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Chapter 1
A1
Glencoe Precalculus
DATE
Before you begin Chapter 1
Functions from a Calculus Perspective
Anticipation Guide
PERIOD
A
A
D
D
A
D
3. Even functions are symmetric with respect to the y-axis.
4. If a function is continuous, you can trace its graph without
lifting your pencil.
5. A horizontal compression is an example of a rigid
transformation.
6. The greatest integer function f(x) is defined as the greatest
integer greater than or equal to x.
7. To decompose a function is to write it as two or more simpler
functions.
8. The graph of a relation and its inverse are symmetric about
the x-axis.
After you complete Chapter 1
10. A one-to-one function passes the horizontal line test.
Chapter 1
3
Chapter Resources
9/30/09 1:57:16 PM
Answers
Glencoe Precalculus
• For those statements that you mark with a D, use a piece of paper to write an example
of why you disagree.
• Did any of your opinions about the statements change from the first column?
D
A
D
2. To find the y-intercept of a function, substitute 0 for y and
solve for x.
9. The inverse of y = 3x is y = – 3x.
A
STEP 2
A or D
1. The range of a function is the set of all possible output
values.
Statement
• Reread each statement and complete the last column by entering an A or a D.
Step 2
STEP 1
A, D, or NS
• Write A or D in the first column OR if you are not sure whether you agree or disagree,
write NS (Not Sure).
• Decide whether you Agree (A) or Disagree (D) with the statement.
• Read each statement.
Step 1
1
NAME
0ii_004_PCCRMC01_893802.indd Sec1:3
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
DATE
Functions
Study Guide and Intervention
PERIOD
Describe x > 18 using set-builder notation and interval notation.
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
005_026_PCCRMC01_893802.indd 5
Chapter 1
{x | x < -11 or x ≥ 1, x ∈ };
(-∞, -11) ∪ [1, ∞)
5. x < -11 or x ≥ 1
{x | x > -8.8, x ∈ }; (-8.8, ∞)
3. x > -8.8
{x | x ≥ 17, x ∈ }
1. {17, 18, 19, 20, …}
5
{x | x ≤ -7, x ∈ }
6. {…, -10, -9, -8, -7}
Lesson 1-1
9/30/09 1:59:41 PM
Glencoe Precalculus
{x | 5 < x < 15, x ∈ }; (5, 15)
4. 5 < x < 15
{x | x ≤ -2, x ∈ }; (-∞, -2]
2. x ≤ -2
Write each set of numbers in set-builder and interval notation, if possible.
Exercises
Use parentheses on the left because 18 is not included in the set. Use
parentheses with infinity since it never ends.
Interval notation: (18, ∞)
The vertical line | means “such that.” The symbol ∈ means “is an element
of.” Read the expression as the set of all x such that x is greater than 18 and
x is an element of the set of real numbers.
Set-builder notation: {x | x > 18, x ∈ }
The set includes all numbers that are greater than 18 but are not equal to 18.
Example
Another way is to use interval notation. With interval notation, you use
brackets if an endpoint is included and parentheses if an endpoint is not
included. Use ∞ to indicate positive infinity and -∞ to indicate negative
infinity.
One way to describe a subset of the real numbers is to use set-builder
notation. With set-builder notation, you choose a variable, list the
properties of the variable, and tell to which set of numbers the variable
belongs.
The set of real numbers
includes the rationals , irrationals , integers , wholes , and naturals .
Describe Subsets of Real Numbers
1-1
NAME
Answers (Anticipation Guide and Lesson 1-1)
Functions
Study Guide and Intervention
DATE
Find each function value.
⎩
⎨
2x2 - 15 if x ≥ 10
3x if 4 < x < 10, find g(6) and g(10).
Simplify.
Substitute -2 for x.
Original function
A2
3+x
x - 6x
3+x
x - 6x
State the domain of f(x) = −
.
2
9t9 - 4t4 + 3t - 2
Glencoe Precalculus
⎩
⎨
005_026_PCCRMC01_893802.indd 6
Chapter 1
4. If f(x) =
⎧
√2x if x < 3
42 if x ≥ 8
6
2x + 10 if 3 ≤ x < 8, find f(3) and f(8.5). 16; 42
Glencoe Precalculus
2. If h(x) = 9x9 - 4x4 + 3x - 2, find h(t).
⎧ x + 45 if x ≤ -1
3. If g(x) = ⎨
, find g(-5) and g(36). 40; 45
⎩ 81 - x if x > -1
27
1. If f(x) = 5x2 - 4x - 6, find f(3).
Find each function value.
Exercises
Solving x2 - 6x = 0, the excluded values in the domain are x = 0 and x = 6.
The domain is {x | x ≠ 0, 6, x ∈ }.
is zero, the expression is undefined.
When the denominator of −
2
Example 2
Look at the “if ” statements to see that 6 fits into the second rule,
so g(6) = 3(6) or 18.
The value 10 fits into the third rule, so g(10) = 2(10)2 - 15 or 185.
b. If g(x) =
⎧ √x + 1 if x ≤ 4
f(x) = 4x3 + 6x2 + 3x
f(-2) = 4(-2)3 + 6(-2)2 + 3(-2)
= -32 + 24 - 6 or -14
a. If f(x) = 4x3 + 6x2 + 3x, find f(-2).
Example 1
A relation is a rule that relates, or pairs, the
elements in set A with the elements in set B. Set A contains the inputs, or
the domain, and set B contains the outputs, or the range. A function f
from set A to set B is a relation that assigns to each element x in set A
exactly one element y in set B. To evaluate a function, replace the
independent variable with the given value from the domain and simplify.
(continued)
PERIOD
9/30/09 2:00:01 PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Chapter 1
Identify Functions
1-1
NAME
Functions
Practice
DATE
PERIOD
{x | x < 0 or x > 8, x ∈ };
(-∞, 0) ∪ (8, ∞)
4. x < 0 or x > 8
y
function
−8
−4 0
4
8
4
8x
x
not a function
0
y
t + 6t + 9
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
005_026_PCCRMC01_893802.indd 7
Chapter 1
7
⎩
⎨
-75 if x > 11
Lesson 1-1
9/30/09 2:00:16 PM
Glencoe Precalculus
√x - 2 if -2 < x ≤ 11.
⎧ 3x2 + 16 if x < -2
{t | t ≠ -3, t ∈ }
14. Find f(-4) and f(11) for the piecewise function f(x) =
{x | x ≤ - −23 , x ∈ }
12. g(x) = √-3x - 2
State the domain of each function.
2t - 6
13. h(t) = −
2
c. f(a + 1) -3 √
a2 + 2a + 10
c. h(x + 8) x2 + 8x + 1
2
b. f(3a) -9 √
a2 + 1
a. f(4) -15
a. h(-1) 10
b. h(2x) 4x - 16x + 1
11. f(a) = -3 √a2 + 9
not a function
9. x = 5(y - 1)2
7.
10. h(x) = x2 - 8x + 1
Find each function value.
function
8. -x + y = 3x
6.
5. The input value x is a car’s license plate number, and the output value y is
the car’s make and model. function
64; 3
{x | -6.5 < x ≤ 3, x ∈ }; (-6.5, 3]
2. -6.5 < x ≤ 3
Determine whether each relation represents y as a function of x.
{x | x = 2n, n ∈ }
3. all multiples of 2
{x | x ≥ -3, x ∈ }
1. {-3, -2, -1, 0, 1, …}
Write each set of numbers in set-builder and interval notation, if possible.
1-1
NAME
Answers (Lesson 1-1)
Chapter 1
Functions
112
122
114
Michigan
New Mexico
Wisconsin
-54
-50
-51
-60
-17
-27
Low
A3
8
PERIOD
Free
20
15
8
Shipping
Cost ($)
Glencoe Precalculus
9/30/09 2:00:46 PM
Answers
Glencoe Precalculus
R = {f() | 2 ≤ f() ≤ 12, f() ∈ }
b. Write the range in set-builder
notation.
D = { | –2 ≤ ≤ 8, ∈ }
a. Write the relevant domain in
set-builder notation.
5. ELEVATOR An elevator starts with
12 people on a building’s eighth floor.
One person exits to each floor. The lowest
level is two floors below ground level.
The function f(ℓ) = ℓ + 4 gives
the number of people on the elevator
after a person exits to that level.
D = [0, ∞), R = {0, 8, 15, 20}
b. Give the domain and range of the
function.
⎧ 8 if 0 ≤ t ≤ 75
⎢ 15 if 75 < t ≤ 150
c(t) = ⎨
⎢ 20 if 150 < t ≤ 250
⎩ 0 if t > 250
a. Write a piecewise function describing
the shipping cost c in terms of the
total purchase amount t.
250.01 and up
150.01 to 250
75.01 to 150
0 to 75
Total
Purchase ($)
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
005_026_PCCRMC01_893802.indd 8
Chapter 1
⎨
⎧1.15c if 0 < c < 40
t(c) = 1.18c if 40 ≤ c < 100
⎩ 1.2c if c ≥ 100
3. TIPPING A restaurant patron has
decided to leave a 15% tip for meals
costing up to $40, an 18% tip for meals
costing at least $40 but less than $100,
and a 20% tip for meals costing $100 or
more. Write a piecewise function to
describe the total amount t the patron
will pay in terms of the meal cost c.
244 deer; 566 deer
2. DEER A park’s deer population over
five years can be modeled by
f(d) = -3d4 + 43d3 - 185d2 + 350d - 59.
Estimate f(3) and f(5), the populations in
the third and fifth years.
c. Determine whether the relation is a
function. no
D = {110, 112, 114, 118, 122},
R = {–60, –54, –51, –50, -27,
-17}
b. State the domain and range of the
relation.
{(112, –27), (110, –17),
(118, –60), (112, –51),
(122, –50), (114, –54)}
a. State the relation of the data as a set
of ordered pairs.
Source: National Climatic Data Center
110
118
Idaho
112
Delaware
High
State
Alabama
Record Highs and Lows (°F)
DATE
4. SHIPPING The table below shows the
cost of shipping items bought from a
catalog where the cost is based on the
total amount of the purchase.
Word Problem Practice
1. CLIMATE The table shows record high
and low temperatures for selected states.
1-1
NAME
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Enrichment
DATE
PERIOD
0
f (a)
f (b)
y
a
b
2.01
b. a = 1 to b = 1.01
2.001
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005_026_PCCRMC01_893802.indd 9
Chapter 1
9
6. Find the instantaneous rate of change of the function f (x) = 3x 2
as x approaches 3. 18
The value you found in Exercise 5d is the instantaneous rate of
change of the function. Instantaneous rate of change has enormous
importance in calculus.
x
Lesson 1-1
3/22/09 5:49:45 PM
Glencoe Precalculus
c. a = 1 to b = 1.001
d. What value does the average rate of change appear to be
approaching as the value of b gets closer and closer to 1? 2
2.1
a. a = 1 to b = 1.1
5. Find the average rate of change for the function f (x) = x 2 in each interval.
h(x)
(a, f (a))
(b, f (b))
y = f (x )
f (b) - f (a)
slope =
b-a
4. Which of these functions has the greatest average rate of change
between 2 and 3: f (x) = x ; g(x) = x 2; h(x) = x 3 ?
the rate between 4 and 5
3. Which is greater, the average rate of change of
f (x) = x 2 between 0 and 1 or between 4 and 5?
The average rate of change of a function f (x) over an
interval is the amount the function changes per unit
change in x. As shown in the figure at the right, the
average rate of change between x = a and x = b
represents the slope of the line passing through the
two points on the graph of f with abscissas a and b.
change: 24; average rate of change: 12
2. f (x) = x 2 + 6x - 10, from x = 2 to x = 4
change: 15; average rate of change: 3
1. f (x) = 3x - 4, from x = 3 to x = 8
Find the change and the average rate of change of f(x) in the given range.
by the expression − .
f(b) - f(a)
b-a
Between x = a and x = b, the function f (x) changes by f (b) - f (a).
