Preview Extract

Chapter 2
More on Functions
6. a) For x-values from 1 to 4, the y-values increase from 2
to 11. Thus the function is increasing on the interval
(1, 4).
Exercise Set 2.1
1. a) For x-values from โ5 to 1, the y-values increase from
โ3 to 3. Thus the function is increasing on the
interval (โ5, 1).
b) For x-values from 3 to 5, the y-values decrease from
3 to 1. Thus the function is decreasing on the interval (3, 5).
c) For x-values from 1 to 3, y is 3. Thus the function
is constant on (1, 3).
2. a) For x-values from 1 to 3, the y-values increase from 1
to 2. Thus, the function is increasing on the interval
(1, 3).
b) For x-values from โ5 to 1, the y-values decrease
from 4 to 1. Thus the function is decreasing on the
interval (โ5, 1).
c) For x-values from 3 to 5, y is 2. Thus the function
is constant on (3, 5).
3. a) For x-values from โ3 to โ1, the y-values increase
from โ4 to 4. Also, for x-values from 3 to 5, the
y-values increase from 2 to 6. Thus the function is
increasing on (โ3, โ1) and on (3, 5).
b) For x-values from โ1 to 1, the y-values decrease
from 6 to 2. Also, for x-values from 4 to โ, the yvalues decrease from 11 to โโ. Thus the function
is decreasing on (โ1, 1) and on (4, โ).
c) For x-values from โโ to โ1, y is 3. Thus the function is constant on (โโ, โ1).
7. The x-values extend from โ5 to 5, so the domain is [โ5, 5].
The y-values extend from โ3 to 3, so the range is [โ3, 3].
8. Domain: [โ5, 5]; range: [1, 4]
9. The x-values extend from โ5 to โ1 and from 1 to 5, so
the domain is [โ5, โ1] โช [1, 5].
The y-values extend from โ4 to 6, so the range is [โ4, 6].
10. Domain: [โ5, 5]; range: [1, 3]
11. The x-values extend from โโ to โ, so the domain is
(โโ, โ).
The y-values extend from โโ to 3, so the range is (โโ, 3].
12. Domain: (โโ, โ); range: (โโ, 11]
b) For x-values from 1 to 3, the y-values decrease from
3 to 2. Thus the function is decreasing on the interval (1, 3).
13. From the graph we see that a relative maximum value of
the function is 3.25. It occurs at x = 2.5. There is no
relative minimum value.
c) For x-values from โ5 to โ3, y is 1. Thus the function is constant on (โ5, โ3).
The graph starts rising, or increasing, from the left and
stops increasing at the relative maximum. From this point,
the graph decreases. Thus the function is increasing on
(โโ, 2.5) and is decreasing on (2.5, โ).
4. a) For x-values from 1 to 2, the y-values increase from 1
to 2. Thus the function is increasing on the interval
(1, 2).
b) For x-values from โ5 to โ2, the y-values decrease
from 3 to 1. For x-values from โ2 to 1, the y-values
decrease from 3 to 1. And for x-values from 3 to 5,
the y-values decrease from 2 to 1. Thus the function
is decreasing on (โ5, โ2), on (โ2, 1), and on (3, 5).
c) For x-values from 2 to 3, y is 2. Thus the function
is constant on (2, 3).
5. a) For x-values from โโ to โ8, the y-values increase
from โโ to 2. Also, for x-values from โ3 to โ2, the
y-values increase from โ2 to 3. Thus the function
is increasing on (โโ, โ8) and on (โ3, โ2).
b) For x-values from โ8 to โ6, the y-values decrease
from 2 to โ2. Thus the function is decreasing on
the interval (โ8, โ6).
c) For x-values from โ6 to โ3, y is โ2. Also, for xvalues from โ2 to โ, y is 3. Thus the function is
constant on (โ6, โ3) and on (โ2, โ).
14. From the graph we see that a relative minimum value of 2
occurs at x = 1. There is no relative maximum value.
The graph starts falling, or decreasing, from the left and
stops decreasing at the relative minimum. From this point,
the graph increases. Thus the function is increasing on
(1, โ) and is decreasing on (โโ, 1).
15. From the graph we see that a relative maximum value of
the function is 2.370. It occurs at x = โ0.667. We also
see that a relative minimum value of 0 occurs at x = 2.
The graph starts rising, or increasing, from the left and
stops increasing at the relative maximum. From this point
it decreases to the relative minimum and then increases
again. Thus the function is increasing on (โโ, โ0.667)
and on (2, โ). It is decreasing on (โ0.667, 2).
16. From the graph we see that a relative maximum value of
2.921 occurs at x = 3.601. A relative minimum value of
0.995 occurs at x = 0.103.
c 2016 Pearson Education, Inc.
Copyright
58
Chapter 2: More on Functions
The graph starts decreasing from the left and stops decreasing at the relative minimum. From this point it increases to the relative maximum and then decreases again.
Thus the function is increasing on (0.103, 3.601) and is decreasing on (โโ, 0.103) and on (3.601, โ).
17.
20.
y
5
4
3
2
1
y
โ5 โ4 โ3 โ2 โ1
โ1
5
โ2
4
โ3
3
โ4
f (x ) = x 2
2
1
2
3
4
5
x
f (x ) = | x + 3 | โ 5
โ5
1
โ5 โ4 โ3 โ2 โ1
โ1
1
2
3
4
5
Increasing: (โ3, โ)
x
Decreasing: (โโ, โ3)
โ2
โ3
โ4
Maxima: none
โ5
Minimum: โ5 at x = โ3
The function is increasing on (0, โ) and decreasing on
(โโ, 0). We estimate that the minimum is 0 at x = 0.
There are no maxima.
21.
y
5
4
3
18.
y
2
1
5
โ5 โ4 โ3 โ2 โ1
f (x ) = 4 โ x 2
4
3
โ2
2
โ1
1
2
3
4
5
x
f (x ) = x 2 โ 6x + 10
โ3
1
โ5 โ4 โ3 โ2 โ1
โ1
โ4
1
2
3
4
5
x
โ5
โ2
The function is decreasing on (โโ, 3) and increasing on
(3, โ). We estimate that the minimum is 1 at x = 3.
There are no maxima.
โ3
โ4
โ5
Increasing: (โโ, 0)
22.
Decreasing: (0, โ)
y
10 2
f (x ) = โ
x โ 8x โ 9
8
Maximum: 4 at x = 0
6
4
Minima: none
2
19.
y
โ10 โ8 โ6 โ4 โ2
5
2
4
6
8 10
x
โ4
f (x ) = 5 โ | x |
4
โ2
โ6
3
โ8
2
โ10
1
โ5 โ4 โ3 โ2 โ1
โ1
1
2
3
4
5
Increasing: (โโ, โ4)
x
Decreasing: (โ4, โ)
โ2
โ3
Maximum: 7 at x = โ4
โ4
โ5
Minima: none
The function is increasing on (โโ, 0) and decreasing on
(0, โ). We estimate that the maximum is 5 at x = 0.
There are no minima.
23. If x = the length of the rectangle, in meters, then the
480 โ 2x
, or 240 โ x. We use the formula Area =
width is
2
length ร width:
A(x) = x(240 โ x), or
A(x) = 240x โ x2
c 2016 Pearson Education, Inc.
Copyright
Exercise Set 2.1
59
24. Let h = the height of the scarf, in inches. Then the length
of the base = 2h โ 7.
1
A(h) = (2h โ 7)(h)
2
7
A(h) = h2 โ h
2
25. We use the Pythagorean theorem.
[h(d)]2 + 35002 = d2
[h(d)]2 = d2 โ 35002
โ
h(d) = d2 โ 35002
We considered only the positive square root since distance
must be nonnegative.
26. After t minutes, the balloon has risen 120t ft. We use the
Pythagorean theorem.
[d(t)]2 = (120t)2 + 4002
d(t) = (120t)2 + 4002
We considered only the positive square root since distance
must be nonnegative.
27. Let w = the width of the rectangle. Then the
40 โ 2w
, or 20 โ w. Divide the rectangle into
length =
2
quadrants as shown below.
1
3d
2
= s
7
12
1
s
ยท 3d = 7 ยท
12
2
7
sd
=
4
2
4 7
d = ยท , so
s 2
14
d(s) =
.
s
30. The volume of the tank is the sum of the volume of a sphere
with radius r and a right circular cylinder with radius r
and height 6 ft.
4
V (r) = ฯr3 + 6ฯr2
3
31. a) After 4 pieces of ๏ฌoat line, each of length x ft, are
used for the sides perpendicular to the beach, there
remains (240โ4x) ft of ๏ฌoat line for the side parallel
to the beach. Thus we have a rectangle with length
240 โ 4x and width x. Then the total area of the
three swimming areas is
A(x) = (240 โ 4x)x, or 240x โ 4×2 .
b) The length of the sides labeled x must be positive
and their total length must be less than 240 ft, so
4x < 240, or x < 60. Thus the domain is
{x|0 < x < 60}, or (0, 60).
c) We see from the graph that the maximum value of
the area function on the interval (0, 60) appears to
be 3600 when x = 30. Thus the dimensions that
yield the maximum area are 30 ft by 240 โ 4 ยท 30, or
240 โ 120, or 120 ft.
20 โ w
w
In each quadrant there are two congruent triangles. One
triangle is part of the rhombus and both are part of the
rectangle. Thus, in each quadrant the area of the rhombus
is one-half the area of the rectangle. Then, in total, the
area of the rhombus is one-half the area of the rectangle.
1
A(w) = (20 โ w)(w)
2
w2
A(w) = 10w โ
2
28. Let w = the width, in feet. Then the length =
or 23 โ w.
A(w) = (23 โ w)w
46 โ 2w
,
2
A(w) = 23w โ w2
29. We will
in
use similar triangles, expressing all distances
s
1
ft, and d yd = 3d ft We
feet. 6 in. = ft, s in. =
2
12
have
32. a) If the length = x feet, then the width = 24 โ x feet.
A(x) = x(24 โ x)
A(x) = 24x โ x2
b) The length of the rectangle must be positive and
less than 24 ft, so the domain of the function is
{x|0 < x < 24}, or (0, 24).
c) We see from the graph that the maximum value of
the area function on the interval (0, 24) appears to
be 144 when x = 12. Then the dimensions that yield
the maximum area are length = 12 ft and width =
24 โ 12, or 12 ft.
33. a) When a square with sides of length x is cut from
each corner, the length of each of the remaining sides
of the piece of cardboard is 12 โ 2x. Then the dimensions of the box are x by 12 โ 2x by 12 โ 2x. We
use the formula Volume = length ร width ร height
to ๏ฌnd the volume of the box:
V (x) = (12 โ 2x)(12 โ 2x)(x)
V (x) = (144 โ 48x + 4×2 )(x)
V (x) = 144x โ 48×2 + 4×3
This can also be expressed as V (x) = 4x(x โ 6)2 , or
V (x) = 4x(6 โ x)2 .
c 2016 Pearson Education, Inc.
Copyright
60
Chapter 2: More on Functions
b) The length of the sides of the square corners that
are cut out must be positive and less than half the
length of a side of the piece of cardboard. Thus, the
domain of the function is {x|0 < x < 6}, or (0, 6).
c) We see from the graph that the maximum value of
the area function on the interval (0, 6) appears to
be 128 when x = 2. When x = 2, then 12 โ 2x =
12 โ 2 ยท 2 = 8, so the dimensions that yield the
maximum volume are 8 cm by 8 cm by 2 cm.
๏ฃฑ1
๏ฃฒ x,
for x < 0,
39. f (x) = 2
๏ฃณ
x + 3, for x โฅ 0
1
x for
2
inputs x less than 0. Then graph f (x) = x + 3 for inputs
x greater than or equal to 0.
We create the graph in two parts. Graph f (x) =
y
4
34. a) V (x) = 8x(14 โ 2x), or 112x โ 16×2
14
b) The domain is x0 < x <
, or
2
{x|0 < x 1
Since โ4 โค 1, g(โ4) = โ4 + 4 = 0.
x
โซบ4
๏ฃฑ 1
๏ฃฒ โ x + 2, for x โค 0,
3
40. f (x) =
๏ฃณ
x โ 5,
for x > 0
Since 0 โค 1, g(0) = 0 + 4 = 4.
y
Since 1 โค 1, g(1) = 1 + 4 = 5.
4
Since 3 > 1, g(3) = 8 โ 3 = 5.
2
๏ฃฑ
๏ฃด
๏ฃฒ 3,
36. f (x) =
โซบ4 โซบ2
for x โค โ2,
1
๏ฃด
๏ฃณ x + 6, for x > โ2
2
4
x
โซบ4
๏ฃฑ 3
๏ฃฒ โ x + 2, for x < 4,
4
41. f (x) =
๏ฃณ
โ1,
for x โฅ 4
We create the graph in two parts. Graph
3
f (x) = โ x + 2 for inputs x less than 4. Then graph
4
f (x) = โ1 for inputs x greater than or equal to 4.
f (โ5) = 3
f (โ2) = 3
1
f (0) = ยท 0 + 6 = 6
2
1
f (2) = ยท 2 + 6 = 7
2
37. h(x) =
2
โซบ2
โ3x โ 18, for x < โ5,
1,
for โ5 โค x < 1,
x + 2,
for x โฅ 1
y
4
Since โ5 is in the interval [โ5, 1), h(โ5) = 1.
Since 0 is in the interval [โ5, 1), h(0) = 1.
โซบ4 โซบ2
Since 1 โฅ 1, h(1) = 1 + 2 = 3.
2
4
โซบ2
Since 4 โฅ 1, h(4) = 4 + 2 = 6.
x
โซบ4
๏ฃฑ
โ5x โ 8, for x 4
42. h(x) =
2x โ 1, for x < 2
2 โ x, for x โฅ 2
y
Since โ4 4, f (6) = 10 โ 2 ยท 6 = โ2.
4
2
โซบ4 โซบ2
2
โซบ2
โซบ4
c 2016 Pearson Education, Inc.
Copyright
4
x
Exercise Set 2.1
61
๏ฃฑ
2
๏ฃด
๏ฃฒ x โ 9 , for x = โ3,
x+3
46. f (x) =
๏ฃด
๏ฃณ
5,
for x = โ3
๏ฃฑ
๏ฃด
x + 1, for x โค โ3,
๏ฃด
๏ฃด
๏ฃด
๏ฃด
๏ฃด
๏ฃฒ
for โ3 < x < 4
43. f (x) = โ1,
๏ฃด
๏ฃด
๏ฃด
๏ฃด
1
๏ฃด
๏ฃด x,
for x โฅ 4
๏ฃณ
2
y
We create the graph in three parts. Graph f (x) = x + 1
for inputs x less than or equal to โ3. Graph f (x) = โ1
for inputs greater than โ3 and less than 4. Then graph
1
f (x) = x for inputs greater than or equal to 4.
2
4
2
โซบ4 โซบ2
y
โซบ6
4
๏ฃฑ
2,
๏ฃด
๏ฃฒ
โซบ4 โซบ2
2
4
โซบ2
47. f (x) =
x
for x = 5,
x โ 25
๏ฃด
๏ฃณ
, for x = 5
xโ5
When x = 5, the denominator of (x2 โ 25)/(x โ 5) is
nonzero so we can simplify:
x
โซบ4
2
(x + 5)(x โ 5)
x2 โ 25
=
= x + 5.
xโ5
xโ5
Thus, f (x) = x + 5, for x = 5.
4,
for x โค โ2,
x + 1, for โ2 < x < 3
โx,
for x โฅ 3
The graph of this part of the function consists of a line
with a โholeโ at the point (5, 10), indicated by an open
dot. At x = 5, we have f (5) = 2, so the point (5, 2) is
plotted below the open dot.
y
4
2
โซบ4 โซบ2
4
โซบ4
2
44. f (x) =
2
โซบ2
2
4
โซบ2
y
x
8
โซบ4
2
โซบ8 โซบ4
๏ฃฑ1
๏ฃด
๏ฃด x โ 1, for x 1
4
8
x
โซบ4
โซบ8
1
xโ1
2
for inputs less than 0. Graph g(x) = 3 for inputs greater
than or equal to 0 and less than or equal to 1. Then graph
g(x) = โ2x for inputs greater than 1.
We create the graph in three parts. Graph g(x) =
๏ฃฑ
2
๏ฃด
๏ฃฒ x + 3x + 2 , for x = โ1,
x+1
48. f (x) =
๏ฃด
๏ฃณ
7,
for x = โ1
y
8
y
6
4
4
2
2
โซบ4 โซบ2
2
4
x
โซบ4 โซบ2
โซบ2
โซบ4
c 2016 Pearson Education, Inc.
Copyright
2
4
x
62
Chapter 2: More on Functions
49. f (x) = [[x]]
1
[[x]] โ 2
2
This function can be de๏ฌned by a piecewise function with
an in๏ฌnite number of statements:
๏ฃฑ
๏ฃด
๏ฃด.
๏ฃด
๏ฃด
๏ฃด.
๏ฃด
๏ฃด
๏ฃด
๏ฃด
.
๏ฃด
๏ฃด
๏ฃด
๏ฃด
๏ฃด โ2 12 , for โ1 โค x < 0,
๏ฃด
๏ฃด
๏ฃฒ โ2,
for 0 โค x < 1,
f (x) = โ1 1 , for 1 โค x < 2,
๏ฃด
2
๏ฃด
๏ฃด
for 2 โค x < 3,
๏ฃด โ1,
๏ฃด
๏ฃด
๏ฃด
.
๏ฃด
๏ฃด
๏ฃด
๏ฃด
.
๏ฃด
๏ฃด
๏ฃด
๏ฃด
๏ฃด.
๏ฃณ
52. f (x) =
See Example 9.
50. f (x) = 2[[x]]
This function can be de๏ฌned by a piecewise function with
an in๏ฌnite
๏ฃฑ number of statements:
๏ฃด
๏ฃด
.
๏ฃด
๏ฃด
๏ฃด
๏ฃด
.
๏ฃด
๏ฃด
๏ฃด
๏ฃด
.
๏ฃด
๏ฃด
๏ฃด
๏ฃด
โ4, for โ2 โค x < โ1,
๏ฃด
๏ฃด
๏ฃฒ โ2, for โ1 โค x < 0,
f (x) = 0,
for 0 โค x < 1,
๏ฃด
๏ฃด
๏ฃด
2,
for 1 โค x < 2,
๏ฃด
๏ฃด
๏ฃด
๏ฃด
.
๏ฃด
๏ฃด
๏ฃด
๏ฃด
๏ฃด.
๏ฃด
