerator approaches a nonzero value, then the overall fraction increases without bound. If we consider the reciprocal values of the sequence above, we see that
-
- \frac{0.9}{0.1} = 9, \frac{0.99}{0.01} = 99, \frac{0.999}{0.001} = 999
- .
- Since the denominator gets closer and closer to 0 and the numerator stays away from 0, the quotients increase without bound.
-
-
-
- These two behaviors show how the zeros and vertical asympototes of a rational function r(x) = \frac{p(x)}{q(x)} arise: where the numerator p(x) is zero and the denominator q(x) is nonzero, the function r will have a zero; and where q(x) is zero and p(x) is nonzero, the function will have a vertical asymptote. What we must be careful of is the special situation where both the numerator p(x) and q(x) are simultaneously zero. Indeed, if the numerator and denominator of a fraction both approach 0, different behavior can arise. For instance, consider the sequence
-
- \frac{0.2}{0.1} = 2, \frac{0.02}{0.01} = 2, \frac{0.002}{0.001} = 2
- .
- In this situation, both the numerator and denominator are approaching 0, but the overall fraction's value is always 2. This is very different from the two sequences we considered above. In Example, we explore similar behavior in the context of a particular rational function.
-
-
-
-
-
- Consider the rational function r(x) = \frac{x^2 - 1}{x^2 - 3x - 4} from Preview Activity, whose numerator is p(x) = x^2 - 1 and whose denominator is q(x) = x^2 - 3x - 4. Explain why the graph of r generated by Desmos or another computational device is incorrect, and also identify the locations of any zeros and vertical asymptotes of r.
-
-
-
-
- It is helpful with any rational function to factor the numerator and denominator. We note that p(x) = x^2 - 1 = (x-1)(x+1) and q(x) = x^2 - 3x - 4 = (x+1)(x-4). The domain of r is thus the set of all real numbers except x = -1 and x = 4, the set of all points where q(x) \ne 0.
-
-
-
- Knowing that r is not defined at x = -1, it is natural to study the graph of r near that value. Plotting the function in Desmos, we get a result similar to the one shown in Figure, which appears to show no unusual behavior at x = -1, and even that r(-1) is defined. If we zoom in on that point, as shown in Figure, the technology still fails to visually demonstrate the fact that r(-1) is not defined. This is because graphing utilities sample functions at a finite number of points and then connect the resulting dots to generate the curve we see.
-
-
-
-
-
A plot of r(x) = \frac{x^2 - 1}{x^2 - 3x - 4}.
-
A plot of r(x) = \frac{x^2 - 1}{x^2 - 3x - 4}.
-
-
-
-
Zooming in on r(x) near x = -1.
-
Zooming in on r(x) near x = -1.
-
-
-
-
How the graph of r(x) should actually appear near x = -1.
-
How the graph of r(x) should actually appear near x = -1.
-
-
-
-
-
- We know from our algebraic work with the denominator, q(x) = (x+1)(x-4), that r is not defined at x = -1. While the denominator q gets closer and closer to 0 as x approaches -1, so does the numerator, since p(x) = (x-1)(x+1). If we consider values close but not equal to x = -1, we see results in Table.
-
- In the table, we see that both the numerator and denominator get closer and closer to 0 as x gets closer and closer to -1, but that their quotient appears to be getting closer and closer to y = 0.4. Indeed, we see this behavior in the graph of r, though the graphing utility misses the fact that r(-1) is actually not defined. A precise graph of r near x = -1 should look like the one presented in Figure, where we see an open circle at the point (-1, 0.4) that demonstrates that r(-1) is not defined, and that r does not have a vertical asymptote or zero at x = -1.
-
-
-
- Finally, we also note that p(1) = 0 and q(1) = -6, so at x = 1, r(x) has a zero (its numerator is zero and its denominator is not). In addition, q(4) = 0 and p(4) = 15 (its denominator is zero and its numerator is not), so r(x) has a vertical asymptote at x = 4. These features are accurately represented by the original Desmos graph shown in Figure.
-
-
-
-
-
- In the situation where a rational function is undefined at a point but does not have a vertical asymptote there, we'll say that the graph of the function has a hole. rational functionhole In calculus, we use limit notation to identify a hole in a function's graph. Indeed, having shown in Example that the value of r(x) gets closer and closer to 0.4 as x gets closer and closer to -1, we naturally write
-
- \lim_{x \to -1} r(x) = 0.4
-
- as a shorthand way to represent the behavior of r (similar to how we've written limits involving \infty). This fact, combined with r(-1) being undefined, tells us that near x = -1 the graph approaches a value of 0.4 but has to have a hole at the point (-1,0.4), as shown in Figure. Because we'll encounter similar behavior with other functions, we formally define limit notation as follows.
-
-
-
-
-
- Let a and L be finite real numbers, and let r be a function defined near x = a, but not necessarily at x = a itself. If we can make the value of r(x) as close to the number L as we like by taking x sufficiently close (but not equal) to a, then we write
-
- \lim_{x \to a} r(x) = L
-
- and say that the limit of r as x approaches a is L.
-
-
-
-
-
- The key observations regarding zeros, vertical asymptotes, and holes in Example apply to any rational function.
-
-
-
- Features of a rational function
-
- Let r(x) = \frac{p(x)}{q(x)} be a rational function.
-
-
-
-
-
-
- If p(a) = 0 and q(a) \ne 0, then r(a) = 0, so r has a zero at x = a.
-
-
-
-
- If q(a) = 0 and p(a) \ne 0, then r(a) is undefined and r has a vertical asymptote at x = a.
-
-
-
-
- If p(a) = 0 and q(a) = 0 and we can show that there is a finite number L such that
-
- \lim_{x \to a} r(x) = L
- ,
- then r(a) is not defined and r has a hole at the point (a,L).It is possible for both p(a) = 0 and q(a) = 0 and for r to still have a vertical asymptote at x = a. We explore this possibility further in Exercise.
-
-
-
-
-
-
-
-
-
-
-
- Sign charts and finding formulas for rational functions
-
-
- Just like with polynomial functions, we can use sign charts to describe the behavior of rational functions. The only significant difference for their use in this context is that we not only must include all x-values where the rational function r(x) = 0, but also all x-values at which the function r is not defined. This is because it is possible for a rational function to change sign at a point that lies outside its domain, such as when the function has a vertical asymptote.
-
-
-
-
-
- Construct a sign chart for the function q(x) = \frac{(x-2)(x^2-9)}{(x-3)(x-1)^2}. Then, graph the function q and compare the graph and sign chart.
-
-
-
-
- First, we fully factor q and identify the x-values that are not in its domain. Since x^2-9 = (x-3)(x+3), we see that
-
- q(x) = \frac{(x-2)(x-3)(x+3)}{(x-3)(x-1)^2}
- .
- From the denominator, we observe that q is not defined at x = 3 and x = 1 since those values make the factors x - 3 = 0 or (x-1)^2 = 0. Thus, the domain of q is the set of all real numbers except x = 1 and x = 3. From the numerator, we see that both x = 2 and x = -3 are zeros of q since these values make the numerator zero while the denominator is nonzero. We expect that q will have a hole at x = 3 since this x-value is not in the domain and it makes both the numerator and denominator 0. Indeed, computing values of q for x near x = 3 suggests that
-
- \lim_{x \to 3} q(x) = 1.5
- ,
- and thus q does not change sign at x = 3.
-
-
-
- Thus, we have three different x-values to place on the sign chart: x = -3, x = 1, and x = 2. We now analyze the sign of each of the factors in q(x) = \frac{(x-2)(x-3)(x+3)}{(x-3)(x-1)^2} on the various intervals. For x \lt -3, (x-2) \lt 0, (x-3) \lt 0, (x+3) \lt 0, and (x-1)^2 \gt 0. Thus, for x \lt -3, the sign of q is
-
- \frac{- - -}{- +} = +
-
- since there are an even number of negative terms in the quotient.
-
-
-
- On the interval -3 \lt x \lt 1, (x-2) \lt 0, (x-3) \lt 0, (x+3) \gt 0, and (x-1)^2 \gt 0. Thus, for these x-values, the sign of q is
-
- \frac{- - +}{- +} = -
- .
-
-
-
- Using similar reasoning, we can complete the sign chart shown in Figure. A plot of the function q, as seen in Figure, shows behavior that matches the sign function, as well as the need to manually identify the hole at (3, 1.5), which is missed by the graphing software.
-
-
-
-
-
The sign chart for q.
-
The sign chart for q.
-
-
-
-
A plot of q.
-
A plot of q.
-
-
-
-
-
- In both the sign chart and the figure, we see that q changes sign at each of its zeros, x = -3 and x = 2, and that it does not change as it passes by its vertical asymptote at x = 1. The reason q doesn't change sign at the asympotote is because of the repeated factor of (x-1)^2 which is always positive.
-
-
-
-
-
- To find a formula for a rational function with certain properties, we can reason in ways that are similar to our work with polynomials. Since the rational function must have a polynomial expression in both the numerator and denominator, by thinking about where the numerator and denominator must be zero, we can often generate a formula whose graph will satisfy the desired properties.
-
-
-
-
-
-
-
- Summary
-
-
-
- If a rational function r(x) = \frac{p(x)}{q(x)} has the properties that p(a) = 0 and q(a) = 0 and
-
- \lim_{x \to a} r(x) = L
- ,
- then r has a hole at the point (a,L). This behavior can occur when there is a matching factor of (x-a) in both p and q.
-
-
-
-
- For a rational function r(x) = \frac{p(x)}{q(x)}, we determine where the function has zeros and where it has vertical asymptotes by considering where the numerator and denominator are 0. In particular, if p(a) = 0 and q(a) \ne 0, then r(a) = 0, so r has a zero at x = a. And if q(a) = 0 and p(a) \ne 0, then r(a) is undefined and r has a vertical asymptote at x = a.
-
-
-
-
- By writing a rational function's numerator in factored form, we can generate a sign chart for the function that takes into account all of the zeros and vertical asymptotes of the function, which are the only points where the function can possibly change sign. By testing x-values in various intervals between zeros and/or vertical asymptotes, we can determine where the rational function is positive and where the function is negative.
-
+ What does it mean to say that a rational function has a hole at a certain point, and what algebraic structure leads to such behavior?
+
+
+
+
+ How do we determine where a rational function has zeros and where it has vertical asymptotes?
+
+
+
+
+ What does a sign chart reveal about the behavior of a rational function and how do we develop a sign chart from a given formula?
+
+
+
+
+
+ Introduction
+
+ Because any rational function is the ratio of two polynomial functions, it's natural to ask questions about rational functions similar to those we ask about polynomials. With polynomials, it is often helpful to know where the function's value is zero. In a rational function r(x) = \frac{p(x)}{q(x)}, we are curious to know where both p(x) = 0 and where q(x) = 0.
+
+
+
+ Connected to these questions, we want to understand both where a rational function's output value is zero, as well as where the function is undefined. In addition, from the behavior of simple rational power functions such as \frac{1}{x}, we expect that rational functions may not only have horizontal asymptotes (as investigated in Section), but also vertical asymptotes. At first glance, these questions about zeros and vertical asymptotes of rational functions may appear to be elementary ones whose answers simply depend on where the numerator and denominator of the rational function are zero. But in fact, rational functions often admit very subtle behavior that can escape the human eye and the graph generated by a computer.
+
+
+
+
+
+
+
+
+ When a rational function has a <q>hole</q>
+
+
+ Two important features of any rational function r(x) = \frac{p(x)}{q(x)} are any zeros and vertical asymptotes the function may have. These aspects of a rational function are closely connected to where the numerator and denominator, respectively, are zero. At the same time, a subtle related issue can lead to radically different behavior. To understand why, we first remind ourselves of a few key facts about fractions that involve 0. Because we are working with a function, we'll think about fractions whose numerator and denominator are approaching particular values.
+
+
+
+ If the numerator of a fraction approaches 0 while the denominator approaches a nonzero value, then the overall fraction values will approach zero. For instance, consider the sequence of values
+
+ \frac{0.1}{0.9} = 0.111111\cdots, \frac{0.01}{0.99} = 0.010101\cdots, \frac{0.001}{0.999} = 0.001001\cdots
+ .
+ Because the numerator gets closer and closer to 0 and the denominator stays away from 0, the quotients tend to 0.
+
+
+
+ Similarly, if the denominator of a fraction approaches 0 while the numerator approaches a nonzero value, then the overall fraction increases without bound. If we consider the reciprocal values of the sequence above, we see that
+
+ \frac{0.9}{0.1} = 9, \frac{0.99}{0.01} = 99, \frac{0.999}{0.001} = 999
+ .
+ Since the denominator gets closer and closer to 0 and the numerator stays away from 0, the quotients increase without bound.
+
+
+
+ These two behaviors show how the zeros and vertical asympototes of a rational function r(x) = \frac{p(x)}{q(x)} arise: where the numerator p(x) is zero and the denominator q(x) is nonzero, the function r will have a zero; and where q(x) is zero and p(x) is nonzero, the function will have a vertical asymptote. What we must be careful of is the special situation where both the numerator p(x) and q(x) are simultaneously zero. Indeed, if the numerator and denominator of a fraction both approach 0, different behavior can arise. For instance, consider the sequence
+
+ \frac{0.2}{0.1} = 2, \frac{0.02}{0.01} = 2, \frac{0.002}{0.001} = 2
+ .
+ In this situation, both the numerator and denominator are approaching 0, but the overall fraction's value is always 2. This is very different from the two sequences we considered above. In Example, we explore similar behavior in the context of a particular rational function.
+
+
+
+
+
+ Consider the rational function r(x) = \frac{x^2 - 1}{x^2 - 3x - 4} from Preview Activity, whose numerator is p(x) = x^2 - 1 and whose denominator is q(x) = x^2 - 3x - 4. Explain why the graph of r generated by Desmos or another computational device is incorrect, and also identify the locations of any zeros and vertical asymptotes of r.
+
+
+
+
+ It is helpful with any rational function to factor the numerator and denominator. We note that p(x) = x^2 - 1 = (x-1)(x+1) and q(x) = x^2 - 3x - 4 = (x+1)(x-4). The domain of r is thus the set of all real numbers except x = -1 and x = 4, the set of all points where q(x) \ne 0.
+
+
+
+ Knowing that r is not defined at x = -1, it is natural to study the graph of r near that value. Plotting the function in Desmos, we get a result similar to the one shown in Figure, which appears to show no unusual behavior at x = -1, and even that r(-1) is defined. If we zoom in on that point, as shown in Figure, the technology still fails to visually demonstrate the fact that r(-1) is not defined. This is because graphing utilities sample functions at a finite number of points and then connect the resulting dots to generate the curve we see.
+
+
+
+
+
A plot of r(x) = \frac{x^2 - 1}{x^2 - 3x - 4}.
+
A plot of r(x) = \frac{x^2 - 1}{x^2 - 3x - 4}.
+
+
+
+
Zooming in on r(x) near x = -1.
+
Zooming in on r(x) near x = -1.
+
+
+
+
How the graph of r(x) should actually appear near x = -1.
+
How the graph of r(x) should actually appear near x = -1.
+
+
+
+
+
+ We know from our algebraic work with the denominator, q(x) = (x+1)(x-4), that r is not defined at x = -1. While the denominator q gets closer and closer to 0 as x approaches -1, so does the numerator, since p(x) = (x-1)(x+1). If we consider values close but not equal to x = -1, we see results in Table.
+
+ In the table, we see that both the numerator and denominator get closer and closer to 0 as x gets closer and closer to -1, but that their quotient appears to be getting closer and closer to y = 0.4. Indeed, we see this behavior in the graph of r, though the graphing utility misses the fact that r(-1) is actually not defined. A precise graph of r near x = -1 should look like the one presented in Figure, where we see an open circle at the point (-1, 0.4) that demonstrates that r(-1) is not defined, and that r does not have a vertical asymptote or zero at x = -1.
+
+
+
+ Finally, we also note that p(1) = 0 and q(1) = -6, so at x = 1, r(x) has a zero (its numerator is zero and its denominator is not). In addition, q(4) = 0 and p(4) = 15 (its denominator is zero and its numerator is not), so r(x) has a vertical asymptote at x = 4. These features are accurately represented by the original Desmos graph shown in Figure.
+
+
+
+
+
+ In the situation where a rational function is undefined at a point but does not have a vertical asymptote there, we'll say that the graph of the function has a hole. rational functionhole In calculus, we use limit notation to identify a hole in a function's graph. Indeed, having shown in Example that the value of r(x) gets closer and closer to 0.4 as x gets closer and closer to -1, we naturally write
+
+ \lim_{x \to -1} r(x) = 0.4
+
+ as a shorthand way to represent the behavior of r (similar to how we've written limits involving \infty). This fact, combined with r(-1) being undefined, tells us that near x = -1 the graph approaches a value of 0.4 but has to have a hole at the point (-1,0.4), as shown in Figure. Because we'll encounter similar behavior with other functions, we formally define limit notation as follows.
+
+
+
+
+
+ Let a and L be finite real numbers, and let r be a function defined near x = a, but not necessarily at x = a itself. If we can make the value of r(x) as close to the number L as we like by taking x sufficiently close (but not equal) to a, then we write
+
+ \lim_{x \to a} r(x) = L
+
+ and say that the limit of r as x approaches a is L.
+
+
+
+
+
+ The key observations regarding zeros, vertical asymptotes, and holes in Example apply to any rational function.
+
+
+
+ Features of a rational function
+
+ Let r(x) = \frac{p(x)}{q(x)} be a rational function.
+
+
+
+
+
+
+ If p(a) = 0 and q(a) \ne 0, then r(a) = 0, so r has a zero at x = a.
+
+
+
+
+ If q(a) = 0 and p(a) \ne 0, then r(a) is undefined and r has a vertical asymptote at x = a.
+
+
+
+
+ If p(a) = 0 and q(a) = 0 and we can show that there is a finite number L such that
+
+ \lim_{x \to a} r(x) = L
+ ,
+ then r(a) is not defined and r has a hole at the point (a,L).It is possible for both p(a) = 0 and q(a) = 0 and for r to still have a vertical asymptote at x = a. We explore this possibility further in Exercise.
+
+
+
+
+
+
+
+
+
+
+
+ Sign charts and finding formulas for rational functions
+
+
+ Just like with polynomial functions, we can use sign charts to describe the behavior of rational functions. The only significant difference for their use in this context is that we not only must include all x-values where the rational function r(x) = 0, but also all x-values at which the function r is not defined. This is because it is possible for a rational function to change sign at a point that lies outside its domain, such as when the function has a vertical asymptote.
+
+
+
+
+
+ Construct a sign chart for the function q(x) = \frac{(x-2)(x^2-9)}{(x-3)(x-1)^2}. Then, graph the function q and compare the graph and sign chart.
+
+
+
+
+ First, we fully factor q and identify the x-values that are not in its domain. Since x^2-9 = (x-3)(x+3), we see that
+
+ q(x) = \frac{(x-2)(x-3)(x+3)}{(x-3)(x-1)^2}
+ .
+ From the denominator, we observe that q is not defined at x = 3 and x = 1 since those values make the factors x - 3 = 0 or (x-1)^2 = 0. Thus, the domain of q is the set of all real numbers except x = 1 and x = 3. From the numerator, we see that both x = 2 and x = -3 are zeros of q since these values make the numerator zero while the denominator is nonzero. We expect that q will have a hole at x = 3 since this x-value is not in the domain and it makes both the numerator and denominator 0. Indeed, computing values of q for x near x = 3 suggests that
+
+ \lim_{x \to 3} q(x) = 1.5
+ ,
+ and thus q does not change sign at x = 3.
+
+
+
+ Thus, we have three different x-values to place on the sign chart: x = -3, x = 1, and x = 2. We now analyze the sign of each of the factors in q(x) = \frac{(x-2)(x-3)(x+3)}{(x-3)(x-1)^2} on the various intervals. For x \lt -3, (x-2) \lt 0, (x-3) \lt 0, (x+3) \lt 0, and (x-1)^2 \gt 0. Thus, for x \lt -3, the sign of q is
+
+ \frac{- - -}{- +} = +
+
+ since there are an even number of negative terms in the quotient.
+
+
+
+ On the interval -3 \lt x \lt 1, (x-2) \lt 0, (x-3) \lt 0, (x+3) \gt 0, and (x-1)^2 \gt 0. Thus, for these x-values, the sign of q is
+
+ \frac{- - +}{- +} = -
+ .
+
+
+
+ Using similar reasoning, we can complete the sign chart shown in Figure. A plot of the function q, as seen in Figure, shows behavior that matches the sign function, as well as the need to manually identify the hole at (3, 1.5), which is missed by the graphing software.
+
+
+
+
+
The sign chart for q.
+
The sign chart for q.
+
+
+
+
A plot of q.
+
A plot of q.
+
+
+
+
+
+ In both the sign chart and the figure, we see that q changes sign at each of its zeros, x = -3 and x = 2, and that it does not change as it passes by its vertical asymptote at x = 1. The reason q doesn't change sign at the asympotote is because of the repeated factor of (x-1)^2 which is always positive.
+
+
+
+
+
+ To find a formula for a rational function with certain properties, we can reason in ways that are similar to our work with polynomials. Since the rational function must have a polynomial expression in both the numerator and denominator, by thinking about where the numerator and denominator must be zero, we can often generate a formula whose graph will satisfy the desired properties.