The average rate of change of f (x) between x = a and x = b is defined
Rates of Change
1-1
NAME
Answers (Lesson 1-1)
PERIOD
Analyzing Graphs of Functions and Relations
Study Guide and Intervention
DATE
A4
−20
−12
0
8
y
2
(2, -16)
8x
g(x) = x - 4x - 12
4
D = {x | x ∈ }, R = {y | y ≥ -16,
y ∈ }
y-intercept: -12
zeros: -2 and 6
x2 - 4x - 12 = (x + 2)(x - 6);
x = -2 or x = 6
g(0) = 02 - 4(0) - 12 = -12
−6
Glencoe Precalculus
005_026_PCCRMC01_893802.indd 10
Chapter 1
1.
10
2.
−8
−4
−4 0
8
y
8x
Glencoe Precalculus
D = {x | x ∈ }, R = {y | y ≤ 9,
y ∈ }
y-intercept: 8
zeros: -2 and 4
8 + 2x - x2 = -(x + 2)(x - 4);
x = -2 or x = 4
g(0) = 8 + 2(0) - 02 = 8
−8
(0, 8)
g(x) = 8 + 2x - x 2
Use the graph of g to find the domain and range of the function and to approximate
its y-intercept and zero(s). Then find its y-intercept and zeros algebraically.
Exercises
To find the zeros algebraically, let f(x) = 0 and solve for x.
-x2 - 1.5x + 4.5 = 0
-1(x + 3)(x - 1.5) = 0
x = -3 or x = 1.5
To find the y-intercept algebraically, find f(0).
f(0) = -(0)2 - 1.5(0) + 4.5 = 4.5
Example
Use the graph of f to find the domain
and range of the function and to approximate the
y
y-intercept and zero(s). Then confirm the estimate
(-0.75, 5.0625)
f(x) = -x 2 - 1.5x + 4.5
algebraically.
The graph is not bounded on the left or right, so the
domain is the set of all real numbers.
0
x
{x | x ∈ }
The graph does not extend above 5.0625 or f(-0.75), so
the range is all real numbers less than or equal to 5.0625.
{y | y ≤ 5.0625, y ∈ }
The y-intercept is the point where the graph intersects the y-axis. It appears
to be 4.5. Likewise, the zeros are the x-coordinates of the points where the
graph crosses the x-axis. They seem to occur at -3 and 1.5.
By looking at the graph of a function,
you can determine the function’s domain and range and estimate the x- and
y-intercepts. The x-intercepts of the graph of a function are also called the
zeros of the function because these input values give an output of 0.
9/30/09 2:57:57 PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Chapter 1
Analyzing Function Graphs
1-2
NAME
DATE
(continued)
PERIOD
Analyzing Graphs of Functions and Relations
Study Guide and Intervention
Replacing y with -y produces
an equivalent equation.
Replacing x with -x produces
an equivalent equation.
Replacing x with -x and y with
-y produces an equivalent
equation.
For every (x, y) on the graph,
(x, -y) is also on the graph.
For every (x, y) on the graph,
(-x, y) is also on the graph.
For every (x, y) on the graph,
(-x, -y) is also on the graph.
x-axis
y-axis
origin
3
3
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
005_026_PCCRMC01_893802.indd 11
Chapter 1
symmetric with respect to y-axis
5
5
even; −
=−
x4
(-x)4
x
5
3. g(x) = −
4
neither; 4(-x) + 1 = -4x + 1
1. f(x) = 4x3 + 1
11
Lesson 1-2
10/1/09 9:22:40 AM
Glencoe Precalculus
odd; (-x)3 - 6(-x) = -x3 + 6x
symmetric with respect to origin
4. g(x) = x3 - 6x
even;
(-x)4 -10(-x)2 + 9 = x4 - 10x2 + 9
symmetric with respect to y-axis
2. g(x) = x4 - 10x2 + 9
GRAPHING CALCULATOR Graph each function. Analyze the graph
to determine whether each function is even, odd, or neither. Confirm
algebraically. If odd or even, describe the symmetry of the graph of
the function.
Exercises
Example
GRAPHING CALCULATOR Graph f(x) = -x3 + 2x.
Analyze the graph to determine whether the function is even, odd, or
neither. Confirm algebraically. If odd or even, describe the
symmetry of the graph of the function.
From the graph, it appears that the function is symmetric to
the origin.
Confirm: f(-x) = -(-x)3 + 2(-x) = x3 - 2x = -f(x)
[-10, 10] scl: 1 by [-10, 10] scl: 1
The function is odd because f(-x) = -f(x).
Functions that are symmetric with respect to the y-axis are even functions,
and for every x in the domain, f(-x) = f(x). Functions that are symmetric
with respect to the origin are odd functions and for every x in the domain,
f(-x) = -f(x).
Algebraic Test
Description
Symmetric with
respect to…
A graph of a relation that is symmetric to the
x-axis and/or the y-axis has line symmetry. A graph of a relation that is
symmetric to the origin has point symmetry.
Symmetry of Graphs
1-2
NAME
Answers (Lesson 1-2)
A5
−8
0
y
4
8x
y = h(x)
−4
4
−8
0
4
8
−8
x
6.
−8
4
-2
y=−
x
-2
origin; -y = −
-x
−4
y
−4
−8 −4 0
y = −x2
y
y = -0.5x 5 - 3
8x
2
12
Glencoe Precalculus
9/30/09 2:01:21 PM
Answers
Glencoe Precalculus
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
005_026_PCCRMC01_893802.indd 12
Chapter 1
x
1
1
even; f(-x) = −
=−
= f(x); symmetric with respect to the y-axis
2
2
(-x)
4x
y-axis;
y = -0.5(-x)2 - 3
y = -0.5(x)2 - 3
1
7. Graph g(x) = −
using a graphing calculator. Analyze the graph to
x2
determine whether the function is even, odd, or neither. Confirm
algebraically. If odd or even, describe the symmetry of the graph of the
function.
5.
−4
−2 0
4 f(x) = 4x - 1 - 4
y
D = (-∞, 4],
R = (-∞, 3]
Use the graph of each equation to test for symmetry with respect to
the x-axis, y-axis, and the origin. Support the answer numerically.
Then confirm algebraically.
3
4 √
0 - 1 - 4 = 4 √
-1 - 4 = 4(-1) - 4 = -8;
y = -8
3
3
3
0 = 4 √
x - 1 - 4; 4 = 4 √
x - 1 ; 1 = √
x - 1,
1 = x - 1; 2 = x
3
−8
4
−4
8x
y = h(x)
3.
−2
4
D = [-4, 3],
R = [-6, 5]
−8
y
−4
0
4
8
4. Use the graph of the function to find its
y-intercept and zeros. Then find these values
algebraically. y-int: -8, zero: 2; f(0) =
2.
6
4
2
y
16
14
12 f(x) = 2|x - 3| + 1
10
−4−3−2−10 1 2 3 4 5 6 7 8 x
Use the graph of h to find the domain and range of each function.
12; 5; 9
PERIOD
Analyzing Graphs of Functions and Relations
Practice
1. Use the graph of the function shown to estimate
f(-2.5), f(1), and f(7). Then confirm the estimates
algebraically. Round to the nearest hundredth, if
necessary.
1-2
DATE
4
6
Years since 2000
2
8
Merchandise Cost ($)
10 20 30 40 50 60 70 80 90 100 110
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005_026_PCCRMC01_893802.indd 13
Chapter 1
b. Use the graph to estimate the shipping
costs for a $245 package. $10
a. State the domain and range of the
function. D = (0, ∞),
R = {4, 7, 10}
0
1
2
3
4
5
6
7
8
9
10
11
12
2. SHIPPING Shipping costs are shown in
the piecewise function below.
c. In what year did the number of
visitors first exceed 20,000? 2007
b. Find the estimated number of visitors
in 2006 algebraically. 16,650
a. Use the graph to estimate the
number of visitors to the park
in 2006. 16,700
0
15
16
17
18
19
20
21
22
DATE
PERIOD
13
0
Setting
−18,000
−12,000
−6000
y
20
40 x
Lesson 1-2
3/22/09 5:50:16 PM
Glencoe Precalculus
f. Is the function even, odd, or neither?
Explain how you know.
Neither; f(-x) is not equal to
f(x) or -f(x).
15,575 units3
e. Find the volume when the machine is
set to 20 algebraically.
d. Estimate the setting for a volume of
9000. 16
16,000 units3
c. Use the graph to estimate the volume
with a machine setting of 20.
D = [0, ∞), R = [0, ∞)
b. State the relevant domain and range
of the function.
D = (-∞, ∞), R = (-∞, ∞)
a. State the domain and range of the
function.
−40
6000
12,000
18,000
Volume of Product
3. FACTORY The function relating a
machine setting x to the volume of the
product being built is modeled by the
function y = x3 + 15x2 + 75x + 75. The
least setting on the machine is 0.
Analyzing Graphs of Functions and Relations
Word Problem Practice
1. PARK The approximate numbers of
annual visitors to a park from 2000
through 2008 can be modeled using
v(x) = 0.05x3 - 0.51x2 + 1.81x + 13.35,
where x represents the number of years
since 2000.
Park Data
1-2
NAME
Number of Visitors (thousands)
NAME
Shipping Cost ($)
Chapter 1
Volume
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Answers (Lesson 1-2)
Enrichment
DATE
A6
Glencoe Precalculus
005_026_PCCRMC01_893802.indd 14
Chapter 1
14
7. A cube has 6 two-fold axes of symmetry. In the space at the right,
draw one of these axes.
3; each axis passes through the centers of a
pair of opposite faces.
6. How many four-fold axes of symmetry does a cube have?
Use a die to help you locate them.
Solid figures can also have rotational symmetry. For example, the
axis drawn through the cube in the illustration is a four-fold axis of
symmetry because the cube can be rotated about this axis into four
different positions that are exactly alike.
9 planes: 3 planes are parallel to pairs of
opposite faces and the other 6 pass through pairs
of opposite edges.
5. a cube
4 planes all passing through the top vertex:
2 planes are parallel to a pair of opposite edges
of the base and the other 2 cut along the
diagonals of the square base.
4. a square pyramid
an infinite number of planes passing through
the central axis, plus one plane cutting the
center of the axis at right angles
3. a soup can
an infinite number of planes; each plane passes
through the center.
2. a tennis ball
3 planes of symmetry; each plane is parallel to
a pair of opposite faces.
1. a brick
Determine the number of planes of symmetry for each object and
describe the planes.
Glencoe Precalculus
Similar to the symmetry of a two-dimensional graph, three-dimensional objects
can display symmetry.
A solid figure that can be superimposed, point for
point, on its mirror image has a plane of symmetry.
A symmetrical solid object may have a finite or
infinite number of planes of symmetry. The chair
in the illustration at the right has just one plane of
symmetry; the doughnut has infinitely many planes
of symmetry, three of which are shown.
PERIOD
9/30/09 3:18:50 PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Chapter 1
Symmetry in Three-Dimensional Figures
1-2
NAME
Graphing Calculator Activity
DATE
PERIOD
keys to
keys to
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
005_026_PCCRMC01_893802.indd 15
Chapter 1
15
Lesson 1-2
9/30/09 2:01:45 PM
Glencoe Precalculus
Sample answer: You could use the ASK option within the TBLSET menu
to input various values for x along with the opposites of those values.
