College algebra 7th edition by blitzer solution manual

Chapter 2
Functions and Graphs
Copyright © 2018 Pearson Education, Inc. 201
Section 2.1
Check Point Exercises
1. The domain is the set of all first components: {0, 10,
20, 30, 42}. The range is the set of all second
components: {9.1, 6.7, 10.7, 13.2, 21.7}.
2. a. The relation is not a function since the two
ordered pairs (5, 6) and (5, 8) have the same
first component but different second
components.
b. The relation is a function since no two ordered
pairs have the same first component and
different second components.
3. a. 2 6
6 2
x y
y x
+ =
= −
For each value of x, there is one and only one
value for y, so the equation defines y as a
function of x.
b. 2 2
2 2
2
1
1
1
x y
y x
y x
+ =
= −
= ± −
Since there are values of x (all values between –
1 and 1 exclusive) that give more than one
value for y (for example, if x = 0, then
2
1 0 1y = ± − = ± ), the equation does not
define y as a function of x.
4. a. 2
( 5) ( 5) 2( 5) 7
25 ( 10) 7
42
f − = − − − +
= − − +
=
b. 2
2
2
( 4) ( 4) 2( 4) 7
8 16 2 8 7
6 15
f x x x
x x x
x x
+ = + − + +
= + + − − +
= + +
c. 2
2
2
( ) ( ) 2( ) 7
( 2 ) 7
2 7
f x x x
x x
x x
− = − − − +
= − − +
= + +
5. x ( ) 2f x x= ( ),x y
-2 –4 ( )2, 4− −
-1 –2 ( )1, 2− −
0 0 ( )0,0
1 2 ( )1,2
2 4 ( )2,4
x ( ) 2 3g x x= − ( ),x y
-2 ( ) 2( 2) 32 7g − − =− = − ( )2, 7− −
-1 ( ) 2( 1) 31 5g − −− = = − ( )1, 5− −
0 ( ) 2(0) 30 3g −= = − ( )0, 3−
1 ( ) 2(1) 31 1g −= = − ( )1, 1−
2 ( ) 2(2) 32 1g −= = ( )2,1
The graph of g is the graph of f shifted down 3
units.

Chapter 2 Functions and Graphs
202 Copyright © 2018 Pearson Education, Inc.
6. The graph (a) passes the vertical line test and is
therefore is a function.
The graph (b) fails the vertical line test and is
therefore not a function.
The graph (c) passes the vertical line test and is
therefore is a function.
The graph (d) fails the vertical line test and is
therefore not a function.
7. a. (5) 400f =
b. 9x = , (9) 100f =
c. The minimum T cell count in the asymptomatic
stage is approximately 425.
8. a. domain: { } [ ]2 1 or 2,1 .x x− ≤ ≤ −
range: { } [ ]0 3 or 0,3 .y y≤ ≤
b. domain: { } ( ]2 1 or 2,1 .x x− < ≤ −
range: { } [ )1 2 or 1,2 .y y− ≤ < −
c. domain: { } [ )3 0 or 3,0 .x x− ≤ < −
range: { }3, 2, 1 .y y = − − −
Concept and Vocabulary Check 2.1
1. relation; domain; range
2. function
3. f; x
4. true
5. false
6. x; 6x +
7. ordered pairs
8. more than once; function
9. [0,3) ; domain
10. [1, )∞ ; range
11. 0; 0; zeros
12. false
Exercise Set 2.1
1. The relation is a function since no two ordered pairs
have the same first component and different second
components. The domain is {1, 3, 5} and the range is
{2, 4, 5}.
2. The relation is a function because no two ordered
pairs have the same first component and different
second components The domain is {4, 6, 8} and the
range is {5, 7, 8}.
3. The relation is not a function since the two ordered
pairs (3, 4) and (3, 5) have the same first component
but different second components (the same could be
said for the ordered pairs (4, 4) and (4, 5)). The
domain is {3, 4} and the range is {4, 5}.
4. The relation is not a function since the two ordered
pairs (5, 6) and (5, 7) have the same first component
but different second components (the same could be
said for the ordered pairs (6, 6) and (6, 7)). The
domain is {5, 6} and the range is {6, 7}.
second components The domain is
{3, 4, 5, 7} and the range is {–2, 1, 9}.
second components The domain is
{–2, –1, 5, 10} and the range is {1, 4, 6}.
7. The relation is a function since there are no same first
components with different second components. The
domain is {–3, –2, –1, 0} and the range is {–3, –2, –
1, 0}.
8. The relation is a function since there are no ordered
pairs that have the same first component but different
second components. The domain is {–7, –5, –3, 0}
and the range is {–7, –5, –3, 0}.
9. The relation is not a function since there are ordered
pairs with the same first component and different
second components. The domain is {1} and the range
is {4, 5, 6}.
10. The relation is a function since there are no two
ordered pairs that have the same first component and
different second components. The domain is
{4, 5, 6} and the range is {1}.

Section 2.1 Basics of Functions and Their Graphs
11. 16
16
x y
y x
+ =
= −
Since only one value of y can be obtained for each
value of x, y is a function of x.
12. 25
25
x y
y x
+ =
= −
13. 2
2
16
16
x y
y x
+ =
= −
14. 2
2
25
25
x y
y x
+ =
= −
15. 2 2
2 2
2
16
16
16
x y
y x
y x
+ =
= −
= ± −
If x = 0, 4.y = ±
Since two values, y = 4 and y = – 4, can be obtained
for one value of x, y is not a function of x.
16. 2 2
2 2
2
25
25
25
If 0, 5.
x y
y x
y x
x y
+ =
= −
= ± −
= = ±
Since two values, y = 5 and y = –5, can be obtained
for one value of x, y is not a function of x.
17. 2
x y
y x
=
= ±
If x = 1, 1.y = ±
for x = 1, y is not a function of x.
18. 4 2
4 2
If 1, then 2.
x y
y x x
x y
=
= ± = ±
= = ±
for x = 1, y is not a function of x.
19. 4y x= +
20. 4y x= − +
21. 3
3
3
8
8
8
x y
y x
y x
+ =
= −
= −
22.
3
3
3
27
27
27
x y
y x
y x
+ =
= −
= −
23. 2 1xy y+ =
( )2 1
1
2
y x
y
x
+ =
=
+
24. 5 1xy y− =
( )5 1
1
5
y x
y
x
− =
=
−
25. 2x y− =
2
2
y x
y x
− = − +
= −
26. 5x y− =
5
5
y x
y x
− = − +
= −

27. a. f(6) = 4(6) + 5 = 29
b. f(x + 1) = 4(x + 1) + 5 = 4x + 9
c. f(–x) = 4(–x) + 5 = – 4x + 5
28. a. f(4) = 3(4) + 7 = 19
b. f(x + 1) = 3(x + 1) + 7 = 3x + 10
c. f(–x) = 3(–x) + 7 = –3x + 7
29. a. 2
( 1) ( 1) 2( 1) 3
1 2 3
2
g − = − + − +
= − +
=
b. 2
2
2
( 5) ( 5) 2( 5) 3
10 25 2 10 3
12 38
g x x x
x x x
x x
+ = + + + +
= + + + + +
= + +
c. 2
2
( ) ( ) 2( ) 3
2 3
g x x x
x x
− = − + − +
= − +
30. a. 2
( 1) ( 1) 10( 1) 3
1 10 3
8
g − = − − − −
= + −
=
b. 2
2
2
( 2) ( 2) 10(8 2) 3
4 4 10 20 3
6 19
g x x
x x x
x x
+ = + − + −
= + + − − −
= − −
c. 2
2
( ) ( ) 10( ) 3
10 3
g x x x
x x
− = − − − −
= + −
31. a. 4 2
(2) 2 2 1
16 4 1
13
h = − +
= − +
=
b. 4 2
( 1) ( 1) ( 1) 1
1 1 1
1
h − = − − − +
= − +
=
c. 4 2 4 2
( ) ( ) ( ) 1 1h x x x x x− = − − − + = − +
d. 4 2
4 2
(3 ) (3 ) (3 ) 1
81 9 1
h a a a
a a
= − +
= − +
32. a. 3
(3) 3 3 1 25h = − + =
b. 3
( 2) ( 2) ( 2) 1
8 2 1
5
h − = − − − +
= − + +
= −
c. 3 3
( ) ( ) ( ) 1 1h x x x x x− = − − − + = − + +
d. 3
3
(3 ) (3 ) (3 ) 1
27 3 1
h a a a
a a
= − +
= − +
33. a. ( 6) 6 6 3 0 3 3f − = − + + = + =
b. (10) 10 6 3
16 3
4 3
7
f = + +
= +
= +
=
c. ( 6) 6 6 3 3f x x x− = − + + = +
34. a. (16) 25 16 6 9 6 3 6 3f = − − = − = − = −
b. ( 24) 25 ( 24) 6
49 6
7 6 1
f − = − − −
= −
= − =
c. (25 2 ) 25 (25 2 ) 6
2 6
f x x
x
− = − − −
= −
35. a.
2
2
4(2) 1 15
(2)
42
f
−
= =
b.
2
2
4( 2) 1 15
( 2)
4( 2)
f
− −
− = =
−
c.
2 2
2 2
4( ) 1 4 1
( )
( )
x x
f x
x x
− − −
− = =
−
36. a.
3
3
4(2) 1 33
(2)
82
f
+
= =
b.
3
3
4( 2) 1 31 31
( 2)
8 8( 2)
f
− + −
− = = =
−−
c.
3 3
3 3
4( ) 1 4 1
( )
( )
x x
f x
x x
− + − +
− = =
− −
3
3
4 1
or
x
x
−

37. a.
6
(6) 1
6
f = =
b.
6 6
( 6) 1
6 6
f
− −
− = = = −
−
c.
2 2
2
22
( ) 1
r r
f r
rr
= = =
38. a.
5 3 8
(5) 1
5 3 8
f
+
= = =
+
b.
5 3 2 2
( 5) 1
5 3 2 2
f
− + −
− = = = = −
− + − −
c.
9 3
( 9 )
9 3
x
f x
x
− − +
− − =
− − +
6 1, if 6
1,if 66
x x
xx
− − < −
= = 
− > −− − 
39. x ( )f x x= ( ),x y
−2 ( )2 2f − = − ( )2, 2− −
−1 ( )1 1f − = − ( )1, 1− −
0 ( )0 0f = ( )0,0
1 ( )1 1f = ( )1,1
2 ( )2 2f = ( )2,2
x ( ) 3g x x= + ( ),x y
−2 ( )2 2 3 1g − = − + = ( )2,1−
−1 ( )1 1 3 2g − = − + = ( )1,2−
0 ( )0 0 3 3g = + = ( )0,3
1 ( )1 1 3 4g = + = ( )1,4
2 ( )2 2 3 5g = + = ( )2,5
The graph of g is the graph of f shifted up 3 units.
40. x ( )f x x= ( ),x y
−2 ( )2 2f − = − ( )2, 2− −
−1 ( )1 1f − = − ( )1, 1− −
0 ( )0 0f = ( )0,0
1 ( )1 1f = ( )1,1
2 ( )2 2f = ( )2,2
x ( ) 4g x x= − ( ),x y
−2 ( )2 2 4 6g − = − − = − ( )2, 6− −
−1 ( )1 1 4 5g − = − − = − ( )1, 5− −
0 ( )0 0 4 4g = − = − ( )0, 4−
1 ( )1 1 4 3g = − = − ( )1, 3−
2 ( )2 2 4 2g = − = − ( )2, 2−
The graph of g is the graph of f shifted down 4 units.
41. x ( ) 2f x x= − ( ),x y
–2 ( ) ( )2 2 2 4f − = − − = ( )2,4−
–1 ( ) ( )1 2 1 2f − = − − = ( )1,2−
0 ( ) ( )0 2 0 0f = − = ( )0,0
1 ( ) ( )1 2 1 2f = − = − ( )1, 2−
2 ( ) ( )2 2 2 4f = − = − ( )2, 4−
x ( ) 2 1g x x= − − ( ),x y
–2 ( ) ( )2 2 1 32g − = − − =− ( )2,3−
–1 ( ) ( )1 2 1 11g − = − − =− ( )1,1−
0 ( ) ( )0 2 1 10g = − − = − ( )0, 1−
1 ( ) ( )1 2 1 31g = − − = − ( )1, 3−
2 ( ) ( )2 2 2 1 5g = − − = − ( )2, 5−