๏ฃด
๏ฃด
๏ฃด
๏ฃณ.
6
2
2
4
2
โซบ2
x
โซบ4
53. From the graph we see that the domain is (โโ, โ) and
the range is (โโ, 0) โช [3, โ).
57. From the graph we see that the domain is (โโ, โ) and
the range is {y|y โค โ2 or y = โ1 or y โฅ 2}.
58. Domain: (โโ, โ); range: (โโ, โ3] โช (โ1, 4]
51. f (x) = 1 + [[x]]
This function can be de๏ฌned by a piecewise function with
an in๏ฌnite
๏ฃฑ number of statements:
๏ฃด
.
๏ฃด
๏ฃด
๏ฃด
๏ฃด
.
๏ฃด
๏ฃด
๏ฃด
๏ฃด
.
๏ฃด
๏ฃด
๏ฃด
๏ฃด
๏ฃด โ1, for โ2 โค x < โ1,
๏ฃด
๏ฃด
๏ฃฒ 0,
for โ1 โค x < 0,
f (x) = 1,
for
0 โค x < 1,
๏ฃด
๏ฃด
๏ฃด
for 1 โค x < 2,
๏ฃด 2,
๏ฃด
๏ฃด
๏ฃด
๏ฃด
.
๏ฃด
๏ฃด
๏ฃด
๏ฃด
.
๏ฃด
๏ฃด
๏ฃด
๏ฃด
๏ฃณ.
y
59. From the graph we see that the domain is (โโ, โ) and
the range is {โ5, โ2, 4}. An equation for the function is:
f (x) =
โ2, for x 2
60. Domain: (โโ, โ); range: {y|y = โ3 or y โฅ 0}
โ3, for x < 0,
g(x) =
x, for x โฅ 0
61. From the graph we see that the domain is (โโ, โ) and
the range is (โโ, โ1] โช [2, โ). Finding the slope of each
segment and using the slope-intercept or point-slope formula, we ๏ฌnd that an equation for the function is:
x, for x โค โ1,
2, for โ1 2
This can also be expressed as follows:
g(x) =
4
2
2
โซบ4
โซบ4 โซบ2
56. Domain: (โ, โ); range: (โโ, 3)
x
f(x) โซฝ 2ๅxๅ
โซบ2
h(x) โซฝ q ๅxๅ โซบ 2
2
55. From the graph we see that the domain is (โโ, โ) and
the range is [โ1, โ).
4
โซบ4
4
54. Domain: (โโ, โ); range: (โ5, โ)
y
โซบ4
y
4
x
g(x) โซฝ 1 โซน ๅxๅ
g(x) =
x, for x โค โ1,
2, for โ1 < x < 2,
x, for x โฅ 2
c 2016 Pearson Education, Inc.
Copyright
Exercise Set 2.1
63
62. Domain: (โโ, โ); range: {y|y = โ2 or y โฅ 0}. An
equation for the function is:
|x|, for x < 3,
h(x) =
โ2, for x โฅ 3
This can also be expressed as follows:
โx, for x โค 0,
x, for 0 < x < 3,
โ2, for x โฅ 3
It can also be expressed as follows:
h(x) =
h(x) =
68.
โx, for x < 0,
x, for 0 โค x < 3,
โ2, for x โฅ 3
63. From the graph we see that the domain is [โ5, 3] and the
range is (โ3, 5). Finding the slope of each segment and
using the slope-intercept or point-slope formula, we ๏ฌnd
that an equation for the function is:
h(x) =
x + 8, for โ5 โค x < โ3,
3,
for โ3 โค x โค 1,
3x โ 6, for 1 < x โค 3
64. Domain: [โ4, โ); range: [โ2, 4]
โ2x โ 4, for โ4 โค x โค โ1,
x โ 1,
for โ1 < x < 2,
2,
for x โฅ 2
This can also be expressed as:
f (x) =
f (x) =
y โ y1 = m(x โ x1 )
1
y โ 1 = โ [x โ (โ1)]
8
1
y โ 1 = โ (x + 1)
8
1
1
yโ1 = โ xโ
8
8
1
7
y = โ x+
8
8
2x โ 9y + 1 = 0
2x + 1 = 9y
1
2
x+ = y
9
9
1
2
Slope: ; y-intercept: 0,
9
9
69. a) The function C(t) can be de๏ฌned piecewise.
๏ฃฑ
3, for 0 < t < 1,
๏ฃด
๏ฃด
๏ฃด
๏ฃด
6, for 1 โค t < 2,
๏ฃด
๏ฃด
๏ฃด
๏ฃฒ 9, for 2 โค t < 3,
C(t) = .
๏ฃด
๏ฃด
๏ฃด
.
๏ฃด
๏ฃด
๏ฃด
๏ฃด
๏ฃณ.
We graph this function.
โ2x โ 4, for โ4 โค x < โ1,
x โ 1,
for โ1 โค x 0.
a) f (2) = 4 ยท 2 โ 5 ยท 2 = 4 ยท 8 โ 5 ยท 2 = 32 โ 10 = 22
3
b) f (โ2) = 4(โ2) โ 5(โ2) = 4(โ8) โ 5(โ2) = โ32 +
10 = โ22
3
c) f (a) = 4a3 โ 5a
d) f (โa) = 4(โa)3 โ 5(โa) = 4(โa3 ) โ 5(โa) =
โ4a3 + 5a
67. First ๏ฌnd the slope of the given line.
70. If [[x + 2]] = โ3, then โ3 โค x + 2 < โ2, or
โ5 โค x < โ4. The possible inputs for x are
{x| โ 5 โค x < โ4}.
71. If [[x]]2 = 25, then [[x]] = โ5 or [[x]] = 5. For
โ5 โค x < โ4, [[x]] = โ5. For 5 โค x < 6, [[x]] = 5.
Thus, the possible inputs for x are
{x| โ 5 โค x < โ4 or 5 โค x 0 for all values of x in its domain, the domain of F/G
is [2, 9].
44. (F + G)(x) = F (x) + G(x)
y
10
37. The domain of G/F is the set of numbers in the domains of
both F and G (See Exercise 35.), excluding those for which
F = 0. Since F (3) = 0, the domain of G/F is [2, 3) โช (3, 9].
8
38.
2
FG
6
4
y
2
4
6
8
10
6
8
10
x
8
6
45.
4
F G
2
2
39.
4
6
y
6
8
10
4
x
2
2
y
โซบ2
GโซบF
2
4
x
6
46.
G F
4
2
y
4
2
4
6
8
10
x
FG
2
2
2
4
4
6
8
2
10
x
4
40.
y
6
47. a) P (x) = R(x) โ C(x) = 60x โ 0.4×2 โ (3x + 13) =
F G
4
60x โ 0.4×2 โ 3x โ 13 = โ0.4×2 + 57x โ 13
2
b) R(100) = 60ยท100โ0.4(100)2 = 6000โ0.4(10, 000) =
2
4
6
8
10
x
6000 โ 4000 = 2000
2
C(100) = 3 ยท 100 + 13 = 300 + 13 = 313
4
P (100) = R(100) โ C(100) = 2000 โ 313 = 1687
41. From the graph, we see that the domain of F is [0, 9] and
the domain of G is [3, 10]. The domain of F + G is the set
of numbers in the domains of both F and G. This is [3, 9].
42. The domain of F โ G and F G is the set of numbers in the
domains of both F and G. (See Exercise 41.) This is [3, 9].
The domain of F/G is the set of numbers in the domains
of both F and G, excluding those for which G = 0. Since
G > 0 for all values of x in its domain, the domain of F/G
is [3, 9].
43. The domain of G/F is the set of numbers in the domains
of both F and G (See Exercise 41.), excluding those for
which F = 0. Since F (6) = 0 and F (8) = 0, the domain
of G/F is [3, 6) โช (6, 8) โช (8, 9].
48. a) P (x) = 200x โ x2 โ (5000 + 8x) =
200x โ x2 โ 5000 โ 8x = โx2 + 192x โ 5000
b) R(175) = 200(175) โ 1752 = 4375
C(175) = 5000 + 8 ยท 175 = 6400
P (175) = R(175) โ C(175) = 4375 โ 6400 = โ2025
(We could also use the function found in part (a) to
๏ฌnd P (175).)
49. f (x) = 3x โ 5
f (x + h) = 3(x + h) โ 5 = 3x + 3h โ 5
3x + 3h โ 5 โ (3x โ 5)
f (x + h) โ f (x)
=
h
h
3x + 3h โ 5 โ 3x + 5
=
h
3h
=
=3
h
c 2016 Pearson Education, Inc.
Copyright
70
50.
Chapter 2: More on Functions
f (x) = 4x โ 1
f (x + h) โ f (x)
4(x + h) โ 1 โ (4x โ 1)
=
=
h
h
4h
4x + 4h โ 1 โ 4x + 1
=
=4
h
h
51. f (x) = 6x + 2
f (x + h) = 6(x + h) + 2 = 6x + 6h + 2
6x + 6h + 2 โ (6x + 2)
f (x + h) โ f (x)
=
h
h
6x + 6h + 2 โ 6x โ 2
=
h
6h
=6
=
h
55. f (x) =
1
3x
f (x + h) =
1
3(x + h)
1
1
โ
f (x + h) โ f (x)
3(x + h) 3x
=
h
h
x
1 x+h
1
ยท โ
ยท
3(x + h) x 3x x + h
=
h
x+h
x
โ
3x(x + h) 3x(x + h)
=
h
xโxโh
x โ (x + h)
3x(x + h)
3x(x + h)
=
=
h
h
โh
โh
1
3x(x + h)
=
ยท
=
h
3x(x + h) h
52. f (x) = 5x + 3
5(x + h) + 3 โ (5x + 3)
f (x + h) โ f (x)
=
=
h
h
5h
5x + 5h + 3 โ 5x โ 3
=
=5
h
h
1
x+1
3
1
1
1
f (x + h) = (x + h) + 1 = x + h + 1
3
3
3
1
1
1
x+ h+1โ
x+1
f (x + h) โ f (x)
3
3
3
=
h
h
1
1
1
x+ h+1โ xโ1
3
3
3
=
h
1
h
1
= 3 =
h
3
โ1 ยท h/
โh
=
3x(x + h) ยท h
3x(x + h) ยท h/
1
โ1
, or โ
=
3x(x + h)
3x(x + h)
=
53. f (x) =
1
54. f (x) = โ x + 7
2
1
1
โ (x + h) + 7 โ โ x + 7
f (x + h) โ f (x)
2
= 2
=
h
h
1
1
1
1
โ h
โ xโ h+7+ โ7
1
2
2
2
2
=
=โ
h
h
2
56. f (x) =
1
2x
1
1
1
x 1 x+h
โ
ยท โ ยท
f (x+h)โf (x) 2(x+h) 2x 2(x+h) x 2x x+h
=
=
=
h
h
h
x+h
xโxโh
โh
x
โ
2x(x + h) 2x(x + h)
2x(x + h)
2x(x + h)
=
=
=
h
h
h
1
โ1
1
โh
ยท =
, or โ
2x(x + h) h
2x(x + h)
2x(x + h)
57. f (x) = โ
1
4x
f (x + h) = โ
1
4(x + h)
f (x + h) โ f (x)
=
h
โ
1
โ
4(x + h)
h
โ
1
4x
x
1
x+h
1
ยท โ โ
ยท
โ
4(x + h) x
4x
x+h
=
h
x+h
x
+
โ
4x(x + h) 4x(x + h)
=
h
h
โx + x + h
4x(x + h)
4x(x + h)
=
=
h
h
=
c 2016 Pearson Education, Inc.
Copyright
1
h/ ยท1
1
h
ยท =
=
4x(x+h) h 4x(x+h)ยทh/ 4x(x+h)
Exercise Set 2.2
58. f (x) = โ
71
1
x
63. f (x) = 3×2 โ 2x + 1
1
1
โ
โ โ
f (x + h) โ f (x)
x+h
x
=
=
h
h
1 x+h
1 x
x
x+h
ยท โ โ
ยท
โ
โ
+
x+h x
x x+h
x(x+h) x(x+h)
=
=
h
h
h
โx + x + h
h
1
1
x(x + h)
x(x + h)
=
=
ยท =
h
h
x(x + h) h
x(x + h)
f (x + h) = 3(x + h)2 โ 2(x + h) + 1 =
3(x2 + 2xh + h2 ) โ 2(x + h) + 1 =
3×2 + 6xh + 3h2 โ 2x โ 2h + 1
f (x) = 3×2 โ 2x + 1
f (x + h) โ f (x)
=
h
(3×2 + 6xh + 3h2 โ2xโ2h + 1)โ(3×2 โ 2x + 1)
=
h
3×2 + 6xh + 3h2 โ 2x โ 2h + 1 โ 3×2 + 2x โ 1
=
h
h(6x + 3h โ 2)
6xh + 3h2 โ 2h
=
=
h
hยท1
h 6x + 3h โ 2
ยท
= 6x + 3h โ 2
h
1
59. f (x) = x2 + 1
f (x + h) = (x + h)2 + 1 = x2 + 2xh + h2 + 1
x2 + 2xh + h2 + 1 โ (x2 + 1)
f (x + h) โ f (x)
=
h
h
=
=
x2 + 2xh + h2 + 1 โ x2 โ 1
h
2xh + h
h
64. f (x) = 5×2 + 4x
f (x+h)โf (x) (5×2+10xh+5h2+4x+4h)โ(5×2+4x)
=
=
h
h
10xh + 5h2 + 4h
= 10x + 5h + 4
h
2
h(2x + h)
h
h 2x + h
= ยท
h
1
= 2x + h
=
65. f (x) = 4 + 5|x|
f (x + h) = 4 + 5|x + h|
4 + 5|x + h| โ (4 + 5|x|)
f (x + h) โ f (x)
=
h
h
60. f (x) = x2 โ 3
(x + h)2 โ 3 โ (x2 โ 3)
f (x + h) โ f (x)
=
=
h
h
x + 2xh + h โ 3 โ x + 3
2xh + h
h(2x + h)
=
=
=
h
h
h
2x + h
2
2
2
2
61. f (x) = 4 โ x2
f (x + h) = 4 โ (x + h)2 = 4 โ (x2 + 2xh + h2 ) =
4 โ x2 โ 2xh โ h2
f (x + h) โ f (x)
4 โ x2 โ 2xh โ h2 โ (4 โ x2 )
=
h
h
4 โ x2 โ 2xh โ h2 โ 4 + x2
=
h
h/(โ2x โ h)
โ2xh โ h2
=
=
h
h/
= โ2x โ h
โ2xh โ h2
2 โ x2 โ 2xh โ h2 โ 2 + x2
=
=
h
h
h(โ2x โ h)
= โ2x โ h
h
4 + 5|x + h| โ 4 โ 5|x|
h
=
5|x + h| โ 5|x|
h
66. f (x) = 2|x| + 3x
f (x+h)โf (x) (2|x+h|+3x+3h)โ(2|x|+3x)
=
=
h
h
2|x + h| โ 2|x| + 3h
h
67. f (x) = x3
f (x + h) = (x + h)3 = x3 + 3×2 h + 3xh2 + h3
f (x) = x3
x3 + 3×2 h + 3xh2 + h3 โ x3
f (x + h) โ f (x)
=
=
h
h
2
2
3
2
2
3x h + 3xh + h
h(3x + 3xh + h )
=
=
h
hยท1
2
2
h 3x + 3xh + h
ยท
= 3×2 + 3xh + h2
h
1
62. f (x) = 2 โ x2
2 โ (x + h)2 โ (2 โ x2 )
f (x + h) โ f (x)
=
=
h
h
=
68.
f (x) = x3 โ 2x
f (x+h)โf (x) (x+h)3 โ2(x+h)โ(x3 โ2x)
=
=
h
h
x3 + 3×2 h + 3xh2 + h3 โ2x โ 2hโx3 +2x
=
h
3×2 h+3xh2 + h3 โ2h h(3×2 +3xh + h2 โ2)
=
=
h
h
3×2 + 3xh + h2 โ 2
c 2016 Pearson Education, Inc.
Copyright
72
Chapter 2: More on Functions
69. f (x) =
xโ4
x+3
72.
y
4
x+hโ4 xโ4
โ
f (x + h) โ f (x)
x
= +h+3 x+3 =
h
h
2x y 4
2
4 2
2
4
x
2
x+hโ4 xโ4
โ
x + h + 3 x + 3 ยท (x + h + 3)(x + 3) =
h
(x + h + 3)(x + 3)
4
(x + h โ 4)(x + 3) โ (x โ 4)(x + h + 3)
=
h(x + h + 3)(x + 3)
73. Graph x โ 3y = 3.
First we ๏ฌnd the x- and y-intercepts.
x2+hxโ4x+3x+3hโ12โ(x2+hx+3xโ4xโ4hโ12)
=
h(x+h+3)(x+3)
xโ3ยท0 = 3
x=3
x2 + hx โ x + 3h โ 12 โ x2 โ hx + x + 4h + 12
=
h(x + h + 3)(x + 3)
h
7
7h
= ยท
=
h(x + h + 3)(x + 3)
h (x + h + 3)(x + 3)
7
(x + h + 3)(x + 3)
The x-intercept is (3, 0).
0 โ 3y = 3
โ3y = 3
y = โ1
The y-intercept is (0, โ1).
We ๏ฌnd a third point as a check. We let x = โ3 and solve
for y.
x
70. f (x) =
2โx
x+h
x
โ
f (x + h) โ f (x)
2 โ (x + h) 2 โ x
=
=
h
h
โ3 โ 3y = 3
โ3y = 6
y = โ2
(x + h)(2 โ x) โ x(2 โ x โ h)
(2 โ x โ h)(2 โ x)
=
h
Another point on the graph is (โ3, โ2). We plot the points
and draw the graph.
y
2x โ x2 + 2h โ hx โ 2x + x2 + hx
(2 โ x โ h)(2 โ x)
=
h
4
2
2h
(2 โ x โ h)(2 โ x)
=
h
4 2
x 3y 3
2
4
2
4
x
2
4
1
2
2h
ยท =
(2 โ x โ h)(2 โ x) h
(2 โ x โ h)(2 โ x)
74.
y
71. Graph y = 3x โ 1.
4
We ๏ฌnd some ordered pairs that are solutions of the equation, plot these points, and draw the graph.
When x = โ1, y = 3(โ1) โ 1 = โ3 โ 1 = โ4.
When x = 0, y = 3 ยท 0 โ 1 = 0 โ 1 = โ1.
2
4 2
2
4
x
y x2 1
When x = 2, y = 3 ยท 2 โ 1 = 6 โ 1 = 5.
x
y
y
โ1 โ4
4
0
โ1
2
2
5
4 2
75. Answers may vary; f (x) =
76. The domain of h + f , h โ f , and hf consists of all numbers
that are in the domain of both h and f , or {โ4, 0, 3}.
2
4
2
4
1
1
, g(x) =
x+7
xโ3
y 3x 1
x
The domain of h/f consists of all numbers that are in the
domain of both h and f , excluding any for which the value
of f is 0, or {โ4, 0}.
7
, and the domain of g(x)
77. The domain of h(x) is xx =
3
7
is {x|x = 3}, so and 3 are not in the domain of (h/g)(x).
3
We must also exclude the value of x for which g(x) = 0.
c 2016 Pearson Education, Inc.
Copyright
Exercise Set 2.3
x4 โ 1
=0
5x โ 15
x4 โ 1 = 0
73
17.
Multiplying by 5x โ 15
x4 = 1
x = ยฑ1
Then the domain of (h/g)(x) is
7
xx = and x = 3 and x = โ1 and x = 1 , or
3
7 7
โช
, 3 โช (3, โ).
(โโ, โ1) โช (โ1, 1) โช 1,
3
3
Exercise Set 2.3
1.
(g โฆ f )(x) = g(f (x)) = g(x + 3) = x + 3 โ 3 = x
The domain of f and of g is (โโ, โ), so the domain of
f โฆ g and of g โฆ f is (โโ, โ).
5
4 5
18. (f โฆ g)(x) = f
x = ยท x=x
4
5 4
5 4
4
x = ยท x=x
(g โฆ f )(x) = g
5
4 5
The domain of f and of g is (โโ, โ), so the domain of
f โฆ g and of g โฆ f is (โโ, โ).
19.
(f โฆ g)(โ1) = f (g(โ1)) = f ((โ1)2 โ 2(โ1) โ 6) =
f (1 + 2 โ 6) = f (โ3) = 3(โ3) + 1 = โ9 + 1 = โ8
2.
(g โฆ f )(โ2) = g(f (โ2)) = g(3(โ2) + 1) = g(โ5) =
(h โฆ f )(1) = h(f (1)) = h(3 ยท 1 + 1) = h(3 + 1) =
20.
h(4) = 43 = 64
4.
1
1
1
1 3
=g h
=g
=
=g
(g โฆ h)
2
2
2
8
1 2 1
1 1
399
โ6 =
โ โ6=โ
โ2
8
8
64 4
64
5.
(g โฆ f )(5) = g(f (5)) = g(3 ยท 5 + 1) = g(15 + 1) =
6.
g(16) = 162 โ 2 ยท 16 โ 6 = 218
1
1
1
1 2
=f g
=f
โ6 =
โ2
(f โฆ g)
3
3
3
3
1 2
59
59
56
f
โ โ 6 =f โ
=3 โ
+ 1=โ
9 3
9
9
3
7.
(f โฆ h)(โ3) = f (h(โ3)) = f ((โ3)3 ) = f (โ27) =
3(โ27) + 1 = โ81 + 1 = โ80
8.
(h โฆ g)(3) = h(g(3)) = h(32 โ 2 ยท 3 โ 6) =
h(9 โ 6 โ 6) = h(โ3) = (โ3)3 = โ27
9.
(g โฆ g)(โ2) = g(g(โ2)) = g((โ2)2 โ 2(โ2) โ 6) =
g(4 + 4 โ 6) = g(2) = 22 โ 2 ยท 2 โ 6 = 4 โ 4 โ 6 = โ6
10.
(g โฆ g)(3) = g(g(3)) = g(32 โ 2 ยท 3 โ 6) = g(9 โ 6 โ 6) =
g(โ3) = (โ3)2 โ 2(โ3) โ 6 = 9 + 6 โ 6 = 9
11. (h โฆ h)(2) = h(h(2)) = h(23 ) = h(8) = 83 = 512
12. (h โฆ h)(โ1) = h(h(โ1)) = h((โ1)3 ) = h(โ1) = (โ1)3 = โ1
13.
(f โฆ f )(โ4) = f (f (โ4)) = f (3(โ4) + 1) = f (โ12 + 1) =
f (โ11) = 3(โ11) + 1 = โ33 + 1 = โ32
14.
(f โฆ f )(1) = f (f (1)) = f (3 ยท 1 + 1) = f (3 + 1) = f (4) =
3 ยท 4 + 1 = 12 + 1 = 13
15. (h โฆ h)(x) = h(h(x)) = h(x3 ) = (x3 )3 = x9
16.
(f โฆ f )(x) = f (f (x)) = f (3x + 1) = 3(3x + 1) + 1 =
9x + 3 + 1 = 9x + 4
(f โฆ g)(x) = f (g(x)) = f (3×2 โ2xโ1) = 3×2 โ2xโ1+1 =
3×2 โ 2x
(g โฆ f )(x) = g(f (x)) = g(x+1) = 3(x+1)2 โ2(x+1)โ1 =
3(x2 +2x+1)โ2(x+1)โ1 = 3×2 +6x+3โ2xโ2โ1 =
3×2 + 4x
The domain of f and of g is (โโ, โ), so the domain of
f โฆ g and of g โฆ f is (โโ, โ).
(โ5)2 โ 2(โ5) โ 6 = 25 + 10 โ 6 = 29
3.
(f โฆ g)(x) = f (g(x)) = f (x โ 3) = x โ 3 + 3 = x
(f โฆ g)(x) = f (x2 + 5) = 3(x2 + 5) โ 2 = 3×2 + 15 โ 2 =
3×2 + 13
(g โฆ f )(x) = g(3xโ2) = (3xโ2)2 +5 = 9×2 โ12x+4+5 =
9×2 โ 12x + 9
The domain of f and of g is (โโ, โ), so the domain of
f โฆ g and of g โฆ f is (โโ, โ).
21.
(f โฆ g)(x) = f (g(x)) = f (4xโ3) = (4xโ3)2 โ3 =