Then check if the y-values stay the same (for an even function) and if the
y-values go to the opposite value (for an odd function).
6. Explain how you could use the ASK option in TBLSET to determine the
relationship between f(x) and f(-x) for a given function.
Sample answer: If the function is even, the graph should be symmetric
with respect to the y-axis. If the function is odd, the graph should be
symmetric with respect to the origin. The graph of f(x) = x8 is symmetric
to the y-axis.
5. How could you use symmetry to help you graph an even or odd
function? Give an example.
f(-x) = (-x)8 - 3(-x)4 + 2(-x)2 + 2 = x8 - 3x4 + 2x2 + 2 = f(x);
f(-x) = (-x)7 + 4(-x)5 - (-x)3 = -x7 - 4x5 + x3 = -f(x)
4. Verify your conjectures algebraically.
even; odd
3. Identify the functions in Exercises 1 and 2 as odd, even, or neither
based on your observations of their graphs.
GRAPH . Then press TRACE and use the
move the cursor along the graph.
A possible window setting is [-10, 10] scl: 1 by [-10, 10] scl: 1.
Press Y=
7 + 4
5 —
3
2. f(x) = x7 + 4x5 - x3
+ 2 GRAPH . Then press TRACE and use the
move the cursor along the graph.
A possible window setting is [-10, 10] scl: 1 by [-10, 10] scl: 1.
Press Y=
8 — 3
4 + 2
1. f(x) = x8 - 3x4 + 2x2 + 2
Graph each function to determine how f(x) and f(2x) are related.
Exercises
You can use the TRACE function to investigate the symmetry of a function.
• Graph the function.
• Use the TRACE function to observe the relationship between
points of the graph having opposite x-coordinates.
• Use this information to determine the relationship between f(x)
and f(-x).
Identifying Odd and Even Functions
1-2
NAME
Answers (Lesson 1-2)
Chapter 1
DATE
Continuity, End Behavior, and Limits
Study Guide and Intervention
PERIOD
A7
6.998
1.99
1.999
2.001
2.01
2.1
x
7.002
7.02
7.2
y = f(x)
-999.5
0.999
x
100.5
1000.5
1.01
10.5
y = f(x)
1.001
1.1
The function has infinite discontinuity
at x = 1.
-99.5
-9.5
y = f(x)
0.99
0.9
x
The function is not defined at x = 1
because it results in a denominator of 0.
The tables show that for values of x
approaching 1 from the left, f(x)
becomes increasingly more negative. For
values approaching 1 from the right,
f(x) becomes increasingly more positive.
x -1
2x
;x=1
b. f(x) = −
2
16
Glencoe Precalculus
9/30/09 3:02:50 PM
Answers
Glencoe Precalculus
so the function is continuous.
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
005_026_PCCRMC01_893802.indd 16
Chapter 1
so the function is not continuous;
it has jump discontinuity.
x → 4+
lim f(x) = 39 and lim f(x) = 39,
x → 4-
x → 2+
lim f(x) = 1 and lim f(x) = 5 ,
x → 2–
Determine whether each function is continuous at the given x-value.
Justify your answer using the continuity test. If discontinuous,
identify the type of discontinuity as infinite, jump, or removable.
⎧ 2x + 1 if x > 2
1. f(x) = ⎨
;x=2
2. f(x) = x2 + 5x + 3; x = 4 f(4) = 39
⎩ x - 1 if x ≤ 2
Exercises
The function is continuous at x = 2.
x→2
(3) lim f(x) = 7 and f(2) = 7.
x→2
The tables show that y approaches 7
as x approaches 2 from both sides.
It appears that lim f(x) = 7.
6.8
6.98
1.9
y = f(x)
x
(1) f(2) = 7, so f(2) exists.
(2) Construct a table that shows values for
f(x) for x-values approaching 2 from the
left and from the right.
a. f(x) = 2|x| + 3; x = 2
Example
Determine whether each function is continuous at the given
x-value. Justify using the continuity test. If discontinuous, identify the type of
discontinuity as infinite, jump, or removable.
Functions that are not continuous are discontinuous. Graphs that are
discontinuous can exhibit infinite discontinuity, jump discontinuity,
or removable discontinuity (also called point discontinuity).
x→c
(3) The function value that f(x) approaches from each side of c is f(c); in
other words, lim f(x) = f(c).
x→c
(2) f(x) approaches the same function value to the left and right of c; in other
words, lim f(x) exists.
(1) f(x) is defined at c; in other words, f(c) exists.
A function f(x) is continuous at x = c if it satisfies the
following conditions.
Continuity
1-3
NAME
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
DATE
Continuity, End Behavior, and Limits
Study Guide and Intervention
(continued)
PERIOD
−4
-100
-999,998
-10
-998
10
1002
0
2
100
1,000,002
1000
4
8
x→∞
See students’ work.
f(x) = -∞; lim f(x) = -∞
lim
x → -∞
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
005_026_PCCRMC01_893802.indd 17
17
16x
5x
x -2
2
x→∞
Lesson 1-3
3/22/09 5:50:38 PM
Glencoe Precalculus
4x
f(x) = x 3 + 2
f(x) = 5; lim f(x) = 5
8
f(x) =
y
See students’ work.
x → -∞
−8
lim
−16 −8 0
4
8
−8
4x
2.
−4
2
f(x) = -x 4 - 2x
y
−4
−4 −2 0
Chapter 1
1.
−8
−4
y
1,000,000,002
Use the graph of each function to describe its end behavior. Support
the conjecture numerically.
Exercises
4
8
−2 0
As x −∞, f(x) -∞. As x ∞, f(x) ∞. This supports the conjecture.
-1000
-999,999,998
x
f(x)
Construct a table of values to investigate function values as |x| increases.
x→∞
As x increases without bound, the y-values increase
without bound. It appears the limit is positive infinity:
lim f(x) = ∞.
x → -∞
Example
Use the graph of f(x) = x3 + 2 to describe
its end behavior. Support the conjecture numerically.
As x decreases without bound, the y-values also
decrease without bound. It appears the limit is negative
infinity: lim f(x) = -∞.
The f(x) values may approach negative infinity, positive infinity, or a specific value.
x→∞
Right-End Behavior (as x becomes more and more positive): lim f(x)
x → -∞
Left-End Behavior (as x becomes more and more negative): lim f(x)
The end behavior of a function describes how the function behaves at
either end of the graph, or what happens to the value of f(x) as x increases or decreases
without bound. You can use the concept of a limit to describe end behavior.
End Behavior
1-3
NAME
Answers (Lesson 1-3)
Continuity, End Behavior, and Limits
Practice
PERIOD
3
No; the function has a removable
discontinuity at x = -1 and infinite
discontinuity at x = -2.
x+1
x + 3x + 2
4. f(x) = −
; at x = -1 and x = -2
2
No; the function is infinitely
discontinuous at x = -4.
x+4
x -2
2. f(x) = −
; at x = -4
A8
[-3, -2], [0, 1]
6. g(x) = x4 + 10x - 6; [-3, 2]
lim
−8
−4
f(x) = x 2 - 4x - 5 4
8
x→∞
x → -∞
See students’ work.
f(x) = -2; lim f(x) = -2
lim
8x
Glencoe Precalculus
005_026_PCCRMC01_893802.indd 18
Chapter 1
18
the resistance? Resistance decreases and approaches zero.
constant but the current keeps increasing in the circuit, what happens to
I
E
. If the voltage remains
voltage E, and current I in a circuit as R = −
Glencoe Precalculus
x→∞
See students’ work.
x → -∞
0 4
f(x) = ∞; lim f(x) = ∞
−8
8.
−4
16x
-6x
3x - 5
−4
8
f(x) =
−2
−16 −8 0
2
y
4
9. ELECTRONICS Ohm’s Law gives the relationship between resistance R,
7.
y
Use the graph of each function to describe its end behavior. Support
the conjecture numerically.
[-5, -4], [-1, 0], [0, 1]
5. f(x) = x3 + 5x2 - 4; [-6, 2]
Determine between which consecutive integers the real zeros of
each function are located on the given interval.
Yes; the function is defined at
x = -1, the function approaches
1 as x approaches 1 from
both sides; f(1) = 1.
3. f(x) = x3 - 2x + 2; at x = 1
Yes; the function is defined at
x = -1, the function approaches
2
-−
as x approaches -1 from
3
2
both sides; f(-1) = -−
.
3x
2
1. f(x) = - −
; at x = -1
2
Determine whether each function is continuous at the given
x-value(s). Justify using the continuity test. If discontinuous,
identify the type of discontinuity as infinite, jump, or removable.
1-3
DATE
9/30/09 2:01:58 PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Chapter 1
lim f(x) = -∞;
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
005_026_PCCRMC01_893802.indd 19
Chapter 1
8
−16
−8
−16 −8 0
y
8
16x
c. Graph the function to verify your
conclusion from part b.
x = 0; infinite.
b. Is the function continuous? Justify the
answer using the continuity test. If
discontinuous, explain your reasoning
and identify the type of discontinuity
as infinite, jump, or removable.
No; because f(0) does not exist,
f(x) is discontinuous at
x→5
function is defined when
x = 5, lim f(x) = 10.
a. Determine whether the function is
continuous at x = 5. Justify the
answer using the continuity test.
Yes; because f(5) = 10, the
side of the base.
x
250
f(x) = −
, where x is the length of one
2
2. GEOMETRY The height of a rectangular
prism with a square base and a volume
of 250 cubic units can be modeled by
x→∞
lim f(x) = -∞
x → -∞
DATE
19
PERIOD
Day
2
4
Stock
6
See students’ work.
Lesson 1-3
9/30/09 3:01:32 PM
Glencoe Precalculus
x→∞
lim f(x) = ∞; lim f(x) = -∞
x → -∞
Use the graph to describe the end
behavior of the function. Support your
conjecture numerically.
0
6
12
24
4. STOCK The average price of a share of
a certain stock x days after a company
restructuring is modeled by
f(x) = -0.15x3 + 1.4x2 - 1.8x + 15.29.
No, x will not be negative
because the fewest number of
people is 0.
b. Are there any points of discontinuity
in the relevant domain? Explain.
a. Graph the function using a graphing
calculator. Use the graph to identify
and describe any points of
discontinuity. infinite
discontinuity at x = -25
3. TRIP The per-person cost of a guided
climbing expedition can be modeled by
600
f(x) = −
, where x is the number of
x + 25
people on the trip.
Continuity, End Behavior, and Limits
Word Problem Practice
1. HOUSING According to the U.S.
Census Bureau, the approximate percent
of Americans who owned a home
from 1900 to 2000 can be modeled by
h(x) = -0.0009x4 - 0.09x3 + 1.54x2 4.12x + 47.37, where x is the number of
decades since 1900. Graph the function
on a graphing calculator. Describe the
end behavior.
1-3
NAME
Price per Share ($)
NAME
Answers (Lesson 1-3)
Chapter 1
Enrichment
DATE
PERIOD
A9
[ )
20
0
1
2
1
f (x)
1
x
Glencoe Precalculus
9/30/09 2:02:25 PM
Answers
Glencoe Precalculus
1
2
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
005_026_PCCRMC01_893802.indd 20
Chapter 1
1
No; it is discontinuous at x = −
.
2
6. Is the function given in Exercise 5 continuous on the
interval [0, 1]? If not, where is the function discontinuous?
⎨
5. In the space at the right, sketch the graph
of the function f(x) defined as follows.
⎧1
1
− if x ∈ 0, −
2
2
f(x) =
1
1 if x ∈ −, 1
2
⎩
4. What notation is used in the selection to express the fact that a number x is
contained in the interval I?
x∈I
3. What mathematical term makes sense in this sentence?
If f(x) is not ____ at x0, it is said to be discontinuous at x0. continuous
The first interval is ∅ and the others reduce to the point a = b.