The graph of g is the graph of f shifted down 1 unit.
43. x ( ) 2
f x x= ( ),x y
−2 ( ) ( )
2
2 2 4f − = − = ( )2,4−
−1 ( ) ( )
2
1 1 1f − = − = ( )1,1−
0 ( ) ( )
2
0 0 0f = = ( )0,0
1 ( ) ( )
2
1 1 1f = = ( )1,1
2 ( ) ( )
2
2 2 4f = = ( )2,4
x ( ) 2
1g x x= + ( ),x y
−2 ( ) ( )
2
2 2 1 5g − = − + = ( )2,5−
−1 ( ) ( )
2
1 11 2g − = +− = ( )1,2−
0 ( ) ( )
2
0 0 1 1g = + = ( )0,1
1 ( ) ( )
2
1 1 1 2g = =+ ( )1,2
2 ( ) ( )
2
2 2 1 5g = =+ ( )2,5
The graph of g is the graph of f shifted up 1 unit.
44. x ( ) 2
f x x= ( ),x y
−2 ( ) ( )
2
2 2 4f − = − = ( )2,4−
−1 ( ) ( )
2
1 1 1f − = − = ( )1,1−
0 ( ) ( )
2
0 0 0f = = ( )0,0
1 ( ) ( )
2
1 1 1f = = ( )1,1
2 ( ) ( )
2
2 2 4f = = ( )2,4
x ( ) 2
2g x x= − ( ),x y
−2 ( ) ( )
2
2 2 2 2g − = − − = ( )2,2−
−1 ( ) ( )
2
1 1 2 1g − = − − = − ( )1, 1− −
0 ( ) ( )
2
0 0 2 2g = − = − ( )0, 2−
1 ( ) ( )
2
1 1 2 1g = − = − ( )1, 1−
2 ( ) ( )
2
2 2 2 2g = − = ( )2,2
42. x ( ) 2f x x= − ( ),x y
–2 ( ) ( )2 2 2 4f − = − − = ( )2,4−
–1 ( ) ( )1 2 1 2f − = − − = ( )1,2−
0 ( ) ( )0 2 0 0f = − = ( )0,0
1 ( ) ( )1 2 1 2f = − = − ( )1, 2−
2 ( ) ( )2 2 2 4f = − = − ( )2, 4−
x ( ) 2 3g x x= − + ( ),x y
–2 ( ) ( )2 2 3 72g − = − + =− ( )2,7−
–1 ( ) ( )1 2 3 51g − = − + =− ( )1,5−
0 ( ) ( )0 2 3 30g = − + = ( )0,3
1 ( ) ( )1 2 3 11g = − + = ( )1,1
2 ( ) ( )2 2 2 3 1g = − + = − ( )2, 1−

45. x ( )f x x= ( ),x y
2− ( )2 2 2f − = − = ( )2,2−
1− ( )1 1 1f − = − = ( )1,1−
0 ( )0 0 0f = = ( )0,0
1 ( )1 1 1f = = ( )1,1
2 ( )2 2 2f = = ( )2,2
x ( ) 2g x x= − ( ),x y
2− ( )2 2 2 0g − = − − = ( )2,0−
1− ( )1 1 2 1g − = − − = − ( )1, 1− −
0 ( )0 0 2 2g = − = − ( )0, 2−
1 ( )1 1 2 1g = − = − ( )1, 1−
2 ( )2 2 2 0g = − = ( )2,0
46. x ( )f x x= ( ),x y
2− ( )2 2 2f − = − = ( )2,2−
1− ( )1 1 1f − = − = ( )1,1−
0 ( )0 0 0f = = ( )0,0
1 ( )1 1 1f = = ( )1,1
2 ( )2 2 2f = = ( )2,2
x ( ) 1g x x= + ( ),x y
2− ( )2 2 1 3g − = − + = ( )2,3−
1− ( )1 1 1 2g − = − + = ( )1,2−
0 ( )0 0 1 1g = + = ( )0,1
1 ( )1 1 1 2g = + = ( )1,2
2 ( )2 2 1 3g = + = ( )2,3
The graph of g is the graph of f shifted up 1 unit.
47. x ( ) 3
f x x= ( ),x y
2−
( ) ( )
3
2 2 8f − = − = − ( )2, 8− −
1−
( ) ( )
3
1 1 1f − = − = − ( )1, 1− −
0 ( ) ( )
3
0 0 0f = = ( )0,0
1 ( ) ( )
3
1 1 1f = = ( )1,1
2 ( ) ( )
3
2 2 8f = = ( )2,8
x ( ) 3
2g x x= + ( ),x y
2− ( ) ( )
3
2 2 2 6g − = − + = − ( )2, 6− −
1− ( ) ( )
3
1 1 2 1g − = − + = ( )1,1−
0 ( ) ( )
3
0 0 2 2g = + = ( )0,2
1 ( ) ( )
3
1 1 2 3g = + = ( )1,3
2 ( ) ( )
3
2 2 2 10g = + = ( )2,10

48. x ( ) 3
f x x= ( ),x y
2− ( ) ( )
3
2 2 8f − = − = − ( )2, 8− −
1− ( ) ( )
3
1 1 1f − = − = − ( )1, 1− −
0 ( ) ( )
3
0 0 0f = = ( )0,0
1 ( ) ( )
3
1 1 1f = = ( )1,1
2 ( ) ( )
3
2 2 8f = = ( )2,8
x ( ) 3
1g x x= − ( ),x y
2− ( ) ( )
3
2 2 1 9g − = − − = − ( )2, 9− −
1− ( ) ( )
3
1 1 1 2g − = − − = − ( )1, 2− −
0 ( ) ( )
3
0 0 1 1g = − = − ( )0, 1−
1 ( ) ( )
3
1 1 1 0g = − = ( )1,0
2 ( ) ( )
3
2 2 1 7g = − = ( )2,7
49. x ( ) 3f x = ( ),x y
2− ( )2 3f − = ( )2,3−
1− ( )1 3f − = ( )1,3−
0 ( )0 3f = ( )0,3
1 ( )1 3f = ( )1,3
2 ( )2 3f = ( )2,3
x ( ) 5g x = ( ),x y
2− ( )2 5g − = ( )2,5−
1− ( )1 5g − = ( )1,5−
0 ( )0 5g = ( )0,5
1 ( )1 5g = ( )1,5
2 ( )2 5g = ( )2,5
50. x ( ) 1f x = − ( ),x y
2− ( )2 1f − = − ( )2, 1− −
1− ( )1 1f − = − ( )1, 1− −
0 ( )0 1f = − ( )0, 1−
1 ( )1 1f = − ( )1, 1−
2 ( )2 1f = − ( )2, 1−
x ( ) 4g x = ( ),x y
2− ( )2 4g − = ( )2,4−
1− ( )1 4g − = ( )1,4−
0 ( )0 4g = ( )0,4
1 ( )1 4g = ( )1,4
2 ( )2 4g = ( )2,4

51. x ( )f x x= ( ),x y
0 ( )0 0 0f = = ( )0,0
1 ( )1 1 1f = = ( )1,1
4 ( )4 4 2f = = ( )4,2
9 ( )9 9 3f = = ( )9,3
x ( ) 1g x x= − ( ),x y
0 ( )0 0 1 1g = − = − ( )0, 1−
1 ( )1 1 1 0g = − = ( )1,0
4 ( )4 4 1 1g = − = ( )4,1
9 ( )9 9 1 2g = − = ( )9,2
52. x ( )f x x= ( ),x y
0 ( )0 0 0f = = ( )0,0
1 ( )1 1 1f = = ( )1,1
4 ( )4 4 2f = = ( )4,2
9 ( )9 9 3f = = ( )9,3
x ( ) 2g x x= + ( ),x y
0 ( )0 0 2 2g = + = ( )0,2
1 ( )1 1 2 3g = + = ( )1,3
4 ( )4 4 2 4g = + = ( )4,4
9 ( )9 9 2 5g = + = ( )9, 5
53. x ( )f x x= ( ),x y
0 ( )0 0 0f = = ( )0,0
1 ( )1 1 1f = = ( )1,1
4 ( )4 4 2f = = ( )4,2
9 ( )9 9 3f = = ( )9,3
x ( ) 1g x x= − ( ),x y
1 ( )1 1 1 0g = − = ( )1,0
2 ( )2 2 1 1g = − = ( )2,1
5 ( )5 5 1 2g = − = ( )5,2
10 ( )10 10 1 3g = − = ( )10,3
The graph of g is the graph of f shifted right 1 unit.
54. x ( )f x x= ( ),x y
0 ( )0 0 0f = = ( )0,0
1 ( )1 1 1f = = ( )1,1
4 ( )4 4 2f = = ( )4,2
9 ( )9 9 3f = = ( )9,3
x ( ) 2g x x= + ( ),x y
–2 ( )2 2 2 0g − = − + = ( )2,0−
–1 ( )1 1 2 1g − = − + = ( )1,1−
2 ( )2 2 2 2g = + = ( )2,2
7 ( )7 7 2 3g = + = ( )7,3
The graph of g is the graph of f shifted left 2 units.

55. function
56. function
57. function
58. not a function
59. not a function
60. not a function
61. function
62. not a function
63. function
64. function
65. ( )2 4f − = −
66. (2) 4f = −
67. ( )4 4f =
68. ( 4) 4f − =
69. ( )3 0f − =
70. ( 1) 0f − =
71. ( )4 2g − =
72. ( )2 2g = −
73. ( )10 2g − =
74. (10) 2g = −
75. When ( )2, 1.x g x= − =
76. When 1, ( ) 1.x g x= = −
77. a. domain: ( , )−∞ ∞
b. range: [ 4, )− ∞
c. x-intercepts: –3 and 1
d. y-intercept: –3
e. ( 2) 3 and (2) 5f f− = − =
78. a. domain: (–∞, ∞)
b. range: (–∞, 4]
d. y-intercept: 3
e. ( 2) 3 and (2) 5f f− = = −
79. a. domain: ( , )−∞ ∞
b. range: [1, )∞
c. x-intercept: none
d. y-intercept: 1
e. ( 1) 2 and (3) 4f f− = =
80. a. domain: (–∞, ∞)
b. range: [0, ∞)
c. x-intercept: –1
d. y-intercept: 1
e. f(–4) = 3 and f(3) = 4
81. a. domain: [0, 5)
b. range: [–1, 5)
c. x-intercept: 2
e. f(3) = 1
82. a. domain: (–6, 0]
b. range: [–3, 4)
c. x-intercept: –3.75
e. f(–5) = 2
83. a. domain: [0, )∞
b. range: [1, )∞
d. y-intercept: 1
e. f(4) = 3

84. a. domain: [–1, ∞)
b. range: [0, ∞)
d. y-intercept: 1
e. f(3) = 2
85. a. domain: [–2, 6]
b. range: [–2, 6]
c. x-intercept: 4
d. y-intercept: 4
e. f(–1) = 5
86. a. domain: [–3, 2]
b. range: [–5, 5]
c. x-intercept:
1
2
−
d. y-intercept: 1
e. f(–2) = –3
87. a. domain: ( , )−∞ ∞
b. range: ( , 2]−∞ −
e. f(–4) = –5 and f(4) = –2
88. a. domain: (–∞, ∞)
b. range: [0, ∞)
c. x-intercept: { }0x x ≤
d. y-intercept: 0
e. f(–2) = 0 and f(2) = 4
89. a. domain: ( , )−∞ ∞
b. range: (0, )∞
d. y-intercept: 1.5
e. f(4) = 6
90. a. domain: ( ,1) (1, )−∞ ∞
b. range: ( ,0) (0, )−∞ ∞
d. y-intercept: 1−
e. f(2) = 1
91. a. domain: {–5, –2, 0, 1, 3}
b. range: {2}
d. y-intercept: 2
e. ( 5) (3) 2 2 4f f− + = + =
92. a. domain: {–5, –2, 0, 1, 4}
b. range: {–2}
e. ( 5) (4) 2 ( 2) 4f f− + = − + − = −
93. ( ) ( )
( )( ) ( ) ( ) ( )
2
1 3 1 5 3 5 2
1 2 2 2 4
4 2 4 10
g
f g f
= − = − = −
= − = − − − +
= + + =
94. ( ) ( )
( )( ) ( ) ( ) ( )
2
1 3 1 5 3 5 8
1 8 8 8 4
64 8 4 76
g
f g f
− = − − = − − = −
− = − = − − − +
= + + =
95. ( ) ( ) ( )
( )
2
3 1 6 6 6 4
3 1 36 6 6 4
4 36 1 4
2 36 4
34 4
38
− − − − + ÷ − ⋅
= + − + ÷ − ⋅
= − + − ⋅
= − + −
= − + −
= −
96. ( ) ( )
2
4 1 3 3 3 6
4 1 9 3 3 6
3 9 1 6
3 9 6 6 6 0
− − − − − + − ÷ ⋅−
= − + − + − ÷ ⋅−
= − − + − ⋅−
= − + = − + =
97. ( ) ( )
( ) ( )
3 3
3 3 3
5 ( 5)
5 5 2 2
f x f x
x x x x
x x x x x x
− −
= − + − − − + −
= − − − − − + = − −

