From a5fa6b559ad98d360731a011d7ecbca1d136a0cb Mon Sep 17 00:00:00 2001
From: David Austin
Date: Fri, 10 Jul 2026 19:57:00 -0400
Subject: [PATCH 1/6] =?UTF-8?q?Add=20solutions=20to=20Chapter=201=20activi?=
=?UTF-8?q?ties=20and=20previews=20(sections=201.2=E2=80=931.9)?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
Co-Authored-By: Claude Sonnet 4.6
---
publication/publication.ptx | 2 +-
.../activities/act-changing-aroc-graphs.xml | 32 +++++++--
.../act-changing-aroc-population.xml | 53 +++++++++++++--
.../activities/act-changing-aroc-trends.xml | 41 ++++++++++--
.../act-changing-combining-arithmetic.xml | 51 ++++++++++++--
.../act-changing-combining-context.xml | 42 ++++++++++--
.../act-changing-combining-piecewise.xml | 32 ++++++++-
.../act-changing-composite-aroc.xml | 34 ++++++++--
...ct-changing-composite-crickets-celsius.xml | 22 +++++-
.../act-changing-composite-tables-graphs.xml | 57 +++++++++++++---
.../act-changing-functions-is-it.xml | 36 ++++++++--
...ging-functions-spherical-tank-draining.xml | 55 +++++++++++++--
.../act-changing-functions-spherical-tank.xml | 52 ++++++++++++--
.../act-changing-inverse-Dolbear.xml | 33 +++++++--
.../act-changing-inverse-does-it.xml | 43 ++++++++++--
.../act-changing-inverse-rainfall.xml | 50 ++++++++++++--
.../act-changing-linear-Kilimanjaro.xml | 43 ++++++++++--
.../act-changing-linear-finding-eqs.xml | 46 +++++++++++--
.../act-changing-linear-in-context.xml | 41 ++++++++++--
.../act-changing-quadratic-falling-ball.xml | 46 +++++++++++--
.../act-changing-quadratic-parameters.xml | 45 +++++++++++--
.../act-changing-quadratic-properties.xml | 51 ++++++++++++--
.../act-changing-transformations-combined.xml | 37 ++++++++--
...-changing-transformations-translations.xml | 34 +++++++++-
...-changing-transformations-vert-stretch.xml | 40 +++++++++--
source/previews/PA-changing-aroc.xml | 67 +++++++++++++++++--
source/previews/PA-changing-combining.xml | 48 +++++++++++--
source/previews/PA-changing-composite.xml | 32 +++++++--
source/previews/PA-changing-inverse-F-C.xml | 40 +++++++++--
source/previews/PA-changing-linear-3-ex.xml | 50 ++++++++++++--
source/previews/PA-changing-quadratic.xml | 37 ++++++++--
.../PA-changing-transformations-quadratic.xml | 34 ++++++++--
32 files changed, 1176 insertions(+), 150 deletions(-)
diff --git a/publication/publication.ptx b/publication/publication.ptx
index 6fcbe811..01f7f032 100644
--- a/publication/publication.ptx
+++ b/publication/publication.ptx
@@ -20,7 +20,7 @@
-
+
diff --git a/source/activities/act-changing-aroc-graphs.xml b/source/activities/act-changing-aroc-graphs.xml
index cc8c872d..f2a65a85 100755
--- a/source/activities/act-changing-aroc-graphs.xml
+++ b/source/activities/act-changing-aroc-graphs.xml
@@ -21,7 +21,7 @@
- f is a function defined on [-1,7] such that f(1) = 4 and AV_{[1,3]} = -2.
+
f is a function defined on [-1,7] such that f(1) = 4 and AV_{[1,3]} = -2.
ADD ALT TEXT TO THIS IMAGE
@@ -30,7 +30,15 @@
-
+
+
+ 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).
+
+
@@ -43,7 +51,15 @@
-
+
+
+ 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).
+
+
@@ -56,7 +72,15 @@
-
+
+
+ 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].