+
+
+
+
+
+
+
+ Summary
+
+
+
+ If a rational function r(x) = \frac{p(x)}{q(x)} has the properties that p(a) = 0 and q(a) = 0 and
+
+ \lim_{x \to a} r(x) = L
+ ,
+ then r has a hole at the point (a,L). This behavior can occur when there is a matching factor of (x-a) in both p and q.
+
+
+
+
+ For a rational function r(x) = \frac{p(x)}{q(x)}, we determine where the function has zeros and where it has vertical asymptotes by considering where the numerator and denominator are 0. In particular, if p(a) = 0 and q(a) \ne 0, then r(a) = 0, so r has a zero at x = a. And if q(a) = 0 and p(a) \ne 0, then r(a) is undefined and r has a vertical asymptote at x = a.
+
+
+
+
+ By writing a rational function's numerator in factored form, we can generate a sign chart for the function that takes into account all of the zeros and vertical asymptotes of the function, which are the only points where the function can possibly change sign. By testing x-values in various intervals between zeros and/or vertical asymptotes, we can determine where the rational function is positive and where the function is negative.
+
- How can we determine key information about a rational function from its algebraic structure?
-
-
-
-
- Why are rational functions important?
-
-
-
-
-
- Introduction
-
- The average rate of change of a function on an interval always involves a ratio. Indeed, for a given function f that interests us near t = 2, we can investigate its average rate of change on intervals near this value by considering
-
- AV_{[2,2+h]} = \frac{f(2+h)-f(2)}{h}
- .
- Suppose, for instance, that f meausures the height of a falling ball at time t and is given by f(t) = -16t^2 + 32t + 48, which happens to be a polynomial function of degree 2. For this particular function, its average rate of change on [2,2+h] is
-
-
- AV_{[2,2+h]} &= \frac{f(2+h)-f(2)}{h}
-
-
- &= \frac{-16(2+h)^2 + 32(2+h) + 48 - (-16 \cdot 4 + 32 \cdot 2 + 48)}{h}
-
-
- &= \frac{-64 - 64h - 16h^2 + 64 + 32h + 48 - (48)}{h}
-
-
- &= \frac{-32h - 16h^2}{h}
-
- .
- Structurally, we observe that AV_{[2,2+h]} is a ratio of the two functions -32h - 16h^2 and h. Moreover, both the numerator and the denominator of the expression are themselves polynomial functions of the variable h. Note that we may be especially interested in what occurs as h \to 0, as these values will tell us the average velocity of the moving ball on shorter and shorter time intervals starting at t = 2. At the same time, AV_{[2,2+h]} is not defined for h = 0.
-
-
-
- Ratios of polynomial functions arise in several different important circumstances. Sometimes we are interested in what happens when the denominator approaches 0, which makes the overall ratio undefined. In other situations, we may want to know what happens in the long term and thus consider what happens when the input variable increases without bound.
-
- The functions AV_{[2,2+h]} = \frac{-32h - 16h^2}{h} and A(q) = \frac{5000000 + 2500q}{q} are both examples of rational functions, since each is a ratio of polynomial functions. Formally, we have the following definition.
-
-
-
- rational function
-
-
- A function r is rational provided that it is possible to write r as the ratio of two polynomials, p and q. That is, r is rational provided that for some polynomial functions p and q, we have
-
- r(x) = \frac{p(x)}{q(x)}
- .
-
-
-
-
-
- Like with polynomial functions, we are interested in such natural questions as
-
-
-
- What is the long range behavior of a given rational function?
-
-
-
-
- What is the domain of a given rational function?
-
-
-
-
- How can we determine where a given rational function's value is 0?
-
-
-
-
-
-
- We begin by focusing on the long-range behavior of rational functions. It's important first to recall our earlier work with power functions of the form p(x) = x^{-n} where n = 1, 2, \ldots. For such functions, we know that p(x) = \frac{1}{x^n} where n \gt 0 and that
-
- \lim_{x \to \infty} \frac{1}{x^n} = 0
-
- since x^n increases without bound as x \to \infty. The same is true when x \to -\infty: \lim_{x \to -\infty} \frac{1}{x^n} = 0. Thus, any time we encounter a quantity such as \frac{1}{x^3}, this quantity will approach 0 as x increases without bound, and this will also occur for any constant numerator. For instance,
-
- \lim_{x \to \infty} \frac{2500}{x^2} = 0
-
- since 2500 times a quantity approaching 0 will still approach 0 as x increases.
-
-
-
-
-
-
-
- We summarize and generalize the results of Activity and Activity as follows.
-
-
-
- rational functionlong-term behavior
- The long-term behavior of a rational function
-
- Let p and q be polynomial functions so that r(x) = \frac{p(x)}{q(x)} is a rational function. Suppose that p has degree n with leading term a_n x^n and q has degree m with leading term b_m x^m for some nonzero constants a_n and b_m. There are three possibilities (n \lt m, n = m, and n \gt m) that result in three different behaviors of r:
-
-
-
-
-
-
- if n \lt m, then the degree of the numerator is less than the degree of the denominator, and thus
-
- \lim_{n \to \infty} r(x) = \lim_{n \to \infty} \frac{a_n x^n + \cdots + a_0}{b_m x^m + \cdots + b_0} = 0
- ,
- which tells us that y = 0 is a horizontal asymptote of r;
-
-
-
-
- if n = m, then the degree of the numerator equals the degree of the denominator, and thus
-
- \lim_{n \to \infty} r(x) = \lim_{n \to \infty} \frac{a_n x^n + \cdots + a_0}{b_n x^n + \cdots + b_0} = \frac{a_n}{b_n}
- ,
- which tells us that y = \frac{a_n}{b_n} (the ratio of the coefficients of the highest order terms in p and q) is a horizontal asymptote of r;
-
-
-
-
- if n \gt m, then the degree of the numerator is greater than the degree of the denominator, and thus
-
- \lim_{n \to \infty} r(x) = \lim_{n \to \infty} \frac{a_n x^n + \cdots + a_0}{b_m x^m + \cdots + b_0} = \pm \infty
- ,
- (where the sign of the limit depends on the signs of a_n and b_m) which tells us that r is does not have a horizontal asymptote.
-
-
-
-
-
- In both situations (a) and (b), the value of \lim_{x \to -\infty} r(x) is identical to \lim_{x \to \infty} r(x).
-
-
-
-
-
-
- The domain of a rational function
-
- Because a rational function can be written in the form r(x) = \frac{p(x)}{q(x)} for some polynomial functions p and q, we have to be concerned about the possibility that a rational function's denominator is zero. Since polynomial functions always have their domain as the set of all real numbers, it follows that any rational function is only undefined at points where its denominator is zero.
-
-
-
- rational functiondomain
- The domain of a rational function
-
- Let p and q be polynomial functions so that r(x) = \frac{p(x)}{q(x)} is a rational function. The domain of r is the set of all real numbers except those for which q(x) = 0.
-
-
-
-
-
-
- Determine the domain of the function r(x) = \frac{5x^3 + 17x^2 - 9x + 4}{2x^3 - 6x^2 - 8x}.
-
-
-
-
- To find the domain of any rational function, we need to determine where the denominator is zero. The best way to find these values exactly is to factor the denominator. Thus, we observe that
-
- 2x^3 - 6x^2 - 8x = 2x(x^2 - 3x - 4) = 2x(x+1)(x-4)
- .
- By the Zero Product Property, it follows that the denominator of r is zero at x = 0, x = -1, and x = 4. Hence, the domain of r is the set of all real numbers except -1, 0, and 4.
-
-
-
-
-
- We note that when it comes to determining the domain of a rational function, the numerator is irrelevant: all that matters is where the denominator is 0.
-
- Rational functions arise naturally in the study of the average rate of change of a polynomial function, leading to expressions such as
-
- AV_{[2,2+h]} = \frac{-32h - 16h^2}{h}
- .
- We will study several subtle issues that correspond to such functions further in Section. For now, we will focus on a different setting in which rational functions play a key role.
-
-
-
- In Section, we encountered a class of problems where a key quantity was modeled by a polynomial function. We found that if we considered a container such as a cylinder with fixed surface area, then the volume of the container could be written as a polynomial of a single variable. For instance, if we consider a circular cylinder with surface area 10 square feet, then we know that
-
- S = 10 = 2\pi r^2 + 2\pi r h
-
- and therefore h = \frac{10 - 2\pi r^2}{2 \pi r}. Since the cylinder's volume is V = \pi r^2 h, it follows that
-
- V = \pi r^2 h = \pi r^2 \left( \frac{10 - 2\pi r^2}{2 \pi r} \right) = r(5 - \pi r^2)
- ,
- which is a polynomial function of r.
-
-
-
- What happens if we instead fix the volume of the container and ask about how surface area can be written as a function of a single variable?
-
-
-
-
-
- Suppose we want to construct a circular cylinder that holds 20 cubic feet of volume. How much material does it take to build the container? How can we state the amount of material as a function of a single variable?
-
-
-
-
- Neglecting any scrap, the amount of material it takes to construct the container is its surface area, which we know to be
-
- S = 2\pi r^2 + 2\pi r h
- .
- Because we want the volume to be fixed, this results in a constraint equation that enables us to relate r and h. In particular, since
-
- V = 20 = \pi r^2 h
- ,
- it follows that we can solve for h and get h = \frac{20}{\pi r^2}. Substituting this expression for h in the equation for surface area, we find that
-
- S = 2\pi r^2 + 2\pi r \cdot \frac{20}{\pi r^2} = 2 \pi r^2 + \frac{40}{r}
- .
- Getting a common denominator, we can also write S in the form
-
- S(r) = \frac{2 \pi r^3 + 40}{r}
-
- and thus we see that S is a rational function of r. Because of the physical context of the problem and the fact that the denominator of S is r, the domain of S is the set of all positive real numbers.
-
-
-
-
-
-
-
-
-
- Summary
-
-
-
-
- A rational function is a function whose formula can be written as the ratio of two polynomial functions. For instance, r(x) = \frac{7x^3 - 5x + 16}{-4x^4 + 2x^3 - 11x + 3} is a rational function.
-
-
-
-
- Two aspects of rational functions are straightforward to determine for any rational function. Given r(x) = \frac{p(x)}{q(x)} where p and q are polynomials, the domain of r is the set of all real numbers except any values of x for which q(x) = 0. In addition, we can determine the long-range behavior of r by examining the highest order terms in p and q:
-
-
-
-
- if the degree of p is less than the degree of q, then r has a horizontal asymptote at y = 0;
-
-
-
-
- if the degree of p equals the degree of q, then r has a horizontal asymptote at y = \frac{a_n}{b_n}, where a_n and b_n are the leading coefficients of p and q respectively;
-
-
-
-
- and if the degree of p is greater than the degree of q, then r does not have a horizontal asymptote.
-
-
-
-
-
-
-
- Two reasons that rational functions are important are that they arise naturally when we consider the average rate of change on an interval whose length varies and when we consider problems that relate the volume and surface area of three-dimensional containers when one of those two quantities is constrained.
-
+ How can we determine key information about a rational function from its algebraic structure?
+
+
+
+
+ Why are rational functions important?
+
+
+
+
+
+ Introduction
+
+ The average rate of change of a function on an interval always involves a ratio. Indeed, for a given function f that interests us near t = 2, we can investigate its average rate of change on intervals near this value by considering
+
+ AV_{[2,2+h]} = \frac{f(2+h)-f(2)}{h}
+ .
+ Suppose, for instance, that f meausures the height of a falling ball at time t and is given by f(t) = -16t^2 + 32t + 48, which happens to be a polynomial function of degree 2. For this particular function, its average rate of change on [2,2+h] is
+
+
+ AV_{[2,2+h]} &= \frac{f(2+h)-f(2)}{h}
+
+
+ &= \frac{-16(2+h)^2 + 32(2+h) + 48 - (-16 \cdot 4 + 32 \cdot 2 + 48)}{h}
+
+
+ &= \frac{-64 - 64h - 16h^2 + 64 + 32h + 48 - (48)}{h}
+
+
+ &= \frac{-32h - 16h^2}{h}
+
+ .
+ Structurally, we observe that AV_{[2,2+h]} is a ratio of the two functions -32h - 16h^2 and h. Moreover, both the numerator and the denominator of the expression are themselves polynomial functions of the variable h. Note that we may be especially interested in what occurs as h \to 0, as these values will tell us the average velocity of the moving ball on shorter and shorter time intervals starting at t = 2. At the same time, AV_{[2,2+h]} is not defined for h = 0.
+
+
+
+ Ratios of polynomial functions arise in several different important circumstances. Sometimes we are interested in what happens when the denominator approaches 0, which makes the overall ratio undefined. In other situations, we may want to know what happens in the long term and thus consider what happens when the input variable increases without bound.
+
+ The functions AV_{[2,2+h]} = \frac{-32h - 16h^2}{h} and A(q) = \frac{5000000 + 2500q}{q} are both examples of rational functions, since each is a ratio of polynomial functions. Formally, we have the following definition.
+
+
+
+ rational function
+
+
+ A function r is rational provided that it is possible to write r as the ratio of two polynomials, p and q. That is, r is rational provided that for some polynomial functions p and q, we have
+
+ r(x) = \frac{p(x)}{q(x)}
+ .
+
+
+
+
+
+ Like with polynomial functions, we are interested in such natural questions as
+
+
+
+ What is the long range behavior of a given rational function?
+
+
+
+
+ What is the domain of a given rational function?
+
+
+
+
+ How can we determine where a given rational function's value is 0?
+
+
+
+
+
+
+ We begin by focusing on the long-range behavior of rational functions. It's important first to recall our earlier work with power functions of the form p(x) = x^{-n} where n = 1, 2, \ldots. For such functions, we know that p(x) = \frac{1}{x^n} where n \gt 0 and that
+
+ \lim_{x \to \infty} \frac{1}{x^n} = 0
+
+ since x^n increases without bound as x \to \infty. The same is true when x \to -\infty: \lim_{x \to -\infty} \frac{1}{x^n} = 0. Thus, any time we encounter a quantity such as \frac{1}{x^3}, this quantity will approach 0 as x increases without bound, and this will also occur for any constant numerator. For instance,
+
+ \lim_{x \to \infty} \frac{2500}{x^2} = 0
+
+ since 2500 times a quantity approaching 0 will still approach 0 as x increases.
+
+
+
+
+
+
+
+ We summarize and generalize the results of Activity and Activity as follows.
+
+
+
+ rational functionlong-term behavior
+ The long-term behavior of a rational function
+
+ Let p and q be polynomial functions so that r(x) = \frac{p(x)}{q(x)} is a rational function. Suppose that p has degree n with leading term a_n x^n and q has degree m with leading term b_m x^m for some nonzero constants a_n and b_m. There are three possibilities (n \lt m, n = m, and n \gt m) that result in three different behaviors of r:
+
+
+
+
+
+
+ if n \lt m, then the degree of the numerator is less than the degree of the denominator, and thus
+
+ \lim_{n \to \infty} r(x) = \lim_{n \to \infty} \frac{a_n x^n + \cdots + a_0}{b_m x^m + \cdots + b_0} = 0
+ ,
+ which tells us that y = 0 is a horizontal asymptote of r;
+
+
+
+
+ if n = m, then the degree of the numerator equals the degree of the denominator, and thus
+
+ \lim_{n \to \infty} r(x) = \lim_{n \to \infty} \frac{a_n x^n + \cdots + a_0}{b_n x^n + \cdots + b_0} = \frac{a_n}{b_n}
+ ,
+ which tells us that y = \frac{a_n}{b_n} (the ratio of the coefficients of the highest order terms in p and q) is a horizontal asymptote of r;
+
+
+
+
+ if n \gt m, then the degree of the numerator is greater than the degree of the denominator, and thus
+
+ \lim_{n \to \infty} r(x) = \lim_{n \to \infty} \frac{a_n x^n + \cdots + a_0}{b_m x^m + \cdots + b_0} = \pm \infty
+ ,
+ (where the sign of the limit depends on the signs of a_n and b_m) which tells us that r is does not have a horizontal asymptote.
+
+
+
+
+
+ In both situations (a) and (b), the value of \lim_{x \to -\infty} r(x) is identical to \lim_{x \to \infty} r(x).
+
+
+
+
+
+
+ The domain of a rational function
+
+ Because a rational function can be written in the form r(x) = \frac{p(x)}{q(x)} for some polynomial functions p and q, we have to be concerned about the possibility that a rational function's denominator is zero. Since polynomial functions always have their domain as the set of all real numbers, it follows that any rational function is only undefined at points where its denominator is zero.
+
+
+
+ rational functiondomain
+ The domain of a rational function
+
+ Let p and q be polynomial functions so that r(x) = \frac{p(x)}{q(x)} is a rational function. The domain of r is the set of all real numbers except those for which q(x) = 0.
+
+
+
+
+
+
+ Determine the domain of the function r(x) = \frac{5x^3 + 17x^2 - 9x + 4}{2x^3 - 6x^2 - 8x}.
+
+
+
+
+ To find the domain of any rational function, we need to determine where the denominator is zero. The best way to find these values exactly is to factor the denominator. Thus, we observe that
+
+ 2x^3 - 6x^2 - 8x = 2x(x^2 - 3x - 4) = 2x(x+1)(x-4)
+ .
+ By the Zero Product Property, it follows that the denominator of r is zero at x = 0, x = -1, and x = 4. Hence, the domain of r is the set of all real numbers except -1, 0, and 4.
+
+
+
+
+
+ We note that when it comes to determining the domain of a rational function, the numerator is irrelevant: all that matters is where the denominator is 0.
+
+ Rational functions arise naturally in the study of the average rate of change of a polynomial function, leading to expressions such as
+
+ AV_{[2,2+h]} = \frac{-32h - 16h^2}{h}
+ .
+ We will study several subtle issues that correspond to such functions further in Section. For now, we will focus on a different setting in which rational functions play a key role.
+
+
+
+ In Section, we encountered a class of problems where a key quantity was modeled by a polynomial function. We found that if we considered a container such as a cylinder with fixed surface area, then the volume of the container could be written as a polynomial of a single variable. For instance, if we consider a circular cylinder with surface area 10 square feet, then we know that
+
+ S = 10 = 2\pi r^2 + 2\pi r h
+
+ and therefore h = \frac{10 - 2\pi r^2}{2 \pi r}. Since the cylinder's volume is V = \pi r^2 h, it follows that
+
+ V = \pi r^2 h = \pi r^2 \left( \frac{10 - 2\pi r^2}{2 \pi r} \right) = r(5 - \pi r^2)
+ ,
+ which is a polynomial function of r.
+
+
+
+ What happens if we instead fix the volume of the container and ask about how surface area can be written as a function of a single variable?
+
+
+
+
+
+ Suppose we want to construct a circular cylinder that holds 20 cubic feet of volume. How much material does it take to build the container? How can we state the amount of material as a function of a single variable?
+
+
+
+
+ Neglecting any scrap, the amount of material it takes to construct the container is its surface area, which we know to be
+
+ S = 2\pi r^2 + 2\pi r h
+ .
+ Because we want the volume to be fixed, this results in a constraint equation that enables us to relate r and h. In particular, since
+
+ V = 20 = \pi r^2 h
+ ,
+ it follows that we can solve for h and get h = \frac{20}{\pi r^2}. Substituting this expression for h in the equation for surface area, we find that
+
+ S = 2\pi r^2 + 2\pi r \cdot \frac{20}{\pi r^2} = 2 \pi r^2 + \frac{40}{r}
+ .
+ Getting a common denominator, we can also write S in the form
+
+ S(r) = \frac{2 \pi r^3 + 40}{r}
+
+ and thus we see that S is a rational function of r. Because of the physical context of the problem and the fact that the denominator of S is r, the domain of S is the set of all positive real numbers.
+
+
+
+
+
+
+
+
+
+ Summary
+
+
+
+
+ A rational function is a function whose formula can be written as the ratio of two polynomial functions. For instance, r(x) = \frac{7x^3 - 5x + 16}{-4x^4 + 2x^3 - 11x + 3} is a rational function.
+
+
+
+
+ Two aspects of rational functions are straightforward to determine for any rational function. Given r(x) = \frac{p(x)}{q(x)} where p and q are polynomials, the domain of r is the set of all real numbers except any values of x for which q(x) = 0. In addition, we can determine the long-range behavior of r by examining the highest order terms in p and q:
+
+
+
+
+ if the degree of p is less than the degree of q, then r has a horizontal asymptote at y = 0;
+
+
+
+
+ if the degree of p equals the degree of q, then r has a horizontal asymptote at y = \frac{a_n}{b_n}, where a_n and b_n are the leading coefficients of p and q respectively;
+
+
+
+
+ and if the degree of p is greater than the degree of q, then r does not have a horizontal asymptote.
+
+
+
+
+
+
+
+ Two reasons that rational functions are important are that they arise naturally when we consider the average rate of change on an interval whose length varies and when we consider problems that relate the volume and surface area of three-dimensional containers when one of those two quantities is constrained.
+
- How can we use inverse trigonometric functions to determine missing angles in right triangles?
-
-
-
-
- What situations require us to use technology to evaluate inverse trigonometric functions?
-
-
-
-
-
- Introduction
-
- In our earlier work in Section and Section, we observed that in any right triangle, if we know the measure of one additional angle and the length of one additional side, we can determine all of the other parts of the triangle. With the inverse trigonometric functions that we developed in Section, we are now also able to determine the missing angles in any right triangle where we know the lengths of two sides.