16×2 โ 24x + 9 โ 3 = 16×2 โ 24x + 6
(g โฆ f )(x) = g(f (x)) = g(x2 โ3) = 4(x2 โ3)โ3 =
4×2 โ 12 โ 3 = 4×2 โ 15
The domain of f and of g is (โโ, โ), so the domain of
f โฆ g and of g โฆ f is (โโ, โ).
22.
(f โฆ g)(x) = f (2x โ 7) = 4(2x โ 7)2 โ (2x โ 7) + 10 =
4(4×2 โ 28x + 49) โ (2x โ 7) + 10 =
16×2 โ 112x + 196 โ 2x + 7 + 10 = 16×2 โ 114x + 213
(g โฆ f )(x) = g(4×2 โ x + 10) = 2(4×2 โ x + 10) โ 7 =
8×2 โ 2x + 20 โ 7 = 8×2 โ 2x + 13
The domain of f and of g is (โโ, โ), so the domain of
f โฆ g and of g โฆ f is (โโ, โ).
1
4
4
=
=
=
23. (f โฆ g)(x) = f (g(x)) = f
1
5
x
1โ5ยท
1โ
x
x
x
4
4x
=4ยท
=
xโ5
xโ5
xโ5
x
4
1
=
=
(g โฆ f )(x) = g(f (x)) = g
4
1 โ 5x
1 โ 5x
1 โ 5x
1 โ 5x
=
1ยท
4
4
1
The domain of f is xx =
and the domain of g is
5
{x|x = 0}. Consider the domain of f โฆ g. Since 0 is not in
1
the domain of g, 0 is not in the domain of f โฆ g. Since
5
1
is not in the domain of f , we know that g(x) cannot be .
5
1
We ๏ฌnd the value(s) of x for which g(x) = .
5
c 2016 Pearson Education, Inc.
Copyright
74
Chapter 2: More on Functions
1
1
=
x
5
5=x
Multiplying by 5x
Thus 5 is also not in the domain of f โฆ g. Then the domain
of f โฆg is {x|x = 0 and x = 5}, or (โโ, 0)โช(0, 5)โช(5, โ).
1
Now consider the domain of g โฆ f . Recall that is not in
5
the domain of f , so it is not in the domain of g โฆ f . Now 0
is not in the domain of g but f (x) is never 0, so the domain
1 1
1
, or โ โ,
โช
,โ .
of g โฆ f is xx =
5
5
5
24.
1
=
2x + 1
2x + 1
6
=6ยท
=
1
1
2x + 1
6(2x + 1), or 12x + 6
6
1
1
1
=
=
=
=
(g โฆ f )(x) = g
6
12
12 + x
x
+1
2ยท +1
x
x
x
x
x
=
1ยท
12 + x
12 + x
The domain of f is {x|x = 0} and the domain of g
1
is xx = โ . Consider the domain of f โฆ g. Since
2
1
1
โ is not in the domain of g, โ is not in the domain
2
2
of f โฆ g. Now 0 is not in the domain of f but g(x)
1
is never 0, so the domain of f โฆ g is xx = โ , or
2
1 1
โช โ ,โ .
โ โ, โ
2
2
Now consider the domain of g โฆ f . Since 0 is not in the
domain of f , then 0 is not in the domain of g โฆ f . Also,
1
since โ is not in the domain of g, we ๏ฌnd the value(s) of
2
1
x for which f (x) = โ .
2
1
6
=โ
x
2
โ12 = x
Then the domain of g โฆ f is xx = โ12 and x = 0 , or
(f โฆ g)(x) = f
(โโ, โ12) โช (โ12, 0) โช (0, โ).
x+7
25. (f โฆ g)(x) = f (g(x)) = f
=
3
x+7
โ7=x+7โ7=x
3
3
(3x โ 7) + 7
=
(g โฆ f )(x) = g(f (x)) = g(3x โ 7) =
3
3x
=x
3
The domain of f and of g is (โโ, โ), so the domain of
f โฆ g and of g โฆ f is (โโ, โ).
26.
(f โฆ g)(x) = f (1.5x + 1.2) =
4
2
(1.5x + 1.2) โ
3
5
4
x + 0.8 โ = x
5
4
2
4
2
xโ
= 1.5 x โ
+ 1.2 =
(g โฆ f )(x) = g
3
5
3
5
x โ 1.2 + 1.2 = x
The domain of f and of g is (โโ, โ), so the domain of
f โฆ g and of g โฆ f is (โโ, โ).
โ
โ
27. (f โฆ g)(x) = f (g(x)) = f ( x) = 2 x + 1
โ
(g โฆ f )(x) = g(f (x)) = g(2x + 1) = 2x + 1
The domain of f is (โโ, โ) and the domain of g is
{x|x โฅ 0}. Thus the domain of f โฆ g is {x|x โฅ 0}, or
[0, โ).
Now consider the domain of g โฆf . There are no restrictions
on the domain of f , but the domain of g is {x|x โฅ 0}. Since
1
1
f (x) โฅ 0 for x โฅ โ , the domain of g โฆ f is xx โฅ โ ,
2
2
1
or โ , โ .
2
โ
28. (f โฆ g)(x) = f (2 โ 3x) = 2 โ 3x
โ
โ
(g โฆ f )(x) = g( x) = 2 โ 3 x
The domain of f is {x|x โฅ 0} and the domain of g is
2
(โโ, โ). Since g(x) โฅ 0 when x โค , the domain of f โฆ g
3
2
is โ โ, .
3
Now consider the domain of g โฆ f . Since the domain of f
is {x|x โฅ 0} and the domain of g is (โโ, โ), the domain
of g โฆ f is {x|x โฅ 0}, or [0, โ).
29.
(f โฆ g)(x) = f (g(x)) = f (0.05) = 20
(g โฆ f )(x) = g(f (x)) = g(20) = 0.05
The domain of f and of g is (โโ, โ), so the domain of
f โฆ g and of g โฆ f is (โโ, โ).
โ
30. (f โฆ g)(x) = ( 4 x)4 = x
โ
4
(g โฆ f )(x) = x4 = |x|
The domain of f is (โโ, โ) and the domain of g is
{x|x โฅ 0}, so the domain of f โฆ g is {x|x โฅ 0}, or [0, โ).
Now consider the domain of g โฆf . There are no restrictions
on the domain of f and f (x) โฅ 0 for all values of x, so the
domain is (โโ, โ).
31.
(f โฆ g)(x) = f (g(x)) = f (x2 โ 5) =
โ
โ
x2 โ 5 + 5 = x2 = |x|
โ
(g โฆ f )(x) = g(f (x)) = g( x + 5) =
โ
( x + 5)2 โ 5 = x + 5 โ 5 = x
The domain of f is {x|x โฅ โ5} and the domain of g is
(โโ, โ). Since x2 โฅ 0 for all values of x, then x2 โ5 โฅ โ5
for all values of x and the domain of g โฆ f is (โโ, โ).
Now consider the domain of f โฆg. There are no restrictions
on the domain of g, so the domain of f โฆ g is the same as
the domain of f , {x|x โฅ โ5}, or [โ5, โ).
โ
32. (f โฆ g)(x) = ( 5 x + 2)5 โ 2 = x + 2 โ 2 = x
โ
โ
5
(g โฆ f )(x) = 5 x5 โ 2 + 2 = x5 = x
The domain of f and of g is (โโ, โ), so the domain of
f โฆ g and of g โฆ f is (โโ, โ).
c 2016 Pearson Education, Inc.
Copyright
Exercise Set 2.3
33.
75
โ
โ
(f โฆ g)(x) = f (g(x)) = f ( 3 โ x) = ( 3 โ x)2 + 2 =
1โx
= โ1
x
1 โ x = โx Multiplying by x
3โx+2=5โx
(g โฆ f )(x) = g(f (x)) = g(x2 + 2) =
โ
โ
3 โ x2 โ 2 = 1 โ x2
3 โ (x2 + 2) =
The domain of f is (โโ, โ) and the domain of g is
{x|x โค 3}, so the domain of f โฆ g is {x|x โค 3}, or (โโ, 3].
Now consider the domain of g โฆf . There are no restrictions
on the domain of f and the domain of g is {x|x โค 3}, so
we ๏ฌnd the values of x for which f (x) โค 3. We see that
x2 + 2 โค 3 for โ1 โค x โค 1, so the domain of g โฆ f is
{x| โ 1 โค x โค 1}, or [โ1, 1].
โ
โ
34. (f โฆ g)(x) = f ( x2 โ 25) = 1 โ ( x2 โ 25)2 =
1 โ (x2 โ 25) = 1 โ x2 + 25 = 26 โ x2
1=0
(g โฆ f )(x) = g(1 โ x2 ) = (1 โ x2 )2 โ 25 =
โ
โ
1 โ 2×2 + x4 โ 25 = x4 โ 2×2 โ 24
1
+2
x
โ
2
=
1
xโ2
2x โ 3
1 + 2x โ 4
x
โ
2
= xโ2
=
1
1
xโ2
xโ2
2x โ 3 x โ 2
ยท
= 2x โ 3
=
xโ2
1
The domain of f is {x|x = 2} and the domain of g is
{x|x = 0}, so 0 is not in the domain of f โฆ g. We ๏ฌnd the
value of x for which g(x) = 2.
x+2
=2
x
x + 2 = 2x
1
(g โฆ f )(x) = g
xโ2
The domain of f is (โโ, โ) and the domain of g is
{x|x โค โ5 or x โฅ 5}, so the domain of f โฆ g is
{x|x โค โ5 or x โฅ 5}, or (โโ, โ5] โช [5, โ).
Now consider the domain of g โฆ f . There are no restrictions on the domain of f and the domain of g is
{x|x โค โ5 or x โฅ 5}, so we ๏ฌnd the values of x for
2
which f (x) โค
โ โ5 or f (x)
โ โฅ 5. We 2see that 1 โ x โค โ5
when x โค โ 6 or x โฅ 6 and 1 โ xโโฅ 5 has no solution,
โ
so the domain
of g โฆ f is {x|x โค โ 6 or x โฅ 6}, or
โ
โ
(โโ, โ 6] โช [ 6, โ).
1
35. (f โฆ g)(x) = f (g(x)) = f
=
1+x
1โ
1
1+x
1
1+x
1+xโ1
= 1+x =
1
1+x
1+
Then the domain of f โฆ g is (โโ, 0) โช (0, 2) โช (2, โ).
Now consider the domain of g โฆ f . Since the domain of f
is {x|x = 2}, we know that 2 is not in the domain of g โฆ f .
Since the domain of g is {x|x = 0}, we ๏ฌnd the value of x
for which f (x) = 0.
1
=0
xโ2
1=0
1
1
=
=
x+1โx
1โx
x
x
1
x
=1ยท =x
1
1
x
The domain of f is {x|x = 0} and the domain of g is
{x|x = โ1}, so we know that โ1 is not in the domain
of f โฆ g. Since 0 is not in the domain of f , values of x
for which g(x) = 0 are not in the domain of f โฆ g. But
g(x) is never 0, so the domain of f โฆ g is {x|x = โ1}, or
(โโ, โ1) โช (โ1, โ).
Now consider the domain of g โฆ f . Recall that 0 is not in
the domain of f . Since โ1 is not in the domain of g, we
know that g(x) cannot be โ1. We ๏ฌnd the value(s) of x
for which f (x) = โ1.
2=x
1+x
x
ยท
=x
1+x
1
1โx
=
(g โฆ f )(x) = g(f (x)) = g
x
False equation
We see that there are no values of x for which f (x) = โ1,
so the domain of g โฆ f is {x|x = 0}, or (โโ, 0) โช (0, โ).
1
36. (f โฆ g)(x) = f x + 2 =
x+2
x
โ2
x
1
1
=
=
x + 2 โ 2x
โx + 2
x
x
x
x
x
=
, or
= 1ยท
โx + 2
โx + 2
2โx
We get a false equation, so there are no such values. Then
the domain of g โฆ f is (โโ, 2) โช (2, โ).
37.
(f โฆ g)(x) = f (g(x)) = f (x + 1) =
(x + 1)3 โ 5(x + 1)2 + 3(x + 1) + 7 =
x3 + 3×2 + 3x + 1 โ 5×2 โ 10x โ 5 + 3x + 3 + 7 =
x3 โ 2×2 โ 4x + 6
(g โฆ f )(x) = g(f (x)) = g(x3 โ 5×2 + 3x + 7) =
x3 โ 5×2 + 3x + 7 + 1 = x3 โ 5×2 + 3x + 8
The domain of f and of g is (โโ, โ), so the domain of
f โฆ g and of g โฆ f is (โโ, โ).
c 2016 Pearson Education, Inc.
Copyright
76
38.
Chapter 2: More on Functions
(g โฆ f )(x) = x3 + 2×2 โ 3x โ 9 โ 1 =
b) r =
x3 + 2×2 โ 3x โ 10
(g โฆ f )(x) = (x โ 1)3 + 2(x โ 1)2 โ 3(x โ 1) โ 9 =
x3 โ 3×2 + 3x โ 1 + 2×2 โ 4x + 2 โ 3x + 3 โ 9 =
x3 โ x2 โ 4x โ 5
The domain of f and of g is (โโ, โ), so the domain of
f โฆ g and of g โฆ f is (โโ, โ).
h
2
2
h
h
h + 2ฯ
2
2
ฯh2
2
S(h) = ฯh +
2
3 2
S(h) = ฯh
2
S(h) = 2ฯ
53. f (x) = (t โฆ s)(x) = t(s(x)) = t(x โ 3) = x โ 3 + 4 = x + 1
39. h(x) = (4 + 3x)5
This is 4 + 3x to the 5th power. The most obvious answer
is f (x) = x5 and g(x) = 4 + 3x.
โ
40. f (x) = 3 x, g(x) = x2 โ 8
1
(x โ 2)4
This is 1 divided by (x โ 2) to the 4th power. One obvious
1
answer is f (x) = 4 and g(x) = x โ 2. Another possibility
x
1
is f (x) = and g(x) = (x โ 2)4 .
x
We have f (x) = x + 1.
54. The manufacturer charges m + 6 per drill. The chain store
sells each drill for 150%(m + 6), or 1.5(m + 6), or 1.5m + 9.
Thus, we have P (m) = 1.5m + 9.
41. h(x) =
55. Equations (a) โ (f ) are in the form y = mx + b, so we can
read the y-intercepts directly from the equations. Equa2
tions (g) and (h) can be written in this form as y = x โ 2
3
and y = โ2x + 3, respectively. We see that only equation (c) has y-intercept (0, 1).
1
42. f (x) = โ , g(x) = 3x + 7
x
56. None (See Exercise 55.)
43. f (x) =
57. If a line slopes down from left to right, its slope is negative.
The equations y = mx + b for which m is negative are (b),
(d), (f), and (h). (See Exercise 55.)
xโ1
, g(x) = x3
x+1
58. The equation for which |m| is greatest is the equation with
the steepest slant. This is equation (b). (See Exercise 55.)
44. f (x) = |x|, g(x) = 9×2 โ 4
2 + x3
2 โ x3
โ
4
46. f (x) = x , g(x) = x โ 3
45. f (x) = x6 , g(x) =
โ
59. The only equation that has (0, 0) as a solution is (a).
60. Equations (c) and (g) have the same slope. (See Exercise 55.)
xโ5
x+2
โ
โ
48. f (x) = 1 + x, g(x) = 1 + x
47. f (x) =
x, g(x) =
61. Only equations (c) and (g) have the same slope and di๏ฌerent y-intercepts. They represent parallel lines.
62. The only equations for which the product of the slopes is
โ1 are (a) and (f).
49. f (x) = x3 โ 5×2 + 3x โ 1, g(x) = x + 2
50. f (x) = 2×5/3 + 5×2/3 , g(x) = x โ 1, or
f (x) = 2×5 + 5×2 , g(x) = (x โ 1)1/3
51. a) Use the distance formula, distance = rate ร
time. Substitute 3 for the rate and t for time.
r(t) = 3t
b) Use the formula for the area of a circle.
63. Only the composition (c โฆ p)(a) makes sense. It represents
the cost of the grass seed required to seed a lawn with area
a.
64. Answers may vary. One example is f (x) = 2x + 5 and
xโ5
. Other examples are found in Exercises 17,
g(x) =
2
18, 25, 26, 32 and 35.
A(r) = ฯr2
c) (A โฆ r)(t) = A(r(t)) = A(3t) = ฯ(3t)2 = 9ฯt2
Chapter 2 Mid-Chapter Mixed Review
This function gives the area of the ripple in terms
of time t.
1. The statement is true. See page 100 in the text.
52. a) h = 2r
S(r) = 2ฯr(2r) + 2ฯr2
2. The statement is false. See page 113 in the text.
S(r) = 4ฯr2 + 2ฯr2
3. The statement is true. See Example 2 in Section 2.3 in the
text, for instance.
S(r) = 6ฯr
2
4. a) For x-values from 2 to 4, the y-values increase from 2
to 4. Thus the function is increasing on the interval
(2, 4).
c 2016 Pearson Education, Inc.
Copyright
Chapter 2 Mid-Chapter Mixed Review
77
b) For x-values from โ5 to โ3, the y-values decrease
from 5 to 1. Also, for x-values from 4 to 5, the yvalues decrease from 4 to โ3. Thus the function is
decreasing on (โ5, โ3) and on (4, 5).
c) For x-values from โ3 to โ1, y is 3. Thus the function is constant on (โ3, โ1).
5. From the graph we see that a relative maximum value of
6.30 occurs at x = โ1.29. We also see that a relative
minimum value of โ2.30 occurs at x = 1.29.
The graph starts rising, or increasing, from the left and
stops increasing at the relative maximum. From this point
it decreases to the relative minimum and then increases
again. Thus the function is increasing on (โโ, โ1.29)
and on (1.29, โ). It is decreasing on (โ1.29, 1.29).
6. The x-values extend from โ5 to โ1 and from 2 to 5, so
the domain is [โ5, โ1] โช [2, 5]. The y-values extend from
โ3 to 5, so the range is [โ3, 5].
1
7. A(h) = (h + 4)h
2
h2
1
+ 2h
A(h) = h2 + 2h, or
2
2
๏ฃฑ
x โ 5, for x โค โ3,
๏ฃด
๏ฃด
๏ฃด
๏ฃด
๏ฃฒ 2x + 3, for โ3 0,
๏ฃด
๏ฃณ 2 x,
Since โ5 โค โ3, f (โ5) = โ5 โ 5 = โ10.
Since โ3 โค โ3, f (โ3) = โ3 โ 5 = โ8.
Since โ3 0, f (6) = ยท 6 = 3.
2
x + 2, for x < โ4,
9. g(x) =
โx,
for x โฅ โ4
We create the graph in two parts. Graph g(x) = x + 2
for inputs less than โ4. Then graph g(x) = โx for inputs
greater than or equal to โ4.
12.
(g โ f )(3) = g(3) โ f (3)
= (32 + 4) โ (3 ยท 3 โ 1)
= 9 + 4 โ (9 โ 1)
= 9+4โ9+1
=5
13.
1
g
1
3
=
(g/f )
1
3
f
3
2
1
+4
3
=
1
3ยท โ1
3
1
+4
9
=
1โ1
37
= 9
0
1
does not exist.
Since division by 0 is not de๏ฌned, (g/f )
3
14. f (x) = 2x + 5, g(x) = โx โ 4
a) The domain of f and of g is the set of all real numbers, or (โโ, โ). Then the domain of f + g, f โ g,
f g, and f f is also (โโ, โ).
For f /g we must exclude โ4 since g(โ4) = 0. Then
the domain of f /g is (โโ, โ4) โช (โ4, โ).
5
5
= 0.
For g/f we must exclude โ since f โ
2
2
Then the domain of g/f is
5
5
โช โ ,โ .
โ โ, โ
2
2
b) (f +g)(x) = f (x)+g(x) = (2x+5)+(โxโ4) = x+1
y
(f โ g)(x) = f (x) โ g(x) = (2x + 5) โ (โx โ 4) =
2x + 5 + x + 4 = 3x + 9
4