+
+
diff --git a/source/activities/act-changing-aroc-population.xml b/source/activities/act-changing-aroc-population.xml
index 0b76d780..87f0a5a3 100755
--- a/source/activities/act-changing-aroc-population.xml
+++ b/source/activities/act-changing-aroc-population.xml
@@ -79,7 +79,24 @@
-
+
+
+ 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.}
+
+
+
@@ -89,7 +106,11 @@
-
+
+
+ The units on AV_{[1990,2010]} for both counties are people per year.
+
+
@@ -99,7 +120,12 @@
-
+
+
+ Between 1990 and 2010, the population of Ottawa County grew by an average of
+ approximately 3801.65 people per year.
+
+
@@ -110,7 +136,17 @@
-
+
+
+ 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.
+
+
@@ -120,7 +156,14 @@
-
+
+
+ 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.}
+
+
diff --git a/source/activities/act-changing-aroc-trends.xml b/source/activities/act-changing-aroc-trends.xml
index 2a546740..b6e9b551 100755
--- a/source/activities/act-changing-aroc-trends.xml
+++ b/source/activities/act-changing-aroc-trends.xml
@@ -30,7 +30,16 @@
-
+
+
+ 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].
+
+
@@ -43,7 +52,18 @@
-
+
+
+ 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.
+
+
@@ -73,7 +93,14 @@
-
+
+
+ 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].
+
+
@@ -86,7 +113,13 @@
-
+
+
+ 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].
+
+
diff --git a/source/activities/act-changing-combining-arithmetic.xml b/source/activities/act-changing-combining-arithmetic.xml
index fc4829ff..e978222e 100755
--- a/source/activities/act-changing-combining-arithmetic.xml
+++ b/source/activities/act-changing-combining-arithmetic.xml
@@ -37,7 +37,12 @@
-
+
+
+ From the graphs, f(0) = \frac{5}{2} and g(0) = 1, so
+ (f+g)(0) = \frac{5}{2} + 1 = \frac{7}{2}.
+
+
@@ -47,7 +52,12 @@
-
+
+
+ From the graphs, f(1) = 4 and g(1) = 3, so
+ (g-f)(1) = 3 - 4 = -1.
+
+
@@ -57,7 +67,12 @@
-
+
+
+ From the graphs, f(-1) = 3 and g(-1) = 3, so
+ (f \cdot g)(-1) = 3 \cdot 3 = 9.
+
+
@@ -67,7 +82,12 @@
-
+
+
+ \left(\frac{f}{g}\right)(x) is undefined wherever g(x) = 0.
+ From the graph, g(x) = 0 at x = -2 and x = 3.
+
+
@@ -77,7 +97,14 @@
-
+
+
+ (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\}.
+
+
@@ -87,7 +114,19 @@
-
+
+
+ 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}.
+
+
diff --git a/source/activities/act-changing-combining-context.xml b/source/activities/act-changing-combining-context.xml
index f6370f69..67a4d1d9 100755
--- a/source/activities/act-changing-combining-context.xml
+++ b/source/activities/act-changing-combining-context.xml
@@ -34,7 +34,12 @@
-
+
+
+ At a speed of 60 miles per hour, the car consumes 0.04 gallons of fuel
+ for each mile traveled.
+
+
@@ -47,7 +52,13 @@
-
+
+
+ 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.
+
+
@@ -60,7 +71,14 @@
-
+
+
+ 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.
+
+
@@ -70,7 +88,14 @@
-
+
+
+ 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.
+
+
@@ -84,7 +109,14 @@
-
+
+
+ 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.
+
+
diff --git a/source/activities/act-changing-combining-piecewise.xml b/source/activities/act-changing-combining-piecewise.xml
index 6933598b..c477c918 100755
--- a/source/activities/act-changing-combining-piecewise.xml
+++ b/source/activities/act-changing-combining-piecewise.xml
@@ -37,7 +37,17 @@
-
+
+
+ For x \lt 0, use p(x) = -(x+2)^2 + 2:
+ p(-4) = -(-2)^2 + 2 = -2;
+ p(-2) = -(0)^2 + 2 = 2.
+ For x \ge 0, use p(x) = \frac{1}{2}(x-2)^2 + 1:
+ p(0) = \frac{1}{2}(4) + 1 = 3;
+ p(2) = 0 + 1 = 1;
+ p(4) = \frac{1}{2}(4) + 1 = 3.