-
-
-
- While the original trigonometric functions take a particular angle as input and provide an output that can be viewed as the ratio of two sides of a right triangle, the inverse trigonometric functions take an input that can be viewed as a ratio of two sides of a right triangle and produce the corresponding angle as output. Indeed, it's imperative to remember that statements such as
-
- \arccos(x) = \theta \ \text{ and } \ \cos(\theta) = x
-
- say the exact same thing from two different perspectives, and that we read \arccos(x) as the angle whose cosine is x.
-
- Like the trigonometric functions themselves, there are a handful of important values of the inverse trigonometric functions that we can determine exactly without the aid of a computer. For instance, we know from the unit circle (Figure) that \arcsin(-\frac{\sqrt{3}}{2}) = -\frac{\pi}{3}, \arccos(-\frac{\sqrt{3}}{2}) = \frac{5\pi}{6}, and \arctan(-\frac{1}{\sqrt{3}}) = -\frac{\pi}{6}. In these evaluations, we have to be careful to remember that the range of the arccosine function is [0,\pi], while the range of the arcsine function is [-\frac{\pi}{2},\frac{\pi}{2}] and the range of the arctangent function is (-\frac{\pi}{2},\frac{\pi}{2}), in order to ensure that we choose the appropriate angle that results from the inverse trigonometric function.
-
-
-
- In addition, there are many other values at which we may wish to know the angle that results from an inverse trigonometric function. To determine such values, we use a computational device (such as Desmos) in order to evaluate the function.
-
-
-
-
-
-
- Consider the right triangle pictured in Figure and assume we know that the vertical leg has length 1 and the hypotenuse has length 3. Let \alpha be the angle opposite the known leg. Determine exact and approximate values for all of the remaining parts of the triangle.
-
-
-
-
A right triangle with one known leg and known hypotenuse.
-
A right triangle with one known leg and known hypotenuse.
-
-
-
-
-
-
-
- Because we know the hypotenuse and the side opposite \alpha, we observe that \sin(\alpha) = \frac{1}{3}. Rewriting this statement using inverse function notation, we have equivalently that \alpha = \arcsin(\frac{1}{3}), which is the exact value of \alpha. Since this is not one of the known special angles on the unit circle, we can find a numerical estimate of \alpha using a computational device. Entering arcsin(1/3) in Desmos, we find that \alpha \approx 0.3398 radians. Note well: whatever device we use, we need to be careful to use degree or radian mode as dictated by the problem we are solving. We will always work in radians unless stated otherwise.
-
-
-
- We can now find the remaining leg's length and the remaining angle's measure. If we let x represent the length of the horizontal leg, by the Pythagorean Theorem we know that
- x^2 + 1^2 = 3^2,
- and thus x^2 = 8 so x = \sqrt{8} \approx 2.8284. Calling the remaining angle \beta, since \alpha + \beta = \frac{\pi}{2}, it follows that
-
- \beta = \frac{\pi}{2} - \arcsin \left(\frac{1}{3}\right) \approx 1.2310
- .
-
- Now that we have developed the (restricted) sine, cosine, and tangent functions and their respective inverses, in any setting in which we have a right triangle together with one side length and any one additional piece of information (another side length or a non-right angle measurement), we can determine all of the remaining pieces of the triangle. In the activities that follow, we explore these possibilities in a variety of different applied contexts.
-
-
-
-
-
-
-
-
-
-
-
- Summary
-
-
-
-
-
-
-
-
- Anytime we know two side lengths in a right triangle, we can use one of the inverse trigonometric functions to determine the measure of one of the non-right angles. For instance, if we know the values of \text{opp} and \text{adj} in Figure, then since
-
- \tan(\theta) = \frac{\text{opp}}{\text{adj}}
- ,
- it follows that \theta = \arctan(\frac{\text{opp}}{\text{adj}}).
-
-
- If we instead know the hypotenuse and one of the two legs, we can use either the arcsine or arccosine function accordingly.
-
-
-
-
Finding an angle from knowing the legs in a right triangle.
-
Finding an angle from knowing the legs in a right triangle.
-
-
-
-
-
-
-
- For situations other than angles or ratios that involve the 16 special points on the unit circle, technology is required in order to evaluate inverse trignometric functions. For instance, from the unit circle we know that \arccos(\frac{1}{2}) = \frac{\pi}{3} (exactly), but if we want to know \arccos(\frac{1}{3}), we have to estimate this value using a computational device such as Desmos. We note that \arccos(\frac{1}{3}) is the exact value of the angle whose cosine is \frac{1}{3}.
-
+ How can we use inverse trigonometric functions to determine missing angles in right triangles?
+
+
+
+
+ What situations require us to use technology to evaluate inverse trigonometric functions?
+
+
+
+
+
+ Introduction
+
+ In our earlier work in Section and Section, we observed that in any right triangle, if we know the measure of one additional angle and the length of one additional side, we can determine all of the other parts of the triangle. With the inverse trigonometric functions that we developed in Section, we are now also able to determine the missing angles in any right triangle where we know the lengths of two sides.
+
+
+
+ While the original trigonometric functions take a particular angle as input and provide an output that can be viewed as the ratio of two sides of a right triangle, the inverse trigonometric functions take an input that can be viewed as a ratio of two sides of a right triangle and produce the corresponding angle as output. Indeed, it's imperative to remember that statements such as
+
+ \arccos(x) = \theta \ \text{ and } \ \cos(\theta) = x
+
+ say the exact same thing from two different perspectives, and that we read \arccos(x) as the angle whose cosine is x.
+
+ Like the trigonometric functions themselves, there are a handful of important values of the inverse trigonometric functions that we can determine exactly without the aid of a computer. For instance, we know from the unit circle (Figure) that \arcsin(-\frac{\sqrt{3}}{2}) = -\frac{\pi}{3}, \arccos(-\frac{\sqrt{3}}{2}) = \frac{5\pi}{6}, and \arctan(-\frac{1}{\sqrt{3}}) = -\frac{\pi}{6}. In these evaluations, we have to be careful to remember that the range of the arccosine function is [0,\pi], while the range of the arcsine function is [-\frac{\pi}{2},\frac{\pi}{2}] and the range of the arctangent function is (-\frac{\pi}{2},\frac{\pi}{2}), in order to ensure that we choose the appropriate angle that results from the inverse trigonometric function.
+
+
+
+ In addition, there are many other values at which we may wish to know the angle that results from an inverse trigonometric function. To determine such values, we use a computational device (such as Desmos) in order to evaluate the function.
+
+
+
+
+
+
+ Consider the right triangle pictured in Figure and assume we know that the vertical leg has length 1 and the hypotenuse has length 3. Let \alpha be the angle opposite the known leg. Determine exact and approximate values for all of the remaining parts of the triangle.
+
+
+
+
A right triangle with one known leg and known hypotenuse.
+
A right triangle with one known leg and known hypotenuse.
+
+
+
+
+
+
+
+ Because we know the hypotenuse and the side opposite \alpha, we observe that \sin(\alpha) = \frac{1}{3}. Rewriting this statement using inverse function notation, we have equivalently that \alpha = \arcsin(\frac{1}{3}), which is the exact value of \alpha. Since this is not one of the known special angles on the unit circle, we can find a numerical estimate of \alpha using a computational device. Entering arcsin(1/3) in Desmos, we find that \alpha \approx 0.3398 radians. Note well: whatever device we use, we need to be careful to use degree or radian mode as dictated by the problem we are solving. We will always work in radians unless stated otherwise.
+
+
+
+ We can now find the remaining leg's length and the remaining angle's measure. If we let x represent the length of the horizontal leg, by the Pythagorean Theorem we know that
+ x^2 + 1^2 = 3^2,
+ and thus x^2 = 8 so x = \sqrt{8} \approx 2.8284. Calling the remaining angle \beta, since \alpha + \beta = \frac{\pi}{2}, it follows that
+
+ \beta = \frac{\pi}{2} - \arcsin \left(\frac{1}{3}\right) \approx 1.2310
+ .
+
+ Now that we have developed the (restricted) sine, cosine, and tangent functions and their respective inverses, in any setting in which we have a right triangle together with one side length and any one additional piece of information (another side length or a non-right angle measurement), we can determine all of the remaining pieces of the triangle. In the activities that follow, we explore these possibilities in a variety of different applied contexts.
+
+
+
+
+
+
+
+
+
+
+
+ Summary
+
+
+
+
+
+
+
+
+ Anytime we know two side lengths in a right triangle, we can use one of the inverse trigonometric functions to determine the measure of one of the non-right angles. For instance, if we know the values of \text{opp} and \text{adj} in Figure, then since
+
+ \tan(\theta) = \frac{\text{opp}}{\text{adj}}
+ ,
+ it follows that \theta = \arctan(\frac{\text{opp}}{\text{adj}}).
+
+
+ If we instead know the hypotenuse and one of the two legs, we can use either the arcsine or arccosine function accordingly.
+
+
+
+
Finding an angle from knowing the legs in a right triangle.
+
Finding an angle from knowing the legs in a right triangle.
+
+
+
+
+
+
+
+ For situations other than angles or ratios that involve the 16 special points on the unit circle, technology is required in order to evaluate inverse trignometric functions. For instance, from the unit circle we know that \arccos(\frac{1}{2}) = \frac{\pi}{3} (exactly), but if we want to know \arccos(\frac{1}{3}), we have to estimate this value using a computational device such as Desmos. We note that \arccos(\frac{1}{3}) is the exact value of the angle whose cosine is \frac{1}{3}.
+
- Is it possible for a periodic function that fails the Horizontal Line Test to have an inverse?
-
-
-
-
- For the restricted cosine, sine, and tangent functions, how do we define the corresponding arccosine, arcsine, and arctangent functions?
-
-
-
-
- What are the key properties of the arccosine, arcsine, and arctangent functions?
-
-
-
-
-
- Introduction
-
- In our prior work with inverse functions, we have seen several important principles, including
-
-
-
- A function f has an inverse function
- if and only if there exists a function g that undoes the work of f. Such
- a function g has the properties that
- g(f(x)) = x for each x in the domain of f,
- and f(g(y)) = y for each y in the range of f.
- We call g the inverse of f,
- and write g = f^{-1}.
-
-
-
-
-
- A function f has an inverse function if and only if the graph of f passes the
- .
-
-
-
-
-
- When f has an inverse,
- we know that writing y = f(t)
- and t = f^{-1}(y)
- say the exact same thing,
- but from two different perspectives.
-
-
-
-
-
-
- The trigonometric functions f(t) = \sin(t),
- g(t) = \cos(t), and h(t) = \tan(t) are periodic, so each fails the horizontal line test,
- and thus these functions on their full domains do not have inverse functions.
- At the same time, it is reasonable to think about changing perspective and viewing angles as outputs in certain restricted settings.
- For instance, we may want to say both
-
- \frac{\sqrt{3}}{2} = \cos\left(\frac{\pi}{6}\right) \ \ \ \mbox{and} \ \ \ \frac{\pi}{6} = \cos^{-1}\left(\frac{\sqrt{3}}{2}\right)
-
- depending on the context in which we are considering the relationship between the angle and side length.
-
-
-
- It's also important to understand why the issue of finding an angle in terms of a known value of a trigonometric function is important.
- Suppose we know the following information about a right triangle:
- one leg has length 2.5, and the hypotenuse has length 4.
- If we let \theta be the angle opposite the side of length 2.5, it follows that \sin(\theta) = \frac{2.5}{4}.
- We naturally want to use the inverse of the sine function to solve the most recent equation for \theta. But the sine function does not have an inverse function, so how can we address this situation?
-
-
-
- While the original trigonometric functions f(t) = \sin(t),
- g(t) = \cos(t), and
- h(t) = \tan(t) do not have inverse functions,
- it turns out that we can consider restricted versions of them that do
- have corresponding inverse functions. We thus investigate how we can think differently about the
- trigonometric functions so that we can discuss inverses in a meaningful way.
-
-
-
-
-
-
-
- The arccosine function
-
- For the cosine function restricted to the domain [0,\pi] that we considered in Preview Activity, the function is strictly decreasing on its domain and thus passes the Horizontal Line Test. Therefore, this restricted version of the cosine function has an inverse function; we will call this inverse function the arccosine function.
-
- Let y = g(t) = \cos(t) be defined on the domain [0,\pi], and observe g : [0,\pi] \to [-1,1]. For any real number y that satisfies -1 \le y \le 1, the arccosine of y, denoted
-
- \arccos(y)
-
- is the angle t satisfying 0 \le t \le \pi such that \cos(t) = y.
-
-
-
-
-
- Note particularly that the output of the arccosine function is an angle. In addition, recall that in the context of the unit circle, an angle measured in radians and the corresponding arc length along the unit circle are numerically equal. This is why we use the arc in arccosine: given a value -1 \le y \le 1, the arccosine function produces the corresponding arc (measured counterclockwise from (1,0)) such that the cosine of that arc is y.
-
-
-
- We recall that for any function with an inverse function, the inverse function reverses the process of the original function. We know that y = \cos(t) can be read as saying y is the cosine of the angle t. Changing perspective and writing the equivalent statement t = \arccos(y), we read this statement as t is the angle whose cosine is y. Just as y = f(t) and t = f^{-1}(y) say the same thing for a function and its inverse in general,
-
- y = \cos(t) \ \text{ and } \ t = \arccos(y)
-
- say the same thing for any angle t that satisfies 0 \le t \le \pi. We also use the equivalent notation t = \cos^{-1}(y) interchangeably with t = \arccos(y). We read t = \cos^{-1}(y) as t is the angle whose cosine is y or t is the inverse cosine of y. Key properties of the arccosine function can be summarized as follows.
-
-
-
- Properties of the arccosine function
-
-
-
-
-
- The restricted cosine function, y = g(t) = \cos(t), is defined on the domain [0,\pi] with range [-1,1]. This function has an inverse function that we call the arccosine function, denoted t = g^{-1}(y) = \arccos(y).
-
-
-
-
- The domain of y = g^{-1}(t) = \arccos(t) is [-1,1] with range [0,\pi].
-
-
-
-
- The arccosine function is always decreasing on its domain.
-
-
-
-
- At right, a plot of the restricted cosine function (in light blue) and its corresponding inverse, the arccosine function (in dark blue).
-
-
-
-
-
-
ADD ALT TEXT TO THIS IMAGE
-
-
-
-
-
- Just as the natural logarithm function allowed us to rewrite exponential equations in an equivalent way (for instance, y = e^t and t = \ln(y) say the exact same thing), the arccosine function allows us to do likewise for certain angles and cosine outputs. For instance, saying \cos(\frac{\pi}{2}) = 0 is the same as writing \frac{\pi}{2} = \arccos(0), which reads \frac{\pi}{2} is the angle whose cosine is 0. Indeed, these relationships are reflected in the plot above, where we see that any point (a,b) that lies on the graph of y = \cos(t) corresponds to the point (b,a) that lies on the graph of y = \arccos(t).
-
-
-
-
-
-
-
- The arcsine function
-
- We can develop an inverse function for a restricted version of the sine function in a similar way. As with the cosine function, we need to choose an interval on which the sine function is always increasing or always decreasing in order to have the function pass the horizontal line test. The standard choice is the domain [-\frac{\pi}{2}, \frac{\pi}{2}] on which f(t) = \sin(t) is increasing and attains all of the values in the range of the sine function. Thus, we consider f(t) = \sin(t) so that f : [-\frac{\pi}{2}, \frac{\pi}{2}] \to [-1,1] and hence define the corresponding arcsine function.
-
-
-
- inverse trigonometric functionsarcsine
-
-
- Let y = f(t) = \sin(t) be defined on the domain [-\frac{\pi}{2},\frac{\pi}{2}], and observe f : [-\frac{\pi}{2},\frac{\pi}{2}] \to [-1,1]. For any real number y that satisfies -1 \le y \le 1, the arcsine of y, denoted
-
- \arcsin(y)
-
- is the angle t satisfying -\frac{\pi}{2} \le t \le \frac{\pi}{2} such that \sin(t) = y.
-
-
-
-
-
-
-
-
-
- The arctangent function
-
-
- Finally, we develop an inverse function for a restricted version of the tangent function. We choose the domain (-\frac{\pi}{2}, \frac{\pi}{2}) on which h(t) = \tan(t) is increasing and attains all of the values in the range of the tangent function.
-
- Let y = h(t) = \tan(t) be defined on the domain (-\frac{\pi}{2},\frac{\pi}{2}), and observe h : (-\frac{\pi}{2},\frac{\pi}{2}) \to (-\infty,\infty). For any real number y, the arctangent of y, denoted
-
- \arctan(y)
-
- is the angle t satisfying -\frac{\pi}{2} \lt t \lt \frac{\pi}{2} such that \tan(t) = y.
-
-
-
-
-
-
-
-
-
- Summary
-
-
-
-
- Any function that fails the Horizontal Line Test cannot have an inverse function. However, for a periodic function that fails the horizontal line test, if we restrict the domain of the function to an interval with no repeated outputs, we then determine a related function that does, in fact, have an inverse function. By choosing such an interval carefully, it is possible for us to develop the inverse functions of the restricted cosine, sine, and tangent functions.
-
-
-
-
- We choose to define the restricted cosine, sine, and tangent functions on the respective domains [0,\pi], [-\frac{\pi}{2}, \frac{\pi}{2}], and (-\frac{\pi}{2}, \frac{\pi}{2}). On each such interval, the restricted function is strictly decreasing (cosine) or strictly increasing (sine and tangent), and thus has an inverse function. The restricted sine and cosine functions each have range [-1,1], while the restricted tangent's range is the set of all real numbers. We thus define the inverse function of each as follows:
-
-
-
-
-
-
- For any y such that -1 \le y \le 1, the arccosine of y (denoted \arccos(y)) is the angle t in the interval [0,\pi] such that \cos(t) = y. That is, t is the angle whose cosine is y.
-
-
-
-
- For any y such that -1 \le y \le 1, the arcsine of y (denoted \arcsin(y)) is the angle t in the interval [-\frac{\pi}{2}, \frac{\pi}{2}] such that \sin(t) = y. That is, t is the angle whose sine is y.
-
-
-
-
- For any real number y, the arctangent of y (denoted \arctan(y)) is the angle t in the interval (-\frac{\pi}{2}, \frac{\pi}{2}) such that \tan(t) = y. That is, t is the angle whose tangent is y.
-
-
-
-
-
-
-
- To discuss the properties of the three inverse trigonometric functions, we plot them on the same axes as their corresponding restricted trigonometric functions. When we do so, we use t as the input variable for both functions simultaneously so that we can plot them on the same coordinate axes.
-
-
-
- The domain of y = g^{-1}(t) = \arccos(t) is [-1,1] with corresponding range [0,\pi], and the arccosine function is always decreasing. These facts correspond to the domain and range of the restricted cosine function and the fact that the restricted cosine function is decreasing on [0,\pi].
-
-
-
-
-
-
The restricted cosine function (in light blue) and its inverse, y = g^{-1}(t) = \arccos(t) (in dark blue).
-
The restricted cosine function (in light blue) and its inverse, y = g^{-1}(t) = \arccos(t) (in dark blue).
-
-
-
-
-
The restricted sine function (in light blue) and its inverse, y = f^{-1}(t) = \arcsin(t) (in dark blue).
-
The restricted sine function (in light blue) and its inverse, y = f^{-1}(t) = \arcsin(t) (in dark blue).
-
-
-
-
-
-
- The domain of y = f^{-1}(t) = \arcsin(t) is [-1,1] with corresponding range [-\frac{\pi}{2}, \frac{\pi}{2}], and the arcsine function is always increasing. These facts correspond to the domain and range of the restricted sine function and the fact that the restricted sine function is increasing on [-\frac{\pi}{2},\frac{\pi}{2}].
-
-
-
- The domain of y = f^{-1}(t) = \arctan(t) is the set of all real numbers with corresponding range (-\frac{\pi}{2}, \frac{\pi}{2}), and the arctangent function is always increasing. These facts correspond to the domain and range of the restricted tangent function and the fact that the restricted tangent function is increasing on (-\frac{\pi}{2},\frac{\pi}{2}).
-
-
-
-
The restricted tangent function (in light blue) and its inverse, y = h^{-1}(t) = \arctan(t) (in dark blue).
-
The restricted tangent function (in light blue) and its inverse, y = h^{-1}(t) = \arctan(t) (in dark blue).
+ Is it possible for a periodic function that fails the Horizontal Line Test to have an inverse?
+
+
+
+
+ For the restricted cosine, sine, and tangent functions, how do we define the corresponding arccosine, arcsine, and arctangent functions?
+
+
+
+
+ What are the key properties of the arccosine, arcsine, and arctangent functions?
+
+
+
+
+
+ Introduction
+
+ In our prior work with inverse functions, we have seen several important principles, including
+
+
+
+ A function f has an inverse function
+ if and only if there exists a function g that undoes the work of f. Such
+ a function g has the properties that
+ g(f(x)) = x for each x in the domain of f,
+ and f(g(y)) = y for each y in the range of f.
+ We call g the inverse of f,
+ and write g = f^{-1}.
+
+
+
+
+
+ A function f has an inverse function if and only if the graph of f passes the
+ .
+
+
+
+
+
+ When f has an inverse,
+ we know that writing y = f(t)
+ and t = f^{-1}(y)
+ say the exact same thing,
+ but from two different perspectives.
+
+
+
+
+
+
+ The trigonometric functions f(t) = \sin(t),
+ g(t) = \cos(t), and h(t) = \tan(t) are periodic, so each fails the horizontal line test,
+ and thus these functions on their full domains do not have inverse functions.
+ At the same time, it is reasonable to think about changing perspective and viewing angles as outputs in certain restricted settings.