(f g)(x) = f (x) ยท g(x) = (2x + 5)(โx โ 4) =
โ2×2 โ 8x โ 5x โ 20 = โ2×2 โ 13x โ 20
2
โซบ4 โซบ2
2
4
x
โซบ2
โซบ4
(f f )(x) = f (x) ยท f (x) = (2x + 5) ยท (2x + 5) =
4×2 + 10x + 10x + 25 = 4×2 + 20x + 25
2x + 5
f (x)
=
(f /g)(x) =
g(x)
โx โ 4
โx โ 4
g(x)
=
f (x)
2x + 5
โ
15. f (x) = x โ 1, g(x) = x + 2
(g/f )(x) =
10.
(f + g)(โ1) = f (โ1) + g(โ1)
= [3(โ1) โ 1] + [(โ1)2 + 4]
= โ3 โ 1 + 1 + 4
=1
11.
(f g)(0) = f (0) ยท g(0)
= (3 ยท 0 โ 1) ยท (02 + 4)
= โ1 ยท 4
a) Any number can be an input for f , so the domain
of f is the set of all real numbers, or (โโ, โ).
The domain of g consists of all values for which x+2
is nonnegative, so we have x + 2 โฅ 0, or x โฅ โ2, or
[โ2, โ). Then the domain of f + g, f โ g, and f g
is [โ2, โ).
= โ4
c 2016 Pearson Education, Inc.
Copyright
78
Chapter 2: More on Functions
The domain of f f is (โโ, โ).
Since g(โ2) = 0, the domain of f /g is (โ2, โ).
Since f (1) = 0, the domain of g/f is [โ2, 1)โช(1, โ).
โ
x+2
โ
(f โ g)(x) = f (x) โ g(x) = x โ 1 โ x + 2
โ
(f g)(x) = f (x) ยท g(x) = (x โ 1) x + 2
b) (f + g)(x) = f (x) + g(x) = x โ 1 +
(f f )(x) = f (x) ยท f (x) = (x โ 1)(x โ 1) =
x2 โ x โ x + 1 = x2 โ 2x + 1
xโ1
f (x)
=โ
(f /g)(x) =
g(x)
x+2
โ
x+2
g(x)
=
(g/f )(x) =
f (x)
xโ1
16. f (x) = 4x โ 3
4(x + h) โ 3 โ (4x โ 3)
f (x + h) โ f (x)
=
=
h
h
4h
4x + 4h โ 3 โ 4x + 3
=
=4
h
h
17. f (x) = 6 โ x2
6 โ (x + h)2 โ (6 โ x2 )
f (x + h) โ f (x)
=
=
h
h
6โ(x2 +2xh+h2 )โ6+x2 6โx2 โ2xhโh2 โ6 + x2
=
=
h
h
h/(โ2x โ h)
โ2xh โ h2
=
= โ2x โ h
h
h/ ยท 1
18. (f โฆ g)(1) = f (g(1)) = f (1 + 1) = f (1 + 1) = f (2) =
5 ยท 2 โ 4 = 10 โ 4 = 6
3
19. (g โฆ h)(2) = g(h(2)) = g(22 โ 2 ยท 2 + 3) = g(4 โ 4 + 3) =
g(3) = 33 + 1 = 27 + 1 = 28
20. (f โฆ f )(0) = f (f (0)) = f (5 ยท 0 โ 4) = f (โ4) = 5(โ4) โ 4 =
โ20 โ 4 = โ24
21. (h โฆ f )(โ1) = h(f (โ1)) = h(5(โ1) โ 4) = h(โ5 โ 4) =
h(โ9) = (โ9)2 โ 2(โ9) + 3 = 81 + 18 + 3 = 102
1
22. (f โฆ g)(x) = f (g(x)) = f (6x + 4) = (6x + 4) = 3x + 2
2
1
1
x = 6 ยท x + 4 = 3x + 4
(g โฆ f )(x) = g(f (x)) = g
2
2
The domain of f and g is (โโ, โ), so the domain of f โฆ g
and g โฆ f is (โโ, โ).
โ
โ
23. (f โฆ g)(x) = f (g(x)) = f ( x) = 3 x + 2
โ
(g โฆ f )(x) = g(f (x)) = g(3x + 2) = 3x + 2
The domain of f is (โโ, โ) and the domain of g is [0, โ).
Consider the domain of f โฆ g. Since any number can be an
input for f , the domain of f โฆ g is the same as the domain
of g, [0, โ).
Now consider the domain of g โฆ f . Since the inputs of g
2
must be nonnegative, we must have 3x+2 โฅ 0, or x โฅ โ .
3
2
Thus the domain of g โฆ f is โ , โ .
3
24. The graph of y = (h โ g)(x) will be the same as the graph
of y = h(x) moved down b units.
25. Under the given conditions, (f + g)(x) and (f /g)(x) have
di๏ฌerent domains if g(x) = 0 for one or more real numbers
x.
26. If f and g are linear functions, then any real number can
be an input for each function. Thus, the domain of f โฆ g =
the domain of g โฆ f = (โโ, โ).
27. This approach is not valid. Consider Exercise 23 in
Exercise Set 2.3 in the text, for example. Since
4x
(f โฆ g)(x) =
, an examination of only this composed
xโ5
function would lead to the incorrect conclusion that the
domain of f โฆ g is (โโ, 5) โช (5, โ). However, we must also
exclude from the domain of f โฆ g those values of x that
are not in the domain of g. Thus, the domain of f โฆ g is
(โโ, 0) โช (0, 5) โช (5, โ).
Exercise Set 2.4
1. If the graph were folded on the x-axis, the parts above and
below the x-axis would not coincide, so the graph is not
symmetric with respect to the x-axis.
If the graph were folded on the y-axis, the parts to the
left and right of the y-axis would coincide, so the graph is
symmetric with respect to the y-axis.
If the graph were rotated 180โฆ , the resulting graph would
not coincide with the original graph, so it is not symmetric
with respect to the origin.
2. If the graph were folded on the x-axis, the parts above and
below the x-axis would not coincide, so the graph is not
symmetric with respect to the x-axis.
If the graph were folded on the y-axis, the parts to the
left and right of the y-axis would coincide, so the graph is
symmetric with respect to the y-axis.
If the graph were rotated 180โฆ , the resulting graph would
not coincide with the original graph, so it is not symmetric
with respect to the origin.
3. If the graph were folded on the x-axis, the parts above and
below the x-axis would coincide, so the graph is symmetric
with respect to the x-axis.
If the graph were folded on the y-axis, the parts to the left
and right of the y-axis would not coincide, so the graph is
not symmetric with respect to the y-axis.
If the graph were rotated 180โฆ , the resulting graph would
not coincide with the original graph, so it is not symmetric
with respect to the origin.
4. If the graph were folded on the x-axis, the parts above and
below the x-axis would not coincide, so the graph is not
symmetric with respect to the x-axis.
If the graph were folded on the y-axis, the parts to the left
and right of the y-axis would not coincide, so the graph is
not symmetric with respect to the y-axis.
c 2016 Pearson Education, Inc.
Copyright
Exercise Set 2.4
79
If the graph were rotated 180โฆ , the resulting graph would
coincide with the original graph, so it is symmetric with
respect to the origin.
The last equation is not equivalent to the original equation,
so the graph is not symmetric with respect to the origin.
8.
y
5. If the graph were folded on the x-axis, the parts above and
below the x-axis would not coincide, so the graph is not
symmetric with respect to the x-axis.
10
8
6
If the graph were folded on the y-axis, the parts to the left
and right of the y-axis would not coincide, so the graph is
not symmetric with respect to the y-axis.
2
โ10 โ8 โ6 โ4 โ2
โ2
If the graph were rotated 180โฆ , the resulting graph would
coincide with the original graph, so it is symmetric with
respect to the origin.
8
10
x
โ8
The graph is not symmetric with respect to the x-axis, the
y-axis, or the origin.
Test algebraically for symmetry with respect to the x-axis:
y = |x + 5|
Original equation
โy = |x + 5|
Replacing y by โy
y = โ|x + 5| Simplifying
The last equation is not equivalent to the original equation,
so the graph is not symmetric with respect to the x-axis.
Test algebraically for symmetry with respect to the y-axis:
y
y = |x + 5|
5
Original equation
y = | โ x + 5| Replacing x by โx
4
3
The last equation is not equivalent to the original equation,
so the graph is not symmetric with respect to the y-axis.
2
1
1
2
3
4
5
x
Test algebraically for symmetry with respect to the origin:
y = |x + 5|
โ2
โ3
y = โ| โ x + 5| Simplifying
โ5
The last equation is not equivalent to the original equation,
so the graph is not symmetric with respect to the origin.
The graph is symmetric with respect to the y-axis. It is
not symmetric with respect to the x-axis or the origin.
Test algebraically for symmetry with respect to the x-axis:
y = |x| โ 2
Original equation
โy = |x| โ 2
Replacing y by โy
y = โ|x| + 2
9.
y
5
4
5y = 4x + 5
Simplifying
1
โ5 โ4 โ3 โ2 โ1
โ1
Test algebraically for symmetry with respect to the y-axis:
y = |x| โ 2
Original equation
y = | โ x| โ 2
Replacing x by โx
y = |x| โ 2
Simplifying
2
3
4
5
x
โ4
โ5
Test algebraically for symmetry with respect to the origin:
Original equation
Replacing x by โx and
y by โy
Simplifying
1
โ2
โ3
The last equation is equivalent to the original equation, so
the graph is symmetric with respect to the y-axis.
โy = | โ x| โ 2
3
2
The last equation is not equivalent to the original equation,
so the graph is not symmetric with respect to the x-axis.
y = |x| โ 2
Original equation
โy = | โ x + 5| Replacing x by โx and y by โy
y = |x | โ 2
โ4
y = โ|x| + 2
6
โ10
If the graph were rotated 180โฆ , the resulting graph would
coincide with the original graph, so it is symmetric with
respect to the origin.
โy = |x| โ 2
4
โ6
If the graph were folded on the y-axis, the parts to the
left and right of the y-axis would coincide, so the graph is
symmetric with respect to the y-axis.
โ5 โ4 โ3 โ2 โ1
โ1
2
โ4
6. If the graph were folded on the x-axis, the parts above and
below the x-axis would coincide, so the graph is symmetric
with respect to the x-axis.
7.
y = |x + 5 |
4
The graph is not symmetric with respect to the x-axis, the
y-axis, or the origin.
Test algebraically for symmetry with respect to the x-axis:
5y = 4x + 5
Original equation
5(โy) = 4x + 5
Replacing y by โy
โ5y = 4x + 5
Simplifying
5y = โ4x โ 5
The last equation is not equivalent to the original equation,
so the graph is not symmetric with respect to the x-axis.
c 2016 Pearson Education, Inc.
Copyright
80
Chapter 2: More on Functions
Test algebraically for symmetry with respect to the y-axis:
5y = 4x + 5
Original equation
5y = 4(โx) + 5
Replacing x by โx
5y = โ4x + 5
Simplifying
11.
y
5
4
3
2
The last equation is not equivalent to the original equation,
so the graph is not symmetric with respect to the y-axis.
1
โ5 โ4 โ3 โ2 โ1
โ1
Test algebraically for symmetry with respect to the origin:
5y = 4x + 5
Original equation
Replacing x by โx
and
y by โy
4
x
5
5y = 2 x 2 โ 3
โ5
The graph is symmetric with respect to the y-axis. It is
not symmetric with respect to the x-axis or the origin.
Simplifying
Test algebraically for symmetry with respect to the x-axis:
The last equation is not equivalent to the original equation,
so the graph is not symmetric with respect to the origin.
5y = 2×2 โ 3
Original equation
5(โy) = 2x โ 3
Replacing y by โy
2
โ5y = 2×2 โ 3
Simplifying
5y = โ2x + 3
2
y
The last equation is not equivalent to the original equation,
so the graph is not symmetric with respect to the x-axis.
5
4
3
โ4
5y = 4x โ 5
10.
2
โ3
5(โy) = 4(โx) + 5
โ5y = โ4x + 5
โ2
1
2x โ 5 = 3y
3
Test algebraically for symmetry with respect to the y-axis:
2
1
โ5 โ4 โ3 โ2 โ1
โ1
1
2
3
4
5
x
โ2
5y = 2×2 โ 3
Original equation
5y = 2(โx)2 โ 3
Replacing x by โx
5y = 2×2 โ 3
โ3
โ4
The last equation is equivalent to the original equation, so
the graph is symmetric with respect to the y-axis.
โ5
The graph is not symmetric with respect to the x-axis, the
y-axis, or the origin.
Test algebraically for symmetry with respect to the origin:
Test algebraically for symmetry with respect to the x-axis:
5(โy) = 2(โx) โ 3
2x โ 5 = 3y
Original equation
2x โ 5 = 3(โy)
Replacing y by โy
โ2x + 5 = 3y
5y = 2×2 โ 3
The last equation is not equivalent to the original equation,
so the graph is not symmetric with respect to the origin.
12.
y
2x โ 5 = 3y
Original equation
2(โx) โ 5 = 3y
Replacing x by โx
5
Simplifying
3
4
2
The last equation is not equivalent to the original equation,
so the graph is not symmetric with respect to the y-axis.
1
โ5 โ4 โ3 โ2 โ1
โ1
Test algebraically for symmetry with respect to the origin:
2(โx) โ 5 = 3(โy)
โ2x โ 5 = โ3y
Simplifying
5y = โ2×2 + 3
Simplifying
Test algebraically for symmetry with respect to the y-axis:
2x โ 5 = 3y
Replacing x by โx and
y by โy
โ5y = 2×2 โ 3
The last equation is not equivalent to the original equation,
so the graph is not symmetric with respect to the x-axis.
โ2x โ 5 = 3y
Original equation
2
โ2
1
2
3
4
5
x
x 2 + 4 = 3y
โ3
Original equation
โ4
Replacing x by โx and
y by โy
โ5
Simplifying
The graph is symmetric with respect to the y-axis. It is
not symmetric with respect to the x-axis or the origin.
The last equation is not equivalent to the original equation,
so the graph is not symmetric with respect to the origin.
Test algebraically for symmetry with respect to the x-axis:
2x + 5 = 3y
x2 + 4 = 3y
Original equation
x2 + 4 = 3(โy)
Replacing y by โy
โx โ 4 = 3y
2
c 2016 Pearson Education, Inc.
Copyright
Simplifying
Exercise Set 2.4
81
The last equation is equivalent to the original equation, so
the graph is symmetric with respect to the origin.
The last equation is not equivalent to the original equation,
so the graph is not symmetric with respect to the x-axis.
Test algebraically for symmetry with respect to the y-axis:
x2 + 4 = 3y Original equation
14.
y
5
Replacing x by โx
(โx)2 + 4 = 3y
4
2
x + 4 = 3y
The last equation is equivalent to the original equation, so
the graph is symmetric with respect to the y-axis.
Test algebraically for symmetry with respect to the origin:
Original equation
x2 + 4 = 3y
2
(โx) + 4 = 3(โy)
x2 + 4 = โ3y
4
y=โโ
x
3
2
1
โ5 โ4 โ3 โ2 โ1
โ1
1
2
3
4
5
x
โ2
โ3
โ4
Replacing x by โx and
y by โy
โ5
Simplifying
The graph is not symmetric with respect to the x-axis or
the y-axis. It is symmetric with respect to the origin.
The last equation is not equivalent to the original equation,
so the graph is not symmetric with respect to the origin.
Test algebraically for symmetry with respect to the x-axis:
4
y=โ
Original equation
x
4
โy = โ
Replacing y by โy
x
4
Simplifying
y=
x
The last equation is not equivalent to the original equation,
so the graph is not symmetric with respect to the x-axis.
โx โ 4 = 3y
2
13.
y
5
4
1
y=โ
x
3
2
1
โ5 โ4 โ3 โ2 โ1
โ1
1
2
3
4
5
x
โ2
โ3
โ4
โ5
The graph is not symmetric with respect to the x-axis or
the y-axis. It is symmetric with respect to the origin.
Test algebraically for symmetry with respect to the x-axis:
1
Original equation
y=
x
1