98. ( ) ( )
( ) ( ) ( )2 2
2 2
3 7 3 7
3 7 3 7
6
f x f x
x x x x
x x x x
x
− −
= − − − + − − +
= + + − + −
=
99. a. {(Iceland, 9.7), (Finland, 9.6), (New Zealand,
9.6), (Denmark, 9.5)}
b. Yes, the relation is a function because each
country in the domain corresponds to exactly
one corruption rating in the range.
c. {(9.7, Iceland), (9.6, Finland), (9.6,
New Zealand), (9.5, Denmark)}
d. No, the relation is not a function because 9.6 in
the domain corresponds to two countries in the
range, Finland and New Zealand.
100. a. {(Bangladesh, 1.7), (Chad, 1.7), (Haiti, 1.8),
(Myanmar, 1.8)}
b. Yes, the relation is a function because each
country in the domain corresponds to exactly
one corruption rating in the range.
c. {(1.7, Bangladesh), (1.7, Chad), (1.8, Haiti),
(1.8, Myanmar)}
d. No, the relation is not a function because 1.7 in
the domain corresponds to two countries in the
range, Bangladesh and Chad.
101. a. (70) 83f = which means the chance that a 60-
year old will survive to age 70 is 83%.
b. (70) 76g = which means the chance that a 60-
c. Function f is the better model.
102. a. (90) 25f = which means the chance that a 60-
b. (90) 10g = which means the chance that a 60-
c. Function f is the better model.
103. a. 2
(30) 0.01(30) (30) 60 81G = − + + =
In 2010, the wage gap was 81%. This is
represented as (30,81) on the graph.
b. (30)G underestimates the actual data shown by
the bar graph by 2%.
104. a. 2
(10) 0.01(10) (10) 60 69G = − + + =
In 1990, the wage gap was 69%. This is
represented as (10,69) on the graph.
b. (10)G underestimates the actual data shown by
the bar graph by 2%.
105. ( ) 100,000 100C x x= +
(90) 100,000 100(90) $109,000C = + =
It will cost $109,000 to produce 90 bicycles.
106. ( ) 22,500 3200V x x= −
(3) 22,500 3200(3) $12,900V = − =
After 3 years, the car will be worth $12,900.
107. ( )
( )
40 40
30
40 40
30
30 30 30
80 40
60 60
120
60
2
T x
x x
T
= +
+
= +
+
= +
=
=
If you travel 30 mph going and 60 mph returning,
your total trip will take 2 hours.
108. ( ) 0.10 0.60(50 )S x x x= + −
(30) 0.10(30) 0.60(50 30) 15S = + − =
When 30 mL of the 10% mixture is mixed with 20
mL of the 60% mixture, there will be 15 mL of
sodium-iodine in the vaccine.
109. – 117. Answers will vary.
118. makes sense
119. does not make sense; Explanations will vary.
Sample explanation: The parentheses used in
function notation, such as ( ),f x do not imply
multiplication.

Sample explanation: The domain is the number of
years worked for the company.
Sample explanation: This would not be a function
because some elements in the domain would
correspond to more than one age in the range.
122. false; Changes to make the statement true will vary.
A sample change is: The domain is [ 4,4].−
A sample change is: The range is [ )2,2 .−
124. true
A sample change is: (0) 0.8f =
126. ( ) 3( ) 7 3 3 7
( ) 3 7
f a h a h a h
f a a
+ = + + = + +
= +
( ) ( )
( ) ( )
3 3 7 3 7
3 3 7 3 7 3
3
f a h f a
h
a h a
h
a h a h
h h
+ −
+ + − +
=
+ + − −
= = =
127. Answers will vary.
An example is {(1,1),(2,1)}
128. It is given that ( ) ( ) ( )f x y f x f y+ = + and (1) 3f = .
To find (2)f , rewrite 2 as 1 + 1.
(2) (1 1) (1) (1)
3 3 6
f f f f= + = +
= + =
Similarly:
(3) (2 1) (2) (1)
6 3 9
f f f f= + = +
= + =
(4) (3 1) (3) (1)
9 3 12
f f f f= + = +
= + =
While ( ) ( ) ( )f x y f x f y+ = + is true for this function,
it is not true for all functions. It is not true
for ( ) 2
f x x= , for example.
129. ( )1 3 4 2
1 3 12 2
3 13 2
13
13
x x
x x
x x
x
x
− + − =
− + − =
− =
− = −
=
The solution set is {13}.
130.
( )
3 4
5
5 2
3 4
10 10 10 5
5 2
2 6 5 20 50
3 14 50
3 36
12
x x
x x
x x
x
x
x
− −
− =
− −   
− =   
   
− − + =
− + =
− =
= −
The solution set is {–12}.
131. Let x = the number of deaths by snakes, in
thousands, in 2014
Let x + 661 = the number of deaths by
mosquitoes, in thousands, in 2014
Let x + 106 = the number of deaths by snails, in
thousands, in 2014
( ) ( )661 106 1049
661 106 1049
3 767 1049
3 282
94
x x x
x x x
x
x
x
+ + + + =
+ + + + =
+ =
=
=
94, thousand deaths by snakes
661 755, thousand deaths by mosquitoes
106 200, thousand deaths by snails
x
x
x
=
+ =
+ =
132. ( ) 20 0.40( 60)
(100) 20 0.40(100 60)
20 0.40(40)
20 16
36
C t t
C
= + −
= + −
= +
= +
=
For 100 calling minutes, the monthly cost is $36.
133.
134. 2 2
2 2 2
2 2 2
2 2 2
2
2( ) 3( ) 5 (2 3 5)
2( 2 ) 3 3 5 2 3 5
2 4 2 3 3 5 2 3 5
2 2 4 2 3 3 3 5 5
4 2 3
x h x h x x
x xh h x h x x
x xh h x h x x
x x xh h x x h
xh h h
+ + + + − + +
= + + + + + − − −
= + + + + + − − −
= − + + + − + + −
= + +

Section 2.2
1. The function is increasing on the interval ( , 1),−∞ −
decreasing on the interval ( 1,1),− and increasing on
the interval (1, ).∞
2. Test for symmetry with respect to the y-axis.
( )
2
2
2
1
1
1
y x
y x
y x
= −
= − −
= −
The resulting equation is equivalent to the original.
Thus, the graph is symmetric with respect to the
y-axis.
Test for symmetry with respect to the x-axis.
2
2
2
1
1
1
y x
y x
y x
= −
− = −
= − +
The resulting equation is not equivalent to the
original. Thus, the graph is not symmetric with
respect to the x-axis.
Test for symmetry with respect to the origin.
( )
2
2
2
2
1
1
1
1
y x
y x
y x
y x
= −
− = − −
− = −
= − +
respect to the origin.
( )
5 3
35
5 3
y x
y x
y x
=
= −
= −
respect to the y-axis.
( )
5 3
5 3
5 3
5 3
y x
y x
y x
y x
=
− =
− =
= −
( ) ( )
5 3
5 3
5 3
5 3
y x
y x
y x
y x
=
− = −
− = −
=
origin.
4. a. The graph passes the vertical line test and is
therefore the graph of a function. The graph is
symmetric with respect to the y-axis. Therefore,
the graph is that of an even function.
b. The graph passes the vertical line test and is
neither symmetric with respect to the y-axis nor
the origin. Therefore, the graph is that of a
function which is neither even nor odd.
c. The graph passes the vertical line test and is
symmetric with respect to the origin. Therefore,
the graph is that of an odd function.
5. a. 2 2
( ) ( ) 6 6 ( )f x x x f x− = − + = + =
The function is even. The graph is symmetric
with respect to the y-axis.
b. 3 3
( ) 7( ) ( ) 7 ( )g x x x x x f x− = − − − = − + = −
The function is odd. The graph is symmetric
with respect to the origin.
c. 5 5
( ) ( ) 1 1h x x x− = − + = − +
The function is neither even nor odd. The graph
is neither symmetric to the y-axis nor the origin.
6.
20 if 0 60
( )
20 0.40( 60) if 60
t
C t
t t
≤ ≤
= 
+ − >
a. Since 0 40 60≤ ≤ , (40) 20C =
With 40 calling minutes, the cost is $20.
This is represented by ( )40,20 .
b. Since 80 60> ,
(80) 20 0.40(80 60) 28C = + − =
With 80 calling minutes, the cost is $28.
This is represented by ( )80,28 .

Section 2.2 More on Functions and Their Graphs
7.
8. a. 2
( ) 2 5f x x x= − + +
2
2 2
2 2
( ) 2( ) ( ) 5
2( 2 ) 5
2 4 2 5
f x h x h x h
x xh h x h
x xh h x h
+ = − + + + +
= − + + + + +
= − − − + + +
b.
( ) ( )f x h f x
h
+ −
( )
( )
2 2 2
2 2 2
2
2 4 2 5 2 5
2 4 2 5 2 5
4 2
4 2 1
4 2 1, 0
x xh h x h x x
h
x xh h x h x x
h
xh h h
h
h x h
h
x h h
− − − + + + − − + +
=
− − − + + + + − −
=
− − +
=
− − +
=
= − − + ≠
1. 2( )f x< ; 2( )f x> ; 2( )f x=
2. maximum; minimum
3. y-axis
4. x-axis
5. origin
6. ( )f x ; y-axis
7. ( )f x− ; origin
8. piecewise
9. less than or equal to x; 2; 3− ; 0
10. difference quotient; x h+ ; ( )f x ; h; h
11. false
12. false
Exercise Set 2.2
1. a. increasing: ( 1, )− ∞
b. decreasing: ( , 1)−∞ −
c. constant: none
2. a. increasing: (–∞, –1)
b. decreasing: (–1, ∞)
c. constant: none
3. a. increasing: (0, )∞
b. decreasing: none
c. constant: none
4. a. increasing: (–1, ∞)
b. decreasing: none
c. constant: none
5. a. increasing: none
b. decreasing: (–2, 6)
c. constant: none
6. a. increasing: (–3, 2)
b. decreasing: none
c. constant: none
7. a. increasing: ( , 1)−∞ −
b. decreasing: none
c. constant: ( 1, )− ∞
8. a. increasing: (0, ∞)
b. decreasing: none
c. constant: (–∞, 0)
9. a. increasing: ( , 0) or (1.5, 3)−∞
b. decreasing: (0,1.5) or (3, )∞
c. constant: none

10. a. increasing: ( 5, 4) or ( 2,0) or (2,4)− − −
b. decreasing: ( 4, 2) or (0,2) or (4,5)− −
c. constant: none
11. a. increasing: (–2, 4)
b. decreasing: none
c. constant: ( , 2) or (4, )−∞ − ∞
12. a. increasing: none
b. decreasing: (–4, 2)
c. constant: ( , 4) or (2, )−∞ − ∞
13. a. x = 0, relative maximum = 4
b. x = −3, 3, relative minimum = 0
14. a. x = 0, relative maximum = 2
b. x = −3, 3, relative minimum = –1
15. a. x = −2, relative maximum = 21
b. x = 1, relative minimum = −6
16. a. x =1, relative maximum = 30
b. x = 4, relative minimum = 3
( )
2
2
2
6
6
6
y x
y x
y x
= +
= − +
= +
y-axis.
2
2
2
6
6
6
y x
y x
y x
= +
− = +
= − −
( )
2
2
2
2
6
6
6
6
y x
y x
y x
y x
= +
− = − +
− = +
= − −
( )
2
2
2
2
2
2
y x
y x
y x
= −
= − −
= −
y-axis.
2
2
2
2
2
2
y x
y x
y x
= −
− = −
= − +
( )
2
2
2
2
2
2
2
2
y x
y x
y x
y x
= −
− = − −
− = −
= − +
2
2
2
6
6
6
x y
x y
x y
= +
− = +
= − −
( )
2
2
2
6
6
6
x y
x y
x y
= +
= − +
= +
x-axis.

( )
2
2
2
2
6
6
6
6
x y
x y
x y
x y
= +
− = − +
− = +
= − −
2
2
2
2
2
2
x y
x y
x y
= −
− = −
= − +
( )
2
2
2
2
2
2
x y
x y
x y
= −
= − −
= −
x-axis.
( )
2
2
2
2
2
2
2
2
x y
x y
x y
x y
= −
− = − −
− = −
= − +
( )
2 2
22
2 2
6
6
6
y x
y x
y x
= +
= − +
= +
Thus, the graph is symmetric with respect to the y-axis.
( )
2 2
2 2
2 2
6
6
6
y x
y x
y x
= +
− = +
= +
Thus, the graph is symmetric with respect to the x-axis.
( ) ( )
2 2
2 2
2
6
6
6
y x
y x
y x
= +
− = − +
= +
Thus, the graph is symmetric with respect to the origin.
( )
2 2
22
2 2
2
2
2
y x
y x
y x
= −
= − −
= −
y-axis.
( )
2 2
2 2
2 2
2
2
2
y x
y x
y x
= −
− = −
= −
x-axis.
( ) ( )
2 2
2 2
2
2
2
2
y x
y x
y x
= −
− = − −
= −
origin.
( )
2 3
2 3
2 3
y x
y x
y x
= +
= − +
= − +
2 3
2 3
2 3
y x
y x
y x
= +
− = +
= − −

( )
2 3
2 3
2 3
y x
y x
y x
= +
− = − +
= −
( )
2 5
2 5
2 5
y x
y x
y x
= +
= − +
= − +
2 5
2 5
2 5
y x
y x
y x
= +
− = +
= − −
( )
2 5
2 5
2 5
y x
y x
y x
= +
− = − +
= −
( )
2 3
2 3
2 3
2
2
2
x y
x y
x y
− =
− − =
− =
Thus, the graph is symmetric with respect to the y-
axis.
( )
2 3
32
2 3
2
2
2
x y
x y
x y
− =
− − =
+ =
( ) ( )
2 3
2 3
2 3
2
2
2
x y
x y
x y
− =
− − − =
+ =
( )
3 2
3 2
3 2
5
5
5
x y
x y
x y
− =
− − =
− − =
( )
3 2
23
3 2
5
5
5
x y
x y
x y
− =
− − =
− =
Thus, the graph is symmetric with respect to the x-
axis.
( ) ( )
3 2
3 2
3 2
5
5
5
x y
x y
x y
− =
− − − =
− − =
( )
2 2
2 2
2 2
100
100
100
x y
x y
x y
+ =
− + =
+ =
axis.
( )
2 2
22
2 2
100
100
100
x y
x y
x y
+ =
+ − =
+ =
axis.