+ For instance, we may want to say both
+
+ \frac{\sqrt{3}}{2} = \cos\left(\frac{\pi}{6}\right) \ \ \ \mbox{and} \ \ \ \frac{\pi}{6} = \cos^{-1}\left(\frac{\sqrt{3}}{2}\right)
+
+ depending on the context in which we are considering the relationship between the angle and side length.
+
+
+
+ It's also important to understand why the issue of finding an angle in terms of a known value of a trigonometric function is important.
+ Suppose we know the following information about a right triangle:
+ one leg has length 2.5, and the hypotenuse has length 4.
+ If we let \theta be the angle opposite the side of length 2.5, it follows that \sin(\theta) = \frac{2.5}{4}.
+ We naturally want to use the inverse of the sine function to solve the most recent equation for \theta. But the sine function does not have an inverse function, so how can we address this situation?
+
+
+
+ While the original trigonometric functions f(t) = \sin(t),
+ g(t) = \cos(t), and
+ h(t) = \tan(t) do not have inverse functions,
+ it turns out that we can consider restricted versions of them that do
+ have corresponding inverse functions. We thus investigate how we can think differently about the
+ trigonometric functions so that we can discuss inverses in a meaningful way.
+
+
+
+
+
+
+
+
+ The arccosine function
+
+ For the cosine function restricted to the domain [0,\pi] that we considered in Preview Activity, the function is strictly decreasing on its domain and thus passes the Horizontal Line Test. Therefore, this restricted version of the cosine function has an inverse function; we will call this inverse function the arccosine function.
+
+ Let y = g(t) = \cos(t) be defined on the domain [0,\pi], and observe g : [0,\pi] \to [-1,1]. For any real number y that satisfies -1 \le y \le 1, the arccosine of y, denoted
+
+ \arccos(y)
+
+ is the angle t satisfying 0 \le t \le \pi such that \cos(t) = y.
+
+
+
+
+
+ Note particularly that the output of the arccosine function is an angle. In addition, recall that in the context of the unit circle, an angle measured in radians and the corresponding arc length along the unit circle are numerically equal. This is why we use the arc in arccosine: given a value -1 \le y \le 1, the arccosine function produces the corresponding arc (measured counterclockwise from (1,0)) such that the cosine of that arc is y.
+
+
+
+ We recall that for any function with an inverse function, the inverse function reverses the process of the original function. We know that y = \cos(t) can be read as saying y is the cosine of the angle t. Changing perspective and writing the equivalent statement t = \arccos(y), we read this statement as t is the angle whose cosine is y. Just as y = f(t) and t = f^{-1}(y) say the same thing for a function and its inverse in general,
+
+ y = \cos(t) \ \text{ and } \ t = \arccos(y)
+
+ say the same thing for any angle t that satisfies 0 \le t \le \pi. We also use the equivalent notation t = \cos^{-1}(y) interchangeably with t = \arccos(y). We read t = \cos^{-1}(y) as t is the angle whose cosine is y or t is the inverse cosine of y. Key properties of the arccosine function can be summarized as follows.
+
+
+
+ Properties of the arccosine function
+
+
+
+
+
+ The restricted cosine function, y = g(t) = \cos(t), is defined on the domain [0,\pi] with range [-1,1]. This function has an inverse function that we call the arccosine function, denoted t = g^{-1}(y) = \arccos(y).
+
+
+
+
+ The domain of y = g^{-1}(t) = \arccos(t) is [-1,1] with range [0,\pi].
+
+
+
+
+ The arccosine function is always decreasing on its domain.
+
+
+
+
+ At right, a plot of the restricted cosine function (in light blue) and its corresponding inverse, the arccosine function (in dark blue).
+
+
+
+
+
+
ADD ALT TEXT TO THIS IMAGE
+
+
+
+
+
+ Just as the natural logarithm function allowed us to rewrite exponential equations in an equivalent way (for instance, y = e^t and t = \ln(y) say the exact same thing), the arccosine function allows us to do likewise for certain angles and cosine outputs. For instance, saying \cos(\frac{\pi}{2}) = 0 is the same as writing \frac{\pi}{2} = \arccos(0), which reads \frac{\pi}{2} is the angle whose cosine is 0. Indeed, these relationships are reflected in the plot above, where we see that any point (a,b) that lies on the graph of y = \cos(t) corresponds to the point (b,a) that lies on the graph of y = \arccos(t).
+
+
+
+
+
+
+
+ The arcsine function
+
+ We can develop an inverse function for a restricted version of the sine function in a similar way. As with the cosine function, we need to choose an interval on which the sine function is always increasing or always decreasing in order to have the function pass the horizontal line test. The standard choice is the domain [-\frac{\pi}{2}, \frac{\pi}{2}] on which f(t) = \sin(t) is increasing and attains all of the values in the range of the sine function. Thus, we consider f(t) = \sin(t) so that f : [-\frac{\pi}{2}, \frac{\pi}{2}] \to [-1,1] and hence define the corresponding arcsine function.
+
+
+
+ inverse trigonometric functionsarcsine
+
+
+ Let y = f(t) = \sin(t) be defined on the domain [-\frac{\pi}{2},\frac{\pi}{2}], and observe f : [-\frac{\pi}{2},\frac{\pi}{2}] \to [-1,1]. For any real number y that satisfies -1 \le y \le 1, the arcsine of y, denoted
+
+ \arcsin(y)
+
+ is the angle t satisfying -\frac{\pi}{2} \le t \le \frac{\pi}{2} such that \sin(t) = y.
+
+
+
+
+
+
+
+
+
+ The arctangent function
+
+
+ Finally, we develop an inverse function for a restricted version of the tangent function. We choose the domain (-\frac{\pi}{2}, \frac{\pi}{2}) on which h(t) = \tan(t) is increasing and attains all of the values in the range of the tangent function.
+
+ Let y = h(t) = \tan(t) be defined on the domain (-\frac{\pi}{2},\frac{\pi}{2}), and observe h : (-\frac{\pi}{2},\frac{\pi}{2}) \to (-\infty,\infty). For any real number y, the arctangent of y, denoted
+
+ \arctan(y)
+
+ is the angle t satisfying -\frac{\pi}{2} \lt t \lt \frac{\pi}{2} such that \tan(t) = y.
+
+
+
+
+
+
+
+
+
+ Summary
+
+
+
+
+ Any function that fails the Horizontal Line Test cannot have an inverse function. However, for a periodic function that fails the horizontal line test, if we restrict the domain of the function to an interval with no repeated outputs, we then determine a related function that does, in fact, have an inverse function. By choosing such an interval carefully, it is possible for us to develop the inverse functions of the restricted cosine, sine, and tangent functions.
+
+
+
+
+ We choose to define the restricted cosine, sine, and tangent functions on the respective domains [0,\pi], [-\frac{\pi}{2}, \frac{\pi}{2}], and (-\frac{\pi}{2}, \frac{\pi}{2}). On each such interval, the restricted function is strictly decreasing (cosine) or strictly increasing (sine and tangent), and thus has an inverse function. The restricted sine and cosine functions each have range [-1,1], while the restricted tangent's range is the set of all real numbers. We thus define the inverse function of each as follows:
+
+
+
+
+
+
+ For any y such that -1 \le y \le 1, the arccosine of y (denoted \arccos(y)) is the angle t in the interval [0,\pi] such that \cos(t) = y. That is, t is the angle whose cosine is y.
+
+
+
+
+ For any y such that -1 \le y \le 1, the arcsine of y (denoted \arcsin(y)) is the angle t in the interval [-\frac{\pi}{2}, \frac{\pi}{2}] such that \sin(t) = y. That is, t is the angle whose sine is y.
+
+
+
+
+ For any real number y, the arctangent of y (denoted \arctan(y)) is the angle t in the interval (-\frac{\pi}{2}, \frac{\pi}{2}) such that \tan(t) = y. That is, t is the angle whose tangent is y.
+
+
+
+
+
+
+
+ To discuss the properties of the three inverse trigonometric functions, we plot them on the same axes as their corresponding restricted trigonometric functions. When we do so, we use t as the input variable for both functions simultaneously so that we can plot them on the same coordinate axes.
+
+
+
+ The domain of y = g^{-1}(t) = \arccos(t) is [-1,1] with corresponding range [0,\pi], and the arccosine function is always decreasing. These facts correspond to the domain and range of the restricted cosine function and the fact that the restricted cosine function is decreasing on [0,\pi].
+
+
+
+
+
+
The restricted cosine function (in light blue) and its inverse, y = g^{-1}(t) = \arccos(t) (in dark blue).
+
The restricted cosine function (in light blue) and its inverse, y = g^{-1}(t) = \arccos(t) (in dark blue).
+
+
+
+
+
The restricted sine function (in light blue) and its inverse, y = f^{-1}(t) = \arcsin(t) (in dark blue).
+
The restricted sine function (in light blue) and its inverse, y = f^{-1}(t) = \arcsin(t) (in dark blue).
+
+
+
+
+
+
+ The domain of y = f^{-1}(t) = \arcsin(t) is [-1,1] with corresponding range [-\frac{\pi}{2}, \frac{\pi}{2}], and the arcsine function is always increasing. These facts correspond to the domain and range of the restricted sine function and the fact that the restricted sine function is increasing on [-\frac{\pi}{2},\frac{\pi}{2}].
+
+
+
+ The domain of y = f^{-1}(t) = \arctan(t) is the set of all real numbers with corresponding range (-\frac{\pi}{2}, \frac{\pi}{2}), and the arctangent function is always increasing. These facts correspond to the domain and range of the restricted tangent function and the fact that the restricted tangent function is increasing on (-\frac{\pi}{2},\frac{\pi}{2}).
+
+
+
+
The restricted tangent function (in light blue) and its inverse, y = h^{-1}(t) = \arctan(t) (in dark blue).
+
The restricted tangent function (in light blue) and its inverse, y = h^{-1}(t) = \arctan(t) (in dark blue).
- What are the other 3 trigonometric functions and how are they related to the cosine, sine, and tangent functions?
-
-
-
-
- How do the graphs of the secant, cosecant, and cotangent functions behave and how do these graphs compare to the cosine, sine, and tangent functions' graphs?
-
-
-
-
- What is a trigonometric identity and why are identities important?
-
-
-
-
-
- Introduction
-
- The sine and cosine functions, originally defined in the context of a point traversing the unit circle, are also central in right triangle trigonometry. They enable us to find missing information in right triangles in a straightforward way when we know one of the non-right angles and one of the three sides of the triangle, or two of the sides where one is the hypotenuse. In addition, we defined the tangent function in terms of the sine and cosine functions, and the tangent function offers additional options for finding missing information in right triangles. We've also seen how the inverses of the restricted sine, cosine, and tangent functions enable us to find missing angles in a wide variety of settings involving right triangles.
-
-
-
- One of the powerful aspects of trigonometry is that the subject offers us the opportunity to view the same idea from many different perspectives. As one example, we have observed that the functions f(t) = \cos(t) and g(t) = \sin(t+\frac{\pi}{2}) are actually the same function; as another, for t values in the domain (-\frac{\pi}{2}, \frac{\pi}{2}), we know that writing y = \tan(t) is the same as writing t = \arctan(y). Which perspective we choose to take often depends on context and given information.
-
-
-
- While almost every question involving trigonometry can be answered using the sine, cosine, and tangent functions, sometimes it is convenient to use three related functions that are connected to the other three possible arrangements of ratios of sides in right triangles.
-
-
-
- secant functioncosecant functioncotangent function
- The secant, cosecant, and cotangent functions
-
-
-
-
-
- For any real number t for which \cos(t) \ne 0, we define the secant of t, denoted \sec(t), by the rule
-
- \sec(t) = \frac{1}{\cos(t)}
- .
-
-
-
-
- For any real number t for which \sin(t) \ne 0, we define the cosecant of t, denoted \csc(t), by the rule
-
- \csc(t) = \frac{1}{\sin(t)}
- .
-
-
-
-
- For any real number t for which \sin(t) \ne 0, we define the cotangent of t, denoted \cot(t), by the rule
-
- \cot(t) = \frac{\cos(t)}{\sin(t)}
- .
-
-
-
-
-
-
-
-
- Note particularly that like the tangent function, the secant, cosecant, and cotangent are also defined completely in terms of the sine and cosine functions. In the context of a right triangle with an angle \theta, we know how to think of \sin(\theta), \cos(\theta), and \tan(\theta) as ratios of sides of the triangle. We can now do likewise with the other trigonometric functions:
-
- With these three additional trigonometric functions, we now have expressions that address all six possible combinations of two sides of a right triangle in a ratio.
-
-
-
-
-
-
-
- Ratios in right triangles
-
- Because the sine and cosine functions are used to define each of the other four trigonometric functions, it follows that we can translate information known about the other functions back to information about the sine and cosine functions. For example, if we know that in a certain triangle \csc(\alpha) = \frac{5}{3}, it follows that \sin(\alpha) = \frac{3}{5}. From there we can reason in the usual way to determine missing information in the given triangle.
-
-
-
- It's also often possible to view given information in the context of the unit circle. With the earlier given information that \csc(\alpha) = \frac{5}{3}, it's natural to view \alpha as being the angle in a right triangle that lies opposite a leg of length 3 with the hypotenuse being 5, since \csc(\alpha) = \frac{\text{hyp}}{\text{opp}}. The Pythagorean Theorem then tells us the leg adjacent to \alpha has length 4, as seen in \triangle OPQ in Figure.
-
-
-
-
A 3-4-5 right triangle.
-
A 3-4-5 right triangle.
-
-
-
-
- But we could also view \sin(\alpha) = \frac{3}{5} as \sin(\alpha) = \frac{\frac{3}{5}}{1}, and thus think of the right triangle has having hypotenuse 1 and vertical leg \frac{3}{5}. This triangle is similar to the originally considered 3-4-5 right triangle, but can be viewed as lying within the unit circle. The perspective of the unit circle is particularly valuable when ratios such as \frac{\sqrt{3}}{2}, \frac{\sqrt{2}}{2}, and \frac{1}{2} arise in right triangles.
-
-
-
-
-
-
-
- Properties of the secant, cosecant, and cotangent functions
-
- Like the tangent function, the secant, cosecant, and cotangent functions are defined in terms of the sine and cosine functions, so we can determine the exact values of these functions at each of the special points on the unit circle. In addition, we can use our understanding of the unit circle and the properties of the sine and cosine functions to determine key properties of these other trigonometric functions. We begin by investigating the secant function.
-
-
-
- Using the fact that \sec(t) = \frac{1}{\cos(t)}, we note that anywhere \cos(t) = 0, the value of \sec(t) is undefined. We denote such instances in the following table by u. At all other points, the value of the secant function is simply the reciprocal of the cosine function's value. Since |\cos(t)| \le 1 for all t, it follows that |\sec(t)| \ge 1 for all t (for which the secant's value is defined).
-
-
-
- Values of the cosine and secant functions at special points on the unit circle (Quadrants I and II).
-
-
- t
- 0
- \frac{\pi}{6}
- \frac{\pi}{4}
- \frac{\pi}{3}
- \frac{\pi}{2}
- \frac{2\pi}{3}
- \frac{3\pi}{4}
- \frac{5\pi}{6}
- \pi
-
-
- \cos(t)
- 1
- \frac{\sqrt{3}}{2}
- \frac{\sqrt{2}}{2}
- \frac{1}{2}
- 0
- -\frac{1}{2}
- -\frac{\sqrt{2}}{2}
- -\frac{\sqrt{3}}{2}
- -1
-
-
- \sec(t)
- 1
- \frac{2}{\sqrt{3}}
- \sqrt{2}
- 2
- u
- -2
- -\sqrt{2}
- -\frac{2}{\sqrt{3}}
- -1
-
-
-
-
-
- Values of the cosine and secant functions at special points on the unit circle (Quadrants III and IV).
-
-
- t
- \frac{7\pi}{6}
- \frac{5\pi}{4}
- \frac{4\pi}{3}
- \frac{3\pi}{2}
- \frac{5\pi}{3}
- \frac{7\pi}{4}
- \frac{11\pi}{6}
- 2\pi
-
-
- \cos(t)
- -\frac{\sqrt{3}}{2}
- -\frac{\sqrt{2}}{2}
- -\frac{1}{2}
- 0
- \frac{1}{2}
- \frac{\sqrt{2}}{2}
- \frac{\sqrt{3}}{2}
- 0
-
-
- \sec(t)
- -\frac{2}{\sqrt{3}}
- -\sqrt{2}
- -2
- u
- 2
- \sqrt{2}
- \frac{2}{\sqrt{3}}
- 1
-
-
-
-
-
- Table and Table help us identify trends in the secant function. The sign of \sec(t) matches the sign of \cos(t) and thus is positive in Quadrant I, negative in Quadrant II, negative in Quadrant III, and positive in Quadrant IV.
-
-
-
- In addition, we observe that as t-values in the first quadrant get closer to \frac{\pi}{2}, \cos(t) gets closer to 0 (while being always positive). Since the numerator of the secant function is always 1, having its denominator approach 0 (while the denominator remains positive) means that \sec(t) increases without bound as t approaches \frac{\pi}{2} from the left side. Once t is slightly greater than \frac{\pi}{2} in Quadrant II, the value of \cos(t) is negative (and close to zero). This makes the value of \sec(t) decrease without bound (negative and getting further away from 0) for t approaching \frac{\pi}{2} from the right side. We therefore see that p(t) = \sec(t) has a vertical asymptote at t = \frac{\pi}{2}; the periodicity and sign behavior of \cos(t) mean this asymptotic behavior of the secant function will repeat.
-
-
-
- Plotting the data in the table along with the expected asymptotes and connecting the points intuitively, we see the graph of the secant function in Figure.
-
-
-
-
A plot of the secant function with special points that come from the unit circle, plus the cosine function (dotted, in light blue).
-
A plot of the secant function with special points that come from the unit circle, plus the cosine function (dotted, in light blue).
-
-
-
-
- We see from both the table and the graph that the secant function has period P = 2\pi. We summarize our recent work as follows.
-
-
-
- Properties of the secant function
-
- For the function p(t) = \sec(t),
-
-
-
- its domain is the set of all real numbers except t = \frac{\pi}{2} \pm k\pi where k is any whole number;
-
-
-
-
- its range is the set of all real numbers y such that |y| \ge 1;
-
-
-
-
- its period is P = 2\pi.
-
-
-
-
-
-
-
-
-
-
-
-
-
- A few important identities
-
- An identity is an equation that is true for all possible values of x for which the involved quantities are defined. An example of a non-trigonometric identity is
-
- (x+1)^2 = x^2 + 2x + 1
- ,
- since this equation is true for every value of x,
- and the left and right sides of the equation are simply two different-looking but entirely equivalent expressions.
-
-
-
- Trigonometric identities are simply identities that involve trigonometric functions. While there are a large number of such identities one can study, we choose to focus on those that turn out to be most useful in the study of calculus. The most important trigonometric identity is the fundamental trigonometric identity,
- which is a trigonometric restatement of the Pythagorean Theorem.
-
-
-
- fundamental trigonometric identity
- The fundamental trigonometric identity
-
- For any real number \theta,
-
- \cos^2(\theta) + \sin^2(\theta) = 1
- .
-
-
-
-
-
- Identities are important because they enable us to view the same idea from multiple perspectives. For example, the fundamental trigonometric identity allows us to think of \cos^2(\theta) + \sin^2(\theta) as simply 1, or alternatively, to view \cos^2(\theta) as the same quantity as 1 - \sin^2(\theta).
-
-
-
- There are two related Pythagorean identities that involve the tangent, secant,
- cotangent, and cosecant functions, which we can derive from the fundamental trigonometric identity by dividing both sides by either \cos^2(\theta) or \sin^2(\theta). If we divide both sides of Equation by \cos^2(\theta) (and assume that \cos(\theta) \ne 0), we see that
-
- 1 + \frac{\sin^2(\theta)}{\cos^2(\theta)} = \frac{1}{\cos^2(\theta)}
- , or equivalently,
-
- 1 + \tan^2(\theta) = \sec^2(\theta)
- .
- A similar argument dividing by \sin^2(\theta) (while assuming \sin(\theta) \ne 0) shows that
-
- \cot^2(\theta) + 1 = \csc^2(\theta)
- . These identities prove useful in calculus when we develop the formulas for the derivatives of the tangent and cotangent functions.
-
-
-
- In calculus, it is also beneficial to know a couple of other standard identities for sums of angles or double angles. sum of two angles identitydouble angle identity We simply state these identities without justification. For more information about them, see Section 10.4 in College Trigonometry, by Stitz and ZeagerMore information on Stitz and Zeager's free texts can be found at ..
-
-
-
-
-
-
- For all real numbers \alpha and \beta, \cos(\alpha + \beta) = \cos(\alpha) \cos(\beta) - \sin(\alpha) \sin(\beta).
-
-
-
-
- For all real numbers \alpha and \beta, \sin(\alpha + \beta) = \sin(\alpha) \cos(\beta) + \cos(\alpha) \sin(\beta).
-
-
-
-
- For any real number \theta, \cos(2\theta) = \cos^2(\theta) - \sin^2(\theta).
-
-
-
-
- For any real number \theta, \sin(2\theta) = 2\sin(\theta)\cos(\theta).
-
-
-
-
-
-
-
-
-
-
- Summary
-
-
-
-
- The secant, cosecant, and cotangent functions are respectively defined as the reciprocals of the cosine, sine, and tangent functions. That is,
-
- \sec(t) = \frac{1}{\cos(t)}, \csc(t) = \frac{1}{\sin(t)}, \text{ and } \cot(t) = \frac{1}{\tan(t)}
- .