+
+
@@ -47,7 +57,14 @@
-
+
+
+ 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).
+
+
@@ -57,7 +74,16 @@
-
+
+
+ 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.
+
+
diff --git a/source/activities/act-changing-composite-aroc.xml b/source/activities/act-changing-composite-aroc.xml
index ac60b27f..90841075 100755
--- a/source/activities/act-changing-composite-aroc.xml
+++ b/source/activities/act-changing-composite-aroc.xml
@@ -27,7 +27,15 @@
-
+
+
+
+ f(1+h) &= 2(1+h)^2 - 3(1+h) + 1
+ &= 2(1 + 2h + h^2) - 3 - 3h + 1
+ &= 2h^2 + h.
+
+
+
@@ -38,7 +46,12 @@
-
+
+
+ Since f(1) = 2 - 3 + 1 = 0 and f(1+h) = 2h^2 + h:
+ AV_{[1,1+h]} = \frac{f(1+h) - f(1)}{h} = \frac{2h^2 + h}{h} = 2h + 1.
+
+
@@ -49,7 +62,12 @@
-
+
+
+ g(1+h) = \frac{5}{1+h}.
+ This expression cannot be simplified further.
+
+
@@ -60,7 +78,15 @@
-
+
+
+ Since g(1) = 5:
+
+ AV_{[1,1+h]} &= \frac{g(1+h) - g(1)}{h} = \frac{\frac{5}{1+h} - 5}{h}
+ &= \frac{5 - 5(1+h)}{(1+h)h} = \frac{-5h}{(1+h)h} = \frac{-5}{1+h}.
+
+
+
diff --git a/source/activities/act-changing-composite-crickets-celsius.xml b/source/activities/act-changing-composite-crickets-celsius.xml
index ca7ada63..bb4ab7c3 100755
--- a/source/activities/act-changing-composite-crickets-celsius.xml
+++ b/source/activities/act-changing-composite-crickets-celsius.xml
@@ -27,7 +27,11 @@
-
+
+
+ H(N) = G(D(N)) = \frac{5}{9}((40 + 0.25N) - 32) = \frac{5}{9}(8 + 0.25N).
+
+
@@ -37,7 +41,12 @@
-
+
+
+ The function H converts the number of cricket chirps per minute directly
+ into a temperature in degrees Celsius.
+
+
@@ -67,7 +76,14 @@
-
+
+
+ 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.
+
+
diff --git a/source/activities/act-changing-composite-tables-graphs.xml b/source/activities/act-changing-composite-tables-graphs.xml
index ecbddb72..7857ecd8 100755
--- a/source/activities/act-changing-composite-tables-graphs.xml
+++ b/source/activities/act-changing-composite-tables-graphs.xml
@@ -71,7 +71,11 @@
-
+
+
+ From the graph, q(0) = 2 and p(2) = 1, so p(q(0)) = 1.
+
+
@@ -81,7 +85,12 @@
-
+
+
+ From the graph, p(0) = -\frac{1}{2} and q\!\left(-\frac{1}{2}\right) = 2,
+ so q(p(0)) = 2.
+
+
@@ -89,7 +98,11 @@
-
+
+
+ From the graph, p(-1) = -1, so (p \circ p)(-1) = p(p(-1)) = p(-1) = -1.
+
+
@@ -99,7 +112,11 @@
-
+
+
+ From the table, g(2) = 0 and f(0) = 6, so (f \circ g)(2) = 6.
+
+
@@ -109,7 +126,11 @@
-
+
+
+ From the table, f(3) = 4 and g(4) = 2, so (g \circ f)(3) = 2.
+
+
@@ -119,7 +140,12 @@
-
+
+
+ From the table, f(0) = 6, but g(6) is not defined in the table,
+ so g(f(0)) is undefined.
+
+
@@ -129,7 +155,14 @@
-
+
+
+ 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.
+
+
@@ -139,7 +172,15 @@
-
+
+
+ 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.
+
+
diff --git a/source/activities/act-changing-functions-is-it.xml b/source/activities/act-changing-functions-is-it.xml
index 7b4ba376..00ad7f1a 100755
--- a/source/activities/act-changing-functions-is-it.xml
+++ b/source/activities/act-changing-functions-is-it.xml
@@ -37,7 +37,17 @@
-
+
+
+ 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.