Replacing y by โy
โy =
x
1
y=โ
Simplifying
x
The last equation is not equivalent to the original equation,
so the graph is not symmetric with respect to the x-axis.
Test algebraically for symmetry with respect to the y-axis:
1
Original equation
y=
x
1
Replacing x by โx
y=
โx
1
y=โ
Simplifying
x
The last equation is not equivalent to the original equation,
so the graph is not symmetric with respect to the y-axis.
Test algebraically for symmetry with respect to the origin:
1
Original equation
y=
x
1
โy =
Replacing x by โx and y by โy
โx
1
Simplifying
y=
x
Test algebraically for symmetry with respect to the y-axis:
4
Original equation
y=โ
x
4
y=โ
Replacing x by โx
โx
4
Simplifying
y=
x
The last equation is not equivalent to the original equation,
so the graph is not symmetric with respect to the y-axis.
Test algebraically for symmetry with respect to the origin:
4
Original equation
y=โ
x
4
โy = โ
Replacing x by โx and y by โy
โx
4
y=โ
Simplifying
x
The last equation is equivalent to the original equation, so
the graph is symmetric with respect to the origin.
15. Test for symmetry with respect to the x-axis:
5x โ 5y = 0
Original equation
5x โ 5(โy) = 0
Replacing y by โy
5x + 5y = 0
Simplifying
The last equation is not equivalent to the original equation,
so the graph is not symmetric with respect to the x-axis.
Test for symmetry with respect to the y-axis:
5x โ 5y = 0
Original equation
5(โx) โ 5y = 0
Replacing x by โx
โ5x โ 5y = 0
5x + 5y = 0
c 2016 Pearson Education, Inc.
Copyright
Simplifying
82
Chapter 2: More on Functions
The last equation is not equivalent to the original equation,
so the graph is not symmetric with respect to the y-axis.
Test for symmetry with respect to the origin:
5x โ 5y = 0
5(โx) โ 5(โy) = 0
โ5x + 5y = 0
Original equation
18. Test for symmetry with respect to the x-axis:
5y = 7×2 โ 2x
Original equation
5(โy) = 7×2 โ 2x
Replacing y by โy
5y = โ7x + 2x Simplifying
2
Replacing x by โx and
y by โy
The last equation is not equivalent to the original equation,
so the graph is not symmetric with respect to the x-axis.
Simplifying
Test for symmetry with respect to the y-axis:
5x โ 5y = 0
The last equation is equivalent to the original equation, so
the graph is symmetric with respect to the origin.
16. Test for symmetry with respect to the x-axis:
5y = 7×2 โ 2x
Original equation
5y = 7(โx)2 โ 2(โx)
Replacing x by โx
2
5y = 7x + 2x
Simplifying
6x + 7y = 0
Original equation
The last equation is not equivalent to the original equation,
so the graph is not symmetric with respect to the y-axis.
6x + 7(โy) = 0
Replacing y by โy
Test for symmetry with respect to the origin:
6x โ 7y = 0
5y = 7×2 โ 2x
Simplifying
The last equation is not equivalent to the original equation,
so the graph is not symmetric with respect to the x-axis.
Test for symmetry with respect to the y-axis:
6x + 7y = 0
Original equation
6(โx) + 7y = 0
Replacing x by โx
6x โ 7y = 0
6x + 7y = 0
Simplifying
The last equation is not equivalent to the original equation,
so the graph is not symmetric with respect to the origin.
Simplifying
Test for symmetry with respect to the origin:
6(โx) + 7(โy) = 0
โ5y = 7×2 + 2x
Replacing x by โx
and y by โy
5y = โ7×2 โ 2x
The last equation is not equivalent to the original equation,
so the graph is not symmetric with respect to the y-axis.
6x + 7y = 0
Original equation
5(โy) = 7(โx)2 โ 2(โx)
Original equation
19. Test for symmetry with respect to the x-axis:
y = |2x|
Original equation
โy = |2x|
Replacing y by โy
y = โ|2x| Simplifying
Replacing x by โx and
y by โy
The last equation is not equivalent to the original equation,
so the graph is not symmetric with respect to the x-axis.
Simplifying
Test for symmetry with respect to the y-axis:
The last equation is equivalent to the original equation, so
the graph is symmetric with respect to the origin.
17. Test for symmetry with respect to the x-axis:
3×2 โ 2y 2 = 3
Original equation
3×2 โ 2(โy)2 = 3
Replacing y by โy
3×2 โ 2y 2 = 3
Simplifying
y = |2x|
Original equation
y = |2(โx)| Replacing x by โx
y = | โ 2x|
Simplifying
y = |2x|
The last equation is equivalent to the original equation, so
the graph is symmetric with respect to the y-axis.
The last equation is equivalent to the original equation, so
the graph is symmetric with respect to the x-axis.
Test for symmetry with respect to the origin:
Test for symmetry with respect to the y-axis:
y = |2x|
Original equation
3×2 โ 2y 2 = 3
Original equation
โy = |2(โx)| Replacing x by โx and y by โy
โy = | โ 2x| Simplifying
3(โx)2 โ 2y 2 = 3
Replacing x by โx
โy = |2x|
3×2 โ 2y 2 = 3
y = โ|2x|
Simplifying
The last equation is equivalent to the original equation, so
the graph is symmetric with respect to the y-axis.
Test for symmetry with respect to the origin:
3×2 โ 2y 2 = 3
Original equation
3(โx)2 โ 2(โy)2 = 3
Replacing x by โx
and y by โy
3×2 โ 2y 2 = 3
Simplifying
The last equation is equivalent to the original equation, so
the graph is symmetric with respect to the origin.
The last equation is not equivalent to the original equation,
so the graph is not symmetric with respect to the origin.
20. Test for symmetry with respect to the x-axis:
y 3 = 2×2
Original equation
(โy)3 = 2×2
Replacing y by โy
โy = 2x
3
2
Simplifying
y 3 = โ2×2
The last equation is not equivalent to the original equation,
so the graph is not symmetric with respect to the x-axis.
c 2016 Pearson Education, Inc.
Copyright
Exercise Set 2.4
83
Test for symmetry with respect to the y-axis:
y 3 = 2×2
Original equation
y 3 = 2(โx)2
Replacing x by โx
23. Test for symmetry with respect to the x-axis:
Simplifying
y 3 = 2×2
The last equation is equivalent to the original equation, so
the graph is symmetric with respect to the y-axis.
Test for symmetry with respect to the origin:
Original equation
y 3 = 2×2
Replacing x by โx and
y by โy
(โy)3 = 2(โx)2
โy 3 = 2×2
Simplifying
y 3 = โ2×2
The last equation is not equivalent to the original equation,
so the graph is not symmetric with respect to the origin.
21. Test for symmetry with respect to the x-axis:
Original equation
2×4 + 3 = y 2
Replacing y by โy
2×4 + 3 = (โy)2
Simplifying
2×4 + 3 = y 2
The last equation is equivalent to the original equation, so
the graph is symmetric with respect to the x-axis.
Test for symmetry with respect to the y-axis:
2×4 + 3 = y 2 Original equation
Replacing x by โx
2(โx)4 + 3 = y 2
2×4 + 3 = y 2 Simplifying
The last equation is equivalent to the original equation, so
the graph is symmetric with respect to the y-axis.
Test for symmetry with respect to the origin:
Original equation
2×4 + 3 = y 2
4
2
2(โx) + 3 = (โy)
Replacing x by โx
and y by โy
Simplifying
2×4 + 3 = y 2
The last equation is equivalent to the original equation, so
the graph is symmetric with respect to the origin.
22. Test for symmetry with respect to the x-axis:
2y 2 = 5×2 + 12 Original equation
2(โy)2 = 5×2 + 12
Replacing y by โy
2y 2 = 5×2 + 12 Simplifying
The last equation is equivalent to the original equation, so
the graph is symmetric with respect to the x-axis.
Test for symmetry with respect to the y-axis:
Original equation
2y 2 = 5×2 + 12
2y 2 = 5(โx)2 + 12
Replacing x by โx
2y 2 = 5×2 + 12
Simplifying
The last equation is equivalent to the original equation, so
the graph is symmetric with respect to the y-axis.
Test for symmetry with respect to the origin:
Original equation
2y 2 = 5×2 + 12
2(โy)2 = 5(โx)2 + 12
2
2
2y = 5x + 12
Replacing x by โx
and y by โy
Simplifying
The last equation is equivalent to the original equation, so
the graph is symmetric with respect to the origin.
3y 3 = 4×3 + 2
Original equation
3(โy)3 = 4×3 + 2
Replacing y by โy
โ3y = 4x + 2
3
3
Simplifying
3y 3 = โ4×3 โ 2
The last equation is not equivalent to the original equation,
so the graph is not symmetric with respect to the x-axis.
Test for symmetry with respect to the y-axis:
3y 3 = 4×3 + 2
Original equation
3y = 4(โx) + 2
Replacing x by โx
3y 3 = โ4×3 + 2
Simplifying
3
3
The last equation is not equivalent to the original equation,
so the graph is not symmetric with respect to the y-axis.
Test for symmetry with respect to the origin:
3y 3 = 4×3 + 2
Original equation
3(โy)3 = 4(โx)3 + 2
Replacing x by โx
and y by โy
โ3y 3 = โ4×3 + 2
Simplifying
3y = 4x โ 2
3
3
The last equation is not equivalent to the original equation,
so the graph is not symmetric with respect to the origin.
24. Test for symmetry with respect to the x-axis:
3x = |y|
Original equation
3x = | โ y| Replacing y by โy
3x = |y|
Simplifying
The last equation is equivalent to the original equation, so
the graph is symmetric with respect to the x-axis.
Test for symmetry with respect to the y-axis:
3x = |y| Original equation
3(โx) = |y| Replacing x by โx
โ3x = |y| Simplifying
The last equation is not equivalent to the original equation,
so the graph is not symmetric with respect to the y-axis.
Test for symmetry with respect to the origin:
3x = |y|
Original equation
3(โx) = | โ y| Replacing x by โx and y by โy
โ3x = |y|
Simplifying
The last equation is not equivalent to the original equation,
so the graph is not symmetric with respect to the origin.
25. Test for symmetry with respect to the x-axis:
xy = 12
Original equation
x(โy) = 12
Replacing y by โy
โxy = 12
Simplifying
xy = โ12
The last equation is not equivalent to the original equation,
so the graph is not symmetric with respect to the x-axis.
c 2016 Pearson Education, Inc.
Copyright
84
Chapter 2: More on Functions
Test for symmetry with respect to the y-axis:
xy = 12
Original equation
โxy = 12
31. x-axis: Replace y with โy; (0, 4)
y-axis: Replace x with โx; (0, โ4)
Replacing x by โx
xy = โ12
Origin: Replace x with โx and y with โy; (0, 4)
Simplifying
32. x-axis: Replace y with โy; (8, 3)
The last equation is not equivalent to the original equation,
so the graph is not symmetric with respect to the y-axis.
Test for symmetry with respect to the origin:
xy = 12 Original equation
โx(โy) = 12
Replacing x by โx and y by โy
xy = 12
Simplifying
The last equation is equivalent to the original equation, so
the graph is symmetric with respect to the origin.
26. Test for symmetry with respect to the x-axis:
Original equation
xy โ x2 = 3
Replacing y by โy
x(โy) โ x2 = 3
xy + x = โ3
2
Simplifying
The last equation is not equivalent to the original equation,
so the graph is not symmetric with respect to the x-axis.
Test for symmetry with respect to the y-axis:
Original equation
xy โ x2 = 3
โxy โ (โx)2 = 3
xy + x = โ3
2
34. The graph is symmetric with respect to the y-axis, so the
function is even.
35. The graph is symmetric with respect to the origin, so the
function is odd.
36. The graph is not symmetric with respect to either the yaxis or the origin, so the function is neither even nor odd.
37. The graph is not symmetric with respect to either the yaxis or the origin, so the function is neither even nor odd.
38. The graph is not symmetric with respect to either the yaxis or the origin, so the function is neither even nor odd.
39.
f (x) = โ3×3 + 2x
f (โx) = โ3(โx)3 + 2(โx) = 3×3 โ 2x
Simplifying
โf (x) = โ(โ3×3 + 2x) = 3×3 โ 2x
Test for symmetry with respect to the origin:
xy โ x2 = 3 Original equation
xy โ x2 = 3
33. The graph is symmetric with respect to the y-axis, so the
function is even.
Replacing x by โx
The last equation is not equivalent to the original equation,
so the graph is not symmetric with respect to the y-axis.
โx(โy) โ (โx)2 = 3
y-axis: Replace x with โx; (โ8, โ3)
Origin: Replace x with โx and y with โy; (โ8, 3)
f (โx) = โf (x), so f is odd.
40.
f (โx) = 7(โx)3 + 4(โx) โ 2 = โ7×3 โ 4x โ 2
โf (x) = โ(7×3 + 4x โ 2) = โ7×3 โ 4x + 2
Replacing x by โx and
y by โy
f (x) = f (โx), so f is not even.
f (โx) = โf (x), so f is not odd.
Simplifying
The last equation is equivalent to the original equation, so
the graph is symmetric with respect to the origin.
f (x) = 7×3 + 4x โ 2
Thus, f (x) = 7×3 + 4x โ 2 is neither even nor odd.
41.
27. x-axis: Replace y with โy; (โ5, โ6)
f (x) = 5×2 + 2×4 โ 1
f (โx) = 5(โx)2 + 2(โx)4 โ 1 = 5×2 + 2×4 โ 1
y-axis: Replace x with โx; (5, 6)
f (x) = f (โx), so f is even.
Origin: Replace x with โx and y with โy; (5, โ6)
7
,0
28. x-axis: Replace y with โy;
2
7
y-axis: Replace x with โx; โ , 0
2
7
Origin: Replace x with โx and y with โy; โ , 0
2
29. x-axis: Replace y with โy; (โ10, 7)
42.
f (x) = x +
1
x
1
1
f (โx) = โx +
= โx โ
โx
x
1
1
= โx โ
โf (x) = โ x +
x
x
f (โx) = โf (x), so f is odd.
43.
f (x) = x17
f (โx) = (โx)17 = โx17
y-axis: Replace x with โx; (10, โ7)
Origin: Replace x with โx and y with โy; (10, 7)
3
30. x-axis: Replace y with โy; 1, โ
8
3
y-axis: Replace x with โx; โ 1,
8
3
Origin: Replace x with โx and y with โy; โ 1, โ
8
โf (x) = โx17
f (โx) = โf (x), so f is odd.
โ
f (x) = 3 x
44.
โ
โ
f (โx) = 3 โx = โ 3 x
โ
โf (x) = โ 3 x
f (โx) = โf (x), so f is odd.
c 2016 Pearson Education, Inc.
Copyright
Exercise Set 2.5
45.
85
f (x) = x โ |x|
53. If the graph were folded on the x-axis, the parts above and
below the x-axis would coincide, so the graph is symmetric
with respect to the x-axis.
f (โx) = (โx) โ |(โx)| = โx โ |x|
โf (x) = โ(x โ |x|) = โx + |x|