-
-
-
-
- The graph of the cotangent function is similar to the graph of the tangent function, except that it is decreasing on every interval on which it is defined and has vertical asymptotes wherever \tan(t) = 0 and is zero wherever \tan(t) has a vertical asymptote.
-
-
-
- The graphs of the secant and cosecant functions are different from the cosine and sine functions' graphs in several ways, including that their range is the set of all real numbers y such that |y| \ge 1 and they have vertical asymptotes wherever the cosine and sine function, respectively, are zero.
-
-
-
-
- A trigonometric identity is an equation involving trigonometric functions that is true for every value of the variable for which the trigonometric functions are defined. For instance, \tan^2(t) + 1 = \sec^2(t) for every real number t except t = \frac{\pi}{2} \pm k\pi. Identities offer us alternate perspectives on the same function. For instance, the function f(t) = \sec^2(t) - \tan^2(t) is the same (at all points where f is defined) as the function whose value is always 1.
-
+ What are the other 3 trigonometric functions and how are they related to the cosine, sine, and tangent functions?
+
+
+
+
+ How do the graphs of the secant, cosecant, and cotangent functions behave and how do these graphs compare to the cosine, sine, and tangent functions' graphs?
+
+
+
+
+ What is a trigonometric identity and why are identities important?
+
+
+
+
+
+ Introduction
+
+ The sine and cosine functions, originally defined in the context of a point traversing the unit circle, are also central in right triangle trigonometry. They enable us to find missing information in right triangles in a straightforward way when we know one of the non-right angles and one of the three sides of the triangle, or two of the sides where one is the hypotenuse. In addition, we defined the tangent function in terms of the sine and cosine functions, and the tangent function offers additional options for finding missing information in right triangles. We've also seen how the inverses of the restricted sine, cosine, and tangent functions enable us to find missing angles in a wide variety of settings involving right triangles.
+
+
+
+ One of the powerful aspects of trigonometry is that the subject offers us the opportunity to view the same idea from many different perspectives. As one example, we have observed that the functions f(t) = \cos(t) and g(t) = \sin(t+\frac{\pi}{2}) are actually the same function; as another, for t values in the domain (-\frac{\pi}{2}, \frac{\pi}{2}), we know that writing y = \tan(t) is the same as writing t = \arctan(y). Which perspective we choose to take often depends on context and given information.
+
+
+
+ While almost every question involving trigonometry can be answered using the sine, cosine, and tangent functions, sometimes it is convenient to use three related functions that are connected to the other three possible arrangements of ratios of sides in right triangles.
+
+
+
+ secant functioncosecant functioncotangent function
+ The secant, cosecant, and cotangent functions
+
+
+
+
+
+ For any real number t for which \cos(t) \ne 0, we define the secant of t, denoted \sec(t), by the rule
+
+ \sec(t) = \frac{1}{\cos(t)}
+ .
+
+
+
+
+ For any real number t for which \sin(t) \ne 0, we define the cosecant of t, denoted \csc(t), by the rule
+
+ \csc(t) = \frac{1}{\sin(t)}
+ .
+
+
+
+
+ For any real number t for which \sin(t) \ne 0, we define the cotangent of t, denoted \cot(t), by the rule
+
+ \cot(t) = \frac{\cos(t)}{\sin(t)}
+ .
+
+
+
+
+
+
+
+
+ Note particularly that like the tangent function, the secant, cosecant, and cotangent are also defined completely in terms of the sine and cosine functions. In the context of a right triangle with an angle \theta, we know how to think of \sin(\theta), \cos(\theta), and \tan(\theta) as ratios of sides of the triangle. We can now do likewise with the other trigonometric functions:
+
+ With these three additional trigonometric functions, we now have expressions that address all six possible combinations of two sides of a right triangle in a ratio.
+
+
+
+
+
+
+
+
+ Ratios in right triangles
+
+ Because the sine and cosine functions are used to define each of the other four trigonometric functions, it follows that we can translate information known about the other functions back to information about the sine and cosine functions. For example, if we know that in a certain triangle \csc(\alpha) = \frac{5}{3}, it follows that \sin(\alpha) = \frac{3}{5}. From there we can reason in the usual way to determine missing information in the given triangle.
+
+
+
+ It's also often possible to view given information in the context of the unit circle. With the earlier given information that \csc(\alpha) = \frac{5}{3}, it's natural to view \alpha as being the angle in a right triangle that lies opposite a leg of length 3 with the hypotenuse being 5, since \csc(\alpha) = \frac{\text{hyp}}{\text{opp}}. The Pythagorean Theorem then tells us the leg adjacent to \alpha has length 4, as seen in \triangle OPQ in Figure.
+
+
+
+
A 3-4-5 right triangle.
+
A 3-4-5 right triangle.
+
+
+
+
+ But we could also view \sin(\alpha) = \frac{3}{5} as \sin(\alpha) = \frac{\frac{3}{5}}{1}, and thus think of the right triangle has having hypotenuse 1 and vertical leg \frac{3}{5}. This triangle is similar to the originally considered 3-4-5 right triangle, but can be viewed as lying within the unit circle. The perspective of the unit circle is particularly valuable when ratios such as \frac{\sqrt{3}}{2}, \frac{\sqrt{2}}{2}, and \frac{1}{2} arise in right triangles.
+
+
+
+
+
+
+
+ Properties of the secant, cosecant, and cotangent functions
+
+ Like the tangent function, the secant, cosecant, and cotangent functions are defined in terms of the sine and cosine functions, so we can determine the exact values of these functions at each of the special points on the unit circle. In addition, we can use our understanding of the unit circle and the properties of the sine and cosine functions to determine key properties of these other trigonometric functions. We begin by investigating the secant function.
+
+
+
+ Using the fact that \sec(t) = \frac{1}{\cos(t)}, we note that anywhere \cos(t) = 0, the value of \sec(t) is undefined. We denote such instances in the following table by u. At all other points, the value of the secant function is simply the reciprocal of the cosine function's value. Since |\cos(t)| \le 1 for all t, it follows that |\sec(t)| \ge 1 for all t (for which the secant's value is defined).
+
+
+
+ Values of the cosine and secant functions at special points on the unit circle (Quadrants I and II).
+
+
+ t
+ 0
+ \frac{\pi}{6}
+ \frac{\pi}{4}
+ \frac{\pi}{3}
+ \frac{\pi}{2}
+ \frac{2\pi}{3}
+ \frac{3\pi}{4}
+ \frac{5\pi}{6}
+ \pi
+
+
+ \cos(t)
+ 1
+ \frac{\sqrt{3}}{2}
+ \frac{\sqrt{2}}{2}
+ \frac{1}{2}
+ 0
+ -\frac{1}{2}
+ -\frac{\sqrt{2}}{2}
+ -\frac{\sqrt{3}}{2}
+ -1
+
+
+ \sec(t)
+ 1
+ \frac{2}{\sqrt{3}}
+ \sqrt{2}
+ 2
+ u
+ -2
+ -\sqrt{2}
+ -\frac{2}{\sqrt{3}}
+ -1
+
+
+
+
+
+ Values of the cosine and secant functions at special points on the unit circle (Quadrants III and IV).
+
+
+ t
+ \frac{7\pi}{6}
+ \frac{5\pi}{4}
+ \frac{4\pi}{3}
+ \frac{3\pi}{2}
+ \frac{5\pi}{3}
+ \frac{7\pi}{4}
+ \frac{11\pi}{6}
+ 2\pi
+
+
+ \cos(t)
+ -\frac{\sqrt{3}}{2}
+ -\frac{\sqrt{2}}{2}
+ -\frac{1}{2}
+ 0
+ \frac{1}{2}
+ \frac{\sqrt{2}}{2}
+ \frac{\sqrt{3}}{2}
+ 0
+
+
+ \sec(t)
+ -\frac{2}{\sqrt{3}}
+ -\sqrt{2}
+ -2
+ u
+ 2
+ \sqrt{2}
+ \frac{2}{\sqrt{3}}
+ 1
+
+
+
+
+
+ Table and Table help us identify trends in the secant function. The sign of \sec(t) matches the sign of \cos(t) and thus is positive in Quadrant I, negative in Quadrant II, negative in Quadrant III, and positive in Quadrant IV.
+
+
+
+ In addition, we observe that as t-values in the first quadrant get closer to \frac{\pi}{2}, \cos(t) gets closer to 0 (while being always positive). Since the numerator of the secant function is always 1, having its denominator approach 0 (while the denominator remains positive) means that \sec(t) increases without bound as t approaches \frac{\pi}{2} from the left side. Once t is slightly greater than \frac{\pi}{2} in Quadrant II, the value of \cos(t) is negative (and close to zero). This makes the value of \sec(t) decrease without bound (negative and getting further away from 0) for t approaching \frac{\pi}{2} from the right side. We therefore see that p(t) = \sec(t) has a vertical asymptote at t = \frac{\pi}{2}; the periodicity and sign behavior of \cos(t) mean this asymptotic behavior of the secant function will repeat.
+
+
+
+ Plotting the data in the table along with the expected asymptotes and connecting the points intuitively, we see the graph of the secant function in Figure.
+
+
+
+
A plot of the secant function with special points that come from the unit circle, plus the cosine function (dotted, in light blue).
+
A plot of the secant function with special points that come from the unit circle, plus the cosine function (dotted, in light blue).
+
+
+
+
+ We see from both the table and the graph that the secant function has period P = 2\pi. We summarize our recent work as follows.
+
+
+
+ Properties of the secant function
+
+ For the function p(t) = \sec(t),
+
+
+
+ its domain is the set of all real numbers except t = \frac{\pi}{2} \pm k\pi where k is any whole number;
+
+
+
+
+ its range is the set of all real numbers y such that |y| \ge 1;
+
+
+
+
+ its period is P = 2\pi.
+
+
+
+
+
+
+
+
+
+
+
+
+
+ A few important identities
+
+ An identity is an equation that is true for all possible values of x for which the involved quantities are defined. An example of a non-trigonometric identity is
+
+ (x+1)^2 = x^2 + 2x + 1
+ ,
+ since this equation is true for every value of x,
+ and the left and right sides of the equation are simply two different-looking but entirely equivalent expressions.
+
+
+
+ Trigonometric identities are simply identities that involve trigonometric functions. While there are a large number of such identities one can study, we choose to focus on those that turn out to be most useful in the study of calculus. The most important trigonometric identity is the fundamental trigonometric identity,
+ which is a trigonometric restatement of the Pythagorean Theorem.
+
+
+
+ fundamental trigonometric identity
+ The fundamental trigonometric identity
+
+ For any real number \theta,
+
+ \cos^2(\theta) + \sin^2(\theta) = 1
+ .
+
+
+
+
+
+ Identities are important because they enable us to view the same idea from multiple perspectives. For example, the fundamental trigonometric identity allows us to think of \cos^2(\theta) + \sin^2(\theta) as simply 1, or alternatively, to view \cos^2(\theta) as the same quantity as 1 - \sin^2(\theta).
+
+
+
+ There are two related Pythagorean identities that involve the tangent, secant,
+ cotangent, and cosecant functions, which we can derive from the fundamental trigonometric identity by dividing both sides by either \cos^2(\theta) or \sin^2(\theta). If we divide both sides of Equation by \cos^2(\theta) (and assume that \cos(\theta) \ne 0), we see that
+
+ 1 + \frac{\sin^2(\theta)}{\cos^2(\theta)} = \frac{1}{\cos^2(\theta)}
+ , or equivalently,
+
+ 1 + \tan^2(\theta) = \sec^2(\theta)
+ .
+ A similar argument dividing by \sin^2(\theta) (while assuming \sin(\theta) \ne 0) shows that
+
+ \cot^2(\theta) + 1 = \csc^2(\theta)
+ . These identities prove useful in calculus when we develop the formulas for the derivatives of the tangent and cotangent functions.
+
+
+
+ In calculus, it is also beneficial to know a couple of other standard identities for sums of angles or double angles. sum of two angles identitydouble angle identity We simply state these identities without justification. For more information about them, see Section 10.4 in College Trigonometry, by Stitz and ZeagerMore information on Stitz and Zeager's free texts can be found at ..
+
+
+
+
+
+
+ For all real numbers \alpha and \beta, \cos(\alpha + \beta) = \cos(\alpha) \cos(\beta) - \sin(\alpha) \sin(\beta).
+
+
+
+
+ For all real numbers \alpha and \beta, \sin(\alpha + \beta) = \sin(\alpha) \cos(\beta) + \cos(\alpha) \sin(\beta).
+
+
+
+
+ For any real number \theta, \cos(2\theta) = \cos^2(\theta) - \sin^2(\theta).
+
+
+
+
+ For any real number \theta, \sin(2\theta) = 2\sin(\theta)\cos(\theta).
+
+
+
+
+
+
+
+
+
+
+ Summary
+
+
+
+
+ The secant, cosecant, and cotangent functions are respectively defined as the reciprocals of the cosine, sine, and tangent functions. That is,
+
+ \sec(t) = \frac{1}{\cos(t)}, \csc(t) = \frac{1}{\sin(t)}, \text{ and } \cot(t) = \frac{1}{\tan(t)}
+ .
+
+
+
+
+ The graph of the cotangent function is similar to the graph of the tangent function, except that it is decreasing on every interval on which it is defined and has vertical asymptotes wherever \tan(t) = 0 and is zero wherever \tan(t) has a vertical asymptote.
+
+
+
+ The graphs of the secant and cosecant functions are different from the cosine and sine functions' graphs in several ways, including that their range is the set of all real numbers y such that |y| \ge 1 and they have vertical asymptotes wherever the cosine and sine function, respectively, are zero.
+
+
+
+
+ A trigonometric identity is an equation involving trigonometric functions that is true for every value of the variable for which the trigonometric functions are defined. For instance, \tan^2(t) + 1 = \sec^2(t) for every real number t except t = \frac{\pi}{2} \pm k\pi. Identities offer us alternate perspectives on the same function. For instance, the function f(t) = \sec^2(t) - \tan^2(t) is the same (at all points where f is defined) as the function whose value is always 1.
+
- How can we view \cos(\theta) and \sin(\theta) as side lengths in right triangles with hypotenuse 1?
-
-
-
-
- Why can both \cos(\theta) and \sin(\theta) be thought of as ratios of certain side lengths in any right triangle?
-
-
-
-
- What is the minimum amount of information we need about a right triangle in order to completely determine all of its sides and angles?
-
-
-
-
-
- Introduction
-
-
- In Section, we defined the cosine and sine functions as the functions that track the location of a point traversing the unit circle counterclockwise from (1,0). In particular, for a central angle of radian measure t that passes through the point (1,0), we define \cos(t) as the x-coordinate of the point where the other side of the angle intersects the unit circle, and \sin(t) as the y-coordinate of that same point, as pictured in Figure.
-
-
-
- By changing our perspective slightly, we can see that it is equivalent to think of the values of the sine and cosine function as representing the lengths of legs in right triangles. Specifically, given a central angleIn our work with right triangles, we'll often represent the angle by \theta and think of this angle as fixed, as opposed to our previous use of t where we frequently think of t as changing.\theta in standard position standard position (vertex at the origin, and one side on the positive x-axis), if we think of the right triangle with vertices (\cos(\theta),0), (0,0), and (\cos(\theta), \sin(\theta)), then the length of the horizontal leg is \cos(\theta) and the length of the vertical leg is \sin(\theta), as seen in Figure.
-
-
-
-
-
The values of \cos(t) and \sin(t) as coordinates on the unit circle.
-
The values of \cos(t) and \sin(t) as coordinates on the unit circle.
-
-
-
-
The values of \cos(\theta) and \sin(\theta) as the lengths of the legs of a right triangle.
-
The values of \cos(\theta) and \sin(\theta) as the lengths of the legs of a right triangle.
-
-
-
-
-
- This right triangle perspective enables us to use the sine and cosine functions to determine missing information in certain right triangles. The field of mathematics that studies relationships among the angles and sides of triangles is called trigonometry. trigonometry In addition, it's important to recall both the Pythagorean Theorem and the Fundamental Trigonometric Identity. Pythagorean Theoremfundamental trigonometric identity The former states that in any right triangle with legs of length a and b and hypotenuse of length c, it follows a^2 + b^2 = c^2. The latter, which is a special case of the Pythagorean Theorem, says that for any angle \theta, \cos^2(\theta) + \sin^2(\theta) = 1.
-
-
-
-
-
-
-
- The geometry of triangles
-
-
- In the study of functions, linear functions are the simplest of all and form a foundation for our understanding of functions that have other shapes. In the study of geometric shapes (polygons, circles, and more), the simplest figure of all is the triangle, and understanding triangles is foundational to understanding many other geometric ideas. To begin, we list some familiar and important facts about triangles.
-
-
-
-
-
-
- Any triangle has 6 important features: 3 sides and 3 angles.
-
-
-
-
- In any triangle in the Cartesian plane, the sum of the measures of the interior angles is \pi radians (or equivalently, 180^\circ).
-
-
-
-
- In any triangle in the plane, knowing three of the six features of a triangle is often enough information to determine the missing three features.Formally, this idea relies on what are called congruence criteria. For instance, if we know the lengths of all three sides, then the angle measures of the triangle are uniquely determined. This is called the Side-Side-Side Criterion (SSS). You are likely familiar with SSS, as well as SAS (Side-Angle-Side), ASA, and AAS, which are the four standard criteria.
-
-
-
-
-
-
- The situation is especially nice for right triangles, because then we only have five unknown features since one of the angles is \frac{\pi}{2} radians (or 90^\circ), as demonstrated in Figure. If we know one of the two non-right angles,
- then we know the other as well. Moreover, if we know any two sides, we can immediately deduce the third, because of the Pythagorean Theorem. As we saw in Preview Activity, the cosine and sine functions offer additional help in determining missing information in right triangles. Indeed, while the functions \cos(t) and \sin(t) have many important applications in modeling periodic phenomena such as oscillating masses on springs, they also find powerful application in settings involving right triangles, such as in navigation and surveying.
-
-
-
-
The 5 potential unknowns in a right triangle.
-
The 5 potential unknowns in a right triangle.
-
-
-
-
- Because we know the values of the cosine and sine functions from the unit circle, right triangles with hypotentuse 1 are the easiest ones in which to determine missing information. In addition, we can relate any other right triangle to a right triangle with hypotenuse 1 through the concept of similarity. Recall that two triangles are similarsimilar triangles provided that one is a magnification of the other. More precisely, two triangles are similar whenever there is some constant k such that every side in one triangle is k times as long as the corresponding side in the other and the corresponding angles in the two triangles are equal. An important result from geometry tells us that if two triangles are known to have all three of their corresponding angles equal, then it follows that the two triangles are similar, and therefore their corresponding sides must be proportionate to one another.
-
-
-
-
-
-
-
- Ratios of sides in right triangles
-
-
- A right triangle with a hypotenuse of length 1 can be viewed as lying in standard position in the unit circle, with one vertex at the origin and one leg along the positive x-axis. If we let the angle formed by the hypotenuse and the horizontal leg be represented by \theta, then the right triangle with hypotenuse 1 has horizontal leg of length \cos(\theta) and vertical leg of length \sin(\theta). If we now consider a similar right triangle with hypotenuse of length r \ne 1, we can view that triangle as a magnification of a triangle with hypotenuse 1. These observations, combined with our work in Activity, show us that the horizontal legs of the right triangle with hypotenuse r have lengths r\cos(\theta) and r\sin(\theta), as pictured in Figure.
-
-
-
-
The roles of r and \theta in a right triangle.
-
The roles of r and \theta in a right triangle.
-
-
-
-
- From the similar triangles in Figure, we can make an important observation about ratios in right triangles. Because the triangles are similar, the ratios of corresponding sides must be equal, so if we consider the two hypotenuses and the two horizontal legs, we have
-
- \frac{r}{1} = \frac{r\cos(\theta)}{\cos(\theta)}
- .
- If we rearrange Equation by dividing both sides by r and multiplying both sides by \cos(\theta), we see that
-
- \frac{\cos(\theta)}{1} = \frac{r\cos(\theta)}{r}
- .
- From a geometric perspective, Equation tells us that the ratio of the length of the horizontal leg of a right triangle to the length of the hypotenuse of the triangle is always the same (regardless of r) and that the value of that ratio is \cos(\theta), where \theta is the angle adjacent to the horizontal leg. In an analogous way, the equation involving the hypotenuses and vertical legs of the similar triangles is
-
- \frac{r}{1} = \frac{r\sin(\theta)}{\sin(\theta)}
- ,
- which can be rearranged to
-
- \frac{\sin(\theta)}{1} = \frac{r\sin(\theta)}{r}
- .
- Equation shows that the ratio of the length of the vertical leg of a right triangle to the length of the hypotenuse of the triangle is always the same (regardless of r) and that the value of that ratio is \sin(\theta), where \theta is the angle opposite the vertical leg. We summarize these recent observations as follows.
-
-
-
- Ratios in right triangles
-
-
- In a right triangle where one of the non-right angles is \theta, and adj denotes the length of the leg adjacent to \theta, opp the length the side opposite \theta, and hyp the length of the hypotenuse,
-
- \cos(\theta) = \frac{\text{adj}}{\text{hyp}} \text{ and } \sin(\theta) = \frac{\text{opp}}{\text{hyp}}
- .
-
-
-
ADD ALT TEXT TO THIS IMAGE
-
-
-
-
-
-
-
-
-
- Using a ratio involving sine and cosine
-
-
- In Activity, we found that in many cases where we have a right triangle, knowing two additional pieces of information enables us to find the remaining three unknown quantities in the triangle. At this point in our studies, the following general principles hold.