+
+
@@ -47,7 +57,14 @@
-
+
+
+ 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.
+
+
@@ -57,7 +74,13 @@
-
+
+
+ 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.
+
+
@@ -112,7 +135,12 @@
-
+
+
+ The table does not represent a function, because the input values
+ x = 1 and x = 2 each correspond to more than one output value.
+
+
diff --git a/source/activities/act-changing-functions-spherical-tank-draining.xml b/source/activities/act-changing-functions-spherical-tank-draining.xml
index e3b7e09b..2324f4c7 100755
--- a/source/activities/act-changing-functions-spherical-tank-draining.xml
+++ b/source/activities/act-changing-functions-spherical-tank-draining.xml
@@ -34,7 +34,20 @@
-
+
+
+ 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.
+
+
@@ -44,7 +57,14 @@
-
+
+
+ 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].
+
+
@@ -54,7 +74,19 @@
-
+
+
+ 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.
+
+
@@ -76,7 +108,14 @@
-
+
+
+ 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.
+
+
@@ -86,7 +125,13 @@
-
+
+
+ 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.
+
+
diff --git a/source/activities/act-changing-functions-spherical-tank.xml b/source/activities/act-changing-functions-spherical-tank.xml
index f32d3374..12a68ab4 100755
--- a/source/activities/act-changing-functions-spherical-tank.xml
+++ b/source/activities/act-changing-functions-spherical-tank.xml
@@ -33,7 +33,14 @@
-
+
+
+ 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}.
+
+
@@ -43,7 +50,16 @@
-
+
+
+ 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.
+
+
@@ -54,7 +70,19 @@
-
+
+
+
+ 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.
+
+
@@ -64,7 +92,15 @@
-
+
+
+ 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.
+
+
@@ -74,7 +110,13 @@
-
+
+
+ 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.
+
+
diff --git a/source/activities/act-changing-inverse-Dolbear.xml b/source/activities/act-changing-inverse-Dolbear.xml
index 65c5fe6e..61a0ae57 100755
--- a/source/activities/act-changing-inverse-Dolbear.xml
+++ b/source/activities/act-changing-inverse-Dolbear.xml
@@ -27,7 +27,12 @@
-
+
+
+ Subtracting 40 and multiplying by 4:
+ E(F) = N = 4(F - 40).
+
+
@@ -37,7 +42,12 @@
-
+
+
+ The function E takes a temperature in degrees Fahrenheit as its input and
+ outputs the corresponding number of snowy tree cricket chirps per minute.
+
+
@@ -47,7 +57,15 @@
-
+
+
+ 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.
+
+
@@ -57,7 +75,14 @@
-
+
+
+ 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.
+
+
diff --git a/source/activities/act-changing-inverse-does-it.xml b/source/activities/act-changing-inverse-does-it.xml
index fd637c8c..42b460ed 100755
--- a/source/activities/act-changing-inverse-does-it.xml
+++ b/source/activities/act-changing-inverse-does-it.xml
@@ -52,7 +52,12 @@
-
+
+
+ The function f does not have an inverse, because f(1) = 2 = f(4),
+ so f^{-1}(2) would need to equal both 1 and 4 — not a function.
+
+
@@ -87,7 +92,12 @@
-
+
+
+ The function g does have an inverse, since all five output values
+ are distinct. For example, g^{-1}(4) = 0 and g^{-1}(0) = 1.
+
+
@@ -97,7 +107,14 @@
-
+
+
+ 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.
+
+
@@ -107,7 +124,13 @@
-
+
+
+ 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.
+
+
@@ -127,7 +150,17 @@
-
+
+
+ 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.
+
+
diff --git a/source/activities/act-changing-inverse-rainfall.xml b/source/activities/act-changing-inverse-rainfall.xml
index e9a1cb62..77768cb4 100755
--- a/source/activities/act-changing-inverse-rainfall.xml
+++ b/source/activities/act-changing-inverse-rainfall.xml
@@ -30,7 +30,13 @@
-
+
+
+ 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.