( ) ( )
2 2
2 2
2 2
100
100
100
x y
x y
x y
+ =
− + − =
+ =
origin.
( )
2 2
2 2
2 2
49
49
49
x y
x y
x y
+ =
− + =
+ =
axis.
( )
2 2
22
2 2
49
49
49
x y
x y
x y
+ =
+ − =
+ =
x-axis.
( ) ( )
2 2
2 2
2 2
49
49
49
x y
x y
x y
+ =
− + − =
+ =
origin.
( ) ( )
2 2
2 2
2 2
3 1
3 1
3 1
x y xy
x y x y
x y xy
+ =
− + − =
− =
( ) ( )
2 2
22
2 2
3 1
3 1
3 1
x y xy
x y x y
x y xy
+ =
− + − =
− =
x-axis.
( ) ( ) ( )( )
2 2
2 2
2 2
3 1
3 1
3 1
x y xy
x y x y
x y xy
+ =
− − + − − =
+ =
origin.
( ) ( )
2 2
2 2
2 2
5 2
5 2
5 2
x y xy
x y x y
x y xy
+ =
− + − =
− =
( ) ( )
2 2
22
2 2
5 2
5 2
5 2
x y xy
x y x y
x y xy
+ =
− + − =
− =
x-axis.
( ) ( ) ( )( )
2 2
2 2
2 2
5 2
5 2
5 2
x y xy
x y x y
x y xy
+ =
− − + − − =
+ =
origin.
( )
4 3
34
4 3
6
6
6
y x
y x
y x
= +
= − +
= − +
( )
4 3
4 3
4 3
6
6
6
y x
y x
y x
= +
− = +
= +
x-axis.

( ) ( )
4 3
4 3
4 3
6
6
6
y x
y x
y x
= +
− = − +
= − +
( )
5 4
45
5 4
2
2
2
y x
y x
y x
= +
= − +
= +
axis.
( )
5 4
5 4
5 4
5 4
2
2
2
2
y x
y x
y x
y x
= +
− = +
− = +
= − −
( ) ( )
4 3
4 3
4 3
6
6
6
y x
y x
y x
= +
− = − +
= − +
33. The graph is symmetric with respect to the y-axis.
The function is even.
34. The graph is symmetric with respect to the origin.
The function is odd.
36. The graph is not symmetric with respect to the y-axis
or the origin. The function is neither even nor odd.
37. 3
3
3 3
( )
( ) ( ) ( )
( ) ( )
f x x x
f x x x
f x x x x x
= +
− = − + −
− = − − = − +
( ) ( ),f x f x− = − odd function
38. 3
3
3 3
( )
( ) ( ) ( )
( ) ( )
( ) ( ), odd function
f x x x
f x x x
f x x x x x
f x f x
= −
− = − − −
− = − + = − −
− = −
39. 2
2
( )
( ) ( ) ( )
g x x x
g x x x
= +
− = − + −
2
( ) ,g x x x− = − neither
40. 2
2
2
( )
( ) ( ) ( )
( ) , neither
g x x x
g x x x
g x x x
= −
− = − − −
− = +
41. 2 4
2 4
2 4
( )
( ) ( ) ( )
( )
h x x x
h x x x
h x x x
= −
− = − − −
− = −
( ) ( ),h x h x− = even function
42. 2 4
2 4
2 4
( ) 2
( ) 2( ) ( )
( ) 2
( ) ( ), even function
h x x x
h x x x
h x x x
h x h x
= +
− = − + −
− = +
− =
43. 2 4
2 4
2 4
( ) 1
( ) ( ) ( ) 1
( ) 1
f x x x
f x x x
f x x x
= − +
− = − − − +
− = − +
( ) ( ),f x f x− = even function
44. 2 4
2 4
2 4
( ) 2 1
( ) 2( ) ( ) 1
( ) 2 1
( ) ( ), even function
f x x x
f x x x
f x x x
f x f x
= + +
− = − + − +
− = + +
− =
45. 6 2
6 2
6 2
1
( ) 3
5
1
( ) ( ) 3( )
5
1
( ) 3
5
f x x x
f x x x
f x x x
= −
− = − − −
− = −
( ) ( )f x f x− = , even function

46. 3 5
3 5
3 5
3 5
( ) 2 6
( ) 2( ) 6( )
( ) 2 6
( ) (2 6 )
( ) ( ), odd function
f x x x
f x x x
f x x x
f x x x
f x f x
= −
− = − − −
− = − +
− = − −
− = −
47.
( )
2
2
2
2
( ) 1
( ) 1 ( )
( ) 1
1
f x x x
f x x x
f x x x
x x
= −
− = − − −
− = − −
= − −
f(–x) = – f(x), odd function
48. ( ) 2 2
1f x x x= −
( ) ( ) ( )
2 2
1f x x x− = − − −
( ) 2 2
1f x x x− = −
f(–x) = f(x), even function
49. a. domain: ( ),−∞ ∞
b. range: [ )4,− ∞
c. x-intercepts: 1, 7
d. y-intercept: 4
e. ( )4,∞
f. ( )0,4
g. ( ),0−∞
h. 4x =
i. 4y = −
j. ( 3) 4f − =
k. (2) 2f = − and (6) 2f = −
l. neither ; ( )f x x− ≠ , ( )f x x− ≠ −
50. a. domain: ( ),−∞ ∞
b. range: ( ],4−∞
c. x-intercepts: –4, 4
d. y-intercept: 1
e. ( ), 2−∞ − or ( )0,3
f. ( )2,0− or ( )3,∞
g. ( ], 4−∞ − or [ )4,∞
h. 2x = − and 3x =
i. ( 2) 4f − = and (3) 2f =
j. ( 2) 4f − =
k. 4x = − and 4x =
l. neither ; ( )f x x− ≠ , ( )f x x− ≠ −
51. a. domain: ( ],3−∞
b. range: ( ],4−∞
c. x-intercepts: –3, 3
d. (0) 3f =
e. ( ),1−∞
f. ( )1,3
g. ( ], 3−∞ −
h. (1) 4f =
i. 1x =
j. positive; ( 1) 2f − = +
52. a. domain: ( ],6−∞
b. range: ( ],1−∞
c. zeros of f: –3, 3
d. (0) 1f =
e. ( ), 2−∞ −
f. ( )2,6
g. ( )2,2−

h. ( )3,3−
i. 5x = − and 5x =
j. negative; (4) 1f = −
k. neither
l. no; f(2) is not greater than the function values to
the immediate left.
53. a. f(–2) = 3(–2) + 5 = –1
b. f(0) = 4(0) + 7 = 7
c. f(3) = 4(3) + 7 = 19
54. a. f(–3) = 6(–3) – 1 = –19
b. f(0) = 7(0) + 3 = 3
c. f(4) = 7(4) + 3 = 31
55. a. g(0) = 0 + 3 = 3
b. g(–6) = –(–6 + 3) = –(–3) = 3
c. g(–3) = –3 + 3 = 0
56. a. g(0) = 0 + 5 = 5
b. g(–6) = –(–6 + 5) = –(–1) = 1
c. g(–5) = –5 + 5 = 0
57. a.
2
5 9 25 9 16
(5) 8
5 3 2 2
h
− −
= = = =
−
b.
2
0 9 9
(0) 3
0 3 3
h
− −
= = =
− −
c. h(3) = 6
58. a.
2
7 25 49 25 24
(7) 12
7 5 2 2
h
− −
= = = =
−
b.
2
0 25 25
(0) 5
0 5 5
h
− −
= = =
− −
c. h(5) = 10
59. a.
b. range: [ )0,∞
60. a.
b. range: ( ],0−∞
61. a.
b. range: ( ],0 {2}−∞ ∪
62. a.
b. range: ( ],0 {3}−∞ ∪
63. a.
b. range: ( , )−∞ ∞

64. a.
b. range: ( , )−∞ ∞
65. a.
b. range: { 3,3}−
66. a.
b. range: { 4,4}−
67. a.
b. range: [ )0,∞
68. a.
b. range: ( ] [ ),0 3,−∞ ∪ ∞
69. a.
b. range: [ )0,∞
70. a.
b. range: [ )1,− ∞
71.
( ) ( )f x h f x
h
+ −
4( ) 4
4 4 4
4
4
x h x
h
x h x
h
h
h
+ −
=
+ −
=
=
=

72.
( ) ( )f x h f x
h
+ −
7( ) 7
7 7 7
7
7
x h x
h
x h x
h
h
h
+ −
=
+ −
=
=
=
73.
( ) ( )f x h f x
h
+ −
3( ) 7 (3 7)
3 3 7 3 7
3
3
x h x
h
x h x
h
h
h
+ + − +
=
+ + − −
=
=
=
74.
( ) ( )f x h f x
h
+ −
6( ) 1 (6 1)
6 6 1 6 1
6
6
x h x
h
x h x
h
h
h
+ + − +
=
+ + − −
=
=
=
75.
( ) ( )f x h f x
h
+ −
( )
( )
2 2
2 2 2
2
2
2
2
2
x h x
h
x xh h x
h
xh h
h
h x h
h
x h
+ −
=
+ + −
=
+
=
+
=
= +
76.
( ) ( )f x h f x
h
+ −
( )
2 2
2 2 2
2 2 2
2
2( ) 2
2( 2 ) 2
2 4 2 2
4 2
4 2
4 2
x h x
h
x xh h x
h
x xh h x
h
xh h
h
h x h
h
x h
+ −
=
+ + −
=
+ + −
=
+
=
+
=
= +
77.
( ) ( )f x h f x
h
+ −
2 2
2 2 2
2
( ) 4( ) 3 ( 4 3)
2 4 4 3 4 3
2 4
(2 4)
2 4
x h x h x x
h
x xh h x h x x
h
xh h h
h
h x h
h
x h
+ − + + − − +
=
+ + − − + − + −
=
+ −
=
+ −
=
= + −
78.
( ) ( )f x h f x
h
+ −
( )
2 2
2 2 2
2
( ) 5( ) 8 ( 5 8)
2 5 5 8 5 8
2 5
2 5
2 5
x h x h x x
h
x xh h x h x x
h
xh h h
h
h x h
h
x h
+ − + + − − +
=
+ + − − + − + −
=
+ −
=
+ −
=
= + −

79.
( ) ( )f x h f x
h
+ −
( ) ( )
( )
2 2
2 2 2
2
2 1 (2 1)
2 4 2 1 2 1
4 2
4 2 1
4 2 1
x h x h x x
h
x xh h x h x x
h
xh h h
h
h x h
h
x h
+ + + − − + −
=
+ + + + − − − +
=
+ +
=
+ +
=
= + +
80.
( ) ( )f x h f x
h
+ −
( ) ( )
( )
2 2
2 2 2
2
3 5 (3 5)
3 6 3 5 3 5
6 3
6 3 1
6 3 1
x h x h x x
h
x xh h x h x x
h
xh h h
h
h x h
h
x h
+ + + + − + +
=
+ + + + + − − −
=
+ +
=
+ +
=
= + +
81.
( ) ( )f x h f x
h
+ −
( ) ( )
( )
2 2
2 2 2
2
2 4 ( 2 4)
2 2 2 4 2 4
2 2
2 2
2 2
x h x h x x
h
x xh h x h x x
h
xh h h
h
h x h
h
x h
− + + + + − − + +
=
− − − + + + + − −
=
− − +
=
− − +
=
= − − +
82.
( ) ( )f x h f x
h
+ −
( ) ( )
( )
2 2
2 2 2
2
3 1 ( 3 1)
2 3 3 1 3 1
2 3
2 3
2 3
x h x h x x
h
x xh h x h x x
h
xh h h
h
h x h
h
x h
− + − + + − − − +
=
− − − − − + + + −
=
− − −
=
− − −
=
= − − −
83.
( ) ( )f x h f x
h
+ −
( ) ( )
( )
2 2
2 2 2
2
2 5 7 ( 2 5 7)
2 4 2 5 5 7 2 5 7
4 2 5
4 2 5
4 2 5
x h x h x x
h
x xh h x h x x
h
xh h h
h
h x h
h
x h
− + + + + − − + +
=
− − − + + + + − −
=
− − +
=
− − +
=
= − − +
84.
( ) ( )f x h f x
h
+ −
( ) ( )
( )
2 2
2 2 2
2
3 2 1 ( 3 2 1)
3 6 3 2 2 1 3 2 1
6 3 2
6 3 2
6 3 2
x h x h x x
h
x xh h x h x x
h
xh h h
h
h x h
h
x h
− + + + − − − + −
=
− − − + + − + − +
=
− − +
=
− − +
=
= − − +