-
-
-
- Missing information in right triangles
-
- In any right triangle,
-
-
-
-
-
-
- if we know one of the non-right angles and the length of the hypotenuse, we can find both the remaining non-right angle and the lengths of the two legs;
-
-
-
-
- if we know the length of two sides of the triangle, then we can find the length of the other side;
-
-
-
-
- if we know the measure of one non-right angle, then we can find the measure of the remaining angle.
-
-
-
-
-
-
-
- In scenario (1.), all 6 features of the triangle are not only determined, but we are able to find their values. In (2.), the triangle is uniquely determined by the given information, but as in Activity parts (d) and (e), while we know the values of the sine and cosine of the angles in the triangle, we haven't yet developed a way to determine the measures of those angles. Finally, in scenario (3.), the triangle is not uniquely determined, since any magnified version of the triangle will have the same three angles as the given one, and thus we need more information to determine side length.
-
-
-
- We will revisit scenario (2) in our future work. Now, however, we want to consider a situation that is similar to (1), but where it is one leg of the triangle instead of the hypotenuse that is known. We encountered this in Activity part (f): a right triangle where one of the non-right angles is \beta = \frac{\pi}{5} and the leg opposite this angle has length 4.
-
-
-
-
-
-
-
-
- Consider a right triangle in which one of the non-right angles is \beta = \frac{\pi}{5} and the leg opposite \beta has length 4.
-
-
-
- Determine (both exactly and approximately) the measures of all of the remaining sides and angles in the triangle.
-
-
-
-
-
The given right triangle.
-
The given right triangle.
-
-
-
-
-
-
-
-
- From the fact that \beta = \frac{\pi}{5}, it follows that \alpha = \frac{\pi}{2} - \frac{\pi}{5} = \frac{3\pi}{10}. In addition, we know that
-
- \sin\left(\frac{\pi}{5}\right) = \frac{4}{h}
-
- and
-
- \cos\left(\frac{\pi}{5}\right) = \frac{x}{h}
-
-
-
-
- Solving Equation for h, we see that
-
- h = \frac{4}{\sin\left(\frac{\pi}{5}\right)}
- , which is the exact numerical value of h. Substituting this result in
- Equation, we find that
-
- \cos\left(\frac{\pi}{5}\right) = \frac{x}{\frac{4}{\sin\left(\frac{\pi}{5}\right)}}
- .
- Solving this equation for the single unknown x shows that
-
- x = \frac{4 \cos\left(\frac{\pi}{5}\right)}{\sin\left(\frac{\pi}{5}\right)}
- .
- The approximate values of x and h are x \approx 5.506 and h \approx 6.805.
-
-
-
-
-
- Example demonstrates that a ratio of values of the sine and cosine function can be needed in order to determine the value of one of the missing sides of a right triangle, and also that we may need to work with two unknown quantities simultaneously in order to determine both of their values.
-
-
-
-
-
-
-
- Summary
-
-
-
-
- In a right triangle with hypotenuse 1, we can view \cos(\theta) as the length of the leg adjacent to \theta and \sin(\theta) as the length of the leg opposite \theta, as seen in Figure. This is simply a change in perspective achieved by focusing on the triangle as opposed to the unit circle.
-
-
-
-
- Because a right triangle with hypotenuse of length r can be thought of as a scaled version of a right triangle with hypotenuse of length 1, we can conclude that in a right triangle with hypotenuse of length r, the leg adjacent to angle \theta has length r\cos(\theta), and the leg opposite \theta has length r\sin(\theta), as seen in Figure. Moreover, in any right triangle with angle \theta, we know that
-
- \cos(\theta) = \frac{\text{adj}}{\text{hyp}} \text{ and } \sin(\theta) = \frac{\text{opp}}{\text{hyp}}
- .
-
-
-
-
- In a right triangle, there are five additional characteristics: the measures of the two non-right angles and the lengths of the three sides. In general, if we know one of those two angles and one of the three sides, we can determine all of the remaining pieces.
-
+ How can we view \cos(\theta) and \sin(\theta) as side lengths in right triangles with hypotenuse 1?
+
+
+
+
+ Why can both \cos(\theta) and \sin(\theta) be thought of as ratios of certain side lengths in any right triangle?
+
+
+
+
+ What is the minimum amount of information we need about a right triangle in order to completely determine all of its sides and angles?
+
+
+
+
+
+ Introduction
+
+
+ In Section, we defined the cosine and sine functions as the functions that track the location of a point traversing the unit circle counterclockwise from (1,0). In particular, for a central angle of radian measure t that passes through the point (1,0), we define \cos(t) as the x-coordinate of the point where the other side of the angle intersects the unit circle, and \sin(t) as the y-coordinate of that same point, as pictured in Figure.
+
+
+
+ By changing our perspective slightly, we can see that it is equivalent to think of the values of the sine and cosine function as representing the lengths of legs in right triangles. Specifically, given a central angleIn our work with right triangles, we'll often represent the angle by \theta and think of this angle as fixed, as opposed to our previous use of t where we frequently think of t as changing.\theta in standard position standard position (vertex at the origin, and one side on the positive x-axis), if we think of the right triangle with vertices (\cos(\theta),0), (0,0), and (\cos(\theta), \sin(\theta)), then the length of the horizontal leg is \cos(\theta) and the length of the vertical leg is \sin(\theta), as seen in Figure.
+
+
+
+
+
The values of \cos(t) and \sin(t) as coordinates on the unit circle.
+
The values of \cos(t) and \sin(t) as coordinates on the unit circle.
+
+
+
+
The values of \cos(\theta) and \sin(\theta) as the lengths of the legs of a right triangle.
+
The values of \cos(\theta) and \sin(\theta) as the lengths of the legs of a right triangle.
+
+
+
+
+
+ This right triangle perspective enables us to use the sine and cosine functions to determine missing information in certain right triangles. The field of mathematics that studies relationships among the angles and sides of triangles is called trigonometry. trigonometry In addition, it's important to recall both the Pythagorean Theorem and the Fundamental Trigonometric Identity. Pythagorean Theoremfundamental trigonometric identity The former states that in any right triangle with legs of length a and b and hypotenuse of length c, it follows a^2 + b^2 = c^2. The latter, which is a special case of the Pythagorean Theorem, says that for any angle \theta, \cos^2(\theta) + \sin^2(\theta) = 1.
+
+
+
+
+
+
+
+
+ The geometry of triangles
+
+
+ In the study of functions, linear functions are the simplest of all and form a foundation for our understanding of functions that have other shapes. In the study of geometric shapes (polygons, circles, and more), the simplest figure of all is the triangle, and understanding triangles is foundational to understanding many other geometric ideas. To begin, we list some familiar and important facts about triangles.
+
+
+
+
+
+
+ Any triangle has 6 important features: 3 sides and 3 angles.
+
+
+
+
+ In any triangle in the Cartesian plane, the sum of the measures of the interior angles is \pi radians (or equivalently, 180^\circ).
+
+
+
+
+ In any triangle in the plane, knowing three of the six features of a triangle is often enough information to determine the missing three features.Formally, this idea relies on what are called congruence criteria. For instance, if we know the lengths of all three sides, then the angle measures of the triangle are uniquely determined. This is called the Side-Side-Side Criterion (SSS). You are likely familiar with SSS, as well as SAS (Side-Angle-Side), ASA, and AAS, which are the four standard criteria.
+
+
+
+
+
+
+ The situation is especially nice for right triangles, because then we only have five unknown features since one of the angles is \frac{\pi}{2} radians (or 90^\circ), as demonstrated in Figure. If we know one of the two non-right angles,
+ then we know the other as well. Moreover, if we know any two sides, we can immediately deduce the third, because of the Pythagorean Theorem. As we saw in Preview Activity, the cosine and sine functions offer additional help in determining missing information in right triangles. Indeed, while the functions \cos(t) and \sin(t) have many important applications in modeling periodic phenomena such as oscillating masses on springs, they also find powerful application in settings involving right triangles, such as in navigation and surveying.
+
+
+
+
The 5 potential unknowns in a right triangle.
+
The 5 potential unknowns in a right triangle.
+
+
+
+
+ Because we know the values of the cosine and sine functions from the unit circle, right triangles with hypotentuse 1 are the easiest ones in which to determine missing information. In addition, we can relate any other right triangle to a right triangle with hypotenuse 1 through the concept of similarity. Recall that two triangles are similarsimilar triangles provided that one is a magnification of the other. More precisely, two triangles are similar whenever there is some constant k such that every side in one triangle is k times as long as the corresponding side in the other and the corresponding angles in the two triangles are equal. An important result from geometry tells us that if two triangles are known to have all three of their corresponding angles equal, then it follows that the two triangles are similar, and therefore their corresponding sides must be proportionate to one another.
+
+
+
+
+
+
+
+ Ratios of sides in right triangles
+
+
+ A right triangle with a hypotenuse of length 1 can be viewed as lying in standard position in the unit circle, with one vertex at the origin and one leg along the positive x-axis. If we let the angle formed by the hypotenuse and the horizontal leg be represented by \theta, then the right triangle with hypotenuse 1 has horizontal leg of length \cos(\theta) and vertical leg of length \sin(\theta). If we now consider a similar right triangle with hypotenuse of length r \ne 1, we can view that triangle as a magnification of a triangle with hypotenuse 1. These observations, combined with our work in Activity, show us that the horizontal legs of the right triangle with hypotenuse r have lengths r\cos(\theta) and r\sin(\theta), as pictured in Figure.
+
+
+
+
The roles of r and \theta in a right triangle.
+
The roles of r and \theta in a right triangle.
+
+
+
+
+ From the similar triangles in Figure, we can make an important observation about ratios in right triangles. Because the triangles are similar, the ratios of corresponding sides must be equal, so if we consider the two hypotenuses and the two horizontal legs, we have
+
+ \frac{r}{1} = \frac{r\cos(\theta)}{\cos(\theta)}
+ .
+ If we rearrange Equation by dividing both sides by r and multiplying both sides by \cos(\theta), we see that
+
+ \frac{\cos(\theta)}{1} = \frac{r\cos(\theta)}{r}
+ .
+ From a geometric perspective, Equation tells us that the ratio of the length of the horizontal leg of a right triangle to the length of the hypotenuse of the triangle is always the same (regardless of r) and that the value of that ratio is \cos(\theta), where \theta is the angle adjacent to the horizontal leg. In an analogous way, the equation involving the hypotenuses and vertical legs of the similar triangles is
+
+ \frac{r}{1} = \frac{r\sin(\theta)}{\sin(\theta)}
+ ,
+ which can be rearranged to
+
+ \frac{\sin(\theta)}{1} = \frac{r\sin(\theta)}{r}
+ .
+ Equation shows that the ratio of the length of the vertical leg of a right triangle to the length of the hypotenuse of the triangle is always the same (regardless of r) and that the value of that ratio is \sin(\theta), where \theta is the angle opposite the vertical leg. We summarize these recent observations as follows.
+
+
+
+ Ratios in right triangles
+
+
+ In a right triangle where one of the non-right angles is \theta, and adj denotes the length of the leg adjacent to \theta, opp the length the side opposite \theta, and hyp the length of the hypotenuse,
+
+ \cos(\theta) = \frac{\text{adj}}{\text{hyp}} \text{ and } \sin(\theta) = \frac{\text{opp}}{\text{hyp}}
+ .
+
+
+
ADD ALT TEXT TO THIS IMAGE
+
+
+
+
+
+
+
+
+
+ Using a ratio involving sine and cosine
+
+
+ In Activity, we found that in many cases where we have a right triangle, knowing two additional pieces of information enables us to find the remaining three unknown quantities in the triangle. At this point in our studies, the following general principles hold.
+
+
+
+ Missing information in right triangles
+
+ In any right triangle,
+
+
+
+
+
+
+ if we know one of the non-right angles and the length of the hypotenuse, we can find both the remaining non-right angle and the lengths of the two legs;
+
+
+
+
+ if we know the length of two sides of the triangle, then we can find the length of the other side;
+
+
+
+
+ if we know the measure of one non-right angle, then we can find the measure of the remaining angle.
+
+
+
+
+
+
+
+ In scenario (1.), all 6 features of the triangle are not only determined, but we are able to find their values. In (2.), the triangle is uniquely determined by the given information, but as in Activity parts (d) and (e), while we know the values of the sine and cosine of the angles in the triangle, we haven't yet developed a way to determine the measures of those angles. Finally, in scenario (3.), the triangle is not uniquely determined, since any magnified version of the triangle will have the same three angles as the given one, and thus we need more information to determine side length.
+
+
+
+ We will revisit scenario (2) in our future work. Now, however, we want to consider a situation that is similar to (1), but where it is one leg of the triangle instead of the hypotenuse that is known. We encountered this in Activity part (f): a right triangle where one of the non-right angles is \beta = \frac{\pi}{5} and the leg opposite this angle has length 4.
+
+
+
+
+
+
+
+
+ Consider a right triangle in which one of the non-right angles is \beta = \frac{\pi}{5} and the leg opposite \beta has length 4.
+
+
+
+ Determine (both exactly and approximately) the measures of all of the remaining sides and angles in the triangle.
+
+
+
+
+
The given right triangle.
+
The given right triangle.
+
+
+
+
+
+
+
+
+ From the fact that \beta = \frac{\pi}{5}, it follows that \alpha = \frac{\pi}{2} - \frac{\pi}{5} = \frac{3\pi}{10}. In addition, we know that
+
+ \sin\left(\frac{\pi}{5}\right) = \frac{4}{h}
+
+ and
+
+ \cos\left(\frac{\pi}{5}\right) = \frac{x}{h}
+
+
+
+
+ Solving Equation for h, we see that
+
+ h = \frac{4}{\sin\left(\frac{\pi}{5}\right)}
+ , which is the exact numerical value of h. Substituting this result in
+ Equation, we find that
+
+ \cos\left(\frac{\pi}{5}\right) = \frac{x}{\frac{4}{\sin\left(\frac{\pi}{5}\right)}}
+ .
+ Solving this equation for the single unknown x shows that
+
+ x = \frac{4 \cos\left(\frac{\pi}{5}\right)}{\sin\left(\frac{\pi}{5}\right)}
+ .
+ The approximate values of x and h are x \approx 5.506 and h \approx 6.805.
+
+
+
+
+
+ Example demonstrates that a ratio of values of the sine and cosine function can be needed in order to determine the value of one of the missing sides of a right triangle, and also that we may need to work with two unknown quantities simultaneously in order to determine both of their values.
+
+
+
+
+
+
+
+ Summary
+
+
+
+
+ In a right triangle with hypotenuse 1, we can view \cos(\theta) as the length of the leg adjacent to \theta and \sin(\theta) as the length of the leg opposite \theta, as seen in Figure. This is simply a change in perspective achieved by focusing on the triangle as opposed to the unit circle.
+
+
+
+
+ Because a right triangle with hypotenuse of length r can be thought of as a scaled version of a right triangle with hypotenuse of length 1, we can conclude that in a right triangle with hypotenuse of length r, the leg adjacent to angle \theta has length r\cos(\theta), and the leg opposite \theta has length r\sin(\theta), as seen in Figure. Moreover, in any right triangle with angle \theta, we know that
+
+ \cos(\theta) = \frac{\text{adj}}{\text{hyp}} \text{ and } \sin(\theta) = \frac{\text{opp}}{\text{hyp}}
+ .
+
+
+
+
+ In a right triangle, there are five additional characteristics: the measures of the two non-right angles and the lengths of the three sides. In general, if we know one of those two angles and one of the three sides, we can determine all of the remaining pieces.
+
- How is the tangent function defined in terms of the sine and cosine functions?
-
-
-
-
- Why is the graph of the tangent function so different from the graphs of the sine and cosine functions?
-
-
-
-
- What are important applications of the tangent function?
-
-
-
-
-
- Introduction
-
-
-
- In Activity, we determined the distance between two points A and B on opposite sides of a river by knowing a length along one shore of the river and the angle formed between a point downstream and the point on the opposite shore, as pictured in Figure. By first using the cosine of the angle, we determined the value of z and from there were able to use the sine of the angle to find w, the river's width, which turns out to be
-
- w = 50 \cdot \frac{\sin(56.4)}{\cos(56.4)}
- .
-
-
-
-
Finding the width of the river.
-
Finding the width of the river.
-
-
-
-
-
-
- It turns out that we regularly need to evaluate the ratio of the sine and cosine functions at the same angle, so it is convenient to define a new function to be their ratio.
-
-
-
- The tangent function
- tangent function
-
-
- For any real number t for which \cos(t) \ne 0, we define the tangent of t, denoted \tan(t), by
-
- \tan(t) = \frac{\sin(t)}{\cos(t)}
- .
-
-
-
-
-
-
-
-
-
- Two perspectives on the tangent function
-
-
-
-
An angle t in standard position in the unit circle that intercepts an arc from (1,0) to (a,b).
-
An angle t in standard position in the unit circle that intercepts an arc from (1,0) to (a,b).
-
-
-
-
A right triangle with legs adjacent and opposite angle \theta.
-
A right triangle with legs adjacent and opposite angle \theta.
-
-
-
-
-
- Because the tangent function is defined in terms of the two fundamental circular functions by the rule \tan(t) = \frac{\sin(t)}{\cos(t)}, we can use our understanding of the sine and cosine functions to make sense of the tangent function. In particular, we can think of the tangent of an angle from two different perspectives: as an angle in standard position in the unit circle, or as an angle in a right triangle.
-
-
-
- From the viewpoint of Figure, as the point corresponding to angle t traverses the circle and generates the point (a,b), we know \cos(t) = a and \sin(t) = b, and therefore the tangent function tracks the ratio of these two quantities, and is given by
-
- \tan(t) = \frac{\sin(t)}{\cos(t)} = \frac{b}{a}
- .
- From the perspective of any right triangle (not necessarily in the unit circle) with hypotenuse hyp and legs
- adj and opp that are respectively adjacent and opposite the known angle \theta, as seen in Figure, we know that \sin(\theta) = \frac{\text{opp}}{\text{hyp}} and \cos(\theta) = \frac{\text{adj}}{\text{hyp}}. Substituting these expressions for \sin(\theta) and \cos(\theta) in the rule for the tangent function, we see that
-
- \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{\frac{\text{opp}}{\text{hyp}}}{\frac{\text{adj}}{\text{hyp}}} = \frac{\text{opp}}{\text{adj}}
- .
-
-
-
- We typically use the first perspective of tracking the ratio of the y-coordinate to the x-coordinate of a point traversing the unit circle in order to think of the overall behavior and graph of the tangent function, and use the second perspective in a right triangle whenever we are working to determine missing values in a triangle.
-
-
-
-
- Properties of the tangent function
-
-
- Because the tangent function is defined in terms of the sine and cosine functions, its values and behavior are completely determined by those two functions. To begin, we know the value of \tan(t) for every special angle t on the unit circle that we identified for the sine and cosine functions. For instance, we know that
-
- \tan \left( \frac{\pi}{6} \right) = \frac{ \sin \left( \frac{\pi}{6} \right) }{ \cos \left( \frac{\pi}{6} \right) } = \frac{ \frac{1}{2} }{ \frac{\sqrt{3}}{2} } = \frac{1}{\sqrt{3}}
- .
- Executing similar computations for every familiar special angle on the unit circle, we find the results shown in Table and Table. We also note that anywhere \cos(t) = 0, the value of \tan(t) is undefined. We record such instances in the table by writing u.
-
- Additional values of the sine, cosine, and tangent functions at special points on the unit circle.
-
-
- t
- \frac{7\pi}{6}
- \frac{5\pi}{4}
- \frac{4\pi}{3}
- \frac{3\pi}{2}
- \frac{5\pi}{3}
- \frac{7\pi}{4}
- \frac{11\pi}{6}
- 2\pi
-
-
- \sin(t)
- -\frac{1}{2}
- -\frac{\sqrt{2}}{2}
- -\frac{\sqrt{3}}{2}
- -1
- -\frac{\sqrt{3}}{2}
- -\frac{\sqrt{2}}{2}
- -\frac{1}{2}
- 0
-
-
- \cos(t)
- -\frac{\sqrt{3}}{2}
- -\frac{\sqrt{2}}{2}
- -\frac{1}{2}
- 0
- \frac{1}{2}
- \frac{\sqrt{2}}{2}
- \frac{\sqrt{3}}{2}
- 0
-
-
- \tan(t)
- \frac{1}{\sqrt{3}}
- 1
- \sqrt{3}
- u
- -\sqrt{3}
- -1
- -\frac{1}{\sqrt{3}}
- 0
-
-
-
-
-
- Table and Table helps us identify trends in the tangent function. For instance, we observe that the sign of \tan(t) is positive in Quadrant I, negative in Quadrant II, positive in Quadrant III, and negative in Quadrant IV. This holds because the sine and cosine functions have the same sign in the first and third quadrants, and opposite signs in the other two quadrants.
-
-
-
- In addition, we observe that as t-values in the first quadrant get closer to \frac{\pi}{2}, \sin(t) gets closer to 1, while \cos(t) gets closer to 0 (while being always positive). Noting that \frac{\pi}{2} \approx 1.57, we observe that
-
- \tan(1.47) = \frac{\sin(1.47)}{\cos(1.47)} \approx \frac{0.995}{0.101} = 9.887
-
- and
-
- \tan(1.56) = \frac{\sin(1.56)}{\cos(1.56)} \approx \frac{0.9994}{0.0108} = 92.6205
- .