+
+
@@ -40,7 +46,15 @@
-
+
+
+ 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.
+
+
@@ -50,7 +64,15 @@
-
+
+
+ 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.
+
+
@@ -60,7 +82,17 @@
-
+
+
+ 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.
+
+
@@ -70,7 +102,15 @@
-
+
+
+ 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].)
+
+
diff --git a/source/activities/act-changing-linear-Kilimanjaro.xml b/source/activities/act-changing-linear-Kilimanjaro.xml
index ebb32d6f..26838d17 100755
--- a/source/activities/act-changing-linear-Kilimanjaro.xml
+++ b/source/activities/act-changing-linear-Kilimanjaro.xml
@@ -35,7 +35,14 @@
-
+
+
+ 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.
+
+
@@ -45,7 +52,13 @@
-
+
+
+ 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).
+
+
@@ -57,7 +70,13 @@
-
+
+
+ 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.
+
+
@@ -68,7 +87,16 @@
-
+
+
+ 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.
+
+
@@ -80,7 +108,12 @@
-
+
+
+ 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.
+
+
diff --git a/source/activities/act-changing-linear-finding-eqs.xml b/source/activities/act-changing-linear-finding-eqs.xml
index 75ad7cd6..42322d66 100755
--- a/source/activities/act-changing-linear-finding-eqs.xml
+++ b/source/activities/act-changing-linear-finding-eqs.xml
@@ -28,7 +28,11 @@
-
+
+
+ y - (-17) = \frac{3}{7}(x - (-11)), \quad \text{i.e.,} \quad y + 17 = \frac{3}{7}(x+11).
+
+
@@ -38,7 +42,13 @@
-
+
+
+ The slope is m = \frac{-1-5}{3-(-2)} = -\frac{6}{5}. Using the point
+ (3,-1):
+ y + 1 = -\frac{6}{5}(x-3).
+
+
@@ -48,7 +58,13 @@
-
+
+
+ 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).
+
+
@@ -99,7 +115,17 @@
-
+
+
+ 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.
+
+
+
@@ -110,11 +136,19 @@
Plot of a linear function h. -->
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.
+
+
+
diff --git a/source/activities/act-changing-linear-in-context.xml b/source/activities/act-changing-linear-in-context.xml
index f0466841..4c1acaf5 100755
--- a/source/activities/act-changing-linear-in-context.xml
+++ b/source/activities/act-changing-linear-in-context.xml
@@ -28,7 +28,11 @@
-
+
+
+ f(t) = 28750 + 825t.
+
+
@@ -38,7 +42,12 @@
-
+
+
+ The slope is -465 people per year. This means the town's population
+ decreases by 465 people each year.
+
+
@@ -48,7 +57,13 @@
-
+
+
+ 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].
+
+
@@ -58,7 +73,12 @@
-
+
+
+ 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.
+
+
@@ -68,7 +88,18 @@
-
+
+
+ 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.
+
+
diff --git a/source/activities/act-changing-quadratic-falling-ball.xml b/source/activities/act-changing-quadratic-falling-ball.xml
index f10d09cf..f150d14a 100755
--- a/source/activities/act-changing-quadratic-falling-ball.xml
+++ b/source/activities/act-changing-quadratic-falling-ball.xml
@@ -28,7 +28,13 @@
-
+
+
+ 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.
+
+
@@ -49,7 +55,11 @@
-
+
+
+ From the graph, the balloon appears to land at approximately t \approx 3.3 seconds.
+
+
@@ -60,7 +70,18 @@
-
+
+
+ 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.
+
+
@@ -70,7 +91,13 @@
-
+
+
+ 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.}
+
+
@@ -83,7 +110,16 @@
-
+
+
+ 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.
+
+
diff --git a/source/activities/act-changing-quadratic-parameters.xml b/source/activities/act-changing-quadratic-parameters.xml
index 2aa55eed..2d35eb25 100755
--- a/source/activities/act-changing-quadratic-parameters.xml
+++ b/source/activities/act-changing-quadratic-parameters.xml
@@ -31,7 +31,13 @@
-
+
+
+ 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.
+
+
@@ -41,7 +47,13 @@
-
+
+
+ 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.