2. What happens to the four intervals in the first paragraph when a = b?
Only the last inequality can be satisfied.
1. What happens to the four inequalities in the first paragraph when a = b?
Use the selection above to answer these questions.
Suppose I is an interval that is either open, closed, or half-open. Suppose ƒ(x) is a function defined
on I and x0 is a point in I. We say that the function ƒ(x) is continuous at the point x0 if the quantity
⎪ƒ(x) - ƒ(x0)⎥ becomes small as x ∈ I approaches x0.
[a, b) or (a, b] is called half-open or half-closed, and an interval of the form
[a, b] is called closed.
An interval of the form (a, b) is called open, an interval of the form
Throughout this book, the set S, called the domain of definition of a function, will usually be
an interval. An interval is a set of numbers satisfying one of the four inequalities a < x < b,
a ≤ x < b, a < x ≤ b, or a ≤ x ≤ b. In these inequalities, a ≤ b. The usual notations for the intervals
corresponding to the four inequalities are (a, b), [a, b), (a, b], and [a, b], respectively.
The following selection gives a definition of a continuous function as
it might be defined in a college-level mathematics textbook.
Notice that the writer begins by explaining the notation to be used for
various types of intervals. Although a great deal of the notation
is standard, it is a common practice for college authors to explain their
notations. Each author usually chooses the notation he or she wishes to use.
Reading Mathematics
1-3
NAME
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
DATE
PERIOD
x→∞
−4
−8
−2 0
4
8
y
-100
-1 × 1010
-2
-7
-1.5
-0.09
-1
-2
-0.5
-3.5
0
-3
0.5
-2.47
1
-4
1.5
-5.91
2
1
100
1 × 1010
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
005_026_PCCRMC01_893802.indd 21
Chapter 1
Lesson 1-4
9/30/09 2:02:38 PM
Glencoe Precalculus
rel. min. of 0 at x = 0;
rel. max. of 108 at x = -6
2. f(x) = x3 + 9x2
21
abs. min. of -5.03 at x = -0.97 and
at x = 0.97; rel. max. of 0 at x = 0
1. f(x) = 2x6 + 2x4 - 9x2
Use a graphing calculator to approximate to the nearest hundredth the
relative or absolute extrema of each function. State the x-value(s) where
they occur.
Exercises
Because f(-1.5) > f(-2) and f(-1.5) > f(-1), there is a relative maximum in the interval
(-2, -1) near -1.5.
Because f(-0.5) < f(-1) and f(-0.5) < f(0), there is a relative minimum in the interval
(-1, 0) near -0.5.
Because f(0.5) > f(0) and f(0.5) > f(1), there is a relative maximum in the interval
(0, 1) near 0.5.
Because f(1.5) < f(1) and f(1.5) < f(2), there is a relative minimum in the interval
(1, 2) near 1.5.
f(-100) < f(-1.5) and f(100) > f(1.5), which supports the conjecture that
f has no absolute extrema.
f(x)
x
4x
g(x) = x 5 - 4x 3 + 2x - 3
Support Numerically
Choose x-values in half-unit intervals on either side of the estimated x-value for each
extremum, as well as one very small and one very large value for x.
be no absolute extrema.
x → -∞
lim f(x) = -∞ and lim f(x) = ∞, so there appears to
Analyze Graphically
It appears that f(x) has a relative maximum of 0 at
x = -1.5, a relative minimum of -3.5 at x = -0.5,
a relative maximum of -2.5 at x = 0.5, and a relative
minimum of -6 at x = 1.5. It also appears that
Example
Estimate to the nearest 0.5 unit and classify the extrema for the
graph of f(x). Support the answers numerically.
Functions can increase, decrease, or remain
constant over a given interval. The points at which a function changes its increasing or
decreasing behavior are called critical points. A critical point can be a relative minimum,
absolute minimum, relative maximum, or absolute maximum. The general term for
minimum or maximum is extremum or extrema.
Extrema and Average Rates of Change
Study Guide and Intervention
Increasing and Decreasing Behavior
1-4
NAME
Answers (Lesson 1-3 and Lesson 1-4)
PERIOD
(continued)
Extrema and Average Rates of Change
Study Guide and Intervention
DATE
1
= −−−
A10
Evaluate and simplify.
Substitute -1 for x1 and 1 for x2.
Simplify.
Evaluate f(-1) and f(-3).
Substitute -3 for x1 and -1 for x2.
Glencoe Precalculus
005_026_PCCRMC01_893802.indd 22
Chapter 1
-56
5. f(x) = x4 + 8x - 3; [-4, 0]
-14
3. f(x) = x3 + 5x2 - 7x - 4; [-3, -1]
-28
1. f(x) = x4 + 2x3 - x - 1; [-3, -2]
22
7
6. f(x) = -x4 + 8x - 3; [0, 1]
26
Glencoe Precalculus
4. f(x) = x3 + 5x2 - 7x - 4; [1, 3]
0
2. f(x) = x4 + 2x3 - x - 1; [-1, 0]
Find the average rate of change of each function on the given interval.
Exercises
2.5 - (-2.5)
5
= − or −
2
1 - (-1)
f(x2) - f(x1)
f(1) - f(-1)
−
= −
x2 - x1
1 - (-1)
b. [-1, 1]
3
[0.5(-1) + 2(-1)] - [0.5(-3) + 2(-3)]
-1 - (-3)
–2.5 - (-19.5)
17
= − or −
2
-1 - (-3)
3
f(x2) - f(x1)
f(-1) - f(-3)
−
= −
x2 - x1
-1 - (-3)
a. [-3, -1]
Example
Find the average rate of change of f(x) = 0.5x3 + 2x on
each interval.
2
2
1
msec = −
x -x
f(x ) - f(x )
The average rate of change on the interval [x1, x2] is the slope of the secant
line, msec.
The average rate of change between any
two points on the graph of f is the slope of the line through those points. The
line through any two points on a curve is called a secant line.
3/22/09 5:51:08 PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Chapter 1
Average Rate of Change
1-4
NAME
Extrema and Average Rates of Change
Practice
DATE
PERIOD
x
increasing on (-∞, 0);
decreasing on (0, 1.5);
increasing on (1.5, ∞);
See students’ work.
0
y
g(x) = x 5 - 2x 3 + 2x 2
2.
0
y
x
decreasing on (-∞, 0); decreasing
on (0, ∞); See students’ work.
5x
f (x) = 3
4
−4
0
4x
rel. min. of -8.5 at x = -1.5;
rel. max. of -5 at x = 0;
rel. min. of -6 at x = 1;
See students’ work.
−4
8
y
f(x) = x 4 - 3x 2 + x - 5
4.
x
rel. max. of 1 at x = -1;
rel. min. of 0 at x = 0.5;
See students’ work.
0
y
f (x) = x 3 + x 2 - x
-160
7. g(x) = -3x3 - 4x; [2, 6]
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
005_026_PCCRMC01_893802.indd 23
Chapter 1
23
8. PHYSICS The height t seconds after a toy rocket is launched straight up
can be modeled by the function h(t) = -16t2 + 32t + 0.5, where h(t) is in
feet. Find the maximum height of the rocket. 16.5 ft
-132
6. g(x) = x4 + 2x2 - 5; [-4, -2]
Lesson 1-4
3/22/09 5:51:12 PM
Glencoe Precalculus
Find the average rate of change of each function on the given interval.
rel. max. (-1.05, 6.02); rel. min. (1.05, -4.02)
5. GRAPHING CALCULATOR Approximate to the nearest hundredth the
relative or absolute extrema of h(x) = x5 - 6x + 1. State the x-values
where they occur.
3.
Estimate to the nearest 0.5 unit and classify the extrema for the
graph of each function. Support the answers numerically.
1.
Use the graph of each function to estimate intervals to the nearest
0.5 unit on which the function is increasing, decreasing, or constant.
Support the answer numerically.
1-4
NAME
Answers (Lesson 1-4)
A11
h(t)
2
4
6
t
Day Number
2 4 6 8 10 12 14 16 18 20 22 24 26
g(x) = -x 4 + 48x 3 - 822x 2 + 5795x - 7455
24
Glencoe Precalculus
9/30/09 2:03:12 PM
Answers
Glencoe Precalculus
9-inch sides are cut from each
corner, the volume of the box
is 0 because no material
remains.
(9, 0); When squares with
c. What is the relative minimum of
the function? Explain what this
minimum means in the context of
the problem.
3 in.; 432 in
3
b. What value of x maximizes the
volume? What is the maximum
volume?
v(x) = 4x3 - 72x2 + 324x
a. Write a function v(x) where v is the
volume of the box and x is the length
of the side of a square that was cut
from each corner of the cardboard.
4. BOXES A box with no top and a square
base is to be made by taking a piece of
cardboard, cutting equal-sized squares
from the corners and folding up each
side. Suppose the cardboard piece is
square and measures 18 inches on
each side.
-921
c. Day 18 to Day 20
19
b. Day 13 to Day 15
1395
a. Day 2 to Day 6
3. RECREATION For the function in
Exercise 2, find the average rate of
change for each time interval.
PERIOD
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
005_026_PCCRMC01_893802.indd 24
Chapter 1
rel. max. (7, 6897); rel. min. (13, 5857);
rel. max. (16, 5909)
0
1000
2000
3000
4000
5000
6000
7000
8000
2. RECREATION The daily attendance at a
state fair is modeled by g(x) = -x4 + 48x3
- 822x2 + 5795x - 7455, where x is the
number of days since opening. Estimate
to the nearest unit the relative or
absolute extrema and the x-values where
they occur.
24.4 m; See students’ work.
b. Estimate the greatest height reached
by the flare. Support the answer
numerically.
0
6
12
18
24
DATE
Extrema and Average Rates of Change
Word Problem Practice
a. Graph the function.
Attendance
Chapter 1
1. FLARE A lost boater shoots a flare
straight up into the air. The height of the
flare, in meters, can be modeled by
h(t) = -4.9t2 + 20t + 4, where t is the
time in seconds since the flare was
launched.
1-4
NAME
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Enrichment
DATE
PERIOD
k = 19
x
k = −12
0
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
005_026_PCCRMC01_893802.indd 25
Chapter 1
25
Sample answer: They are not functions.
x
Lesson 1-4
10/23/09 4:58:49 PM
Glencoe Precalculus
a. k < -13; b. k = -13;
c. k > -13;
y
d.
8. x2 + 4x + y2 - 6y - k = 0
9. Why would it make no sense to discuss extrema and average rate
of change for the graphs in Exercises 7 and 8?
0
a. k > 20; b. k = 20;
c. k < 20;
y
d.
7. x2 - 4x + y2 + 8y + k = 0
d. Choose a value of k for which the graph is a curve. Then sketch
the curve on the axes provided.
c. will the graph be a curve?
b. will the graph be a point?
a. will the solutions of the equation be imaginary?
no
6. x2 + 4y2 + 4xy + 16 = 0
no
4. x2 + 16 = 0
no
2. x2 - 3x + y2 + 4y = -7
In Exercises 7 and 8, for what values of k :
no
5. x4 + 4y2 + 4 = 0
yes
3. (x + 2)2 + y2 - 6y + 8 = 0
no
1. (x + 3)2 + (y - 2)2 = -4
Determine whether each equation can be graphed on the
real-number plane. Write yes or no.
There are some equations that cannot be graphed on the real-number
coordinate system. One example is the equation x2 - 2x + 2y2 + 8y + 14 = 0.
Completing the squares in x and y gives the equation (x - 1)2 + 2 (y + 2)2 = -5.