85.
( ) ( )f x h f x
h
+ −
( ) ( )
( )
2 2
2 2 2
2
2 3 ( 2 3)
2 4 2 3 2 3
4 2
4 2 1
4 2 1
x h x h x x
h
x xh h x h x x
h
xh h h
h
h x h
h
x h
− + − + + − − − +
=
− − − − − + + + −
=
− − −
=
− − −
=
= − − −
86.
( ) ( )f x h f x
h
+ −
( ) ( )
( )
2 2
2 2 2
2
3 1 ( 3 1)
3 6 3 1 3 1
6 3
6 3 1
6 3 1
x h x h x x
h
x xh h x h x x
h
xh h h
h
h x h
h
x h
− + + + − − − + −
=
− − − + + − + − +
=
− − +
=
− − +
=
= − − +
87.
( ) ( ) 6 6 0
0
f x h f x
h h h
+ − −
= = =
88.
( ) ( ) 7 7 0
0
f x h f x
h h h
+ − −
= = =
89.
( ) ( )f x h f x
h
+ −
( )
1 1
( ) ( )
( )
( )
1
( )
1
( )
x h x
h
x hx
x x h x x h
h
x x h
x x h
h
h
x x h
h
h
x x h h
x x h
−
+=
− +
+
+ +
=
− −
+
=
−
+
=
−
= ⋅
+
−
=
+
90.
( ) ( )f x h f x
h
+ −
( )
1 1
2( ) 2
2 ( ) 2 ( )
2 ( )
1
2 ( )
1
2
x h x
h
x x h
x x h x x h
h
h
x x h
h
h
x x h h
x x h
−
+
=
+
−
+ +
=
−
+
=
−
= ⋅
+
−
=
+
91.
( ) ( )f x h f x
h
+ −
( )
( )
1
x h x
h
x h x x h x
h x h x
x h x
h x h x
h
h x h x
x h x
+ −
=
+ − + +
= ⋅
+ +
+ −
=
+ +
=
+ +
=
+ +
92.
( ) ( )f x h f x
h
+ −
( )
( )
( )
1 1
1 1 1 1
1 1
1 ( 1)
1 1
1 1
1 1
1 1
1
1 1
x h x
h
x h x x h x
h x h x
x h x
h x h x
x h x
h x h x
h
h x h x
x h x
+ − − −
=
+ − − − + − + −
= ⋅
+ − + −
+ − − −
=
+ − + −
+ − − +
=
+ − + −
=
+ − + −
=
+ − + −

93. [ ]
2
( 1.5) ( 0.9) ( ) ( 3) (1) ( )f f f f f fπ π− + − − + − ÷ ⋅ −
[ ] ( )
( )
2
1 0 4 2 2 3
1 16 1 3
1 16 3
18
= + − − + ÷ − ⋅
= − + − ⋅
= − −
= −
94. [ ]
2
( 2.5) (1.9) ( ) ( 3) (1) ( )f f f f f fπ π− − − − + − ÷ ⋅
[ ]
[ ] ( ) ( )
( )( )
2
2
( 2.5) (1.9) ( ) ( 3) (1) ( )
2 ( 2) 3 2 2 4
4 9 1 4
2 9 4
3
f f f f f fπ π− − − − + − ÷ ⋅
= − − − + ÷ − ⋅ −
= − + − −
= − +
= −
95. 30 0.30( 120) 30 0.3 36 0.3 6t t t+ − = + − = −
96. 40 0.30( 200) 40 0.3 60 0.3 20t t t+ − = + − = −
97.
50 if 0 400
( )
50 0.30( 400) if 400
t
C t
t t
≤ ≤
= 
+ − >
98.
60 if 0 450
( )
60 0.35( 450) if 450
t
C t
t t
≤ ≤
= 
+ − >
99. increasing: (25, 55); decreasing: (55, 75)
100. increasing: (25, 65); decreasing: (65, 75)
101. The percent body fat in women reaches a maximum
at age 55. This maximum is 38%.
102. The percent body fat in men reaches a maximum at
age 65. This maximum is 26%.
103. domain: [25, 75]; range: [34, 38]
104. domain: [25, 75]; range: [23, 26]
105. This model describes percent body fat in men.
106. This model describes percent body fat in women.
107. (20,000) 850 0.15(20,000 8500)
2575
T = + −
=
A single taxpayer with taxable income of $20,000
owes $2575.
108. (50,000) 4750 0.25(50,000 34,500)
8625
T = + −
=
A single taxpayer with taxable income of $50,000
owes $8625.
109. 42,449+ 0.33( 174,400)x −
110. 110,016.50+ 0.35( ( 379,150)x x− −
111. (3) 0.93f =
The cost of mailing a first-class letter weighing 3
ounces is $0.93.
112. (3.5) 1.05f =
The cost of mailing a first-class letter weighing 3.5
ounces is $1.05.
113. The cost to mail a letter weighing 1.5 ounces is
$0.65.

114. The cost to mail a letter weighing 1.8 ounces is
$0.65.
115.
125.
The number of doctor visits decreases during
childhood and then increases as you get older.
The minimum is (20.29, 3.99), which means that the
minimum number of doctor visits, about 4, occurs at
around age 20.
126.
Increasing: ( ,1) or (3, )−∞ ∞
Decreasing: (1, 3)
127.
Increasing: (–2, 0) or (2, ∞)
Decreasing: (–∞, –2) or (0, 2)
128.
Increasing: (2, )∞
Decreasing: ( , 2)−∞ −
Constant: (–2, 2)
129.
Increasing: (1, ∞)
Decreasing: (–∞, 1)
130.
Increasing: (0, )∞
Decreasing: ( , 0)−∞
131.
Increasing: (–∞, 0)
Decreasing: (0, ∞)

132. a.
b.
c. Increasing: (0, ∞)
Decreasing: (–∞, 0)
d. ( ) n
f x x= is increasing from (–∞, ∞) when n
is odd.
e.
Sample explanation: It’s possible the graph is not
defined at a.
134. makes sense
135. makes sense
136. makes sense
137. answers will vary
138. answers will vary
139. a. h is even if both f and g are even or if both f and
g are odd.
f and g are both even:
(– ) ( )
(– ) ( )
(– ) ( )
f x f x
h x h x
g x g x
= = =
f and g are both odd:
(– ) – ( ) ( )
(– ) ( )
(– ) – ( ) ( )
f x f x f x
h x h x
g x g x g x
= = = =
b. h is odd if f is odd and g is even or if f is even
and g is odd.
f is odd and g is even:
(– ) – ( ) ( )
(– ) – – ( )
(– ) ( ) ( )
f x f x f x
h x h x
g x g x g x
= = = =
f is even and g is odd:
(– ) ( ) ( )
(– ) – – ( )
(– ) – ( ) ( )
f x f x f x
h x h x
g x g x g x
= = = =
140. Let x = the amount invested at 5%.
Let 80,000 – x = the amount invested at 7%.
0.05 0.07(80,000 ) 5200
0.05 5600 0.07 5200
0.02 5600 5200
0.02 400
20,000
80,000 60,000
x x
x x
x
x
x
x
+ − =
+ − =
− + =
− = −
=
− =
$20,000 was invested at 5% and $60,000 was
invested at 7%.
141. C A Ar= +
( )1
1
C A Ar
C A r
C
A
r
= +
= +
=
+

142. 2
5 7 3 0
5, 7, 3
x x
a b c
− + =
= = − =
2
2
4
2
( 7) ( 7) 4(5)(3)
2(5)
7 49 60
10
7 11
10
7 11
10
7 11
10 10
b b ac
x
a
x
x
x
i
x
x i
− ± −
=
− − ± − −
=
± −
=
± −
=
±
=
= ±
The solution set is
7 11 7 11
, .
10 10 10 10
i
  
+ − 
  
143. 2 1
2 1
4 1 3
3
2 ( 3) 1
y y
x x
− −
= = =
− − − −
144. When 0 :y =
4 3 6 0
4 3(0) 6 0
4 6 0
4 6
3
2
x y
x
x
x
x
− − =
− − =
− =
=
=
The point is
3
,0 .
2
 
 
 
When 0 :x =
4 3 6 0
4(0) 3 6 0
3 6 0
3 6
2
x y
y
y
y
x
− − =
− − =
− − =
− =
= −
The point is ( )0, 2 .−
145. 3 2 4 0
2 3 4
3 4
2
or
3
2
2
x y
y x
x
y
y x
+ − =
= − +
− +
=
= − +
Section 2.3
1. a.
2 4 6
6
4 ( 3) 1
m
− − −
= = =
− − − −
b.
5 ( 2) 7 7
1 4 5 5
m
− −
= = = −
− − −
2. Point-slope form:
1 1( )
( 5) 6( 2)
5 6( 2)
y y m x x
y x
y x
− = −
− − = −
+ = −
Slope-intercept form:
5 6( 2)
5 6 12
6 17
y x
y x
y x
+ = −
+ = −
= −
3.
6 ( 1) 5
5
1 ( 2) 1
m
− − − −
= = = −
− − −
,
so the slope is –5.
Using the point (–2, –1), we get the following point-
slope equation:
1 1( )
( 1) 5[ ( 2)]
1 5( 2)
y y m x x
y x
y x
− = −
− − = − − −
+ = − +
Using the point (–1, –6), we get the following point-
slope equation:
1 1( )
( 6) 5[ ( 1)]
6 5( 1)
y y m x x
y x
y x
− = −
− − = − − −
+ = − +
Solve the equation for y:
1 5( 2)
1 5 10
5 11.
y x
y x
y x
+ = − +
+ = − −
= − −
4. The slope m is 3
5
and the y-intercept is 1, so one
point on the line is (0, 1). We can find a second point
on the line by using the slope 3 Rise
5 Run
:m = = starting at
the point (0, 1), move 3 units up and 5 units to the
right, to obtain the point (5, 4).

Section 2.3 Linear Functions and Slope
5. 3y = is a horizontal line.
6. All ordered pairs that are solutions of 3x = − have a
value of x that is always –3. Any value can be used
for y.
7. 3 6 12 0
6 3 12
3 12
6 6
1
2
2
x y
y x
y x
y x
+ − =
= − +
−
= +
= − +
The slope is
1
2
− and the y-intercept is 2.
8. Find the x-intercept:
3 2 6 0
3 2(0) 6 0
3 6 0
3 6
2
x y
x
x
x
x
− − =
− − =
− =
=
=
Find the y-intercept:
3 2 6 0
3(0) 2 6 0
2 6 0
2 6
3
x y
y
y
y
y
− − =
− − =
− − =
− =
= −
9. First find the slope.
Change in 57.64 57.04 0.6
0.016
Change in 354 317 37
y
m
x
−
= = = ≈
−
Use the point-slope form and then find slope-
intercept form.
1 1( )
57.04 0.016( 317)
57.04 0.016 5.072
0.016 51.968
( ) 0.016 52.0
y y m x x
y x
y x
y x
f x x
− = −
− = −
− = −
= +
= +
Find the temperature at a concentration of 600 parts
per million.
( ) 0.016 52.0
(600) 0.016(600) 52.0
61.6
f x x
f
= +
= +
=
The temperature at a concentration of 600 parts per
million would be 61.6 F.°
1. scatter plot; regression
2. 2 1
2 1
y y
x x
−
−
3. positive
4. negative
5. zero
6. undefined
7. 1 1( )y y m x x− = −

8. y mx b= + ; slope; y-intercept
9. (0,3) ; 2; 5
10. horizontal
11. vertical
12. general
Exercise Set 2.3
1.
10 7 3
;
8 4 4
m
−
= =
−
rises
2.
4 1 3
3;
3 2 1
m
−
= = =
−
rises
3.
2 1 1
;
2 ( 2) 4
m
−
= =
− −
rises
4.
4 3 1
;
2 ( 1) 3
m
−
= =
− −
rises
5.
2 ( 2) 0
0;
3 4 1
m
− −
= = =
− −
horizontal
6.
1 ( 1) 0
0;
3 4 1
m
− − −
= = =
− −
horizontal
7.
1 4 5
5;
1 ( 2) 1
m
− − −
= = = −
− − −
falls
8.
2 ( 4) 2
1;
4 6 2
m
− − −
= = = −
− −
falls
9.
2 3 5
5 5 0
m
− − −
= =
−
undefined; vertical
10.
5 ( 4) 9
3 3 0
m
− −
= =
−
undefined; vertical
11. 1 12, 3, 5;m x y= = =
point-slope form: y – 5 = 2(x – 3);
slope-intercept form: 5 2 6
2 1
y x
y x
− = −
= −
12. point-slope form: y – 3 = 4(x – 1);
1 14, 1, 3;m x y= = =
slope-intercept form: y = 4x – 1
13. 1 16, 2, 5;m x y= = − =
point-slope form: y – 5 = 6(x + 2);
6 17
y x
y x
− = +
= +
14. point-slope form: y + 1 = 8(x – 4);
1 18, 4, 1;m x y= = = −
15. 1 13, 2, 3;m x y= − = − = −
point-slope form: y + 3 = –3(x + 2);
3 9
y x
y x
+ = − −
= − −
16. point-slope form: y + 2 = –5(x + 4);
1 15, 4, 2;m x y= − = − = −
slope-intercept form: y = –5x – 22
17. 1 14, 4, 0;m x y= − = − =
point-slope form: y – 0 = –4(x + 4);
slope-intercept form: 4( 4)
4 16
y x
y x
= − +
= − −
18. point-slope form: y + 3 = –2(x – 0)
1 12, 0, 3;m x y= − = = −
19. 1 1
1
1, , 2;
2
m x y
−
= − = = −
point-slope form:
1
2 1 ;
2
y x
 