+
+
@@ -51,7 +63,13 @@
-
+
+
+ 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.
+
+
@@ -63,7 +81,15 @@
-
+
+
+ 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.
+
+
@@ -74,7 +100,16 @@
-
+
+
+ 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.
+
+
diff --git a/source/activities/act-changing-quadratic-properties.xml b/source/activities/act-changing-quadratic-properties.xml
index 11df9e0b..ea52eed6 100755
--- a/source/activities/act-changing-quadratic-properties.xml
+++ b/source/activities/act-changing-quadratic-properties.xml
@@ -27,7 +27,15 @@
-
+
+
+ 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.
+
+
@@ -37,7 +45,13 @@
-
+
+
+ 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.
+
+
@@ -47,7 +61,16 @@
-
+
+
+ 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.
+
+
@@ -57,7 +80,17 @@
-
+
+
+ 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}.
+
+
+
@@ -67,7 +100,15 @@
-
+
+
+ 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.
+
+
diff --git a/source/activities/act-changing-transformations-combined.xml b/source/activities/act-changing-transformations-combined.xml
index f82caf17..8fb0bb34 100755
--- a/source/activities/act-changing-transformations-combined.xml
+++ b/source/activities/act-changing-transformations-combined.xml
@@ -37,7 +37,14 @@
-
+
+
+ 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.
+
+
@@ -47,7 +54,14 @@
-
+
+
+ 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.
+
+
@@ -57,7 +71,17 @@
-
+
+
+ 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.
+
+
@@ -70,7 +94,12 @@
-
+
+
+ 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.
+
+
diff --git a/source/activities/act-changing-transformations-translations.xml b/source/activities/act-changing-transformations-translations.xml
index a8af2e3b..d801707f 100755
--- a/source/activities/act-changing-transformations-translations.xml
+++ b/source/activities/act-changing-transformations-translations.xml
@@ -37,7 +37,16 @@
-
+
+
+ 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).
+
+
@@ -47,7 +56,16 @@
-
+
+
+ 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).
+
+
@@ -57,7 +75,17 @@
-
+
+
+ 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.
+
+
diff --git a/source/activities/act-changing-transformations-vert-stretch.xml b/source/activities/act-changing-transformations-vert-stretch.xml
index d6f88d94..76e28837 100755
--- a/source/activities/act-changing-transformations-vert-stretch.xml
+++ b/source/activities/act-changing-transformations-vert-stretch.xml
@@ -37,7 +37,15 @@
-
+
+
+ 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}).
+
+
@@ -47,7 +55,15 @@
-
+
+
+ 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}).
+
+
@@ -61,7 +77,14 @@
-
+
+
+ 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.
+
+
@@ -71,7 +94,16 @@
-
+
+
+ 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.
+
+
diff --git a/source/previews/PA-changing-aroc.xml b/source/previews/PA-changing-aroc.xml
index ba44e289..bd30defd 100755
--- a/source/previews/PA-changing-aroc.xml
+++ b/source/previews/PA-changing-aroc.xml
@@ -28,7 +28,15 @@
-
+
+
+ First compute s(1.5) = 64 - 16(1.5-1)^2 = 60 and
+ s(2.5) = 64 - 16(2.5-1)^2 = 28. Then
+
+ AV_{[1.5,2.5]} &= \frac{s(2.5)-s(1.5)}{2.5-1.5} = \frac{28-60}{1} = -32.
+
+
+
@@ -39,7 +47,13 @@
-
+
+
+ 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.
+
+
@@ -60,7 +74,14 @@
-
+
+
+ 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.
+
+
@@ -74,7 +95,17 @@
-
+
+
+ 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.
+
+
@@ -85,7 +116,12 @@
-
+
+
+ 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.
+
+
@@ -96,7 +132,26 @@
-
+
+
+ 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.
+
+
diff --git a/source/previews/PA-changing-combining.xml b/source/previews/PA-changing-combining.xml
index c72e14fb..c96b2d87 100755
--- a/source/previews/PA-changing-combining.xml
+++ b/source/previews/PA-changing-combining.xml
@@ -68,7 +68,12 @@
-
+
+
+ From the table, f(3) = 20 and g(3) = 2, so
+ h(3) = f(3) + g(3) = 20 + 2 = 22.