If the graph were folded on the y-axis, the parts to the left
and right of the y-axis would not coincide, so the graph is
not symmetric with respect to the y-axis.
f (x) = f (โx), so f is not even.
f (โx) = โf (x), so f is not odd.
Thus, f (x) = x โ |x| is neither even nor odd.
1
46.
f (x) = 2
x
1
1
f (โx) =
= 2
(โx)2
x
f (x) = f (โx), so f is even.
47.
f (โx) = 8
If the graph were rotated 180โฆ , the resulting graph would
coincide with the original graph, so it is symmetric with
respect to the origin.
f (x) = f (โx), so f is even.
โ
48.
f (x) = x2 + 1
โ
f (โx) = (โx)2 + 1 = x2 + 1
55. See the answer section in the text.
f (x) = f (โx), so f is even.
f (โx) โ f (x)
f (โx) โ f (โ(โx))
=
,
2
2
f (โx) โ f (x)
f (x) โ f (โx)
=
. Thus,
โO(x) = โ
2
2
O(โx) = โO(x) and O is odd.
56. O(โx) =
y
4
2
4 2
2
2
4
x
57. a), b) See the answer section in the text.
4
50. Let v = the number of volunteers from the University of
Wisconsin – Madison. Then v + 464 = the number of
volunteers from the University of California – Berkeley.
Solve: v + (v + 464) = 6688
v = 3112, so there were 3112 volunteers from the University of Wisconsin – Madison and 3112 + 464, or 3576
volunteers from the University of California – Berkeley.
โ
51.
f (x) = x 10 โ x2
โ
f (โx) = โx 10 โ (โx)2 = โx 10 โ x2
โ
โf (x) = โx 10 โ x2
Since f (โx) = โf (x), f is odd.
52.
f (x) =
f (โx) =
54. If the graph were folded on the x-axis, the parts above and
below the x-axis would not coincide, so the graph is not
symmetric with respect to the x-axis.
If the graph were folded on the y-axis, the parts to the left
and right of the y-axis would not coincide, so the graph is
not symmetric with respect to the y-axis.
f (x) = 8
49.
If the graph were rotated 180โฆ , the resulting graph would
not coincide with the original graph, so it is not symmetric
with respect to the origin.
58. Let f (x) = g(x) = x. Now f and g are odd functions, but
(f g)(x) = x2 = (f g)(โx). Thus, the product is even, so
the statement is false.
59. Let f (x) and g(x) be even functions. Then by de๏ฌnition,
f (x) = f (โx) and g(x) = g(โx). Thus, (f + g)(x) =
f (x) + g(x) = f (โx) + g(โx) = (f + g)(โx) and f + g is
even. The statement is true.
60. Let f (x) be an even function, and let g(x) be an odd function. By de๏ฌnition f (x) = f (โx) and g(โx) = โg(x),
or g(x) = โg(โx). Then f g(x) = f (x) ยท g(x) = f (โx) ยท
[โg(โx)] = โf (โx) ยท g(โx) = โf g(โx), and f g is odd.
The statement is true.
Exercise Set 2.5
x2 + 1
x3 + 1
1. Shift the graph of f (x) = x2 right 3 units.
x2 + 1
(โx)2 + 1
=
3
(โx) + 1
โx3 + 1
y
4
x2 + 1
โf (x) = โ 3
x +1
Since f (x) = f (โx), f is not even.
2
4 2
2
Since f (โx) = โf (x), f is not odd.
2
x2 + 1
is neither even nor odd.
Thus, f (x) = 3
x +1
4
c 2016 Pearson Education, Inc.
Copyright
4
x
f(x) (x 3)2
86
Chapter 2: More on Functions
2. Shift the graph of g(x) = x2 up
1
unit.
2
6. Shift the graph of g(x) =
โ
x right 1 unit.
y
y
4
4
2
2
4 2
2
4 2
x
4
2
x
4
2
2
4
4
g(x) โ๏ฃฅ๏ฃฅ๏ฃฅ๏ฃฅ๏ฃฅ
x 1
1
g(x) x 2
2
1
up 4 units.
x
7. Shift the graph of h(x) =
3. Shift the graph of g(x) = x down 3 units.
y
y
8
4
6
2
4 2
4
2
4
x
1
h(x)
x 4
2
2
4 2
g(x) x 3
2
4
x
2
4. Re๏ฌect the graph of g(x) = x across the x-axis and then
shift it down 2 units.
8. Shift the graph of g(x) =
1
right 2 units.
x
y
y
4
4
2
2
4 2
2
4 2
x
4
2
2
2
g(x) x 2
5. Re๏ฌect the graph of h(x) =
โ
x across the x-axis.
y
h(x) 3x 3
h(x) โ๏ฃฅx
2
4 2
x
9. First stretch the graph of h(x) = x vertically by multiplying each y-coordinate by 3. Then re๏ฌect it across the
x-axis and shift it up 3 units.
y
4
4
1
g(x)
x
2
4
4
4
2
2
4
x
4 2
2
2
2
4
4
4
x
10. First stretch the graph of f (x) = x vertically by multiplying each y-coordinate by 2. Then shift it up 1 unit.
y
4
2
4 2
2
4
2
4
f(x) 2x 1
c 2016 Pearson Education, Inc.
Copyright
x
Exercise Set 2.5
87
11. First shrink the graph of h(x) = |x| vertically by multiply1
ing each y-coordinate by . Then shift it down 2 units.
2
15. Shift the graph of g(x) = x2 left 1 unit and down 1 unit.
y
y
4
4
2
1
h (x) |x|
2
2
4 2
2
2
4
x
2
4
2
4
x
4
2
g(x) (x 1)2 1
4
12. Re๏ฌect the graph of g(x) = |x| across the x-axis and shift
it up 2 units.
16. Re๏ฌect the graph of h(x) = x2 across the x-axis and down
4 units.
y
4 2
y
2
x
4
2
4
4
2
6
4 2
2
x
4
8
2
h(x) x 2 4
4
g(x) x 2
13. Shift the graph of g(x) = x3 right 2 units and re๏ฌect it
across the x-axis.
17. First shrink the graph of g(x) = x3 vertically by multiply1
ing each y-coordinate by . Then shift it up 2 units.
3
y
y
4
4
g (x) (x 2)3
2
2
4
4 2
2
4
2
2
x
2
4
4
x
1 3
g(x) x
2
3
4
18. Re๏ฌect the graph of h(x) = x3 across the y-axis.
14. Shift the graph of f (x) = x3 left 1 unit.
y
y
4
4
2
2
4 2
4 2
2
4
2
x
2
x
4
2
4
4
f(x) (x 1)3
h(x) (x)3
โ
19. Shift the graph of f (x) =
y
4
2
4 2
2
4
x
2
4
c 2016 Pearson Education, Inc.
Copyright
f(x) โ๏ฃฅ๏ฃฅ๏ฃฅ๏ฃฅ๏ฃฅ
x 2
x left 2 units.
88
Chapter 2: More on Functions
20. First shift the graph of f (x) =
โ
x right 1 unit. Shrink it
1
vertically by multiplying each y-coordinate by and then
2
re๏ฌect it across the x-axis.
y
4
29. Think of the graph of f (x) = |x|. Since g(x) =
1
1
x โ 4, the graph of g(x) = x โ 4 is the graph of
f
3
3
2
4 2
2
x
4
2
4
f (x) = |x| stretched horizontally by multiplying each xcoordinate by 3 and then shifted down 4 units.
1
f(x) โ
x 1
2 ๏ฃฅ๏ฃฅ๏ฃฅ๏ฃฅ๏ฃฅ
21. Shift the graph of f (x) =
โ
3
x down 2 units.
y
4
3
f(x) โ๏ฃฅx 2
2
30. Think of the graph of g(x) = x3 . Since
2
2
f (x) = g(x) โ 4, the graph of f (x) = x3 โ 4 is the
3
3
graph of g(x) = x3 shrunk vertically by multiplying each
2
y-coordinate by and then shifted down 4 units.
3
1
31. Think of the graph of g(x) = x2 . Since f (x) = โ g(x โ 5),
4
1
the graph of f (x) = โ (x โ 5)2 is the graph of g(x) = x2
4
shifted right 5 units, shrunk vertically by multiplying each
1
y-coordinate by , and re๏ฌected across the x-axis.
4
2
4 2
1
. Since f (x) = 5 โ g(x), or
x
1
f (x) = โg(x) + 5, the graph of f (x) = 5 โ is the graph
x
1
of g(x) = re๏ฌected across the x-axis and then shifted up
x
5 units.
28. Think of the graph of g(x) =
x
2
4
โ
22. Shift the graph of h(x) = 3 x left 1 unit.
32. Think of the graph of g(x) = x3 . Since f (x) = g(โx) โ 5,
the graph of f (x) = (โx)3 โ 5 is the graph of g(x) = x3
re๏ฌected across the y-axis and shifted down 5 units.
y
4
1
. Since f (x) =
x
1
+ 2 is the graph
g(x + 3) + 2, the graph of f (x) =
x+3
1
of g(x) = shifted left 3 units and up 2 units.
x
โ
34. Think of the graph ofโf (x) = x. Since g(x) = f (โx) +
โ5,
the graph of g(x) = โx + 5 is the graph of f (x) = x
re๏ฌected across the y-axis and shifted up 5 units.
33. Think of the graph of g(x) =
2
4 2
2
4
x
2
4
3
h(x) โ๏ฃฅ๏ฃฅ๏ฃฅ๏ฃฅ๏ฃฅ
x 1
23. Think of the graph of f (x) = |x|. Since
g(x) = f (3x), the graph of g(x) = |3x| is the graph
of f (x) = |x| shrunk horizontally by dividing each x
1
.
coordinate by 3 or multiplying each x-coordinate by
3
โ
1
24. Think of the graph of g(x) = 3 x. Since f (x) = g(x), the
2
โ
1โ
graph of f (x) = 3 x is the graph of g(x) = 3 x shrunk
2
1
vertically by multiplying each y-coordinate by .
2
1
25. Think of the graph of f (x) = . Since h(x) = 2f (x),
x
1
2
the graph of h(x) = is the graph of f (x) = stretched
x
x
vertically by multiplying each y-coordinate by 2.
26. Think of the graph of g(x) = |x|. Since f (x) = g(xโ3)โ4,
the graph of f (x) = |x โ 3| โ 4 is the graph of g(x) = |x|
shifted right 3 units and down 4 units.
โ
27. Think of the graph of g(x)
โ = x. Since f (x) = 3g(x) โโ5,
the graph of f (x) = 3 x โ 5 is the graph of g(x) = x
stretched vertically by multiplying each y-coordinate by 3
and then shifted down 5 units.
35. Think of the graph of f (x) = x2 . Since h(x) = โf (xโ3)+
5, the graph of h(x) = โ(x โ 3)2 + 5 is the graph of f (x) =
x2 shifted right 3 units, re๏ฌected across the x-axis, and
shifted up 5 units.
36. Think of the graph of g(x) = x2 . Since f (x) = 3g(x + 4)โ
3, the graph of f (x) = 3(x+4)2 โ3 is the graph of g(x) = x2
shifted left 4 units, stretched vertically by multiplying each
y-coordinate by 3, and then shifted down 3 units.
37. The graph of y = g(x) is the graph of y = f (x) shrunk
1
vertically by a factor of . Multiply the y-coordinate by
2
1
: (โ12, 2).
2
38. The graph of y = g(x) is the graph of y = f (x) shifted
right 2 units. Add 2 to the x-coordinate: (โ10, 4).
39. The graph of y = g(x) is the graph of y = f (x) re๏ฌected
across the y-axis, so we re๏ฌect the point across the y-axis:
(12, 4).
c 2016 Pearson Education, Inc.
Copyright
Exercise Set 2.5
89
40. The graph of y = g(x) is the graph of y = f (x) shrunk
1
the
horizontally. The x-coordinates of y = g(x) are
4
corresponding x-coordinates of y = f (x), so we divide the
1
: (โ3, 4).
x-coordinate by 4 or multiply it by
4
54. Shape: h(x) = x2
Shift h(x) right 6 units: g(x) = h(x โ 6) = (x โ 6)2
Shift g(x) up 2 units: f (x) = g(x) + 2 = (x โ 6)2 + 2
55. Shape: m(x) = x2
41. The graph of y = g(x) is the graph of y = f (x) shifted
down 2 units. Subtract 2 from the y-coordinate: (โ12, 2).
Turn m(x) upside-down (that is, re๏ฌect it across the xaxis): h(x) = โm(x) = โx2
42. The graph of y = g(x) is the graph of y = f (x) stretched
horizontally. The x-coordinates of y = g(x) are twice the
corresponding x-coordinates of y = f (x), so we multiply
1
: (โ24, 4).
the x-coordinate by 2 or divide it by
2
Shift g(x) up 4 units: f (x) = g(x) + 4 = โ(x โ 3)2 + 4
43. The graph of y = g(x) is the graph of y = f (x) stretched
vertically by a factor of 4. Multiply the y-coordinate by 4:
(โ12, 16).
44. The graph of y = g(x) is the graph y = f (x) re๏ฌected
across the x-axis. Re๏ฌect the point across the x-axis:
(โ12, โ4).
45. g(x) = x2 + 4 is the function f (x) = x2 + 3 shifted up
1 unit, so g(x) = f (x) + 1. Answer B is correct.
46. If we substitute 3x for x in f , we get 9×2 + 3, so
g(x) = f (3x). Answer D is correct.
47. If we substitute x โ 2 for x in f , we get (x โ 2)3 + 3, so
g(x) = f (x โ 2). Answer A is correct.
2
2
48. If we multiply x + 3 by 2, we get 2x + 6, so g(x) = 2f (x).
Answer C is correct.
49. Shape: h(x) = x2
Turn h(x) upside-down (that is, re๏ฌect it across the xaxis): g(x) = โh(x) = โx2
Shift g(x) right 8 units: f (x) = g(x โ 8) = โ(x โ 8)2
โ
50. Shape: h(x) = x
โ
Shift h(x) left 6 units: g(x) = h(x + 6) = x + 6
โ
Shift g(x) down 5 units: f (x) = g(x) โ 5 = x + 6 โ 5
Shift h(x) right 3 units: g(x) = h(x โ 3) = โ(x โ 3)2
56. Shape: h(x) = |x|
Stretch h(x) horizontally by a factor of 2 that is, multiply
1
1
1
: g(x) = h x = x
each x-value by
2
2
2
1
Shift g(x) down 5 units: f (x) = g(x) โ 5 = x โ 5
2
โ
57. Shape: m(x) = x
โ
Re๏ฌect m(x) across the y-axis: h(x) = m(โx) = โx
Shift h(x) left 2 units: g(x) = h(x + 2) = โ(x + 2)
Shift g(x) down 1 unit: f (x) = g(x) โ 1 =
โ(x + 2) โ 1
58. Shape: h(x) =
Re๏ฌect h(x) across the x-axis: g(x) = โh(x) = โ
Shift g(x) up 1 unit: f (x) = g(x) + 1 = โ
1
+1
x
y
(โซบ1, 4)
(โซบ3, 4)
g(x) โซฝ โซบ2f(x)
4
(5, 0)
(โซบ4, 0)
โซบ4 โซบ2
2
4
6 x
โซบ2
โซบ4
Shift h(x) left 7 units: g(x) = h(x + 7) = |x + 7|
Shift g(x) up 2 units: f (x) = g(x) + 2 = |x + 7| + 2
52. Shape: h(x) = x3
Turn h(x) upside-down (that is, re๏ฌect it across the xaxis): g(x) = โh(x) = โx3
โซบ6
(2, โซบ6)
1
60. Each y-coordinate is multiplied by . We plot and connect
2
(โ4, 0), (โ3, โ1), (โ1, โ1), (2, 1.5), and (5, 0).
y
Shift g(x) right 5 units: f (x) = g(x โ 5) = โ(x โ 5)3
1
x
1
x
59. Each y-coordinate is multiplied by โ2. We plot and connect (โ4, 0), (โ3, 4), (โ1, 4), (2, โ6), and (5, 0).
51. Shape: h(x) = |x|
53. Shape: h(x) =
1
x
4
1
Shrink h(x) vertically by a factor of
2
1
:
multiply each function value by
2
1 1
1
1
g(x) = h(x) = ยท , or
2
2 x
2x
(4, 0)
that is,
2 (2, 1.5)
4 2
2
(3, 1) (1, 1)
4
Shift g(x) down 3 units: f (x) = g(x) โ 3 =
(5, 0)
4
1
โ3
2x
c 2016 Pearson Education, Inc.
Copyright
x
g(x) qf(x)
90
Chapter 2: More on Functions
61. The graph is re๏ฌected across the y-axis and stretched horizontally by a factor of 2. That is, each x-coordinate is
1
multiplied by โ2 or divided by โ . We plot and con2
nect (8, 0), (6, โ2), (2, โ2), (โ4, 3), and (โ10, 0).
y
2
(โซบ10, 0)
2
โซบ4
2
4
(2, โซบ2)
6
4
(5, 4)
(1, 4)
x
(4, 4)
12
(1, 13)
g(x) 3f(x 1) 4
(8, 0)
โซบ10 โซบ8 โซบ6 โซบ4 โซบ2
โซบ2
4
(4, 2)
g (x) โซฝ f ๅขโซบqxๅฃ
4
(โซบ4, 3)
y
(2, 2)
8
x
(6, โซบ2)
65. The graph is re๏ฌected across the y-axis so each
x-coordinate is replaced by its opposite.
y
62. The graph is shrunk horizontally by a factor of 2. That
1
.
is, each x-coordinate is divided by 2 or multiplied by
2
We plot and connect (โ2, 0), (โ1.5, โ2), (โ0.5, โ2), (1, 3),
and (2.5, 0).
y
4
2
(1.5, 2)
4
4
(5, 0)
(4, 0)
6 4 2
2
2
4
6
x
(3, 2)
(1, 2)
4
y
(2.5, 0)
(2, 0)
4
2
66. The graph is re๏ฌected across the x-axis so each
y-coordinate is replaced by its opposite.
(1, 3)
2
g(x) f (x )
4
(2, 3)
(1, 2)
x
(3, 2)
g(x) f (x )
4
2
(0.5, 2)
(4, 0)
g(x) f (2x)
6 4 2
(5, 0)
2
4
6
x
2
63. The graph is shifted right 1 unit so each x-coordinate is
increased by 1. The graph is also re๏ฌected across the xaxis, shrunk vertically by a factor of 2, and shifted up 3
1
units. Thus, each y-coordinate is multiplied by โ and
2
then increased by 3. We plot and connect (โ3, 3), (โ2, 4),
(0, 4), (3, 1.5), and (6, 3).