- Because the ratio of numbers closer and closer to 1 divided by numbers closer and closer to 0 (but positive) increases without bound, this means that \tan(t) increases without bound as t approaches \frac{\pi}{2} from the left side. Once t is slightly greater than \frac{\pi}{2} in Quadrant II, the value of \sin(t) stays close to 1, but now the value of \cos(t) is negative (and close to zero). For instance, \cos(1.58) \approx -0.0092. This makes the value of \tan(t) decrease without bound (negative and getting further away from 0) for t approaching \frac{\pi}{2} from the right side, and results in h(t) = \tan(t) having a vertical asymptote at t = \frac{\pi}{2}. The periodicity and sign behaviors of \sin(t) and \cos(t) mean this asymptotic behavior of the tangent function will repeat.
-
-
-
- Plotting the data in the table along with the expected asymptotes and connecting the points intuitively, we see the graph of the tangent function in Figure.
-
-
-
-
A plot of the tangent function together with special points that come from the unit circle.
-
A plot of the tangent function together with special points that come from the unit circle.
-
-
-
-
- We see from Table and Table as well as from Figure that the tangent function has period P = \pi and that the function is increasing on any interval on which it is defined. We summarize our recent work as follows.
-
-
-
- Properties of the tangent function
-
- For the function h(t) = \tan(t),
-
-
-
- its domain is the set of all real numbers except t = \frac{\pi}{2} \pm k\pi where k is any whole number;
-
-
-
-
- its range is the set of all real numbers;
-
-
-
-
- its period is P = \pi;
-
-
-
-
- is increasing on any interval on which the function is defined at every point in the interval.
-
-
-
-
-
-
-
- While the tangent function is an interesting mathematical function for its own sake, its most important applications arise in the setting of right triangles, and for the remainder of this section we will focus on that perspective.
-
-
-
-
-
- Using the tangent function in right triangles
-
- The tangent function offers us an additional choice when working in right triangles with limited information. In the setting where we have a right triangle with one additional known angle, if we know the length of the hypotenuse, we can use either the sine or cosine of the angle to help us easily find the remaining side lengths. But in the setting where we know only the length of one leg, the tangent function now allows us to determine the value of the remaining leg in a similarly straightforward way, and from there the hypotenuse.
-
-
-
-
-
- Use the tangent function to determine the width, w, of the river in Figure. (Note that here we are revisiting the problem in Activity, which we previously solved without using the tangent function.) What other information can we now easily determine?
-
-
-
-
A right triangle with one angle and one leg known.
-
A right triangle with one angle and one leg known.
-
-
-
-
-
- Using the perspective that \tan(\theta) = \frac{\text{opp}}{\text{adj}} in a right triangle, in this context we have
-
- \tan(56.4^\circ) = \frac{w}{50}
-
- and thus w = 50\tan(56.4) is the exact width of the river. Using a computational device, we find that w \approx 75.256.
-
-
- Once we know the river's width, we can use the Pythagorean theorem or the sine function to determine the distance from P to A, at which point all 6 parts of the triangle are known.
-
-
-
-
-
- The tangent function finds a wide range of applications in finding missing information in right triangles where information about one or more legs of the triangle is known.
-
-
-
-
-
-
-
-
-
-
-
- Summary
-
-
-
-
- The tangent function is defined defined to be the ratio of the sine and cosine functions according to the rule
-
- \tan(t) = \frac{\sin(t)}{\cos(t)}
-
- for all values of t for which \cos(t) \ne 0.
-
-
-
-
- The graph of the tangent function differs substantially from the graphs of the sine and cosine functions, primarily because near values where \cos(t) = 0, the ratio of \frac{\sin(t)}{\cos(t)} increases or decreases without bound, producing vertical asymptotes. In addition, while the period of the sine and cosine functions is P = 2\pi, the period of the tangent function is P = \pi due to how the sine and cosine functions repeat the same values (with different signs) as a point traverses the unit circle.
-
-
-
-
- The tangent function finds some of its most important applications in the setting of right triangles where one leg of the triangle is known and one of the non-right angles is known. Computing the tangent of the known angle, say \alpha, and using the fact that
-
- \tan(\alpha) = \frac{\text{opp}}{\text{adj}}
-
- we can then find the missing leg's length in terms of the other and the tangent of the angle.
-
+ How is the tangent function defined in terms of the sine and cosine functions?
+
+
+
+
+ Why is the graph of the tangent function so different from the graphs of the sine and cosine functions?
+
+
+
+
+ What are important applications of the tangent function?
+
+
+
+
+
+ Introduction
+
+
+
+ In Activity, we determined the distance between two points A and B on opposite sides of a river by knowing a length along one shore of the river and the angle formed between a point downstream and the point on the opposite shore, as pictured in Figure. By first using the cosine of the angle, we determined the value of z and from there were able to use the sine of the angle to find w, the river's width, which turns out to be
+
+ w = 50 \cdot \frac{\sin(56.4)}{\cos(56.4)}
+ .
+
+
+
+
Finding the width of the river.
+
Finding the width of the river.
+
+
+
+
+
+
+ It turns out that we regularly need to evaluate the ratio of the sine and cosine functions at the same angle, so it is convenient to define a new function to be their ratio.
+
+
+
+ The tangent function
+ tangent function
+
+
+ For any real number t for which \cos(t) \ne 0, we define the tangent of t, denoted \tan(t), by
+
+ \tan(t) = \frac{\sin(t)}{\cos(t)}
+ .
+
+
+
+
+
+
+
+
+
+
+ Two perspectives on the tangent function
+
+
+
+
An angle t in standard position in the unit circle that intercepts an arc from (1,0) to (a,b).
+
An angle t in standard position in the unit circle that intercepts an arc from (1,0) to (a,b).
+
+
+
+
A right triangle with legs adjacent and opposite angle \theta.
+
A right triangle with legs adjacent and opposite angle \theta.
+
+
+
+
+
+ Because the tangent function is defined in terms of the two fundamental circular functions by the rule \tan(t) = \frac{\sin(t)}{\cos(t)}, we can use our understanding of the sine and cosine functions to make sense of the tangent function. In particular, we can think of the tangent of an angle from two different perspectives: as an angle in standard position in the unit circle, or as an angle in a right triangle.
+
+
+
+ From the viewpoint of Figure, as the point corresponding to angle t traverses the circle and generates the point (a,b), we know \cos(t) = a and \sin(t) = b, and therefore the tangent function tracks the ratio of these two quantities, and is given by
+
+ \tan(t) = \frac{\sin(t)}{\cos(t)} = \frac{b}{a}
+ .
+ From the perspective of any right triangle (not necessarily in the unit circle) with hypotenuse hyp and legs
+ adj and opp that are respectively adjacent and opposite the known angle \theta, as seen in Figure, we know that \sin(\theta) = \frac{\text{opp}}{\text{hyp}} and \cos(\theta) = \frac{\text{adj}}{\text{hyp}}. Substituting these expressions for \sin(\theta) and \cos(\theta) in the rule for the tangent function, we see that
+
+ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{\frac{\text{opp}}{\text{hyp}}}{\frac{\text{adj}}{\text{hyp}}} = \frac{\text{opp}}{\text{adj}}
+ .
+
+
+
+ We typically use the first perspective of tracking the ratio of the y-coordinate to the x-coordinate of a point traversing the unit circle in order to think of the overall behavior and graph of the tangent function, and use the second perspective in a right triangle whenever we are working to determine missing values in a triangle.
+
+
+
+
+ Properties of the tangent function
+
+
+ Because the tangent function is defined in terms of the sine and cosine functions, its values and behavior are completely determined by those two functions. To begin, we know the value of \tan(t) for every special angle t on the unit circle that we identified for the sine and cosine functions. For instance, we know that
+
+ \tan \left( \frac{\pi}{6} \right) = \frac{ \sin \left( \frac{\pi}{6} \right) }{ \cos \left( \frac{\pi}{6} \right) } = \frac{ \frac{1}{2} }{ \frac{\sqrt{3}}{2} } = \frac{1}{\sqrt{3}}
+ .
+ Executing similar computations for every familiar special angle on the unit circle, we find the results shown in Table and Table. We also note that anywhere \cos(t) = 0, the value of \tan(t) is undefined. We record such instances in the table by writing u.
+
+ Additional values of the sine, cosine, and tangent functions at special points on the unit circle.
+
+
+ t
+ \frac{7\pi}{6}
+ \frac{5\pi}{4}
+ \frac{4\pi}{3}
+ \frac{3\pi}{2}
+ \frac{5\pi}{3}
+ \frac{7\pi}{4}
+ \frac{11\pi}{6}
+ 2\pi
+
+
+ \sin(t)
+ -\frac{1}{2}
+ -\frac{\sqrt{2}}{2}
+ -\frac{\sqrt{3}}{2}
+ -1
+ -\frac{\sqrt{3}}{2}
+ -\frac{\sqrt{2}}{2}
+ -\frac{1}{2}
+ 0
+
+
+ \cos(t)
+ -\frac{\sqrt{3}}{2}
+ -\frac{\sqrt{2}}{2}
+ -\frac{1}{2}
+ 0
+ \frac{1}{2}
+ \frac{\sqrt{2}}{2}
+ \frac{\sqrt{3}}{2}
+ 0
+
+
+ \tan(t)
+ \frac{1}{\sqrt{3}}
+ 1
+ \sqrt{3}
+ u
+ -\sqrt{3}
+ -1
+ -\frac{1}{\sqrt{3}}
+ 0
+
+
+
+
+
+ Table and Table helps us identify trends in the tangent function. For instance, we observe that the sign of \tan(t) is positive in Quadrant I, negative in Quadrant II, positive in Quadrant III, and negative in Quadrant IV. This holds because the sine and cosine functions have the same sign in the first and third quadrants, and opposite signs in the other two quadrants.
+
+
+
+ In addition, we observe that as t-values in the first quadrant get closer to \frac{\pi}{2}, \sin(t) gets closer to 1, while \cos(t) gets closer to 0 (while being always positive). Noting that \frac{\pi}{2} \approx 1.57, we observe that
+
+ \tan(1.47) = \frac{\sin(1.47)}{\cos(1.47)} \approx \frac{0.995}{0.101} = 9.887
+
+ and
+
+ \tan(1.56) = \frac{\sin(1.56)}{\cos(1.56)} \approx \frac{0.9994}{0.0108} = 92.6205
+ .
+ Because the ratio of numbers closer and closer to 1 divided by numbers closer and closer to 0 (but positive) increases without bound, this means that \tan(t) increases without bound as t approaches \frac{\pi}{2} from the left side. Once t is slightly greater than \frac{\pi}{2} in Quadrant II, the value of \sin(t) stays close to 1, but now the value of \cos(t) is negative (and close to zero). For instance, \cos(1.58) \approx -0.0092. This makes the value of \tan(t) decrease without bound (negative and getting further away from 0) for t approaching \frac{\pi}{2} from the right side, and results in h(t) = \tan(t) having a vertical asymptote at t = \frac{\pi}{2}. The periodicity and sign behaviors of \sin(t) and \cos(t) mean this asymptotic behavior of the tangent function will repeat.
+
+
+
+ Plotting the data in the table along with the expected asymptotes and connecting the points intuitively, we see the graph of the tangent function in Figure.
+
+
+
+
A plot of the tangent function together with special points that come from the unit circle.
+
A plot of the tangent function together with special points that come from the unit circle.
+
+
+
+
+ We see from Table and Table as well as from Figure that the tangent function has period P = \pi and that the function is increasing on any interval on which it is defined. We summarize our recent work as follows.
+
+
+
+ Properties of the tangent function
+
+ For the function h(t) = \tan(t),
+
+
+
+ its domain is the set of all real numbers except t = \frac{\pi}{2} \pm k\pi where k is any whole number;
+
+
+
+
+ its range is the set of all real numbers;
+
+
+
+
+ its period is P = \pi;
+
+
+
+
+ is increasing on any interval on which the function is defined at every point in the interval.
+
+
+
+
+
+
+
+ While the tangent function is an interesting mathematical function for its own sake, its most important applications arise in the setting of right triangles, and for the remainder of this section we will focus on that perspective.
+
+
+
+
+
+ Using the tangent function in right triangles
+
+ The tangent function offers us an additional choice when working in right triangles with limited information. In the setting where we have a right triangle with one additional known angle, if we know the length of the hypotenuse, we can use either the sine or cosine of the angle to help us easily find the remaining side lengths. But in the setting where we know only the length of one leg, the tangent function now allows us to determine the value of the remaining leg in a similarly straightforward way, and from there the hypotenuse.
+
+
+
+
+
+ Use the tangent function to determine the width, w, of the river in Figure. (Note that here we are revisiting the problem in Activity, which we previously solved without using the tangent function.) What other information can we now easily determine?
+
+
+
+
A right triangle with one angle and one leg known.
+
A right triangle with one angle and one leg known.
+
+
+
+
+
+ Using the perspective that \tan(\theta) = \frac{\text{opp}}{\text{adj}} in a right triangle, in this context we have
+
+ \tan(56.4^\circ) = \frac{w}{50}
+
+ and thus w = 50\tan(56.4) is the exact width of the river. Using a computational device, we find that w \approx 75.256.
+
+
+ Once we know the river's width, we can use the Pythagorean theorem or the sine function to determine the distance from P to A, at which point all 6 parts of the triangle are known.
+
+
+
+
+
+ The tangent function finds a wide range of applications in finding missing information in right triangles where information about one or more legs of the triangle is known.
+
+
+
+
+
+
+
+
+
+
+
+ Summary
+
+
+
+
+ The tangent function is defined defined to be the ratio of the sine and cosine functions according to the rule
+
+ \tan(t) = \frac{\sin(t)}{\cos(t)}
+
+ for all values of t for which \cos(t) \ne 0.
+
+
+
+
+ The graph of the tangent function differs substantially from the graphs of the sine and cosine functions, primarily because near values where \cos(t) = 0, the ratio of \frac{\sin(t)}{\cos(t)} increases or decreases without bound, producing vertical asymptotes. In addition, while the period of the sine and cosine functions is P = 2\pi, the period of the tangent function is P = \pi due to how the sine and cosine functions repeat the same values (with different signs) as a point traverses the unit circle.
+
+
+
+
+ The tangent function finds some of its most important applications in the setting of right triangles where one leg of the triangle is known and one of the non-right angles is known. Computing the tangent of the known angle, say \alpha, and using the fact that
+
+ \tan(\alpha) = \frac{\text{opp}}{\text{adj}}
+
+ we can then find the missing leg's length in terms of the other and the tangent of the angle.
+
+
+
+
+
+
+
+
+
+
+
diff --git a/source/titlepage.xml b/source/titlepage.xml
index 236f7b2d..a0cfd0e3 100644
--- a/source/titlepage.xml
+++ b/source/titlepage.xml
@@ -1,16 +1,16 @@
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/xsl/activity-task.xsl b/xsl/activity-task.xsl
index 4c2b1541..de2028fb 100644
--- a/xsl/activity-task.xsl
+++ b/xsl/activity-task.xsl
@@ -1,107 +1,107 @@
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
- ws-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
- 0
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
- -task-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ ws-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ 0
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ -task-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
\ No newline at end of file
diff --git a/xsl/apc-activity-workbook.xsl b/xsl/apc-activity-workbook.xsl
index 1129812c..8eab3da3 100644
--- a/xsl/apc-activity-workbook.xsl
+++ b/xsl/apc-activity-workbook.xsl
@@ -1,213 +1,213 @@
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
- %entities;
-]>
-
-
-
-
-
-
-
-
-
-
-
-
-
-
- enhanced,frame hidden,interior hidden, sharp corners,
- boxrule=0pt,borderline west={3pt}{0pt}{ActiveBlue},
- runintitlestyle, blockspacingstyle, after title={.\space},
- colback=white,
- coltitle=black,after={\cleardoublepage}
-
-
-
-
-
-
-
-
-
-
-
-
-
- %% Customized to load Palatino fonts
- \usepackage[T1]{fontenc}
- \renewcommand{\rmdefault}{zpltlf} %Roman font for use in math mode
- \usepackage[scaled=.85]{beramono}% used only by \mathtt
- \usepackage[type1]{cabin}%used only by \mathsf
- \usepackage{amsmath,amssymb}%load before newpxmath
- \usepackage[varg,cmintegrals,bigdelims,varbb]{newpxmath}
- \usepackage[scr=rsfso]{mathalfa}
- \usepackage{bm} %load after all math to give access to bold math
- % Now load the otf text fonts using fontspec--wont affect math
- \usepackage[no-math]{fontspec}
- \setmainfont{TeXGyrePagellaX}
- \defaultfontfeatures{Ligatures=TeX,Scale=1,Mapping=tex-text}
- \linespread{1.02}
-
-
-
- %% Used to get WeBWorK logo into margin next to WW exercises
- \usepackage{marginnote}
-
- %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
- % CC icon at bottom of first page of each chapter
- %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
- \newpagestyle{chapopen}{
- \sethead[][][] % even
- {}{}{} % odd
- \setfoot[\includegraphics[height=1pc]{external/images/CC-BY-SA-license.pdf}][][]
-{}{}{\includegraphics[height=1pc]{external/images/CC-BY-SA-license.pdf}}}
- \assignpagestyle{\chapter}{chapopen}
-
- %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
- % Modified from Mitch Keller's chapter handling
- %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
- %%% This is from common
- \definecolor{ActiveBlue}{cmyk}{1, 0.5, 0, 0.35}
- \colorlet{chaptercolor}{ActiveBlue}
- %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
- % Basic paragraph parameters
- %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
- \setlength{\parindent}{0mm}
- \setlength{\parskip}{0.5pc}
- %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
- % In print, trying to reduce color use
- %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
- \hypersetup{colorlinks=true,linkcolor=black,citecolor=black,
- filecolor=black,urlcolor=black}
-
-
-
-
- {\sethead[\textsl{\ifthechapter{\chaptertitlename{}
- \thechapter }{} }][][]
- {}{}{\textsl{\ifthesection{\thesection{} \sectiontitle}
- {} }}
- \setfoot[\thepage][][]
- {}{}{\thepage}}
-
-
-
-
-
- \titleformat{\chapter}[display]
- {\raggedleft\normalfont\color{chaptercolor}\Large}{
- \MakeUppercase{\divisionnameptx}\space
- \rlap{\enskip\resizebox{!}{0.95cm}{\thechapter}
- }}{10pt}{\normalfont\Huge\itshape#1}
- [{\Large\authorsptx}]
- \titleformat{name=\chapter,numberless}[display]
- {\raggedleft\normalfont\color{chaptercolor}\Huge\itshape}{}{0pt}{#1}
- [{\Large\authorsptx}]
- \titlespacing*{\chapter}{0pt}{0pt}{0pt}
-
-
-
- \titleformat{\section}[block]
- {\normalfont\Large\bfseries}{\thesection\space\titleptx}{1em}{}
- [{\large\authorsptx}]
- \titleformat{name=\section,numberless}[block]
- {\normalfont\Large\bfseries}{}{0pt}{#1}
- [{\large\authorsptx}]
- \titlespacing*{\section}{0pt}{3.5ex plus 1ex minus .2ex}{2.3ex plus .2ex}
- \titleformat{\subsection}[block]
- {\normalfont\large\bfseries}{\thesubsection\space\titleptx}{1em}{}
- [{\normalsize\authorsptx}]
- \titleformat{name=\subsection,numberless}[block]
- {\normalfont\large\bfseries}{}{0pt}{#1}
- [{\normalsize\authorsptx}]
- \titlespacing*{\subsection}{0pt}{3.25ex plus 1ex minus .2ex}{1.5ex plus .2ex}
- \titleformat{\subsubsection}[block]