+
+
@@ -78,7 +83,12 @@
-
+
+
+ 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}.
+
+
@@ -88,7 +98,18 @@
-
+
+
+ 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}.
+
+
+
@@ -98,7 +119,12 @@
-
+
+
+ From the table, f(0) = 5 and g(0) = 9, so
+ k(0) = f(0) \cdot g(0) = 5 \cdot 9 = 45.
+
+
@@ -108,7 +134,12 @@
-
+
+
+ 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}.
+
+
@@ -118,7 +149,12 @@
-
+
+
+ Yes: s(x) is undefined wherever q(x) = 0.
+ From the graph, q(-3) = 0, so s(-3) is undefined.
+
+
diff --git a/source/previews/PA-changing-composite.xml b/source/previews/PA-changing-composite.xml
index be75c29b..0408ec83 100755
--- a/source/previews/PA-changing-composite.xml
+++ b/source/previews/PA-changing-composite.xml
@@ -27,7 +27,15 @@
-
+
+
+
+ r(t) &= p(q(t)) = p(t^2 - 1)
+ &= 3(t^2 - 1) - 4
+ &= 3t^2 - 7.
+
+
+
@@ -37,7 +45,14 @@
-
+
+
+ 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.
+
+
@@ -47,7 +62,11 @@
-
+
+
+ q(s(z)) = \left(\frac{1}{z+4}\right)^2 - 1 = \frac{1}{(z+4)^2} - 1.
+
+
@@ -57,7 +76,12 @@
-
+
+
+ One natural decomposition: f(x) = \sqrt{x} and g(t) = 2t^2 + 5,
+ so that f(g(t)) = \sqrt{2t^2 + 5} = h(t).
+
+
diff --git a/source/previews/PA-changing-inverse-F-C.xml b/source/previews/PA-changing-inverse-F-C.xml
index ec046ad7..f98707e9 100755
--- a/source/previews/PA-changing-inverse-F-C.xml
+++ b/source/previews/PA-changing-inverse-F-C.xml
@@ -27,7 +27,13 @@
-
+
+
+ Subtracting 32 from both sides: F - 32 = \frac{9}{5}C.
+ Multiplying both sides by \frac{5}{9}:
+ C = \frac{5}{9}(F-32).
+
+
@@ -35,22 +41,36 @@
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)).
+ Find the simplest expression that you can for the composite function j(C) = h(g(C)).
-
+
+
+
+ j(C) &= h(g(C)) = \frac{5}{9}\!\left(\frac{9}{5}C + 32 - 32\right)
+ &= \frac{5}{9} \cdot \frac{9}{5}C = C.
+
+
+
- Find the simplest expression that you can for the composite function k(F) = g(h(F)).
+ Find the simplest expression that you can for the composite function k(F) = g(h(F)).
-
+
+
+
+ k(F) &= g(h(F)) = \frac{9}{5}\!\left(\frac{5}{9}(F-32)\right) + 32
+ &= (F-32) + 32 = F.
+
+
+
@@ -60,7 +80,15 @@
-
+
+
+ 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.
+
+
diff --git a/source/previews/PA-changing-linear-3-ex.xml b/source/previews/PA-changing-linear-3-ex.xml
index 4c76a343..1c1d1325 100755
--- a/source/previews/PA-changing-linear-3-ex.xml
+++ b/source/previews/PA-changing-linear-3-ex.xml
@@ -24,7 +24,14 @@
-
+
+
+ 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.
+
+
@@ -118,7 +125,15 @@
-
+
+
+ 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.
+
+
@@ -135,7 +150,15 @@
-
+
+
+ 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}.
+
+
@@ -145,12 +168,19 @@
-
+
+
+ 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
+ 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}
@@ -159,7 +189,15 @@
-
+
+
+
+ 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.
+
+
diff --git a/source/previews/PA-changing-quadratic.xml b/source/previews/PA-changing-quadratic.xml
index b0e44115..3950fa1c 100755
--- a/source/previews/PA-changing-quadratic.xml
+++ b/source/previews/PA-changing-quadratic.xml
@@ -25,7 +25,7 @@
- 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.
+ 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.
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