+ = − + 
 
slope-intercept form:
1
2
2
5
2
y x
y x
+ = − −
= − −
20. point-slope form:
1
1( 4);
4
y x+ = − +
1 1
1
1, 4, ;
4
m x y= − = − = −
17
4
y x= − −
21. 1 1
1
, 0, 0;
2
m x y= = =
point-slope form:
1
0 ( 0);
2
y x− = −
1
2
y x=

1
0 ( 0);
3
y x− = −
1 1
1
, 0, 0;
3
m x y= = =
1
3
y x=
23. 1 1
2
, 6, 2;
3
m x y= − = = −
point-slope form:
2
2 ( 6);
3
y x+ = − −
2
2 4
3
2
2
3
y x
y x
+ = − +
= − +
3
4 ( 10);
5
y x+ = − −
1 1
3
, 10, 4;
5
m x y= − = = −
3
2
5
y x= − +
25.
10 2 8
2
5 1 4
m
−
= = =
−
;
point-slope form: y – 2 = 2(x – 1) using
1 1( , ) (1, 2)x y = , or y – 10 = 2(x – 5) using
1 1( , ) (5, 10)x y = ;
slope-intercept form: 2 2 2or
10 2 10,
2
y x
y x
y x
− = −
− = −
=
26.
15 5 10
2
8 3 5
m
−
= = =
−
;
( ) ( )1 1, 3,5x y = ,or y – 15 = 2(x – 8) using
( ) ( )1 1, 8,15x y = ;
27.
3 0 3
1
0 ( 3) 3
m
−
= = =
− −
;
point-slope form: y – 0 = 1(x + 3) using
1 1( , ) ( 3, 0)x y = − , or y – 3 = 1(x – 0) using
1 1( , ) (0, 3)x y = ; slope-intercept form: y = x + 3
28.
2 0 2
1
0 ( 2) 2
m
−
= = =
− −
;
point-slope form: y – 0 = 1(x + 2) using
( ) ( )1 1, 2,0x y = − , or y – 2 = 1(x – 0) using
( ) ( )1 1, 0,2x y = ;
slope-intercept form: y = x + 2
29.
4 ( 1)
1
2 ( 3) 5
m
− − 5
= = =
− −
;
point-slope form: y + 1 = 1(x + 3) using
1 1( , ) ( 3, 1)x y = − − , or y – 4 = 1(x – 2) using
1 1( , ) (2, 4)x y = ; slope-intercept form:
1 3or
4 2
2
y x
y x
y x
+ = +
− = −
= +
30.
1 ( 4) 3
1
1 ( 2) 3
m
− − −
= = =
− −
;
( ) ( )1 1, 2, 4x y = − − , or y + 1 = 1(x – 1) using
( ) ( )1 1, 1, 1x y = −
slope-intercept form: y = x – 2
31.
6 ( 2) 8 4
3 ( 3) 6 3
m
− −
= = =
− −
;
point-slope form:
4
2 ( 3)
3
y x+ = + using
1 1( , ) ( 3, 2)x y = − − , or
4
6 ( 3)
3
y x− = − using
1 1( , ) (3, 6)x y = ;
4
2 4or
3
4
6 4,
3
4
2
3
y
x
y x
y x
+ = +
− = −
= +
32.
2 6 8 4
3 ( 3) 6 3
m
− − −
= = = −
− −
;
point-slope form:
4
6 ( 3)
3
y x− = − + using
( ) ( )1 1, 3,6x y = − , or
4
2 ( 3)
3
y x+ = − − using
( ) ( )1 1, 3, 2x y = − ;
4
2
3
y x= − +

33.
1 ( 1) 0
0
4 ( 3) 7
m
− − −
= = =
− −
;
1 1( , ) ( 3, 1)x y = − − , or y + 1 = 0(x – 4) using
1 1( , ) (4, 1)x y = − ;
slope-intercept form: 1 0,so
1
y
y
+ =
= −
34.
5 ( 5) 0
0
6 ( 2) 8
m
− − −
= = =
− −
;
( ) ( )1 1, 2, 5x y = − − , or y + 5 = 0(x – 6) using
( ) ( )1 1, 6, 5x y = − ;
slope-intercept form: 5 0, so
5
y
y
+ =
= −
35.
0 4 4
1
2 2 4
m
− −
= = =
− − −
;
1 1( , ) (2, 4)x y = , or y – 0 = 1(x + 2) using
1 1( , ) ( 2, 0)x y = − ;
slope-intercept form: 9 2,or
2
y x
y x
− = −
= +
36.
0 ( 3) 3 3
1 1 2 2
m
− −
= = = −
− − −
point-slope form:
3
3 ( 1)
2
y x+ = − − using
( ) ( )1 1, 1, 3x y = − , or
3
0 ( 1)
2
y x− = − + using
( ) ( )1 1, 1,0x y = − ;
3 3
3 , or
2 2
3 3
2 2
y x
y x
+ = − +
= − −
37.
( )1 1
2 2
4 0 4
8
0
m
−
= = =
− −
;
1 1( , ) (0, 4)x y = , or ( )1
2
0 8y x− = + using
( )1
1 1 2
( , ) , 0x y = − ; or ( )1
2
0 8y x− = +
slope-intercept form: 8 4y x= +
38.
2 0 2 1
0 4 4 2
m
− − −
= = =
− −
;
point-slope form:
1
0 ( 4)
2
y x− = − using
( ) ( )1 1, 4,0x y = ,
or
1
2 ( 0)
2
y x+ = − using ( ) ( )1 1, 0, 2x y = − ;
1
2
2
y x= −
39. m = 2; b = 1
40. m = 3; b = 2
41. m = –2; b = 1
42. m = –3; b = 2

43.
3
;
4
m = b = –2
44.
3
; 3
4
m b= = −
45.
3
;
5
m = − b = 7
46.
2
; 6
5
m b= − =
47.
1
; 0
2
m b= − =
48.
1
; 0
3
m b= − =
49.
50.
51.
52.
53.

54.
55.
56.
57. 3 18 0x − =
3 18
6
x
x
=
=
58. 3 12 0x + =
3 12
4
x
x
= −
= −
59. a. 3 5 0
5 3
3 5
x y
y x
y x
+ − =
− = −
= − +
b. m = –3; b = 5
c.
60. a. 4 6 0
6 4
4 6
x y
y x
y x
+ − =
− = −
= − +
b. 4; 6m b= − =
c.
61. a. 2 3 18 0x y+ − =
2 18 3
3 2 18
2 18
3 3
2
6
3
x y
y x
y x
y x
− = −
− = −
= −
− −
= − +
b.
2
;
3
m = − b = 6
c.
62. a. 4 6 12 0x y+ + =
4 12 6
6 4 12
4 12
6 6
2
2
3
x y
y x
y x
y x
+ = −
− = +
= +
− −
= − −

b.
2
;
3
m = − b = –2
c.
63. a. 8 4 12 0
8 12 4
4 8 12
8 12
4 4
2 3
x y
x y
y x
y x
y x
− − =
− =
= −
= −
= −
b. m = 2; b = –3
c.
64. a. 6 5 20 0x y− − =
6 20 5
5 6 20
6 20
5 5
6
4
5
x y
y x
y x
y x
− =
= −
= −
= −
b.
6
; 4
5
m b= = −
c.
65. a. 3 9 0y − =
3 9
3
y
y
=
=
b. 0; 3m b= =
c.
66. a. 4 28 0y + =
4 28
7
y
y
= −
= −
b. 0; 7m b= = −
c.
6 2 12 0
6 2(0) 12 0
6 12 0
6 12
2
x y
x
x
x
x
− − =
− − =
− =
=
=
6 2 12 0
6(0) 2 12 0
2 12 0
2 12
6
x y
y
y
y
y
− − =
− − =
− − =
− =
= −

6 9 18 0
6 9(0) 18 0
6 18 0
6 18
3
x y
x
x
x
x
− − =
− − =
− =
=
=
6 9 18 0
6(0) 9 18 0
9 18 0
9 18
2
x y
y
y
y
y
− − =
− − =
− − =
− =
= −
2 3 6 0
2 3(0) 6 0
2 6 0
2 6
3
x y
x
x
x
x
+ + =
+ + =
+ =
= −
= −
2 3 6 0
2(0) 3 6 0
3 6 0
3 6
2
x y
y
y
y
y
+ + =
+ + =
+ =
= −
= −
3 5 15 0
3 5(0) 15 0
3 15 0
3 15
5
x y
x
x
x
x
+ + =
+ + =
+ =
= −
= −
3 5 15 0
3(0) 5 15 0
5 15 0
5 15
3
x y
y
y
y
y
+ + =
+ + =
+ =
= −
= −
8 2 12 0
8 2(0) 12 0
8 12 0
8 12
8 12
8 8
3
2
x y
x
x
x
x
x
− + =
− + =
+ =
= −
−
=
−
=
8 2 12 0
8(0) 2 12 0
2 12 0
2 12
6
x y
y
y
y
y
− + =
− + =
− + =
− = −
= −

6 3 15 0
6 3(0) 15 0
6 15 0
6 15
6 15
6 6
5
2
x y
x
x
x
x
x
− + =
− + =
+ =
= −
−
=
= −
6 3 15 0
6(0) 3 15 0
3 15 0
3 15
5
x y
y
y
y
y
− + =
− + =
− + =
− = −
=
73. 0
0
a a a
m
b b b
− −
= = = −
−
Since a and b are both positive,
a
b
− is
negative. Therefore, the line falls.
74.
( )
0
0
b b b
m
a a a
− − −
= = = −
− −
b
a
− is
negative. Therefore, the line falls.
75. ( )
0
b c b c
m
a a
+ −
= =
−
The slope is undefined.
The line is vertical.
76. ( )
( )
a c c a
m
a a b b
+ −
= =
− −
a
b
is positive.
Therefore, the line rises.
77. Ax By C
By Ax C
A C
y x
B B
+ =
= − +
= − +
The slope is
A
B
− and the y − intercept is .
C
B
78. Ax By C
Ax C By
A C
x y
B B
= −
+ =
+ =
The slope is
A
B
and the y − intercept is .
C
B
79. 4
3
1 3
4
3
2
6 4
2
2
y
y
y
y
y
−
− =
−
−
− =
−
= −
= −
− =
80.
( )
( )
1 4
3 4 2
1 4
3 4 2
1 4
3 6
6 3 4
6 12 3
18 3
6
y
y
y
y
y
y
y
− −
=
− −
− −
=
+
− −
=
= − −
= − −
= −
− =
81. ( )
( )
( )
3 4 6
4 3 6
3 3
4 2
x f x
f x x
f x x
− =
− = − +
= −

82. ( )
( )
( )
6 5 20
5 6 20
6
4
5
x f x
f x x
f x x
− =
− = − +
= −
83. Using the slope-intercept form for the equation
of a line:
( )1 2 3
1 6
5
b
b
b
− = − +
− = − +
=
84.
( )
3
6 2
2
6 3
3
b
b
b
− = − +
− = − +
− =
85. 1 3 2 4, , ,m m m m
86. 2 1 4 3, , ,b b b b
87. a. First, find the slope using ( )20,38.9 and
( )30,47.8 .
47.8 38.9 8.9
0.89
30 20 10
m
−
= = =
−
Then use the slope and one of the points to
write the equation in point-slope form.
( )
( )
( )
1 1
47.8 0.89 30
or
38.9 0.89 20
y y m x x
y x
y x
− = −
− = −
− = −
b. ( )
( )
47.8 0.89 30
47.8 0.89 26.7
0.89 21.1
0.89 21.1
y x
y x
y x
f x x
− = −
− = −
= +
= +
c. ( )40 0.89(40) 21.1 56.7f = + =
The linear function predicts the percentage of
never married American females, ages 25 – 29,
to be 56.7% in 2020.
88. a. First, find the slope using ( )20,51.7 and
( )30,62.6 .
51.7 62.6 10.9
1.09
20 30 10
m
− −
= = =
− −
Then use the slope and one of the points to
write the equation in point-slope form.
( )
( )
( )
1 1
62.6 1.09 30
or
51.7 1.09 20
y y m x x
y x
y x
− = −
− = −
− = −
b. ( )
( )
62.6 1.09 30
62.6 1.09 32.7
1.09 29.9
1.09 29.9
y x
y x
y x
f x x
− = −
− = −
= +
= +
c. ( )35 1.09(35) 29.9 68.05f = + =
The linear function predicts the percentage of
never married American males, ages 25 – 29, to
be 68.05% in 2015.
89. a.
b.
Change in 74.3 70.0
0.215
Change in 40 20
y
m
x
−
= = =
−
1 1( )
70.0 0.215( 20)
70.0 0.215 4.3
0.215 65.7
( ) 0.215 65.7
y y m x x
y x
y x
y x
E x x
− = −
− = −
− = −
= +
= +
c. ( ) 0.215 65.7
(60) 0.215(60) 65.7
78.6
E x x
E
= +
= +
=
The life expectancy of American men born in
2020 is expected to be 78.6.