y
6
(โซบ2, 4)
(โซบ3, 3)
(9, 1)
โซบ2
โซบ4
4
y
4
(3, 3)
(2, 1) 2
(3, 0)
10 8 6 4 2
(3, 1.5)
2
67. The graph is shifted left 2 units so each x-coordinate is
decreased by 2. It is also re๏ฌected across the x-axis so each
y-coordinate is replaced with its opposite. In addition, the
graph is shifted up 1 unit, so each y-coordinate is then
increased by 1.
h(x) g (x 2) 1
(0, 4)
(1, 3) (6, 3)
2
โซบ4 โซบ2
(2, 3)
4
(1, 0)
2
4
6
x
2
6 x
(7, 3) (4, 3) 4 (0, 3)
g (x) โซฝ โซบqf (x โซบ 1) โซน 3
(5, 5)
6
64. The graph is shifted left 1 unit so each x-coordinate is
decreased by 1. The graph is also re๏ฌected across the
x-axis, stretched vertically by a factor of 3, and shifted
down 4 units. Thus, each y-coordinate is multiplied by โ3
and then decreased by 4. We plot and connect (โ5, โ4),
(โ4, 2), (โ2, 2), (1, โ13), and (4, โ4).
68. The graph is re๏ฌected across the y-axis so each
x-coordinate is replaced with its opposite. It is also shrunk
1
vertically by a factor of , so each y-coordinate is multi2
1
plied by (or divided by 2).
2
y
(7, 3)
4 (0, 0)
(2, 2)
2
8 6 4 2
(5, 1)
2
1
(1, )
2 4
c 2016 Pearson Education, Inc.
Copyright
(2, 2)
(5, 2)
(7, 0)
2 4
6
8 x
1
(1, )
2
1
h(x) g(x)
2
Exercise Set 2.5
91
69. The graph is shrunk horizontally. The x-coordinates of
y = h(x) are one-half the corresponding x-coordinates of
y = g(x).
y
4
7
2
(,
2 0)
7
6
5
(,
2 4)
by a factor of 3 that is, each y-coordinate is multiplied
1
and then shifted down 3 units. This is graph (e).
by
3
1
77. g(x) = f (x + 2)
3
The graph of g(x)
is the graph of f (x) shrunk vertically
h(x) g (2x )
8
(1, 4)
(1, 4)
(,
2 6)
1
(,
2 1)
4 2
1
2
(,
2 1)
4
2
4
by a factor of 3 that is, each y-coordinate is multiplied
1
and then shifted left 2 units. This is graph (c).
by
3
78. g(x) = โf (x + 2)
x
5
(,
2 2)
(0, 0)
70. The graph is shifted right 1 unit, so each x-coordinate is
increased by 1. It is also stretched vertically by a factor
of 2, so each y-coordinate is multiplied by 2 or divided
1
. In addition, the graph is shifted down 3 units, so
by
2
each y-coordinate is decreased by 3.
y
10
(8, 9)
6
(4, 5)
(0, 1) 2
(6, 3)
4
6
8
2
4
(1, 3)
6
8
x
(2, 1)
1
1
(โx)4 + (โx)3 โ 81(โx)2 โ 17 =
4
5
83. The graph of f (x) = x3 โ 3×2 is shifted left 1 unit. A
formula for the transformed function is k(x) = f (x + 1),
or k(x) = (x + 1)3 โ 3(x + 1)2 .
(6, 7)
71. g(x) = f (โx) + 3
The graph of g(x) is the graph of f (x) re๏ฌected across the
y-axis and shifted up 3 units. This is graph (f).
72. g(x) = f (x) + 3
The graph of g(x) is the graph of f (x) shifted up 3 units.
This is graph (h).
73. g(x) = โf (x) + 3
The graph of g(x) is the graph of f (x) re๏ฌected across the
x-axis and shifted up 3 units. This is graph (f).
74. g(x) = โf (โx)
The graph of g(x) is the graph of f (x) re๏ฌected across the
x-axis and the y-axis. This is graph (a).
1
f (x โ 2)
3
The graph of g(x)
is the graph of f (x) shrunk vertically
by a factor of 3 that is, each y-coordinate is multiplied
1
and then shifted right 2 units. This is graph (d).
by
3
75. g(x) =
80. f (โx) =
82. Each y-coordinate of the graph of f (x) = x3 โ 3×2 is mul1
tiplied by . A formula for the transformed function is
2
1
1
h(x) = f (x), or h(x) = (x3 โ 3×2 ).
2
2
2
6 4 2
79. f (โx) = 2(โx)4 โ 35(โx)3 + 3(โx) โ 5 =
2×4 + 35×3 โ 3x โ 5 = g(x)
81. The graph of f (x) = x3 โ 3×2 is shifted up 2 units. A
formula for the transformed function is g(x) = f (x) + 2,
or g(x) = x3 โ 3×2 + 2.
(3, 5)
4
The graph of g(x) is the graph f (x) re๏ฌected across the
x-axis and shifted left 2 units. This is graph (b).
1 4 1 3
x โ x โ 81×2 โ 17 = g(x)
4
5
h(x) 2g (x 1) 3
8
(1, 5)
1
f (x) โ 3
3
The graph of g(x)
is the graph of f (x) shrunk vertically
76. g(x) =
84. The graph of f (x) = x3 โ 3×2 is shifted right 2 units and
up 1 unit. A formula for the transformed function is
t(x) = f (x โ 2) + 1, or t(x) = (x โ 2)3 โ 3(x โ 2)2 + 1.
85. Test for symmetry with respect to the x-axis.
y = 3×4 โ 3 Original equation
โy = 3×4 โ 3
Replacing y by โy
y = โ3x + 3 Simplifying
The last equation is not equivalent to the original equation,
so the graph is not symmetric with respect to the x-axis.
4
Test for symmetry with respect to the y-axis.
Original equation
y = 3×4 โ 3
y = 3(โx)4 โ 3 Replacing x by โx
Simplifying
y = 3×4 โ 3
The last equation is equivalent to the original equation, so
the graph is symmetric with respect to the y-axis.
Test for symmetry with respect to the origin:
y = 3×4 โ 3
โy = 3(โx)4 โ 3 Replacing x by โx and
y by โy
โy = 3×4 โ 3
y = โ3×4 + 3
c 2016 Pearson Education, Inc.
Copyright
Simplifying
92
Chapter 2: More on Functions
The last equation is not equivalent to the original equation,
so the graph is not symmetric with respect to the origin.
86. Test for symmetry with respect to the x-axis.
y 2 = x Original equation
(โy)2 = x Replacing y by โy
1828
y 2 = x Simplifying
The last equation is equivalent to the original equation, so
the graph is symmetric with respect to the x-axis.
Test for symmetry with respect to the y-axis:
y2 = x
Original equation
2ยท
=
n
โ 418
Solve.
1828 = 2n โ 418
2246 = 2n
1123 = n
Check. 2 ยท 1123 = 2246 and 2246 โ 418 = 1828. This is
the number of guns found in 2013, so the answer checks.
y 2 = โx Replacing x by โx
The last equation is not equivalent to the original equation,
so the graph is not symmetric with respect to the y-axis.
Test for symmetry with respect to the origin:
y2 = x
Translate.
Number of
number of guns
guns found was twice
less 418.
found in 2010
in 2013
๏ฃฆ
๏ฃฆ
๏ฃฆ
๏ฃฆ
๏ฃฆ ๏ฃฆ
Original equation
(โy)2 = โx Replacing x by โx and
y by โy
y 2 = โx Simplifying
State. In 2010, 1123 guns were found with airline travelers.
90. Let a = the total number of acres of pumpkins harvested
in Michigan, Ohio, and Illinois in 2012.
Solve: 0.545a = 16, 200
a โ 29, 700 acres
The last equation is not equivalent to the original equation,
so the graph is not symmetric with respect to the origin.
91. Each point for which f (x) < 0 is re๏ฌected across the x-axis.
y
87. Test for symmetry with respect to the x-axis:
4
2x โ 5y = 0 Original equation
(1, 2)
(1, 2)
2
(4, 0)
2x โ 5(โy) = 0 Replacing y by โy
4 2
2x + 5y = 0 Simplifying
2
(4, 2)
4
x
2
The last equation is not equivalent to the original equation,
so the graph is not symmetric with respect to the x-axis.
|f(x)|
4
Test for symmetry with respect to the y-axis:
2x โ 5y = 0 Original equation
92. The graph of y = f (|x|) consists of the points of y = f (x)
for which x โฅ 0 along with their re๏ฌections across the
y-axis.
2(โx) โ 5y = 0 Replacing x by โx
โ2x โ 5y = 0 Simplifying
The last equation is not equivalent to the original equation,
so the graph is not symmetric with respect to the y-axis.
Test for symmetry with respect to the origin:
2x โ 5y = 0 Original equation
2(โx) โ 5(โy) = 0 Replacing x by โx and
y by โy
โ2x + 5y = 0
y
4
(1, 2)
(4, 2)
2
4 2
(1, 2)
2
(4, 2)
4
x
2
f(|x|)
4
2x โ 5y = 0 Simplifying
The last equation is equivalent to the original equation, so
the graph is symmetric with respect to the origin.
93. The graph of y = g(|x|) consists of the points of y = g(x)
for which x โฅ 0 along with their re๏ฌections across the
y-axis.
88. Let p = the number of pages of federal tax rules in 1995.
Then the number of pages in 2014 was p + 84.2% of p, or
1.842p.
Solve: 1.842p = 74, 608
p = 40, 504 pages
y
4
(2, 1) 2
(4, 0)
(2, 1)
4 2
2
89. Familiarize. Let n = the number of guns that were found
with airline travelers in 2010.
c 2016 Pearson Education, Inc.
Copyright
(4, 0)
x
4
2
4
g(|x|)
Exercise Set 2.6
93
94. Each point for which g(x) < 0 is re๏ฌected across the x-axis.
5.
1 = kยท
y
(2, 1) 2
(4, 0)
(2, 1)
4 2
2
The variation constant is 4. The equation of variation is
y = 4x.
(4, 0)
x
4
2
4
(The graph of y = f (x โ 3) is the graph of y = f (x)
shifted right 3 units, so the point (โ1, 5) on y = f (x) is
transformed to the point (โ1 + 3, 5), or (2, 5) on
y = f (x โ 3).)
96. Call the transformed function g(x).
g(5) = 4 โ f (โ3) = 4 โ f (5 โ 8),
g(8) = 4 โ f (0) = 4 โ f (8 โ 8),
and g(11) = 4 โ f (3) = 4 โ f (11 โ 8).
Thus g(x) = 4 โ f (x โ 8), or g(x) = 4 โ |x โ 8|.
Exercise Set 2.6
y = kx
54 = k ยท 12
9
54
= k, or k =
12
2
9
The variation constant is , or 4.5. The equation of vari2
9
ation is y = x, or y = 4.5x.
2
2.
y = kx
0.1 = k(0.2)
1
= k Variation constant
2
1
Equation of variation: y = x, or y = 0.5x.
2
3.
k
x
k
3=
12
36 = k
y=
The variation constant is 36. The equation of variation is
36
.
y=
x
4.
k
x
k
12 =
5
60 = k Variation constant
y=
Equation of variation: y =
k
x
k
0.1 =
0.5
0.05 = k Variation constant
0.05
Equation of variation: y =
x
k
7.
y=
x
k
32 =
1
8
1
ยท 32 = k
8
4=k
6.
|g(x)|
95. f (2 โ 3) = f (โ1) = 5, so b = 5.
1.
1
4
4=k
4
Then
y = kx
60
x
y=
The variation constant is 4. The equation of variation is
4
y= .
x
8.
y = kx
3 = k ยท 33
1
= k Variation constant
11
1
x
Equation of variation: y =
11
9.
y = kx
3
= kยท2
4
1 3
ยท =k
2 4
3
=k
8
3
The variation constant is . The equation of variation is
8
3
y = x.
8
k
10. y =
x
1
k
=
5
35
7 = k Variation constant
7
Equation of variation: y =
x
k
11.
y=
x
k
1.8 =
0.3
0.54 = k
The variation constant is 0.54. The equation of variation
0.54
.
is y =
x
c 2016 Pearson Education, Inc.
Copyright
94
12.
Chapter 2: More on Functions
17. Let F = the number of grams of fat and w = the weight.
y = kx
F = kw
0.9 = k(0.4)
9
= k Variation constant
4
9
Equation of variation: y = x, or y = 2.25x
4
F varies directly as w.
60 = k ยท 120
60
= k, or
120
1
=k
2
13. Let W = the weekly allowance and a = the childโs age.
W = ka
1
w
2
1
F = ยท 180
2
5.50 = k ยท 6
11
=k
12
F =
11
x
12
11
ยท9
W =
12
W = $8.25
F = 90
Substituting
Solving for k
Variation constant
Equation of variation
Substituting
W =
The maximum daily fat intake for a person weighing 180
lb is 90 g.
18.
53 = k ยท 38, 040, 000 Substituting
53
=k
Variation constant
38, 040, 000
53
P
N=
38, 040, 000
53
ยท 26, 060, 000 Substituting
N =
38, 040, 000
N โ 36
14. Let S = the sales tax and p = the purchase price.
S = kp
S varies directly as p.
7.14 = k ยท 119
Substituting
0.06 = k
Variation constant
S = 0.06p
Equation of variation
S = 0.06(21)
Substituting
S โ 1.26
The sales tax is $1.26.
15.
16.
k
r
k
5=
80
400 = r
400
t=
r
400
t=
70
5
40
, or 5 hr
t=
7
7
k
W =
W varies inversely as L.
L
k
Substituting
1200 =
8
9600 = k Variation constant
t=
9600
L
9600
W =
14
W โ 686
W =
Texas has 36 representatives.
k
19. T =
T varies inversely as P .
P
k
Substituting
5=
7
35 = k
Variation constant
35
P
35
T =
10
T = 3.5
T =
Equation of variation
Substituting
It will take 10 bricklayers 3.5 hr to complete the job.
20.
Equation of variation
Substituting
A 14-m beam can support about 686 kg.
N = kP
21.
k
r
k
45 =
600
27, 000 = k
27, 000
t=
r
27, 000
t=
1000
t = 27 min
t=
d = km
d varies directly as m.
40 = k ยท 3 Substituting
40
=k
Variation constant
3
c 2016 Pearson Education, Inc.
Copyright
Exercise Set 2.6
95
40
m
Equation of variation
3
200
40
ยท5=
Substituting
d=
3
3
2
d = 66
3
26.
d=
A 5-kg mass will stretch the spring 66
22.
2
cm.
3
f = kF
y=
0.15 = k(0.1)2
f = 0.042F
f = 0.042(80)
f = 3.36
k
P varies inversely as W .
P =
W
k
Substituting
330 =
3.2
The equation of variation is y = 15×2 .
28.
550W = 1056
y=
Equation of variation
29.
y = kxz
Dividing by 550
W = 1.92
Simplifying
1=k
The equation of variation is y = xz.
kx
30. y =
z
k ยท 12
4=
15
5=k
M varies directly as E.
35.9 = k ยท 95
Substituting
35.9
=k
Variation constant
95
35.9
E
Equation of variation
M =
95
35.9
M =
ยท 100 Substituting
95
M โ 37.8
A 100-lb person would weigh about 37.8 lb on Mars.
k
25.
y= 2
x
k
0.15 =
Substituting
(0.1)2
k
0.15 =
0.01
0.15(0.01) = k
The equation of variation is y =
y=
5x
z
y = kxz 2
31.
105 = k ยท 14 ยท 52
Substituting
105 = 350k
105
=k
350
3
=k
10
3 2
The equation of variation is y =
xz .
10
xz
32. y = k ยท
w
3
2ยท3
= kยท
2
4
1=k
y=
0.0015 = k
Substituting
56 = 56k
Multiplying by W
1056
W =
550
M = kE
2 2
x
3
56 = k ยท 7 ยท 8
Substituting
A tone with a pitch of 550 vibrations per second has a
wavelength of 1.92 ft.
24.
y = kx2
6 = k ยท 32
2
=k
3
Variation constant
1056
P =
W
1056
550 =
W
Substituting
0.15 = 0.01k
0.15
=k
0.01
15 = k
0.042 = k
1056 = k
54
x2
y = kx2
27.
6.3 = k ยท 150
23.
k
x2
k
6= 2
3
54 = k
y=
xz
w
0.0015
.
x2
c 2016 Pearson Education, Inc.
Copyright
96
33.
Chapter 2: More on Functions
xz
wp
3
3 ยท 10
=k
28
7ยท8
30
3
= kยท
28
56
3 56
ยท
=k
28 30
1
=k
5
y=k
The equation of variation is d =
34.
Substitute 72 for d and ๏ฌnd r.
1 2
r
72 =
18
1296 = r2
Substituting
The equation of variation is y =
36 = r
xz
1 xz
, or
.
5 wp
5wp
xz
w2
12
16 ยท 3
= kยท 2
5
5
5
=k
4
y = kยท
A car can travel 36 mph and still stop in 72 ft.
k
38.
W = 2
d
k
220 =
(3978)2
3, 481, 386, 480 = k
3, 481, 386, 480
d2
3, 481, 386, 480
W =
(3978 + 200)2
W โ 199 lb
W =
5xz
5 xz
, or
4 w2
4w2
k
I = 2
d
k
90 = 2 Substituting
5
k
90 =
25
2250 = k
y=
35.
The equation of variation is I =
kR
I
We ๏ฌrst ๏ฌnd k.
39. E =
k ยท 89
3.75 =
213.1
213.1
=k
3.75
89
9โk
2250
.
d2
d = 7.5
The distance from 5 m to 7.5 m is 7.5 โ 5, or 2.5 m, so it is
2.5 m further to a point where the intensity is 40 W/m2 .
222 = k ยท 37.8 ยท 40
37
=k
252
37