For any real numbers x and y, the values of (x - 1)2 and 2(y + 2)2 are
nonnegative. So, their sum cannot be -5. Thus, no real values of x and y
satisfy the equation; only imaginary values can be solutions.
“Unreal” Equations
1-4
NAME
Answers (Lesson 1-4)
TI-Nspire Activity
PERIOD
A12
2
Glencoe Precalculus
005_026_PCCRMC01_893802.indd 26
Chapter 1
-9
7. f(x) = x4 - 3x3 - x2 - 6; [1, 2]
11
5. f(x) = x4 - x3 + 7x; [0, 2]
16
3. f(x) = x + 7x - 11; [4, 5]
17
-−
2
1. f(x) = x3 + 4x2 - 6x - 5; [-4, -2]
4
3
26
6
8. f(x) = x2 - 1; [1, 5]
-33
Glencoe Precalculus
6. f(x) = x4 - 3x3 - x2 - 6; [-2, -1]
-15
4. f(x) = x - x + 7x; [-2, -1]
-7
2. f(x) = x2 + 7x - 11; [-8, -6]
Use the method shown above to find the average rate of change of
each function on the given interval.
Exercises
Step 4: Press b and select MEASUREMENT > SLOPE. Choose the
line. The slope is -10, so the average rate of change for the
interval [-2, 0] is -10.
Step 3: Press b and choose POINTS & LINES > LINE. Connect the
points on the graph.
Step 2: Press b and choose POINTS & LINES >
POINT ON. Place two points anywhere on the graph.
Double-click on each x-coordinate, changing one to -2
and the other to 0. The y-coordinates will update. You
may need to adjust your viewing window to see the points.
Step 1: Add a GRAPHS & GEOMETRY page. Enter
the function rule in the function entry line.
Press / + G to hide the function entry line.
Example
For the function f(x) = x3 + 4x2 - 6x - 5,
find the average rate of change for the interval [-2, 0].
Given a function, you can
draw two points on the function, connect the points with a line, and then
find the slope of that line, giving you the average rate of change for that
interval.
DATE
3/22/09 5:51:32 PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Chapter 1
Finding an Average Rate of Change
1-4
NAME
DATE
PERIOD
Notes
graph is symmetric about the origin
f(x) =
1
f(x) = −
x
f(x) = | x |
f(x) = x
square root function
reciprocal function
absolute value function
greatest integer function
x→∞
lim f(x) = -∞ and lim f(x) = ∞
0
y
x
f(x) = x 3
x → -∞
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
027-042_PCCRMC01_893802.indd 27
Chapter 1
27
decreasing for (-∞, 0) and increasing for (0, ∞)
x→∞
Lesson 1-5
3/22/09 5:51:42 PM
Glencoe Precalculus
to y-axis; even function; continuous; lim f(x) = ∞ and lim f(x) = ∞;
D = {x | x ∈ }, R = {y | y ≥ 0, y ∈ }; x-int: 0; y-int: 0; symmetric with respect
Describe the following characteristics of the graph of the parent function
f(x) = x2 : domain, range, intercepts, symmetry, continuity, end behavior,
and intervals on which the graph is increasing/decreasing.
Exercise
The graph is always increasing, so it is increasing for (-∞, ∞).
x → -∞
As x decreases, y approaches negative infinity, and as
x increases, y approaches positive infinity.
The graph is continuous because it can be traced without
lifting the pencil off the paper.
f(x) = -f(x).
It is symmetric about the origin and it is an odd function:
The graph intersects the origin, so the x-intercept is 0 and
the y-intercept is 0.
The graph confirms that D = {x | x ∈ } and R = {y | y ∈ }.
Example
Describe the following characteristics of the graph of
the parent function f(x) = x3 : domain, range, intercepts, symmetry,
continuity, end behavior, and intervals on which the graph is
increasing/decreasing.
defined as the greatest integer less than
or equal to x; type of step function
graph is V-shaped
graph has two branches
graph is in first quadrant
graph is U-shaped
f(x) = x3
cubic function
√
x
points on graph have coordinates (a, a)
f(x) = x
f(x) = x2
graph is a horizontal line
identity function
constant function
quadratic function
Form
f(x) = c
Parent Function
A parent function is the simplest of the functions in a family.
Parent Functions and Transformations
Study Guide and Intervention
Parent Functions
1-5
NAME
Answers (Lesson 1-4 and Lesson 1-5)
Dilations
Reflections
Translations
…expanded horizontally if 0 < a < 1.
…compressed horizontally if a > 1.
…compressed vertically if 0 < a < 1.
…expanded vertically if a > 1.
…reflected in the y-axis.
…reflected in the x-axis.
…h units left when h < 0.
…h units right when h > 0.
…k units down when k < 0.
…k units up when k > 0.
Parent functions can be transformed to
g(x) = √-x - 1
0
f(x) = √x
y
28
4
8x
Glencoe Precalculus
9/30/09 2:04:21 PM
Answers
Glencoe Precalculus
x
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
027-042_PCCRMC01_893802.indd 28
Chapter 1
y
The graph of g(x) is the graph of the
square function f(x) = x2 expanded
vertically and translated 4 units down.
−8
−8 −4
−8
8x
−4
4
4
8
2. g(x) = 2x - 4
−4
−4 0
y
The graph of g(x) is the graph of the
absolute value function f(x) = |x|
compressed vertically and translated
4 units left.
−8
4
8
1. g(x) = 0.5 ⎪x + 4⎥
2
Identify the parent function f(x) of g(x), and describe how the graphs of g(x) and
f(x) are related. Then graph f(x) and g(x) on the same axes.
Exercises
The graph of g(x) is the graph of the square root
function f(x) = √
x reflected in the y-axis and
then translated one unit down.
Identify the parent function f(x) of g(x) = √
-x - 1, and describe how
the graphs of g(x) and f(x) are related. Then graph f(x) and g(x) on the same axes.
g(x) = f(ax) is the graph
of f(x)…
g(x) = a f(x) is the graph
of f(x)…
g(x) = f(-x) is the graph
of f(x)…
g(x) = -f(x) is the graph
of f(x)…
g(x) = f(x - h) is the graph
of f(x) translated…
g(x) = f(x) + k is the graph
of f(x) translated…
PERIOD
(continued)
Parent Functions and Transformations
create other members in a family of graphs.
Example
DATE
Study Guide and Intervention
Transformations of Parent Functions
1-5
NAME
DATE
f(x)
x
to graph
0
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
027-042_PCCRMC01_893802.indd 29
Chapter 1
0
y
x
-1 if x ≤ -3
5. Graph f(x) = 1 + x if -2 < x ≤ 2.
x if 4 ≤ x ≤ 6
y
g(x)
f(x)
x
g(x)
−8−6−4
−4
−6
−8
8
6
4
2
0
y
y
29
g(x)
0
y
x
f(x)
x
2 4 6 8x
g(x)
Lesson 1-5
10/23/09 4:24:52 PM
Glencoe Precalculus
6. Use the graph of f(x) = x3 to graph
g(x) = ⎪(x + 1)3⎥.
The graph of g(x) is the graph of f(x) = | x |
stretched vertically and translated 2 units left
and 3 units down.
4. Identify the parent function f(x) of g(x) = 2|x + 2| - 3.
Describe how the graphs of g(x) and f(x) are related.
Then graph f(x) and g(x) on the same axes.
g(x) is f(x) reflected in the x-axis,
translated 1 unit right and 1 unit up.
g(x) = -(x - 1)2 + 1
PERIOD
2. Use the graph of f(x) = ⎪x⎥ to graph
g(x) = -|2x|.
3. Describe how the graph of f(x) = x2 and g(x) are
related. Then write an equation for g(x).
0
y g(x)
√
x
Parent Functions and Transformations
Practice
1. Use the graph of f(x) =
g(x) = √
x + 3 + 1.
1-5
NAME
⎧
A13
⎨
Chapter 1
⎩
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Answers (Lesson 1-5)
A14
Glencoe Precalculus
027-042_PCCRMC01_893802.indd 30
Chapter 1
1 2
h(x) = - −
x + 10
10
b. Suppose the same shot was made
from a tee located 10 yards behind the
original tee. Rewrite h(x) to reflect
this change.
h(x) is the graph of f(x)
translated 10 units right,
compressed vertically,
reflected in the x-axis, and then
translated 10 units up.
a. Describe the transformation of the
parent function f(x) = x2 used to graph
h(x).
1 2
x + 2x,
can be modeled by h(x) = - −
10
where h(x) is the distance above the
ground in yards and x is the horizontal
distance from the tee in yards.
2. GOLF The path of the flight of a golf ball
d. Find the minimum perimeter for an
area of 1000 square feet. 126.5 ft
sample answer: they are two
different kinds of rational
functions.
c. Is the function you found in part b a
transformation of f(x)? Explain. No;
A
P(ℓ) = 2ℓ + 2 −
(ℓ)
b. Describe a function of the length that
could be used to find a minimum
perimeter for a given area
a. If the area is 1000 square feet,
describe the transformations of the
1
parent function f(x) = −
x used to graph
w(x). f(x) is expanded vertically.
30
PERIOD
36
39.6
151,751 to 271,050
271,051 and up
30
60
Taxable Income
(thousands)
90 120 150 180 210 240 270 300
Glencoe Precalculus
It is the parent function
expanded vertically.
b. The function g(x) = 1.2 √
x is also
used to approximate the distance to
the horizon. How does the graph of
g(x) compare to the graph of its
parent function?
It is the parent function
compressed horizontally.
a. How does the graph of f(x) compare
to the graph of its parent function?
4. HORIZON The function f(x) = √1.5x
can be used to approximate the distance
to the apparent horizon, or how far a
person can see on a clear day, where
f(x) is the distance in miles and x is the
person’s elevation in feet.
10
20
30
40
50
31
99,601 to 151,750
Source: Information Please Almanac
15
28
41,201 to 99,600
Tax Rate
(%)
0 to 41,200
Limits of Taxable
Income ($)
Income Tax Rates for a Couple
Filing Jointly
3. TAXES Graph the tax rates for the
different incomes by using a step
function.
Parent Functions and Transformations
Word Problem Practice
1. AREA The width w of a rectangular plot
of land with fixed area A is modeled by
A
the function w() = −
, where is the
length.
1-5
DATE
3/22/09 5:52:06 PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Chapter 1
Tax Rate (%)
NAME
Enrichment
DATE
PERIOD
"
"'
0
y
y
8x
−8
−4
4
8
−8
−4
2
y
4
1 2
5. f(x) = −
x -1
8x
y
4
8x
−8
−8
−4
−4 0
4
8
y
4
7. f(x) = √
-x - 4
8x
3
2
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
027-042_PCCRMC01_893802.indd 31
Chapter 1
31
represents the rotated graph? f(x) = - − x + −
1
2
8. The graph of the function f(x) = 2x - 3 is rotated 90°. What function
−8
−4
−8 −4 0
4
6. f(x) = ⎪x3 + 2x - 4⎥
"
"
x
Lesson 1-5
3/22/09 5:52:11 PM
Glencoe Precalculus
Graph each function. Then graph the function after it is rotated 270°.
−8
−8 −4 0
4
8
4. f(x) = x3 - 2
Graph each function. Then graph the function after it is rotated 90°.
To rotate a function, you can plot several image points and then connect them.