90. a.
b.
Change in 79.7 74.7
0.17
Change in 40 10
y
m
x
−
= = ≈
−
1 1( )
74.7 0.17( 10)
74.7 0.17 1.7
0.17 73
( ) 0.17 73
y y m x x
y x
y x
y x
E x x
− = −
− = −
− = −
= +
= +
c. ( ) 0.17 73
(60) 0.17(60) 73
83.2
E x x
E
= +
= +
=
The life expectancy of American women born
in 2020 is expected to be 83.2.
91. (10, 230) (60, 110) Points may vary.
110 230 120
2.4
60 10 50
230 2.4( 10)
230 2.4 24
2.4 254
m
y x
y x
y x
−
= = − = −
−
− = − −
− = − +
= − +
Answers will vary for predictions.
100. Two points are (0,4) and (10,24).
24 4 20
2.
10 0 10
m
−
= = =
−
101. Two points are (0, 6) and (10, –24).
24 6 30
3.
10 0 10
m
− − −
= = = −
−
Check: : 3 6y mx b y x= + = − + .
102. Two points are (0,–5) and (10,–10).
10 ( 5) 5 1
.
10 0 10 2
m
− − − −
= = = −
−
103. Two points are (0, –2) and (10, 5.5).
5.5 ( 2) 7.5 3
0.75 or .
10 0 10 4
m
− −
= = =
−
Check:
3
: 2
4
y mx b y x= + = − .
104. a. Enter data from table.
b.

c. 22.96876741a = −
260.5633751b =
0.8428126855r = −
d.
Sample explanation: Linear functions never change
from increasing to decreasing.
Sample explanation: Since college cost are going
up, this function has a positive slope.
Sample explanation: The slope of line’s whose
equations are in this form can be determined in
several ways. One such way is to rewrite the
equation in slope-intercept form.
108. makes sense
A sample change is: It is possible for m to equal b.
A sample change is: Slope-intercept form is
y mx b= + . Vertical lines have equations of the
form x a= . Equations of this form have undefined
slope and cannot be written in slope-intercept form.
111. true
A sample change is: The graph of 7x = is a vertical
line through the point (7, 0).
113. We are given that the interceptx − is 2− and the
intercept is 4y − . We can use the points
( ) ( )2,0 and 0,4− to find the slope.
( )
4 0 4 4
2
0 2 0 2 2
m
−
= = = =
− − +
Using the slope and one of the intercepts, we can
write the line in point-slope form.
( )
( )( )
( )
1 1
0 2 2
2 2
2 4
2 4
y y m x x
y x
y x
y x
x y
− = −
− = − −
= +
= +
− + =
Find the x– and y–coefficients for the equation of the
line with right-hand-side equal to 12. Multiply both
sides of 2 4x y− + = by 3 to obtain 12 on the right-
hand-side.
( ) ( )
2 4
3 2 3 4
6 3 12
x y
x y
x y
− + =
− + =
− + =
Therefore, the coefficient of x is –6 and the
coefficient of y is 3.
114. We are given that the intercept is 6y − − and the
slope is
1
.
2
So the equation of the line is
1
6.
2
y x= −
We can put this equation in the form ax by c+ = to
find the missing coefficients.
( )
1
6
2
1
6
2
1
2 2 6
2
2 12
2 12
y x
y x
y x
y x
x y
= −
− = −
 
− = − 
 
− = −
− =
Therefore, the coefficient of x is 1 and the
coefficient of y is 2.−
116. Let (25, 40) and (125, 280) be ordered pairs
(M, E) where M is degrees Madonna and E is degrees
Elvis. Then
280 40 240
2.4
125 25 100
m
−
= = =
−
. Using ( ) ( )1 1, 25,40x y = ,
point-slope form tells us that
E – 40 = 2.4 (M – 25) or
E = 2.4 M – 20.

Section 2.4 More on Slope
118. Let x = the number of years after 1994.
714 17 289
17 425
25
x
x
x
− =
− = −
=
Violent crime incidents will decrease to 289 per
100,000 people 25 years after 1994, or 2019.
119.
3 2
1
4 3
3 2
12 12 1
4 3
x x
x x
+ −
≥ +
+ −   
≥ +   
   
3( 3) 4( 2) 12
3 9 4 8 12
3 9 4 4
5
5
x x
x x
x x
x
x
+ ≥ − +
+ ≥ − +
+ ≥ +
≥
≤
The solution set is { } ( ]5 or ,5 .x x ≤ −∞
120. 3 2 6 9 15x + − <
3 2 6 24
3 2 6 24
3 3
2 6 8
8 2 6 8
14 2 2
7 1
x
x
x
x
x
x
+ <
+
<
+ <
− < + <
− < <
− < <
The solution set is { } ( )7 1 or 7,1x x− < < − .
121. Since the slope is the same as the slope of 2 1,y x= +
then 2.m =
( )
( )( )
( )
1 1
1 2 3
1 2 3
1 2 6
2 7
y y m x x
y x
y x
y x
y x
− = −
− = − −
− = +
− = +
= +
122. Since the slope is the negative reciprocal of
1
,
4
−
then 4.m =
( )
( )
1 1
( 5) 4 3
5 4 12
4 17 0
4 17 0
y y m x x
y x
y x
x y
x y
− = −
− − = −
+ = −
− + + =
− − =
123. 2 1
2 1
2 2
( ) ( ) (4) (1)
4 1
4 1
4 1
15
3
5
f x f x f f
x x
− −
=
− −
−
=
−
=
=
Section 2.4
1. The slope of the line 3 1y x= + is 3.
( )
1 1( )
5 3 ( 2)
5 3( 2) point-slope
5 3 6
3 11 slope-intercept
y y m x x
y x
y x
y x
y x
− = −
− = − −
− = +
− = +
= +
2. a. Write the equation in slope-intercept form:
3 12 0
3 12
1
4
3
x y
y x
y x
+ − =
= − +
= − +
The slope of this line is
1
3
− thus the slope of
any line perpendicular to this line is 3.
b. Use 3m = and the point (–2, –6) to write the
equation.
( )
1 1( )
( 6) 3 ( 2)
6 3( 2)
6 3 6
3 0
3 0 general form
y y m x x
y x
y x
y x
x y
x y
− = −
− − = − −
+ = +
+ = +
− + =
− =

3.
Change in 15 11.2 3.8
0.29
Change in 2013 2000 13
y
m
x
−
= = = ≈
−
The slope indicates that the number of U.S. men
living alone increased at a rate of 0.29 million each
year.
The rate of change is 0.29 million men per year.
4. a.
3 3
2 1
2 1
( ) ( ) 1 0
1
1 0
f x f x
x x
− −
= =
− −
b.
3 3
2 1
2 1
( ) ( ) 2 1 8 1
7
2 1 1
f x f x
x x
− − −
= = =
− −
c.
3 3
2 1
2 1
( ) ( ) 0 ( 2) 8
4
0 ( 2) 2
f x f x
x x
− − −
= = =
− − −
5. 2 1
2 1
( ) ( ) (3) (1)
3 1
0.05 0.03
3 1
0.01
f x f x f f
x x
− −
=
− −
−
=
−
=
The average rate of change in the drug's
concentration between 1 hour and 3 hours is 0.01
mg per 100 mL per hour.
1. the same
2. 1−
3.
1
3
− ; 3
4. 2− ;
1
2
5. y; x
6. 2 1
2 1
( ) ( )f x f x
x x
−
−
Exercise Set 2.4
1. Since L is parallel to 2 ,y x= we know it will have
slope 2.m = We are given that it passes through
( )4,2 . We use the slope and point to write the
equation in point-slope form.
( )
( )
1 1
2 2 4
y y m x x
y x
− = −
− = −
Solve for y to obtain slope-intercept form.
( )2 2 4
2 2 8
2 6
y x
y x
y x
− = −
− = −
= −
In function notation, the equation of the line is
( ) 2 6.f x x= −
2. L will have slope 2m = − . Using the point and the
slope, we have ( )4 2 3 .y x− = − − Solve for y to
obtain slope-intercept form.
( )
4 2 6
2 10
2 10
y x
y x
f x x
− = − +
= − +
= − +
3. Since L is perpendicular to 2 ,y x= we know it will
have slope
1
.
2
m = − We are given that it passes
through (2,4). We use the slope and point to write the
equation in point-slope form.
( )
( )
1 1
1
4 2
2
y y m x x
y x
− = −
− = − −
( )
1
4 2
2
1
4 1
2
1
5
2
y x
y x
y x
− = − −
− = − +
= − +
In function notation, the equation of the line is
( )
1
5.
2
f x x= − +

4. L will have slope
1
.
2
m = The line passes through
(–1, 2). Use the slope and point to write the equation
in point-slope form.
( )( )
( )
1
2 1
2
1
2 1
2
y x
y x
− = − −
− = +
1 1
2
2 2
1 1
2
2 2
y x
y x
− = +
= + +
( )
1 5
2 2
1 5
2 2
y x
f x x
= +
= +
5. m = –4 since the line is parallel to
1 14 3; 8, 10;y x x y= − + = − = −
point-slope form: y + 10 = –4(x + 8)
slope-intercept form: y + 10 = –4x – 32
y = –4x – 42
6. m = –5 since the line is parallel to 5 4y x= − + ;
1 12, 7x y= − = − ;
point-slope form: y + 7 = –5(x + 2)
5 17
y x
y x
+ = − −
= − −
7. m = –5 since the line is perpendicular to
1 1
1
6; 2, 3;
5
y x x y= + = = −
point-slope form: y + 3 = –5(x – 2)
5 7
y x
y x
+ = − +
= − +
8. 3m = − since the line is perpendicular to
1
7
3
y x= + ;
1 14, 2x y= − = ;
point-slope form: 2 3( 4)y x− = − +
3 10
y x
y x
− = − −
= − −
9. 2 3 7 0
3 2 7
2 7
3 3
x y
y x
y x
− − =
− = − +
= −
The slope of the given line is
2 2
, so
3 3
m = since the
lines are parallel.
point-slope form:
2
2 ( 2)
3
y x− = +
general form: 2 3 10 0x y− + =
10. 3 2 0
2 3 5
3 5
2 2
x y
y x
y x
− − =
− = − +
= −
3
2
, so
3
2
m = since the
lines are parallel.
point-slope form:
3
3 ( 1)
2
y x− = +
general form: 3 2 9 0x y− + =
11. 2 3 0
2 3
1 3
2 2
x y
y x
y x
− − =
− = − +
= −
1
2
, so m = –2 since the
lines are perpendicular.
point-slope form: ( )7 –2 4y x+ = −
general form: 2 1 0x y+ − =
12. 7 12 0
7 12
1 12
7 7
x y
y x
y x
+ − =
= − +
−
= +
1
7
− , so m = 7 since the
lines are perpendicular.
point-slope form: y + 9 = 7(x – 5)
general form: 7 44 0x y− − =
13.
15 0 15
3
5 0 5
−
= =
−
14.
24 0 24
6
4 0 4
−
= =
−

15.
2 2
5 2 5 (3 2 3) 25 10 (9 6)
5 3 2
20
2
10
+ ⋅ − + ⋅ + − +
=
−
=
=
16.
( ) ( )2 2
6 2 6 (3 2 3) 36 12 9 6 21
7
6 3 3 3
− − − ⋅ − − −
= = =
−
17.
9 4 3 2 1
9 4 5 5
− −
= =
−
18.
16 9 4 3 1
16 9 7 7
− −
= =
−
19. Since the line is perpendicular to 6x = which is a
vertical line, we know the graph of f is a horizontal
line with 0 slope. The graph of f passes through
( )1,5− , so the equation of f is ( ) 5.f x =
20. Since the line is perpendicular to 4x = − which is a
vertical line, we know the graph of f is a horizontal
line with 0 slope. The graph of f passes through
( )2,6− , so the equation of f is ( ) 6.f x =
21. First we need to find the equation of the line with
x − intercept of 2 and y − intercept of 4.− This line
will pass through ( )2,0 and ( )0, 4 .− We use these
points to find the slope.
4 0 4
2
0 2 2
m
− − −
= = =
− −
Since the graph of f is perpendicular to this line, it
will have slope
1
.
2
m = −
Use the point ( )6,4− and the slope
1
2
− to find the
equation of the line.
( )
( )( )
( )
( )
1 1
1
4 6
2
1
4 6
2
1
4 3
2
1
1
2
1
1
2
y y m x x
y x
y x
y x
y x
f x x
− = −
− = − − −
− = − +
− = − −
= − +
= − +
22. First we need to find the equation of the line with
x − intercept of 3 and y − intercept of 9.− This line
will pass through ( )3,0 and ( )0, 9 .− We use these
points to find the slope.
9 0 9
3
0 3 3
m
− − −
= = =
− −
Since the graph of f is perpendicular to this line, it
will have slope
1
.
3
m = −
Use the point ( )5,6− and the slope
1
3
− to find the
equation of the line.
( )
( )( )
( )
( )
1 1
1
6 5
3
1
6 5
3
1 5
6
3 3
1 13
3 3
1 13
3 3
y y m x x
y x
y x
y x
y x
f x x
− = −
− = − − −
− = − +
− = − −
= − +
= − +