Av
252
37
ยท 51v
430 =
252
v โ 57.4 mph
D=
37.
200 = k ยท 602
200 = 3600k
200
=k
3600
1
=k
18
40.
kT
P
k ยท 42
231 =
20
110 = k
V =
110T
P
110 ยท 30
V =
15
V = 220 cm3
Substituting
213.1
89
John Lester would have given up about 98 earned runs if
he had pitched 235 innings.
V =
d = kr2
Multiplying by
9R
.
I
Substitute 3.75 for E and 235 for I and solve for R.
9R
3.75 =
235
235
235
3.75
=R
Multiplying by
9
9
98 โ R
d2 = 56.25
D = kAv
Substituting
The equation of variation is E =
Substitute 40 for I and ๏ฌnd d.
2250
40 = 2
d
40d2 = 2250
36.
1 2
r .
18
41. parallel
42. zero
43. relative minimum
44. odd function
45. inverse variation
c 2016 Pearson Education, Inc.
Copyright
Chapter 2 Review Exercises
97
49. We are told A = kd2 , and we know A = ฯr2 so we have:
46. a) 7xy = 14
2
y=
x
Inversely
kd2 = ฯr2
kd2 = ฯ
b) x โ 2y = 12
x
y = โ6
2
Neither
d 2
d
2
r=
2
ฯd2
4
ฯ
k=
4
kd2 =
c) โ2x + y = 0
Variation constant
Chapter 2 Review Exercises
y = 2x
Directly
1. This statement is true by the de๏ฌnition of the greatest
integer function.
3
d) x = y
4
4
y= x
3
Directly
2. This statement is false. See Example 3 in Section 2.3 in
the text.
3. The graph of y = f (x โ d) is the graph of y = f (x) shifted
right d units, so the statement is true.
e) x = 2
y
1
y= x
2
Directly
4. The graph of y = โf (x) is the re๏ฌection of the graph of
y = f (x) across the x-axis, so the statement is true.
47. Let V represent the volume and p represent the price of a
jar of peanut butter.
V = kp
ฯ
5. a) For x-values from โ4 to โ2, the y-values increase
from 1 to 4. Thus the function is increasing on the
interval (โ4, โ2).
b) For x-values from 2 to 5, the y-values decrease from
4 to 3. Thus the function is decreasing on the interval (2, 5).
V varies directly as p.
2
3
(5) = k(2.89) Substituting
2
3.89ฯ โ k
Variation constant
V = 3.89ฯp
2
ฯ(1.625) (5.25) = 3.89ฯp
Equation of variation
Substituting
3.56 โ p
If cost is directly proportional to volume, the larger jar
should cost $3.56.
c) For x-values from โ2 to 2, y is 4. Thus the function
is constant on the interval (โ2, 2).
6. a) For x-values from โ1 to 0, the y-values increase from
3 to 4. Also, for x-values from 2 to โ, the y-values
increase from 0 to โ. Thus the function is increasing on the intervals (โ1, 0), and (2, โ).
b) For x-values from 0 to 2, the y-values decrease from
4 to 0. Thus, the function is decreasing on the interval (0, 2).
Now let W represent the weight and p represent the price
of a jar of peanut butter.
c) For x-values from โโ to โ1, y is 3. Thus the function is constant on the interval (โโ, โ1).
W = kp
18 = k(2.89)
Substituting
7.
6.23 โ k
Variation constant
W = 6.23p
Equation of variation
5
22 = 6.23p
Substituting
3
y
4
2
3.53 = p
1
If cost is directly proportional to weight, the larger jar
should cost $3.53.
kp2
48. Q = 3
q
Q varies directly as the square of p and inversely as the
cube of q.
โ5 โ4 โ3 โ2 โ1
โ1
โ2
1
2
3
4
5
x
f (x ) = x 2 โ 1
โ3
โ4
โ5
The function is increasing on (0, โ) and decreasing on
(โโ, 0). We estimate that the minimum value is โ1 at
x = 0. There are no maxima.
c 2016 Pearson Education, Inc.
Copyright
98
Chapter 2: More on Functions
8.
y
y
4
5
4
2
f (x ) = 2 โ | x |
3
4 2
2
1
โ5 โ4 โ3 โ2 โ1
โ1
2
4
x
2
1
2
3
4
5
x
4
โ2
โ3
โ4
โ5
The function is increasing on (โโ, 0) and decreasing on
(0, โ). We estimate that the maximum value is 2 at x = 0.
There are no minima.
9. If two sides of the patio are each x feet, then the remaining
side will be (48 โ 2x) ft. We use the formula Area =
length ร width.
๏ฃฑ 3
for x 2
We create the graph in three parts. Graph f (x) = x3 for
inputs less than โ2. Then graph f (x) = |x| for inputs
greater than or equal โto โ2 and less than or equal to 2.
Finally graph f (x) = x โ 1 for inputs greater than 2.
y
A(x) = x(48 โ 2x), or 48x โ 2×2
2
10. The length of the rectangle is 2x. The width is the second
coordinate of the point (x, y) on the circle. The circle has
center (0,โ0) and radius 2, so its equation is x2 + y 2 = 4
and y = 4 โโx2 . Thus the area of the rectangle is given
by A(x) = 2x 4 โ x2 .
4 2
108 = x2 h
108
=h
x2
Now ๏ฌnd the surface area.
S = x2 + 4 ยท x ยท h
108
S(x) = x2 + 4 ยท x ยท 2
x
432
S(x) = x2 +
x
b) x must be positive, so the domain is (0, โ).
c) From the graph, we see that the minimum value of
the function occurs when x = 6 in. For this value
of x,
108
108
108
= 3 in.
h= 2 = 2 =
x
6
36
๏ฃฑ
for x โค โ4,
๏ฃฒ โx,
12. f (x) = 1 x + 1, for x > โ4
๏ฃณ2
4
x
4
6
8
11. a) Let h = the height of the box. Since the volume is
108 in3 , we have:
108 = x ยท x ยท h
2
2
10
12
๏ฃฑ 2
๏ฃฒ x โ 1 , for x = โ1,
14. f (x) =
x+1
๏ฃณ
3,
for x = โ1
x2 โ 1
x+1
for all inputs except โ1. Then graph f (x) = 3 for x = โ1.
We create the graph in two parts. Graph f (x) =
y
4
2
4 2
2
4
x
4
15. f (x) = [[x]]. See Example 9 in Section 2.1 of the text.
We create the graph in two parts. Graph f (x) = โx for
1
inputs less than or equal to โ4. Then graph f (x) = x + 1
2
for inputs greater than โ4.
16. f (x) = [[x โ 3]]
This function could be de๏ฌned by a piecewise function with
an in๏ฌnite number of statements.
c 2016 Pearson Education, Inc.
Copyright
Chapter 2 Review Exercises
๏ฃฑ
๏ฃด
๏ฃด
.
๏ฃด
๏ฃด
๏ฃด
๏ฃด
.
๏ฃด
๏ฃด
๏ฃด
๏ฃด
.
๏ฃด
๏ฃด
๏ฃด
๏ฃด
โ4,
๏ฃด
๏ฃด
๏ฃฒ โ3,
f (x) = โ2,
๏ฃด
๏ฃด
๏ฃด
๏ฃด โ1,
๏ฃด
๏ฃด
๏ฃด
๏ฃด
.
๏ฃด
๏ฃด
๏ฃด
๏ฃด
.
๏ฃด
๏ฃด
๏ฃด
๏ฃด.
๏ฃณ
99
4
, g(x) = 3 โ 2x
x2
a) Division by zero is unde๏ฌned, so the domain of f is
{x|x = 0}, or (โโ, 0) โช (0, โ). The domain of g is
the set of all real numbers, or (โโ, โ).
22. f (x) =
for โ1 โค x < 0,
for 0 โค x < 1,
for 1 โค x < 2,
for 2 โค x < 3,
๏ฃฑ 3
for x 2
Since โ1 is in the interval [โ2, 2], f (โ1) = | โ 1| = 1.
โ
โ
Since 5 > 2, f (5) = 5 โ 1 = 4 = 2.
Since โ2 is in the interval [โ2, 2], f (โ2) = | โ 2| = 2.
Since โ3 0, the graph of y = f (x)+b is the graph of y = f (x)
shifted up b units. Answer C is correct.
1
74. The graph of g(x) = โ f (x) + 1 is the graph of y = f (x)
2
1
shrunk vertically by a factor of , then re๏ฌected across the
2
x-axis, and shifted up 1 unit. The correct graph is B.
c) For x-values from โ2 to 2, y is 2. Thus the function
is constant on the interval (โ2, 2).
2.
y
5
4
f (x ) = 2 โ x 2
3
75. Let f (x) and g(x) be odd functions. Then by de๏ฌnition,
f (โx) = โf (x), or f (x) = โf (โx), and g(โx) = โg(x),
or g(x) = โg(โx). Thus (f + g)(x) = f (x) + g(x) =
โf (โx) + [โg(โx)] = โ[f (โx) + g(โx)] = โ(f + g)(โx)
and f + g is odd.
2
1
โ5 โ4 โ3 โ2 โ1
โ1
1
2
3
4
5
x
โ2
โ3
โ4
76. Re๏ฌect the graph of y = f (x) across the x-axis and then
across the y-axis.
77. f (x) = 4×3 โ 2x + 7
a) f (x) + 2 = 4x โ 2x + 7 + 2 = 4x โ 2x + 9
3
3
b) f (x + 2) = 4(x + 2)3 โ 2(x + 2) + 7
= 4(x3 + 6×2 + 12x + 8) โ 2(x + 2) + 7
= 4×3 + 24×2 + 48x + 32 โ 2x โ 4 + 7
= 4×3 + 24×2 + 46x + 35
c) f (x) + f (2) = 4×3 โ 2x + 7 + 4 ยท 23 โ 2 ยท 2 + 7
= 4×3 โ 2x + 7 + 32 โ 4 + 7
โ5
The function is increasing on (โโ, 0) and decreasing on
(0, โ). The relative maximum is 2 at x = 0. There are no
minima.
3. If b = the length of the base, in inches, then the height =
4b โ 6. We use the formula for the area of a triangle,
1
A = bh.
2
1
A(b) = b(4b โ 6), or
2
A(b) = 2b2 โ 3b
= 4×3 โ 2x + 42
f (x) + 2 adds 2 to each function value; f (x + 2) adds 2 to
each input before the function value is found; f (x) + f (2)
adds the output for 2 to the output for x.
c 2016 Pearson Education, Inc.
Copyright
Chapter 2 Test
105
๏ฃฑ 2
for x 1
1
x+4
2
1
1
1
f (x + h) = (x + h) + 4 = x + h + 4
2
2
2
1
1
1
x+ h+4โ
x+4
f (x + h) โ f (x)
2
2
2
=
h
h
1
1
1
x+ h+4โ xโ4
2
2
2
=
h
1
h
1
1
1 h
1
= 2 = hยท = ยท =
h
2
h
2 h
2
20. f (x) =
y
4
2
4
2
2
4
x
2
4
7 7
7
7
โค 1, f โ
= โ = .
8
8
8
8
โ
โ
Since 5 > 1, f (5) = 5 โ 1 = 4 = 2.
5. Since โ1 โค โ
21. f (x) = 2×2 โ x + 3
f (x+h) = 2(x+h)2 โ(x+h)+3 = 2(x2 +2xh+h2 )โx โ h+3 =
2×2 + 4xh + 2h2 โ x โ h + 3
Since โ4 < โ1, f (โ4) = (โ4)2 = 16.
6.
7.
8.
2×2+4xh+2h2โxโh + 3โ(2×2 โx + 3)
f (x+h)โf (x)
=
h
h
(f + g)(โ6) = f (โ6) + g(โ6) =
(โ6)2 โ 4(โ6) + 3 + 3 โ (โ6) =
โ
โ
36 + 24 + 3 + 3 + 6 = 63 + 9 = 63 + 3 = 66
(f โ g)(โ1) = f (โ1) โ g(โ1) =
(โ1)2 โ 4(โ1) + 3 โ 3 โ (โ1) =
โ
โ
1+4+3โ 3+1=8โ 4=8โ2=6
โ
(f g)(2) = f (2) ยท g(2) = (22 โ 4 ยท 2 + 3)( 3 โ 2) =
โ
(4 โ 8 + 3)( 1) = โ1 ยท 1 = โ1
=
4xh + 2h2 โ h
h
h/(4x + 2h โ 1)
h/
= 4x + 2h โ 1
22.
1 โ4ยท1+3
f (1)
1โ4+3
0
โ
โ
=
=
= โ =0
g(1)
3โ1
2
2
10. Any real number can be an input for f (x) = x2 , so the
domain is the set of real numbers, or (โโ, โ).
โ
11. The domain of g(x) = x โ 3 is the set of real numbers for
which x โ 3 โฅ 0, or x โฅ 3. Thus the domain is {x|x โฅ 3},
or [3, โ).
2×2 +4xh+2h2โxโh+3โ2×2 +xโ3
h
=
2
9. (f /g)(1) =
=
(g โฆ h)(2) = g(h(2)) = g(3 ยท 22 + 2 ยท 2 + 4) =
g(3 ยท 4 + 4 + 4) = g(12 + 4 + 4) = g(20) = 4 ยท 20 + 3 =
80 + 3 = 83
23.
(f โฆ g)(โ1) = f (g(โ1)) = f (4(โ1) + 3) = f (โ4 + 3) =
f (โ1) = (โ1)2 โ 1 = 1 โ 1 = 0
24.
(h โฆ f )(1) = h(f (1)) = h(12 โ 1) = h(1 โ 1) = h(0) =
3 ยท 02 + 2 ยท 0 + 4 = 0 + 0 + 4 = 4
(g โฆ g)(x) = g(g(x)) = g(4x + 3) = 4(4x + 3) + 3 =
12. The domain of f + g is the intersection of the domains of
f and g. This is {x|x โฅ 3}, or [3, โ).
25.
13. The domain of f โ g is the intersection of the domains of
f and g. This is {x|x โฅ 3}, or [3, โ).
โ
26. (f โฆ g)(x) = f (g(x)) = f (x2 + 1) = x2 + 1 โ 5 =
โ
x2 โ 4
โ
โ
(g โฆ f )(x) = g(f (x)) = g( x โ 5) = ( x โ 5)2 + 1 =
14. The domain of f g is the intersection of the domains of f
and g. This is {x|x โฅ 3}, or [3, โ).
15. The domain of f /g is the intersection of the domains of f
and g, excluding those x-values for which g(x) = 0. Since
x โ 3 = 0 when x = 3, the domain is (3, โ).
โ
16. (f + g)(x) = f (x) + g(x) = x2 + x โ 3
โ
17. (f โ g)(x) = f (x) โ g(x) = x2 โ x โ 3
โ
18. (f g)(x) = f (x) ยท g(x) = x2 x โ 3
x2
f (x)
=โ
19. (f /g)(x) =
g(x)
xโ3
16x + 12 + 3 = 16x + 15
xโ5+1=xโ4
27. The inputs for f (x) must be such that x โ 5 โฅ 0, or x โฅ 5.
Then for (f โฆg)(x) we must have g(x) โฅ 5, or x2 +1 โฅ 5, or
x2 โฅ 4. Then the domain of (f โฆg)(x) is (โโ, โ2]โช[2, โ).
Since we can substitute any real number for x in g, the
domain of (g โฆ f )(x) is the same as the domain of f (x),
[5, โ).
28. Answers may vary. f (x) = x4 , g(x) = 2x โ 7
c 2016 Pearson Education, Inc.
Copyright
106
Chapter 2: More on Functions
29. y = x4 โ 2×2
34.
Replace y with โy to test for symmetry with respect to
the x-axis.
โy = x4 โ 2×2
y = โx4 + 2×2
The resulting equation is not equivalent to the original
equation, so the graph is not symmetric with respect to
the x-axis.
Variation constant
Equation of variation: y =
35.
30
x
y = kx
60 = k ยท 12
Replace x with โx to test for symmetry with respect to
the y-axis.
5=k
y = (โx)4 โ 2(โx)2
Variation constant
Equation of variation: y = 5x
y = x โ 2x
4
k
x
k
5=
6
30 = k
y=
2
The resulting equation is equivalent to the original equation, so the graph is symmetric with respect to the y-axis.
36.
Replace x with โx and y with โy to test for symmetry
with respect to the origin.
โy = (โx)4 โ 2(โx)2
kxz 2
w
k(0.1)(10)2
100 =
5
100 = 2k
y=
50 = k
โy = x4 โ 2×2
y=
y = โx4 + 2×2
The resulting equation is not equivalent to the original
equation, so the graph is not symmetric with respect to
the origin.
2x
x2 + 1
2x
2(โx)
=โ 2
f (โx) =
(โx)2 + 1
x +1
f (x) = f (โx), so f is not even.
2x
โf (x) = โ 2
x +1
f (โx) = โf (x), so f is odd.
30. f (x) =
37.
50xz
w
Variation constant
2
Equation of variation
d = kr2
200 = k ยท 602
1
=k
18
1 2
r
d=
18
1
ยท 302
d=
18
d = 50 ft
Variation constant
Equation of variation
38. The graph of g(x) = 2f (x) โ 1 is the graph of y = f (x)
stretched vertically by a factor of 2 and shifted down 1 unit.
The correct graph is C.
31. Shape: h(x) = x2
Shift h(x) right 2 units: g(x) = h(x โ 2) = (x โ 2)2
Shift g(x) down 1 unit: f (x) = (x โ 2) โ 1
2
32. Shape: h(x) = x2
39. Each x-coordinate on the graph of y = f (x) is divided
by
โ3
, 1 , or
3 on the graph of y = f (3x). Thus the point
3
(โ1, 1) is on the graph of f (3x).
Shift h(x) left 2 units: g(x) = h(x + 2) = (x + 2)2
Shift g(x) down 3 units: f (x) = (x + 2)2 โ 3
1
33. Each y-coordinate is multiplied by โ . We plot and con2
nect (โ5, 1), (โ3, โ2), (1, 2) and (4, โ1).
c 2016 Pearson Education, Inc.
Copyright

## Document Preview (50 of 727 Pages)

You are viewing preview pages of the document. Purchase to get full access instantly.

-37%

### Solution Manual for Algebra and Trigonometry, 5th Edition

$18.99 ~~$29.99~~Save:$11.00(37%)

24/7 Live Chat

Instant Download

100% Confidential

Store

##### Henry Lewis

0 (0 Reviews)

## Best Selling

The World Of Customer Service, 3rd Edition Test Bank

$18.99 ~~$29.99~~Save:$11.00(37%)

Chemistry: Principles And Reactions, 7th Edition Test Bank

$18.99 ~~$29.99~~Save:$11.00(37%)

2023-2024 ATI Pediatrics Proctored Exam with Answers (139 Solved Questions)

$18.99 ~~$29.99~~Save:$11.00(37%)

Solution Manual for Designing the User Interface: Strategies for Effective Human-Computer Interaction, 6th Edition

$18.99 ~~$29.99~~Save:$11.00(37%)

Test Bank for Hospitality Facilities Management and Design, 4th Edition

$18.99 ~~$29.99~~Save:$11.00(37%)

Data Structures and Other Objects Using C++ 4th Edition Solution Manual

$18.99 ~~$29.99~~Save:$11.00(37%)