3. Rotate point A'' by 90°. Graph the point. Give the
coordinates of A'''. Then use the result to write a rule for
rotating (x, y) by 270°. (2, -3); (x, y) → (y, -x)
2. Rotate point A' by 90°. Graph the point. Give the
coordinates of A''. Then use the result to write a rule for
rotating (x, y) by 180°. (-3, -2); (x, y) → (-x, -y)
1. Rotate point A by 90° using the rule. Graph the point.
Give the coordinates of A'. (-2, 3)
A rotation is a rigid transformation. A rotation turns a figure about a point
a certain number of degrees. The rotation can be clockwise or
counterclockwise. For this activity, assume all rotations are about the origin
and in the counterclockwise direction. To rotate a point 90° about the origin,
use the rule (x, y) → (-y, x).
Rotations
1-5
NAME
Answers (Lesson 1-5)
Chapter 1
DATE
PERIOD
Function Operations and Composition of Functions
Study Guide and Intervention
A15
f
()
x
The domain of f is (-∞, ∞) and the
domain of g is (−∞, 0) ∪ (0, ∞), so the
domain of (f - g) is (−∞, 0) ∪ (0, ∞).
x
3
=x-−
1
= (x2 - 3) −
(f g)x = f(x) g(x)
x
32
√x
Glencoe Precalculus
9/30/09 2:04:35 PM
Answers
Glencoe Precalculus
x2 + 4x - 7
− ; D = (0, ∞)
√
x
x2 √
x + 4x √
x - 7 √
x ; D = [0, ∞)
x2 + 4x - 7 − √
x ; D = [0, ∞)
x + 4x - 7 + √
x ; D = [0, ∞)
2
2. f(x) = x2 + 4x − 7, g(x) =
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
027-042_PCCRMC01_893802.indd 32
Chapter 1
x -x+2
−
; D = (-∞, 0) ∪ (0, ∞)
x
x3 - x - 2
−
; D = (-∞, 0) ∪ (0, ∞)
x
2x2 - 2
−
;
D
= (-∞, 0) ∪ (0, ∞)
x
x3 - x
−; D = (-∞, 0) ∪ (0, ∞)
2
3
2
1. f(x) = x2 - 1, g(x) = −
Find (f + g)(x), (f - g)(x), (f g)(x), and − (x) for each f(x) and g(x).
g
State the domain of each new function.
Exercises
The domain of f is (-∞, ∞) and the
domain of g is (−∞, 0) ∪ (0, ∞), so the
domain of (f - g) is (−∞, 0) ∪ (0, ∞).
(f - g)x = f(x) - g(x)
1
= x2 - 3 - −
x
b. (f g)(x)
a. (f - g)(x)
x
1
Given f(x) = x2 - 3 and g(x) = −
, find each function and its domain.
is {x | x ≠ -2, x ∈ }.
the denominator of −
g . So, the domain
(f)
The domains of f and g are both
(-∞, ∞), but x = -2 yields a zero in
= − =x-3
x+2
(x - 3)(x + 2)
x+2
2
-x-6
= x−
()
f(x)
(−gf )x = −
g(x)
Example 2
The domains of f and g are both
(-∞, ∞), so the domain of (f + g) is
(-∞, ∞).
= x2 - 4
(f + g)x = f(x) + g(x)
= x2 - x - 6 + x + 2
Example 1
Given f(x) = x2 - x - 6 and g(x) = x + 2, find each
function and its domain.
f
b. −
a. (f + g)(x)
g (x)
Two functions can be added, subtracted,
multiplied, or divided to form a new function. For the new function, the
domain consists of the intersection of the domains of the two functions,
excluding values that make a denominator equal to zero.
Operations with Functions
1-6
NAME
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
DATE
(continued)
PERIOD
Function Operations and Composition of Functions
Study Guide and Intervention
Simplify.
= 3(16x2 + 16x + 4) + 8x + 4 - 1
= 12x + 8x - 2
Simplify.
Substitute 3x2 + 2x - 1 for x in g(x).
Replace f(x) with 3x2 + 2x - 1.
x-2
2x - 5
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
027-042_PCCRMC01_893802.indd 33
Chapter 1
8 - 3x
1
−
;−
; -2
x-2
1
7. f(x) = 2x - 3, g(x) = −
3x2 - 2; 9x2 - 30x + 26; 46
5. f(x) = 3x - 5, g(x) = x2 + 1
125x3; 5x3; 8000
3. f(x) = x3, g(x) = 5x
2x2 - 4x - 7; 4x2 - 5; 9
1. f(x) = 2x + 1, g(x) = x2 - 2x - 4
3x - 4
16
2
33
x - 4; x - 4; 0
8. f(x) = x - 8, g(x) = x + 4
x - 2 x - 2x + 1 14
1
2x - x
1
−
;−
;−
2
2
x-1
1
6. f(x) = −
, g(x) = x2 - 1
Lesson 1-6
3/22/09 5:52:26 PM
Glencoe Precalculus
-2
4 √
x + 3 - 2; √
4x + 1 ; 4 √7
4. f(x) = 4x − 2, g(x) = √x + 3
x
3 - 4x
61
1
−
;−
; -−
2
2
2
1
2. f(x) = 3x2 − 4, g(x) = −
x
For each pair of functions, find [f ◦ g](x), [g ◦ f](x), and [f ◦ g](4).
Exercises
2
= 4(3x + 2x - 1) + 2
2
= g(3x2 + 2x - 1)
[g ◦ f](x) = g(f(x))
Definition of composite functions
Substitute 4x + 2 for x in f(x).
= 3(4x + 2)2 + 2(4x + 2) - 1
= 48x2 + 56x + 15
Replace g(x) with 4x + 2.
= f(4x + 2)
Definition of composite functions
Given f(x) = 3x2 + 2x - 1 and g(x) = 4x + 2, find [f ◦ g](x)
[f ◦ g](x) = f[g(x)]
and [g ◦ f](x).
Example
Given functions f and g, the composite function f ◦ g can be described by the
equation [f ◦ g](x) = f[g(x)]. The domain of f ◦ g includes all x-values in the
domain of g for which g(x) is in the domain of f.
In a function composition, the result of
one function is used to evaluate a second function.
Compositions of Functions
1-6
NAME
Answers (Lesson 1-6)
(f)
2x2 + 5x + 2, D = (-∞, ∞)
1. f(x) = 2x2 + 8 and g(x) = 5x - 6
2
{|
5
}
3
x
−
, D = (-1, ∞)
√
x+1
3
x √
x + 1 , D = [-1, ∞)
x3 - √
x + 1 , D = [-1, ∞)
A16
2
g(x) = x + 1
1
Sample answer: f(x) = −
,
3x
3x +3
1
10. h(x) = −
Glencoe Precalculus
027-042_PCCRMC01_893802.indd 34
Chapter 1
34
Glencoe Precalculus
f(x) = 3x, where x is the cost for one meal; g(x) = 1.18x; g(f(x)) = 3.54x
11. RESTAURANT A group of three restaurant patrons order the same meal
and drink and leave an 18% tip. Determine functions that represent the
cost of all of the meals before tip, the actual tip, and the composition of
the two functions that gives the cost for all of the meals including tip.
Sample answer: f(x) = √
x - 1,
g(x) = 2x - 6
9. h(x) = √2x
- 6 -1
Find two functions f and g such that h(x) = [f ◦ g](x). Neither
function may be the identity function f(x) = x.
1
{x | x ≠ ± √3, x ∈ }; f ◦ g = −
x -3
3x - 2
{x | x ≥ −23 , x ∈ }; f ◦ g = √
x-8
g(x) = x2 + 5
1
8. f(x) = −
12x2 - 16x + 10; 6x2 - 4x + 9; 70
6. f(x) = 3x2 - 2x + 5 and g(x) = 2x - 1
54x3 - 27x2 + 1; 6x3 - 9x2 + 3; 1216
4. f(x) = 2x3 - 3x2 + 1 and g(x) = 3x
g(x) = 3x
7. f(x) = √x
-2
Find f ◦ g.
8x2 - 34x + 34; 4x2 - 10x - 1; 4
5. f(x) = 2x2 - 5x + 1 and g(x) = 2x - 3
x + 2; x + 2; 4
3. f(x) = x + 5 and g(x) = x - 3
For each pair of functions, find [f ◦ g](x), [g ◦ f](x), and [f ◦ g](3).
6
−, D = x x ≠ −
,x∈
2x2 + 8
5x - 6
10x - 12x + 40x - 48,
D = (-∞, ∞)
3
2x - 5x + 14, D = (-∞, ∞)
x3 + √
x + 1 , D = [-1, ∞)
2. f(x) = x3 and g(x) = √
x+1
g(x). State the domain of each new function.
2
PERIOD
Function Operations and Composition of Functions
Practice
DATE
9/30/09 2:04:49 PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Chapter 1
Find (f + g)(x), (f - g)(x), (f · g)(x), and −
g (x) for each f(x) and
1-6
NAME
2
2
x
Sample answer: s(x) = −
+ 6.25;
28
2x
r(x) = √
gives the temperature in degrees Celsius
of the liquid in a beaker after x seconds.
Decompose the function into two
separate functions, s(x) and r(x), so that
s(r(x)) = t(x).
√
2x
28
3. SCIENCE The function t(x) = − + 6.25
s(x) = 0.9x;
s{f[c(h)]} = 10.8 + 0.9h
b. A sale reduces the cost of making
c candles by 10%. Write the sale
function s(x) and the composite
function that gives the cost of candle
making after h hours if materials are
purchased during the sale.
f[c(h)] = 12 + h
a. Write the composite function that
gives the cost of candle making after
h hours.
2. CANDLES A hobbyist makes and sells
candles at a local market. The function
c(h) = 4h gives the number of candles
she has made after h hours. The function
f(c) = 12 + 0.25c gives the cost of making
c candles.
g[f(t)] = 6.25πt ; 314.2 ft
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
027-042_PCCRMC01_893802.indd 35
Chapter 1
DATE
PERIOD
35
Lesson 1-6
10/1/09 10:15:58 PM
Glencoe Precalculus
Sample answer: a(x) = 2x2;
b(x) = x - 3
5. POPULATION The function
p(x) = 2x2 - 12x + 18 predicts the
population of elk in a forest for the years
2010 through 2015 where x is the
number of years since 2000. Decompose
the function into two separate functions,
a(x) and b(x), so that [a ◦ b](x) = p(x) and
a(x) is a quadratic function and b(x) is a
linear function.
b: how much more the second
traveler can spend than the
first
c: $230; how much more the
second traveler can spend on a
7-night trip
a: (g - f)(x) = 15x + 125;
D = {x | x ≥ 0, x ∈ }
d. Repeat parts a–c for (g - f)(x).
combined amount that can be
spent by the travelers on a
7-night trip
c. Find (f + g)(7) and explain what the
value represents. $1560; the
budget of both travelers
b. What does the composite function in
part a represent? the combined
D = {x | x ≥ 0, x ∈ }
a. Find (f + g)(x) and the relevant
domain. (f + g)(x) = 105x + 825;
4. TRAVEL Two travelers are budgeting
money for the same trip. The first
traveler’s budget (in dollars) can be
represented by f(x) = 45x + 350. The
second traveler’s budget (in dollars) can
be represented by g(x) = 60x + 475x is
the number of nights.
Function Operations and Composition of Functions
Word Problem Practice
1. MARCHING BAND Band members
form a circle of radius r when the music
starts. They march outward as they
play. The function f(t) = 2.5t gives the
radius of
the circle in feet after t seconds.
Using g(r) = πr2 for the area of the
circle, write a composite function that
gives the area of the circle after t
seconds.
Then find the area, to the nearest tenth,
after 4 seconds.
1-6
NAME
Answers (Lesson 1-6)
Chapter 1
Enrichment
DATE
A17
PERIOD
36
Glencoe Precalculus
3/22/09 5:52:39 PM
Answers
Glencoe Precalculus
3
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
027-042_PCCRMC01_893802.indd 36
Chapter 1
b. Find the volume after 10 seconds of inflation. Use 3.14 for π. 1436.03 m
32
πt3
V(r(t)) = −
+ 2πt2 + 8πt + −
π
6
3
a. Find V(r(t)).
meters and t is the number of seconds since inflation began.