- {\normalfont\normalsize\bfseries}{\thesubsubsection\space\titleptx}{1em}{}
- [{\small\authorsptx}]
- \titleformat{name=\subsubsection,numberless}[block]
- {\normalfont\normalsize\bfseries}{}{0pt}{#1}
- [{\normalsize\authorsptx}]
- \titlespacing*{\subsubsection}{0pt}{3.25ex plus 1ex minus .2ex}{1.5ex plus .2ex}
-
-
-
-
- bwminimalstyle, runintitlestyle, blockspacingstyle, after title={\space},
-
-
-
- bwminimalstyle, runintitlestyle, blockspacingstyle, after title={\space},
-
-
-
-
-
-
- enhanced,frame hidden,interior hidden,sharp corners,
- blockspacingstyle,boxrule=0pt,left=0pt,right=0pt,
- fonttitle=\large\bfseries,
- borderline north={0.1ex}{0pt}{black},
- toptitle=0.5ex,top=2ex, bottom=0.5ex,
- borderline south={0.1ex}{0pt}{black},coltitle=black,
-
-
-
-
-
-
- skin=enhanced, arc=2ex,
- colback=ActiveBlue!5,colframe=ActiveBlue!75!black,
- colbacktitle=ActiveBlue!20, coltitle=black,
- boxed title style={sharp corners, frame hidden},
- fonttitle=\bfseries, attach boxed title to top
- left={xshift=4mm,yshift=-3mm}, top=3mm,
-
-
-
-
-
-
-
-
-
- (\nolinkurl{
-
-
-
-
-
- })
-
-
-
-
-
- \footnotetext[
-
- ]
- {
-
-
- }%
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ %entities;
+]>
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ enhanced,frame hidden,interior hidden, sharp corners,
+ boxrule=0pt,borderline west={3pt}{0pt}{ActiveBlue},
+ runintitlestyle, blockspacingstyle, after title={.\space},
+ colback=white,
+ coltitle=black,after={\cleardoublepage}
+
+
+
+
+
+
+
+
+
+
+
+
+
+ %% Customized to load Palatino fonts
+ \usepackage[T1]{fontenc}
+ \renewcommand{\rmdefault}{zpltlf} %Roman font for use in math mode
+ \usepackage[scaled=.85]{beramono}% used only by \mathtt
+ \usepackage[type1]{cabin}%used only by \mathsf
+ \usepackage{amsmath,amssymb}%load before newpxmath
+ \usepackage[varg,cmintegrals,bigdelims,varbb]{newpxmath}
+ \usepackage[scr=rsfso]{mathalfa}
+ \usepackage{bm} %load after all math to give access to bold math
+ % Now load the otf text fonts using fontspec--wont affect math
+ \usepackage[no-math]{fontspec}
+ \setmainfont{TeXGyrePagellaX}
+ \defaultfontfeatures{Ligatures=TeX,Scale=1,Mapping=tex-text}
+ \linespread{1.02}
+
+
+
+ %% Used to get WeBWorK logo into margin next to WW exercises
+ \usepackage{marginnote}
+
+ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+ % CC icon at bottom of first page of each chapter
+ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+ \newpagestyle{chapopen}{
+ \sethead[][][] % even
+ {}{}{} % odd
+ \setfoot[\includegraphics[height=1pc]{external/images/CC-BY-SA-license.pdf}][][]
+{}{}{\includegraphics[height=1pc]{external/images/CC-BY-SA-license.pdf}}}
+ \assignpagestyle{\chapter}{chapopen}
+
+ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+ % Modified from Mitch Keller's chapter handling
+ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+ %%% This is from common
+ \definecolor{ActiveBlue}{cmyk}{1, 0.5, 0, 0.35}
+ \colorlet{chaptercolor}{ActiveBlue}
+ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+ % Basic paragraph parameters
+ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+ \setlength{\parindent}{0mm}
+ \setlength{\parskip}{0.5pc}
+ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+ % In print, trying to reduce color use
+ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+ \hypersetup{colorlinks=true,linkcolor=black,citecolor=black,
+ filecolor=black,urlcolor=black}
+
+
+
+
+ {\sethead[\textsl{\ifthechapter{\chaptertitlename{}
+ \thechapter }{} }][][]
+ {}{}{\textsl{\ifthesection{\thesection{} \sectiontitle}
+ {} }}
+ \setfoot[\thepage][][]
+ {}{}{\thepage}}
+
+
+
+
+
+ \titleformat{\chapter}[display]
+ {\raggedleft\normalfont\color{chaptercolor}\Large}{
+ \MakeUppercase{\divisionnameptx}\space
+ \rlap{\enskip\resizebox{!}{0.95cm}{\thechapter}
+ }}{10pt}{\normalfont\Huge\itshape#1}
+ [{\Large\authorsptx}]
+ \titleformat{name=\chapter,numberless}[display]
+ {\raggedleft\normalfont\color{chaptercolor}\Huge\itshape}{}{0pt}{#1}
+ [{\Large\authorsptx}]
+ \titlespacing*{\chapter}{0pt}{0pt}{0pt}
+
+
+
+ \titleformat{\section}[block]
+ {\normalfont\Large\bfseries}{\thesection\space\titleptx}{1em}{}
+ [{\large\authorsptx}]
+ \titleformat{name=\section,numberless}[block]
+ {\normalfont\Large\bfseries}{}{0pt}{#1}
+ [{\large\authorsptx}]
+ \titlespacing*{\section}{0pt}{3.5ex plus 1ex minus .2ex}{2.3ex plus .2ex}
+ \titleformat{\subsection}[block]
+ {\normalfont\large\bfseries}{\thesubsection\space\titleptx}{1em}{}
+ [{\normalsize\authorsptx}]
+ \titleformat{name=\subsection,numberless}[block]
+ {\normalfont\large\bfseries}{}{0pt}{#1}
+ [{\normalsize\authorsptx}]
+ \titlespacing*{\subsection}{0pt}{3.25ex plus 1ex minus .2ex}{1.5ex plus .2ex}
+ \titleformat{\subsubsection}[block]
+ {\normalfont\normalsize\bfseries}{\thesubsubsection\space\titleptx}{1em}{}
+ [{\small\authorsptx}]
+ \titleformat{name=\subsubsection,numberless}[block]
+ {\normalfont\normalsize\bfseries}{}{0pt}{#1}
+ [{\normalsize\authorsptx}]
+ \titlespacing*{\subsubsection}{0pt}{3.25ex plus 1ex minus .2ex}{1.5ex plus .2ex}
+
+
+
+
+ bwminimalstyle, runintitlestyle, blockspacingstyle, after title={\space},
+
+
+
+ bwminimalstyle, runintitlestyle, blockspacingstyle, after title={\space},
+
+
+
+
+
+
+ enhanced,frame hidden,interior hidden,sharp corners,
+ blockspacingstyle,boxrule=0pt,left=0pt,right=0pt,
+ fonttitle=\large\bfseries,
+ borderline north={0.1ex}{0pt}{black},
+ toptitle=0.5ex,top=2ex, bottom=0.5ex,
+ borderline south={0.1ex}{0pt}{black},coltitle=black,
+
+
+
+
+
+
+ skin=enhanced, arc=2ex,
+ colback=ActiveBlue!5,colframe=ActiveBlue!75!black,
+ colbacktitle=ActiveBlue!20, coltitle=black,
+ boxed title style={sharp corners, frame hidden},
+ fonttitle=\bfseries, attach boxed title to top
+ left={xshift=4mm,yshift=-3mm}, top=3mm,
+
+
+
+
+
+
+
+
+
+ (\nolinkurl{
+
+
+
+
+
+ })
+
+
+
+
+
+ \footnotetext[
+
+ ]
+ {
+
+
+ }%
+
+
+
\ No newline at end of file
diff --git a/xsl/apc-common.xsl b/xsl/apc-common.xsl
index da6c9177..585b0e8a 100644
--- a/xsl/apc-common.xsl
+++ b/xsl/apc-common.xsl
@@ -1,162 +1,162 @@
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
- %entities;
-]>
-
-
-
-
-
-
-
- %% Customized to load Palatino fonts
- \usepackage[T1]{fontenc}
- \renewcommand{\rmdefault}{zpltlf} %Roman font for use in math mode
- \usepackage[scaled=.85]{beramono}% used only by \mathtt
- \usepackage[type1]{cabin}%used only by \mathsf
- \usepackage[varg,cmintegrals,bigdelims,varbb]{newpxmath}
- \usepackage[scr=rsfso]{mathalfa}
- \usepackage{bm} %load after all math to give access to bold math
- % Now load the otf text fonts using fontspec--wont affect math
- \usepackage[no-math]{fontspec}
- \setmainfont{TeXGyrePagellaX}
- \defaultfontfeatures{Ligatures=TeX,Scale=1,Mapping=tex-text}
- \linespread{1.02}
- % Deal with a conflict with amssymb and our fonts
- \let\Bbbk\relax
-
-
-
- %% Used to get WeBWorK logo into margin next to WW exercises
- \usepackage{marginnote}
-
- %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
- % CC icon at bottom of first page of each chapter
- %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
- \newpagestyle{chapopen}{
- \sethead[][][] % even
- {}{}{} % odd
- \setfoot[\includegraphics[height=1pc]{external/images/CC-BY-SA-license.pdf}][][]
-{}{}{\includegraphics[height=1pc]{external/images/CC-BY-SA-license.pdf}}}
- \assignpagestyle{\chapter}{chapopen}
-
- %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
- % Modified from Mitch Keller's chapter handling
- %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
- %%% This is from common
- \definecolor{ActiveBlue}{cmyk}{1, 0.5, 0, 0.35}
- \colorlet{chaptercolor}{ActiveBlue}
- %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
- % Basic paragraph parameters
- %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
- \setlength{\parindent}{0mm}
- \setlength{\parskip}{0.5pc}
- %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
- % In print, trying to reduce color use
- %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
- \hypersetup{colorlinks=true,linkcolor=black,citecolor=black,
- filecolor=black,urlcolor=black}
-
-
-
-
-
- {\sethead[\textsl{\ifthechapter{\chaptertitlename{}
- \thechapter }{} \chaptertitle}][][]
- {}{}{\textsl{\ifthesection{\thesection{} \sectiontitle}
- {\sectiontitle} }}
- \setfoot[\thepage][][]
- {}{}{\thepage}}
-
-
-
-
-
- \titleformat{\chapter}[display]
- {\raggedleft\normalfont\color{chaptercolor}\Large}{
- \MakeUppercase{\divisionnameptx}\space
- \rlap{\enskip\resizebox{!}{0.95cm}{\thechapter} \rule{15cm}{0.95cm}
- }}{10pt}{\normalfont\Huge\itshape#1}
- [{\Large\authorsptx}]
- \titleformat{name=\chapter,numberless}[display]
- {\raggedleft\normalfont\color{chaptercolor}\Huge\itshape}{}{0pt}{#1}
- [{\Large\authorsptx}]
- \titlespacing*{\chapter}{0pt}{30pt}{20pt}
-
-
- \titleformat{\section}[block]
-{\normalfont\Large\bfseries}{\thesection\space\titleptx}{1em}{}
- [{\large\authorsptx}]
- \titleformat{name=\section,numberless}[block]
- {\normalfont\Large\bfseries}{}{0pt}{#1}
- [{\large\authorsptx}]
- \titlespacing*{\section}{0pt}{3.5ex plus 1ex minus .2ex}{2.3ex plus .2ex}
- \titleformat{\subsection}[block]
- {\normalfont\large\bfseries}{\thesubsection\space\titleptx}{1em}{}
- [{\normalsize\authorsptx}]
- \titleformat{name=\subsection,numberless}[block]
- {\normalfont\large\bfseries}{}{0pt}{#1}
- [{\normalsize\authorsptx}]
- \titlespacing*{\subsection}{0pt}{3.25ex plus 1ex minus .2ex}{1.5ex plus .2ex}
- \titleformat{\subsubsection}[block]
- {\normalfont\normalsize\bfseries}{\thesubsubsection\space\titleptx}{1em}{}
- [{\small\authorsptx}]
- \titleformat{name=\subsubsection,numberless}[block]
- {\normalfont\normalsize\bfseries}{}{0pt}{#1}
- [{\normalsize\authorsptx}]
- \titlespacing*{\subsubsection}{0pt}{3.25ex plus 1ex minus .2ex}{1.5ex plus .2ex}
-
-
-
-
- enhanced,frame hidden,interior hidden, sharp corners,
- boxrule=0pt,borderline west={3pt}{0pt}{ActiveBlue},
- runintitlestyle, blockspacingstyle, after title={.\space},
- colback=white,
- coltitle=black,
-
-
-
-
-
- enhanced,frame hidden,interior hidden,sharp corners,
- blockspacingstyle,boxrule=0pt,left=0pt,right=0pt,
- fonttitle=\large\bfseries,
- borderline north={0.1ex}{0pt}{black},
- toptitle=0.5ex,top=2ex, bottom=0.5ex,
- borderline south={0.1ex}{0pt}{black},coltitle=black,
-
-
-
-
-
-
- breakable, skin=enhanced, arc=2ex,
- colback=ActiveBlue!5,colframe=ActiveBlue!75!black,
- colbacktitle=ActiveBlue!20, coltitle=black,
- boxed title style={sharp corners, frame hidden},
- fonttitle=\bfseries, attach boxed title to top
- left={xshift=4mm,yshift=-3mm}, top=3mm,
-
-
-
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ %entities;
+]>
+
+
+
+
+
+
+
+ %% Customized to load Palatino fonts
+ \usepackage[T1]{fontenc}
+ \renewcommand{\rmdefault}{zpltlf} %Roman font for use in math mode
+ \usepackage[scaled=.85]{beramono}% used only by \mathtt
+ \usepackage[type1]{cabin}%used only by \mathsf
+ \usepackage[varg,cmintegrals,bigdelims,varbb]{newpxmath}
+ \usepackage[scr=rsfso]{mathalfa}
+ \usepackage{bm} %load after all math to give access to bold math
+ % Now load the otf text fonts using fontspec--wont affect math
+ \usepackage[no-math]{fontspec}
+ \setmainfont{TeXGyrePagellaX}
+ \defaultfontfeatures{Ligatures=TeX,Scale=1,Mapping=tex-text}
+ \linespread{1.02}
+ % Deal with a conflict with amssymb and our fonts
+ \let\Bbbk\relax
+
+
+
+ %% Used to get WeBWorK logo into margin next to WW exercises
+ \usepackage{marginnote}
+
+ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+ % CC icon at bottom of first page of each chapter
+ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+ \newpagestyle{chapopen}{
+ \sethead[][][] % even
+ {}{}{} % odd
+ \setfoot[\includegraphics[height=1pc]{external/images/CC-BY-SA-license.pdf}][][]
+{}{}{\includegraphics[height=1pc]{external/images/CC-BY-SA-license.pdf}}}
+ \assignpagestyle{\chapter}{chapopen}
+
+ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+ % Modified from Mitch Keller's chapter handling
+ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+ %%% This is from common
+ \definecolor{ActiveBlue}{cmyk}{1, 0.5, 0, 0.35}
+ \colorlet{chaptercolor}{ActiveBlue}
+ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+ % Basic paragraph parameters
+ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+ \setlength{\parindent}{0mm}
+ \setlength{\parskip}{0.5pc}
+ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+ % In print, trying to reduce color use
+ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+ \hypersetup{colorlinks=true,linkcolor=black,citecolor=black,
+ filecolor=black,urlcolor=black}
+
+
+
+
+
+ {\sethead[\textsl{\ifthechapter{\chaptertitlename{}
+ \thechapter }{} \chaptertitle}][][]
+ {}{}{\textsl{\ifthesection{\thesection{} \sectiontitle}
+ {\sectiontitle} }}
+ \setfoot[\thepage][][]
+ {}{}{\thepage}}
+
+
+
+
+
+ \titleformat{\chapter}[display]
+ {\raggedleft\normalfont\color{chaptercolor}\Large}{
+ \MakeUppercase{\divisionnameptx}\space
+ \rlap{\enskip\resizebox{!}{0.95cm}{\thechapter} \rule{15cm}{0.95cm}
+ }}{10pt}{\normalfont\Huge\itshape#1}
+ [{\Large\authorsptx}]
+ \titleformat{name=\chapter,numberless}[display]
+ {\raggedleft\normalfont\color{chaptercolor}\Huge\itshape}{}{0pt}{#1}
+ [{\Large\authorsptx}]
+ \titlespacing*{\chapter}{0pt}{30pt}{20pt}
+
+
+ \titleformat{\section}[block]
+{\normalfont\Large\bfseries}{\thesection\space\titleptx}{1em}{}
+ [{\large\authorsptx}]
+ \titleformat{name=\section,numberless}[block]
+ {\normalfont\Large\bfseries}{}{0pt}{#1}
+ [{\large\authorsptx}]
+ \titlespacing*{\section}{0pt}{3.5ex plus 1ex minus .2ex}{2.3ex plus .2ex}
+ \titleformat{\subsection}[block]
+ {\normalfont\large\bfseries}{\thesubsection\space\titleptx}{1em}{}
+ [{\normalsize\authorsptx}]
+ \titleformat{name=\subsection,numberless}[block]
+ {\normalfont\large\bfseries}{}{0pt}{#1}
+ [{\normalsize\authorsptx}]
+ \titlespacing*{\subsection}{0pt}{3.25ex plus 1ex minus .2ex}{1.5ex plus .2ex}
+ \titleformat{\subsubsection}[block]
+ {\normalfont\normalsize\bfseries}{\thesubsubsection\space\titleptx}{1em}{}
+ [{\small\authorsptx}]
+ \titleformat{name=\subsubsection,numberless}[block]
+ {\normalfont\normalsize\bfseries}{}{0pt}{#1}
+ [{\normalsize\authorsptx}]
+ \titlespacing*{\subsubsection}{0pt}{3.25ex plus 1ex minus .2ex}{1.5ex plus .2ex}
+
+
+
+
+ enhanced,frame hidden,interior hidden, sharp corners,
+ boxrule=0pt,borderline west={3pt}{0pt}{ActiveBlue},
+ runintitlestyle, blockspacingstyle, after title={.\space},
+ colback=white,
+ coltitle=black,
+
+
+
+
+
+ enhanced,frame hidden,interior hidden,sharp corners,
+ blockspacingstyle,boxrule=0pt,left=0pt,right=0pt,
+ fonttitle=\large\bfseries,
+ borderline north={0.1ex}{0pt}{black},
+ toptitle=0.5ex,top=2ex, bottom=0.5ex,
+ borderline south={0.1ex}{0pt}{black},coltitle=black,
+
+
+
+
+
+
+ breakable, skin=enhanced, arc=2ex,
+ colback=ActiveBlue!5,colframe=ActiveBlue!75!black,
+ colbacktitle=ActiveBlue!20, coltitle=black,
+ boxed title style={sharp corners, frame hidden},
+ fonttitle=\bfseries, attach boxed title to top
+ left={xshift=4mm,yshift=-3mm}, top=3mm,
+
+
+
+
+
+
+
diff --git a/xsl/apc-latex.xsl b/xsl/apc-latex.xsl
index 1724715b..0a2e8aab 100644
--- a/xsl/apc-latex.xsl
+++ b/xsl/apc-latex.xsl
@@ -1,120 +1,120 @@
-
-
-
-
-
-
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-
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-
-
- %entities;
-]>
-
-
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-
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-
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-
-
-
-
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-
-
-
- %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
- % Start sections on new page, just not the first one
- %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
- \let\oldsection\section
- \renewcommand\section{\znewpage\oldsection}
-
- \let\oldchapter\chapter
- \renewcommand\chapter{\clearpage\gdef\znewpage{
- \global\let\znewpage\clearpage}\oldchapter}
-
- \global\let\znewpage\clearpage
-
-
-
-
-
-
-
-
-
-
- \renewcommand\section{\oldsection}
-
-
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- \marginnote{\hspace*{-10ex}\tiny \includegraphics[width=0.25in]{external/images/webwork-logo.png}}
-
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- \Large
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- \large
-
-
-
- \Large
-
-
- \large
-
-
- \normalsize
-
-
- \normalsize
-
-
- PTX:BUG: "solutions" division title does not have a font size
-
-
-
-
-
-
-
-
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-
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-
+
+
+
+
+
+
+
+
+
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+
+
+
+
+
+
+
+ %entities;
+]>
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+ % Start sections on new page, just not the first one
+ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+ \let\oldsection\section
+ \renewcommand\section{\znewpage\oldsection}
+
+ \let\oldchapter\chapter
+ \renewcommand\chapter{\clearpage\gdef\znewpage{
+ \global\let\znewpage\clearpage}\oldchapter}
+
+ \global\let\znewpage\clearpage
+
+
+
+
+
+
+
+
+
+
+ \renewcommand\section{\oldsection}
+
+
+
+
+
+ \marginnote{\hspace*{-10ex}\tiny \includegraphics[width=0.25in]{external/images/webwork-logo.png}}
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ \Large
+
+
+
+ \large
+
+
+
+ \Large
+
+
+ \large
+
+
+ \normalsize
+
+
+ \normalsize
+
+
+ PTX:BUG: "solutions" division title does not have a font size
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/xsl/apc-solution-manual.xsl b/xsl/apc-solution-manual.xsl
index 10aa8958..2b6ea7c7 100644
--- a/xsl/apc-solution-manual.xsl
+++ b/xsl/apc-solution-manual.xsl
@@ -1,50 +1,50 @@
-
-
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-
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- % These override what PreTeXt supplies by default. The only addition
- % is the after clause. If things start looking odd, look earlier in
- % the LaTeX file to see how PreTeXt has redefined these three styles
- % and update latex.preamble.late
- \tcbset{ divisionsolutionstyle/.style={bwminimalstyle,
- runintitlestyle, exercisespacingstyle, after title={\space},
- breakable, parbox=false, after={\clearpage} } }
- \tcbset{ explorationsolutionstyle/.style={bwminimalstyle,
- runintitlestyle, exercisespacingstyle, after
- title={\space},
- breakable, parbox=false ,after={\clearpage}} }
- \tcbset{ activitysolutionstyle/.style={bwminimalstyle,
- runintitlestyle, exercisespacingstyle, after title={\space},
- breakable, parbox=false, after={\clearpage}} }
-
-
-
-
-
+
+
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+
+
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+
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+
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+
+
+
+ % These override what PreTeXt supplies by default. The only addition
+ % is the after clause. If things start looking odd, look earlier in
+ % the LaTeX file to see how PreTeXt has redefined these three styles
+ % and update latex.preamble.late
+ \tcbset{ divisionsolutionstyle/.style={bwminimalstyle,
+ runintitlestyle, exercisespacingstyle, after title={\space},
+ breakable, parbox=false, after={\clearpage} } }
+ \tcbset{ explorationsolutionstyle/.style={bwminimalstyle,
+ runintitlestyle, exercisespacingstyle, after
+ title={\space},
+ breakable, parbox=false ,after={\clearpage}} }
+ \tcbset{ activitysolutionstyle/.style={bwminimalstyle,
+ runintitlestyle, exercisespacingstyle, after title={\space},
+ breakable, parbox=false, after={\clearpage}} }
+
+
+
+
+
diff --git a/xsl/entities.ent b/xsl/entities.ent
index 681f71b5..f03bc63d 100644
--- a/xsl/entities.ent
+++ b/xsl/entities.ent
@@ -1,139 +1,139 @@
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