1
t + 2, where r is in
radius increases at a constant rate r(t) = −
2
3
4 3
πr . The balloon is being inflated so that the
given by V(r) = −
2. The volume V of a spherical weather balloon with radius r is
77.2 swings per minute
b. Find the frequency, to the nearest tenth, of a brass pendulum
at 300°C if the pendulum’s length at 0°C is 15 centimeters.
ℓ(1 + 0.00002C)
f(p(L(ℓ, C, e ))) = −−
300 √
(1 + 0.00002C)
1. a. Find and simplify f(p(L(ℓ, C, e))), an expression for the
frequency of a brass pendulum, e = 0.00002, in terms of its
length, in centimeters at 0°C, and the Celsius temperature.
Finally, the length of a pendulum is a function of its length ℓ at
0° Celsius, the Celsius temperature C, and the coefficient of
expansion e of the material of which the pendulum is made:
L(ℓ, C, e) = ℓ(1 + eC).
In turn, the period of a pendulum is a function of its length L in
centimeters: p(L) = 0.2 √
L.
The frequency f of a pendulum is the number of complete swings the
pendulum makes in 60 seconds. It is a function of the period p of the
pendulum, the number of seconds the pendulum requires to
60
make one complete swing: f(p) = −
p.
Because the area of a square A is explicitly determined by the length
of a side of the square, the area can be expressed as a function of one
variable, the length of a side s: A = f(s) = s2. Physical quantities are
often functions of numerous variables, each of which may itself be a
function of several additional variables. A car’s gas mileage, for
example, is a function of the mass of the car, the type of gasoline
being used, the condition of the engine, and many other factors, each
of which is further dependent on other factors. Finding the value of
such a quantity for specific values of the variables is often easiest by
first finding a single function composed of all the functions and then
substituting for the variables.
Applying Composition of Functions
1-6
NAME
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
DATE
Graphing Calculator Activity
Y=
2
—
1
VARS
11 +
VARS
ENTER
3
12
ENTER
.
+ 2
ENTER
.
PERIOD
Y=
VARS
2
Y=
VARS
2
Y=
VARS
2
1
1
1
3
3
VARS
ENTER
VARS
ENTER
11 ÷
—
3
VARS
ENTER
—
11 ×
—
11
—
12
12
12
+ 2
ENTER
+ 2
ENTER
+ 2
ENTER
ENTER
.
ENTER
.
ENTER
.
. Then enter Y3 by
. Then enter Y3 by
. Then enter Y3 by
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
027-042_PCCRMC01_893802.indd 37
Chapter 1
See students’ work.
37
6. Make conjectures about the functions that are the sum, difference,
product, and quotient of two functions.
Lesson 1-6
3/22/09 5:52:42 PM
Glencoe Precalculus
The Y table shows ERROR when x = -5. The domain does not
include -5.
5. Graph Y1 ÷ Y2 for the functions described in Exercise 4. Use the table
function to compare values. What do you notice about the domain?
See students’ work.
4. Repeat the activity using functions f(x) = x2 - 1 and g(x) = 5 - x as Y1
and Y2, respectively. What do you observe?
pressing
Press
3. Y3 = Y1 ÷ Y2
pressing
Press
2. Y3 = Y1 Y2
pressing
Press
1. Y3 = Y1 - Y2
Use the functions f(x) = 2x - 1 and f(x) = 3x + 2 as Y1 and Y2,
respectively. Use TABLE to observe the results for each definition of
Y3.
Exercises
• Use TABLE to compare the function values for Y1, Y2, and Y3.
Press 2nd [TABLE].
To enter Y3, press
• Enter Y1 + Y2 as the function for Y3.
respectively. Press
Use a graphing calculator to explore the sum of two functions.
• Enter the functions f(x) = 2x - 1 and g(x) = 3x + 2 as Y1 and Y2,
Sum of Two Functions
1-6
NAME
Answers (Lesson 1-6)
Inverse Relations and Functions
Study Guide and Intervention
DATE
2
A18
[-10, 10] scl: 1 by [-10, 10] scl: 1
[-10, 10] scl: 1 by [-10, 10] scl: 1
3
2
Glencoe Precalculus
027-042_PCCRMC01_893802.indd 38
Chapter 1
yes
5. f(x) = -x3 + 6
no
3. f(x) = x - 8x + 6x - 4
yes
1
1. f(x) = −
x
38
no
6. f(x) = -x3 + 2x
yes
1
4. f(x) = −
√
x-4
no
2. f(x) = x - 5
2
Glencoe Precalculus
Graph each function using a graphing calculator, and apply the
horizontal line test to determine whether its inverse function exists.
Write yes or no.
Exercises
No, the inverse function does not exist. It is possible to find
a horizontal line that intersects the graph in more than point.
Therefore, you can conclude that g-1(x) does not exist.
b. g(x) = x + 2x - 5x + 1
4
Yes, the inverse function exists. It is not possible to find
a horizontal line that intersects the graph of f(x) in more
than one point, so you can conclude that f-1(x) exists.
4
1 3
a. f(x) = −
x -3
Graph each function using a graphing calculator, and
apply the horizontal line test to determine whether its inverse
function exists. Write yes or no.
Example
A function has an inverse function if and only if each horizontal line
intersects the graph of the function in at most one point. This is known as the
horizontal line test. If a function passes the horizontal line test, it is said to
be one-to-one because every x-value is matched with exactly one y-value.
Two relations are inverse
relations if and only if one relation contains the element (b, a) whenever the
other relation contains the element (a, b). If the inverse of the function f(x) is
also a function, then the inverse is denoted by f-1(x).
PERIOD
3/22/09 5:52:48 PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Chapter 1
Inverse and One-to-One-Functions
1-7
NAME
DATE
4
1
x=−
y+3
4
1
y=−
x+3
Replace y with f -1(x).
Solve for y.
Exchange x and y.
Replace f(x) with y.
[-10, 10] scl: 1 by [-10, 10] scl: 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
027-042_PCCRMC01_893802.indd 39
Chapter 1
5 + 2x
yes; f -1(x) = −
x ;x≠0
x-2
5
3. f(x) = −
3
Lesson 1-7
3/22/09 5:52:53 PM
Glencoe Precalculus
yes; f -1(x) = √
x - 4 ; no
restrictions
4. f(x) = x3 + 4
no
2. f(x) = (x - 1)2 + 2
39
1
yes; f -1(x) = −
x + 2; no restrictions
2
1. f(x) = 2x - 4
Determine whether f has an inverse function. If it does, find the
inverse function and state any restrictions on the domain.
Exercises
Step 4: There are no restrictions on the domain.
f-1(x) = 4x - 12
y = 4x - 12
Step 3: 4x = y + 12
Step 2:
Step 1: The graph shows the function passes the
horizontal line test, so the inverse function exists.
1
Example
Determine whether f(x) = −
x + 3 has an inverse function. If it does,
4
find the inverse function and state any restrictions on the domain.
If given the graph of a function, you can graph its inverse. Locate points on
f (x), and reflect them in the line y = x. Interchange the x- and y-coordinates,
and connect them with a straight line or smooth curve.
You can verify the inverse function by showing that f f -1(x) = x and that f -1
f(x) = x. In other words, the composition of a function and its inverse is
always the identity function.
Step 4: State any restrictions on the domain.
Step 3: Solve for y. Replace y with f -1(x).
Step 2: Replace f(x) with y and then interchange x and y.
Step 1: Use the horizontal line test to confirm the inverse function exists.
function algebraically.
(continued)
PERIOD
Follow the steps below to find an inverse
Inverse Relations and Functions
Study Guide and Intervention
Find Inverse Functions
1-7
NAME
Answers (Lesson 1-7)
Chapter 1
Inverse Relations and Functions
Practice
DATE
PERIOD
5
yes
x
4. f(x) = −
+9
yes
2. f(x) = - √
x+3-1
A19
(
2
)
2
yes; f (x) = x + 2; x ≥ 0
-1
8. f(x) = √
x-2
2
0
( √
2x + 12 )2
y
gf(x) =
x
√ (2 )
x2
2−
- 6 + 12 = x
2
fg(x) = − - 6 = x
2
x
10. f(x) = −
- 6; x ≥ 0; g(x) = √
2x + 12
40
Glencoe Precalculus
9/30/09 2:05:04 PM
Answers
Glencoe Precalculus
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
027-042_PCCRMC01_893802.indd 40
Chapter 1
t(h) = −, where h is given in feet. Find the inverse of the function. If
4
the water takes 8 seconds to hit the ground, from what height did the
plane drop the water? t -1(h) = 16h2
√
h
the water to travel from the plane to the ground is given by the function
12. FIRE FIGHTING Airplanes are often used to drop water on forest fires in
an effort to stop the spread of the fire. The time in seconds it takes for
11. Use the graph of f(x) to graph f -1(x).
2x + 3 - 3
gf(x) = − = x
2
x-3
fg(x) = 2 −
+3=x
2
x-3
9. f(x) = 2x + 3; g(x) = −
7x + 1
2-x
yes; f -1(x) = − ; x ≠ 2
x+7
2x - 1
6. f(x) = −
Show algebraically that f and g are inverse functions.
no
4
7. f(x) = −
(x - 3)2
yes; f -1(x) = x3 + 1
3
5. f(x) = √
x-1
Determine whether f has an inverse function. If it does, find the
inverse function and state any restrictions on its domain.
yes
3. f(x) = x5 + 5x3
no
1. f(x) = 3⎪x⎥ + 2
Graph each function using a graphing calculator, and apply the
horizontal line test to determine whether its inverse function exists.
Write yes or no.
1-7
NAME
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
)
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
027-042_PCCRMC01_893802.indd 41
4π
9.8x
f -1(x) = −
2
2
x is the pendulum length in meters. Find
the inverse of the function.
√9.8
forth one time is modeled by the function
x
−
, x ≥ 0, where
f(x) = 2π takes for a pendulum to swing back and
0.15
0.15x + 20 - 20
-1
f f(x) = − = x
0.15
3. PENDULUM The time, in seconds, it
x - 20
ff-1(x) = 0.15 −
+ 20 = x
(
c. Verify that f(x) and f-1(x) are inverses.
0.15
x - 20
f -1(x) = −
b. Write an equation for the inverse of
the function.
f(x) = 0.15x + 20
a. Write a function for the amount of
money Claire saves, where x is the
amount of her paycheck.
2. SAVINGS Claire saves 15% of every
paycheck that she earns, plus an
additional $20.
c. If the electrician charges $225, how
many hours did the job last? 3 h
the amount the electrician
charges
b. What does x represent in the inverse
function?
the function. f -1(x) = −
x - 60
55
a. Find an equation for the inverse of
Chapter 1
DATE
41
PERIOD
y
4
8
12
x
x
−
; x = surface area
√
4π
Lesson 1-7
9/30/09 2:05:16 PM
Glencoe Precalculus
No; if you graph the
function, it fails the horizontal
line test.
⎨
6. WAGES During her free time, Daphne
babysits. She charges $4.50 for each
whole hour or any fraction of an hour.
The cost f of x hours of babysitting is
given by
⎧
4.5 x if x = x
f(x) =
.
4.5 x + 1 if x < x
⎩
Does the function have an inverse?
Explain.
4.8 in.
c. A basketball has a surface area of 278
square inches. What is the radius of
the ball? Round to the nearest tenth.
of sphere and f-1(x) = radius of
sphere.
f-1(x) =
b. Find the inverse of the function.
Explain what each variable
represents.
a. What is the relevant domain of the
function? (0, ∞)
5. BASKETBALL The surface area of a
sphere is given by f(x) = 4πx2, where
x is the radius of the sphere.
0
4
8
12
4. SALE The graph represents the flat-rate
shipping cost of an object that is on sale
by a certain percent. Use the graph of the
function to graph its inverse function.
Inverse Relations and Functions
Word Problem Practice
1. ELECTRICIAN The amount an electrician
charges can be modeled by the function
f(x) = 60 + 55x, where x is the number of
hours worked.
1-7
NAME
Answers (Lesson 1-7)
Enrichment
V E
29 6 27
A20
E
L E F T
T
25
Glencoe Precalculus
H
S
26
2
027-042_PCCRMC01_893802.indd 42
I
E
14
Chapter 1
W
13
1
E
H
15
3
F
27
A
16
O
E
28
V
17
4
N
A
29
E
18
5
L
Y
T
19
7
42
R
30
6
31
I
O
20
T
T
32
8
9
22 5
S
33
F
21
9
H
I
E
34
E
22
10
F
36
R
24
G
12
Glencoe Precalculus
L
35
A
23
N
11
26 32 30 23 25 12 15 1
S T R A I G H T
13
W H E N
35 3 21 8
From President Franklin D. Roosevelt’s inaugural address
during the Great Depression; delivered March 4, 1933.
10. The graph of the inverse of a linear
function is a __ line.
9. Two variables are inversely proportional __ their
product is constant.
8. To solve a matrix equation, multiply each side of the
matrix equation on the __ by the inverse matrix.
18
7. If · is a binary operation on set S and
x · e = e · x = x for all x in S, then an identity
element for the operation is __.
24 16 19 10 4
R A T I O
36 7
F Y
20 11 34
O N E
31 33 14
I S E
2
H A L F
17 28
PERIOD
6. The inverse ratio of two numbers is the __ of the
reciprocals of the numbers.
5. If the second coordinate of the inverse of (x, f (x)) is y,
then the first coordinate is read “__ of __.”
4. This is the product of a number and its
multiplicative inverse.
3. The first letter and the last two letters of the meaning
of the symbol f 1 are __ .
2. The inverse of the function 2x is found by
computing __ of x.
1. If a relation contains the element (e, v), then the
inverse of the relation must contain the element
( __ , __ ).
The puzzle on this page is called an acrostic. To solve the puzzle, work
back and forth between the clues and the puzzle box. You may need a
math dictionary to help with some of the clues.
DATE
3/22/09 5:53:08 PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Chapter 1
An Inverse Acrostic
1-7
NAME
Answers (Lesson 1-7)
Chapter 1 Assessment Answer Key
(Lessons 1-1 and 1-2)
Page 43
(Lessons 1-5 and 1-6)
Page 44
{x | x < -2, x }
2.
16
3.
D
D = {x | x > -5, x R = {y | y ≥ -3, y Page 45
0
};
}
x
g(x) is reflection
of f(x) in the x-axis
and translated
2. 2 units down
2
3. x + √x - 2; [0, ∞)
Sample answer:
f(x) = x 2 ,
5. g(x) = x + 5
Quiz 2 (Lessons 1-3 and 1-4)
Quiz 4 (Lesson 1-7)
Page 43
Page 44
discontinuous;
2. infinite
3.
A
lim f(x) = ∞;
2.
H
3.
D
4.
J
5.
D
x+6
yes; f -1(x) = − ;
3
2. no restrictions
6. 532; 763
no
3.
2
7. 2a + 3a + 3
5
yes; f -1(x) = −
;
x-1
4.x ≠ 1
x→-∞
lim f(x) = ∞
B
C
1.
discontinuous;
1. jump
1.
B
4.
odd; symmetric
5. about origin
Mid-Chapter Test
g(x)
1.
1. (-∞, -2)
4.
Quiz 3
Answers
Quiz 1
5.
8
4. x → ∞
8. {
x | x ≥ 11, x }
D = (-5, 2) ∪ (2, ∞);
R = -4 ∪ (-2, ∞)
y
9.
4
abs. min. of -1
5. at x = 1
−8
−4 0
4
−4
x
10.
10.5
−8
6.
Chapter 1
-43
A21
Glencoe Precalculus
Chapter 1 Assessment Answer Key
Vocabulary Test
Page 46
Form 1
Page 47
1.
C
2.
H
3.
A
12.
G
13.
D
14.
H
15.
A
16.
J
17.
D
18.
H
19.
D
20.
J
even
1.
2.
Page 48
one-to-one function
3.
point
4.
J
4.
dilation
5.
B
5.
zeros
6.
G
6.
range
7.
B
7.
8.
decreasing
8. G
continuous
9.
C
9. the simplest of
the functions in a
family
10. combining
functions so the
result of one is
used to evaluate
the second
Chapter 1
10.
11.
G
A
B:
A22
g(x) = x 2 -3
Glencoe Precalculus
Chapter 1 Assessment Answer Key
1.
D
Page 50
12.
14.
J
3.
A
4.
G
5.
D
6.
F
7.
8.
C
J
9.
C
10.
G
11.
D
Chapter 1
J
1.
13.
2.
Form 2B
Page 51
15.
16.
17.
12.
F
13.
C
14.
J
15.
C
16.
F
17.
D
18.
G
19.
A
20.
H
A
H
A
G
D
18.
F
19.
B
20.
J
B:
B
Page 52
2.
F
3.
A
4.
G
5.
B
6.
G
7.
8.
3
A
G
9.
B
10.
H
11.
A
8
−
A23
Answers
Form 2A
Page 49
B:
g(x) = x2 - 2
Glencoe Precalculus
Chapter 1 Assessment Answer Key
Form 2C
Page 53
Page 54
D = (-4, ∞);
1. R = [-3, ∞)
12.
y
x
0
2x2 + 4xh + 2h2 2. x - h
3.
1
-−
6
4.
-15
p(x)
translated right 3
units,
compressed
vertically by a
13. factor of 0.5
y
14.
5.
x-axis
6.
odd
15.
discontinuous;
7. infinite
lim g(x) = -∞;
x → -∞
x
0
16.
lim g(x) = -∞
8. x → ∞
17.
1
(f g)(x) = −
;
x+3
{x|x ≠ 3, -3, x ∈ }
v[f(t)] = 24πt2
1
[g ◦ f](x) = −
2
x - 6x
3
18.
9.
21 more
employees
19.
20.
10.
41 ft/s
11.
42
Chapter 1
B:
A24
f-1(x) = √x + 4
no
x - 40
f(x) = −
0.15
k=6
Glencoe Precalculus
Chapter 1 Assessment Answer Key
Form 2D
Page 55
Page 56
12.
y
p(x)
x
0
2
2. 3a - 12a + 8
3.
-12
4.
-35
1
translated right −
4
unit, compressed
horizontally by a
13. factor of 4
y
14.
5.
origin
6.
odd
x
0
15.
1
;
[f g](x) = −
x-4
Answers
D = (-∞, 3];
R = (-∞, 4]
1.
{x|x ≠ 4, -4, x ∈ }
7.
discontinuous;
jump
lim g(x) = -∞;
x → -∞
lim g(x) = ∞
8. x → ∞
16.
17.
v[f(t)] = 112πt2
1
[g ◦ f](x) = −
2
x + 8x
3
18.
25 more
employees
9.
10.
-61 ft/sec
11.
28
Chapter 1
19.
20.
B:
A25
f -1(x) = √x + 6
no
x - 50
f(x) = −
0.05
k=0
Glencoe Precalculus
Chapter 1 Assessment Answer Key
Form 3
Page 57
Page 58
1. D = (-5, -3] ∪
y
12.
(-2, 1) ∪ (2, ∞);
R = [-2, 1) ∪ 2 ∪
(3, ∞)
2.
32a2 - 36a + 20
3
-7, −
5
3.
-29.2
4.
x
0
1
translated −
left, 4 up,
2
reflected in x-axis,
compressed horizontally
by factor of 6 and
13. vertically by factor of 0.5
y
14.
x
0
origin
5.
(−gf )(x) = 4x
even
6.
discontinuous;
7. removable
lim g(x) = -4;
x → -∞
lim g(x) = -4
8. x → ∞
3
- 2x2;
15. {x | x ≠ 0, x ∈ }
16.
v[f(t)] = 54.675πt2
1
[g ◦ f](x) = −
9x + 21x + 12
4
17.
2
3
18.
f -1(x) = −
+2
x
no
19.
x - 203.5
f(x) = −
-1.14
9.
20 weeks; $640
20.
B:
10.
-38 ft/sec
11.
19.5
Chapter 1
where x is the
number of boxes
of wood screws.
y
0
A26
x
Glencoe Precalculus
Chapter 1 Assessment Answer Key
Page 59, Extended-Response Test
Sample Answers
1a.
1. parent graph
2. transformation
y
y
G (Y) = (x-5) 2
G (Y) = x 2
3. transformation
y
2d. Sample answer: h(x) = x - 5 and
j(x) = 2x2
4. transformation
y
x
= 2x2 - 16x + 31
x
0
0
[g º f](x) = 2(x - 4)2 - 1
= 2(x2 - 8x + 16) - 1
(5, 0)
x
0
2c. no; [f º g](x) = 2x2 - 1 - 4
= 2x2 - 5
0
(5, 0)
G (Y) = -(x-5) 2 -2
x
3a. Sample answer: f(x) = 0.04x +
0.07(1000 - x), where x is number of
small cups
(5, -2)
G (Y) = -(x-5) 2
The parent graph f(x) = x2 is reflected in
the x-axis and translated 5 units right
and 2 units down.
1b. D = {x|x ∈ }; R = {y|y ≤ -2, y ∈ }
x - 70
3b. Sample answer: f –1(x) = −
; x is total
-0.03
cost of order and f-1(x) is number of
small cups
3c. 650 small cups and 350 large cups
1c. There is an absolute maximum at (5, -2).
lim f(x) = -∞ and lim f(x) = -∞
x→∞
Answers
x → -∞
-3 - (-18)
4-1
1d. − = 5
2a. yes; (f g)(x) = (x - 4)(2x2 - 1)
= 2x3 - x - 8x2 + 4
= 2x3 - 8x2 - x + 4
(g f)(x) = (2x2 - 1)(x - 4)
= 2x3 - 8x2 - x + 4
2b. no; (f - g)(x) = x - 4 - (2x2 - 1)
= -2x2 + x - 3
(g - f)(x) = 2x2 - 1 - (x – 4)
= 2x3 - x + 3
Chapter 1
A27
Glencoe Precalculus
Chapter 1 Assessment Answer Key
Standardized Test Practice
Page 60
1.
A
B
C
F
G
H
J
3.
A
B
C
D
4.
F
G
H
J
5.
A
B
C
D
6.
F
G
H
J
A
B
Chapter 1
C
8.
F
G
H
J
9.
A
B
C
D
10.
F
G
H
J
11.
A
B
C
D
12.
F
G
H
J
13.
A
B
C
D
14.
F
G
H
J
D
2.
7.
Page 61
D
A28
Glencoe Precalculus
Chapter 1 Assessment Answer Key
Standardized Test Practice
(continued)
Page 62
y
15.
x
0
= x - 3x
(−gf )(x)
+ 9x - 27
3
16.
2
1
(g ° f)(x) = −
2
x +6
17.
Answers
-3
18.
discontinuous;
19. infinite
20. continuous
discontinuous;
21. removable
22a. f(x) = 350 + 0.08x
22b.
x - 350
y=−
22c.
Chapter 1
0.08
$2250
A29
Glencoe Precalculus
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