23. First put the equation 3 2 4 0x y− − = in slope-
intercept form.
3 2 4 0
2 3 4
3
2
2
x y
y x
y x
− − =
− = − +
= −
The equation of f will have slope
2
3
− since it is
perpendicular to the line above and the same
y − intercept 2.−
So the equation of f is ( )
2
2.
3
f x x= − −
24. First put the equation 4 6 0x y− − = in slope-intercept
form.
4 6 0
4 6
4 6
x y
y x
y x
− − =
− = − +
= −
The equation of f will have slope
1
4
− since it is
perpendicular to the line above and the same
y − intercept 6.−
So the equation of f is ( )
1
6.
4
f x x= − −
25. ( ) 0.25 22p x x= − +
26. ( ) 0.22 3p x x= +
27.
1163 617 546
137
1998 1994 4
m
−
= = ≈
−
There was an average increase of approximately 137
discharges per year.
28.
623 1273 650
130
2006 2001 5
m
− −
= = ≈ −
−
There was an average decrease of approximately 130
29. a. 3 2
( ) 1.1 35 264 557f x x x x= − + +
3 2
(0) 1.1(0) 35(0) 264(0) 557 557f = − + + =
3 2
(4) 1.1(4) 35(4) 264(4) 557 1123.4f = − + + =
1123.4 557
142
4 0
m
−
= ≈
−
b. This overestimates by 5 discharges per year.
30. a. 3 2
( ) 1.1 35 264 557f x x x x= − + +
3 2
(0) 1.1(7) 35(7) 264(7) 557 1067.3f = − + + =
3 2
(12) 1.1(12) 35(12) 264(12) 557 585.8f = − + + =
585.8 1067.3
96
12 7
m
−
= ≈ −
−
b. This underestimates the decrease by 34
37.
1
1
3
3 2
y x
y x
= +
= − −
a. The lines are perpendicular because their slopes
are negative reciprocals of each other. This is
verified because product of their slopes is –1.
b. The lines do not appear to be perpendicular.
c. The lines appear to be perpendicular. The
calculator screen is rectangular and does not
have the same width and height. This causes
the scale of the x–axis to differ from the scale
on the y–axis despite using the same scale in
the window settings. In part (b), this causes the
lines not to appear perpendicular when indeed
they are. The zoom square feature compensates
for this and in part (c), the lines appear to be
perpendicular.
Sample explanation: Perpendicular lines have slopes
with opposite signs.

39. makes sense
Sample explanation: Slopes can be used for
segments of the graph.
41. makes sense
42. Write 0Ax By C+ + = in slope-intercept form.
0Ax By C
By Ax C
By Ax C
B B B
A C
y x
B B
+ + =
= − −
−
= −
= − −
A
B
− .
The slope of any line perpendicular to
0Ax By C+ + = is
B
A
.
43. The slope of the line containing ( )1, 3− and ( )2,4−
has slope
( )4 3 4 3 7 7
2 1 3 3 3
m
− − +
= = = = −
− − − −
Solve 2 0Ax y+ − = for y to obtain slope-intercept
form.
2 0
2
Ax y
y Ax
+ − =
= − +
So the slope of this line is .A−
This line is perpendicular to the line above so its
slope is
3
.
7
Therefore,
3
7
A− = so
3
.
7
A = −
44. 24 3( 2) 5( 12)x x+ + = −
24 3 6 5 60
3 30 5 60
90 2
45
x x
x x
x
x
+ + = −
+ = −
=
=
45. Let x = the television’s price before the
reduction.
0.30 980
0.70 980
980
0.70
1400
x x
x
x
x
− =
=
=
=
Before the reduction the television’s price was
$1400.
46. 2/3 1/3
2 5 3 0x x− − =
Let 1/3
t x= .
2
2 5 3 0
(2 1)( 3) 0
t t
t t
− − =
+ − =
2 1 0 or 3 0
2 = 1
1
= 3
2
t t
t
t t
+ = − =
−
− =
1/3 1/3
3
3
1
3
2
1
3
2
1
27
8
x x
x x
x x
= − =
 
= − = 
 
= − =
The solution set is
1
, 27
8
 
− 
 
.
47. a.
b.
c. The graph in part (b) is the graph in part (a)
shifted down 4 units.
48. a.

Mid-Chapter 2 Check Point
b.
shifted to the right 2 units.
49. a.
b.
reflected across the y-axis.
1. The relation is not a function.
The domain is {1,2}.
The range is { 6,4,6}.−
2. The relation is a function.
The domain is {0,2,3}.
The range is {1,4}.
The domain is { | 2 2}.x x− ≤ <
The range is { | 0 3}.y y≤ ≤
The domain is { | 3 4}.x x− < ≤
The range is { | 1 2}.y y− ≤ ≤
The domain is { 2, 1,0,1,2}.− −
The range is { 2, 1,1,3}.− −
The domain is { | 1}.x x ≤
The range is { | 1}.y y ≥ −
7. 2
5x y+ =
2
5y x= − +
For each value of x, there is one and only one value
for y, so the equation defines y as a function of x.
8. 2
5x y+ =
2
5
5
y x
y x
= −
= ± −
Since there are values of x that give more than one
value for y (for example, if x = 4, then
5 4 1y = ± − = ± ), the equation does not define y as
a function of x.
9. No vertical line intersects the graph in more than one
point. Each value of x corresponds to exactly one
value of y.
10. Domain: ( ),−∞ ∞
11. Range: ( ],4−∞
12. x-intercepts: –6 and 2
13. y-intercept: 3
14. increasing: (–∞, –2)
15. decreasing: (–2, ∞)
16. 2x = −
17. ( 2) 4f − =
18. ( 4) 3f − =
19. ( 7) 2f − = − and (3) 2f = −
20. ( 6) 0f − = and (2) 0f =
21. ( )6,2−
22. (100)f is negative.

23. neither; ( )f x x− ≠ and ( )f x x− ≠ −
24. 2 1
2 1
( ) ( ) (4) ( 4) 5 3
1
4 ( 4) 4 4
f x f x f f
x x
− − − − −
= = = −
− − − +
2
2
2
1
1
1
x y
x y
x y
= +
− = +
= − −
( )
2
2
2
1
1
1
x y
x y
x y
= +
= − +
= +
axis.
( )
2
2
2
2
1
1
1
1
x y
x y
x y
x y
= +
− = − +
− = +
= − −
( )
3
3
3
1
1
1
y x
y x
y x
= −
= − −
= − −
3
3
3
1
1
1
y x
y x
y x
= −
− = −
= − +
( )
3
3
3
3
1
1
1
1
y x
y x
y x
y x
= −
− = − −
− = − −
= +
27.
28.
29.
30.
31.

32.
33.
34.
35.
36. 5 3y x= −
3
5
y x= −
37. 5 20y =
4y =
38.
39. a. 2
2
( ) 2( ) 5
2 5
f x x x
x x
− = − − − −
= − − −
neither; ( )f x x− ≠ and ( )f x x− ≠ −
b.
( ) ( )f x h f x
h
+ −
( )
2 2
2 2 2
2
2( ) ( ) 5 ( 2 5)
2 4 2 5 2 5
4 2
4 2 1
4 2 1
x h x h x x
h
x xh h x h x x
h
xh h h
h
h x h
h
x h
− + + + − − − + −
=
− − − + + − + − +
=
− − +
=
− − +
=
= − − +
40.
30 if 0 200
( )
30 0.40( 200) if 200
t
C x
t t
≤ ≤
= 
+ − >
a. (150) 30C =
b. (250) 30 0.40(250 200) 50C = + − =
41. 1 1( )y y m x x− = −
( )3 2 ( 4)
3 2( 4)
3 2 8
2 5
( ) 2 5
y x
y x
y x
y x
f x x
− = − − −
− = − +
− = − −
= − −
= − −
42.
Change in 1 ( 5) 6
2
Change in 2 ( 1) 3
y
m
x
− −
= = = =
− −
1 1( )y y m x x− = −
( )1 2 2
1 2 4
2 3
( ) 2 3
y x
y x
y x
f x x
− = −
− = −
= −
= −

43. 3 5 0x y− − =
3 5
3 5
y x
y x
− = − +
= −
The slope of the given line is 3, and the lines are
parallel, so 3.m =
1 1( )
( 4) 3( 3)
4 3 9
3 13
( ) 3 13
y y m x x
y x
y x
y x
f x x
− = −
− − = −
+ = −
= −
= −
44. 2 5 10 0x y− − =
5 2 10
5 2 10
5 5 5
2
2
5
y x
y x
y x
− = − +
− −
= +
− − −
= −
2
5
, and the lines are
perpendicular, so
5
.
2
m = −
( )
1 1( )
5
( 3) ( 4)
2
5
3 10
2
5
13
2
5
( ) 13
2
y y m x x
y x
y x
y x
f x x
− = −
− − = − − −
+ = − −
= − −
= − −
45. 1
Change in 0 ( 4) 4
Change in 7 2 5
y
m
x
− −
= = =
−
2
Change in 6 2 4
Change in 1 ( 4) 5
y
m
x
−
= = =
− −
The slope of the lines are equal thus the lines are
parallel.
46. a.
Change in 42 26 16
0.16
Change in 180 80 100
y
m
x
−
= = = =
−
b. For each minute of brisk walking, the
percentage of patients with depression in
remission increased by 0.16%. The rate of
change is 0.16% per minute of brisk walking.
47. 2 1
2 1
( ) ( ) (2) ( 1)
2 ( 1)
f x f x f f
x x
− − −
=
− − −
( ) ( )2 2
3(2) 2 3( 1) ( 1)
2 1
2
− − − − −
=
+
=
Section 2.5
1. Shift up vertically 3 units.
2. Shift to the right 4 units.
3. Shift to the right 1 unit and down 2 units.
4. Reflect about the x-axis.

Section 2.5 Transformations of Functions
5. Reflect about the y-axis.
6. Vertically stretch the graph of ( )f x x= .
7. a. Horizontally shrink the graph of ( )y f x= .
b. Horizontally stretch the graph of ( )y f x= .
8. The graph of ( )y f x= is shifted 1 unit left, shrunk
by a factor of
1
,
3
reflected about the x-axis, then
shifted down 2 units.
9. The graph of 2
( )f x x= is shifted 1 unit right,
stretched by a factor of 2, then shifted up 3 units.
1. vertical; down
2. horizontal; to the right
3. x-axis
4. y-axis
5. vertical; y
6. horizontal; x
7. false
Exercise Set 2.5
1.
2.

3.
4.
5.
6.
7.
8.
9.
10.
11.
12.

13.
14.
15.
16.
17.
18.
19.
20.
21.
22.

23.
24.
25.
26.
27.
28.
29.
30.
31.
32.
33.

34.
35.
36.
37.
38.
39.
40.
41.
42.
43.

44.
45.
46.
47.
48.
49.
50.
51.
52.
53.

54.
55.
56.
57.
58.
59.
60.
61.
62.
63.
64.
65.

66.
67.
68.
69.
70.
71.
72.
73.
74.
75.
76.
77.
78.

79.
80.
81.
82.
83.
84.
85.
86.
87.
88.
89.
90.
91.

92.
93.
94.
95.
96.
97.
98.
99.
100.
101.
102.
103.

104.
105.
106.
107.
108.
109.
110.
111.
112.
113.
114.
115.

116.
117.
118.
119.
120.
121.
122.
123. 2y x= −
124. 3
2y x= − +
125. 2
( 1) 4y x= + −
126. 2 1y x= − +
127. a. First, vertically stretch the graph of ( )f x x= by
the factor 2.9; then shift the result up 20.1 units.
b. ( ) 2.9 20.1f x x= +
(48) 2.9 48 20.1 40.2f = + ≈
The model describes the actual data very well.
c. 2 1
2 1
( ) ( )f x f x
x x
−
−
( ) ( )
(10) (0)
10 0
2.9 10 20.1 2.9 0 20.1
10 0
29.27 20.1
10
0.9
f f−
=
−
+ − +
=
−
−
=
≈
0.9 inches per month
d. 2 1
2 1
( ) ( )f x f x
x x
−
−
( ) ( )
(60) (50)
60 50
2.9 60 20.1 2.9 50 20.1
60 50
42.5633 40.6061
10
0.2
f f−
=
−
+ − +
=
−
−
=
≈
This rate of change is lower than the rate of
change in part (c). The relative leveling off of
the curve shows this difference.

128. a. First, vertically stretch the graph of ( )f x x=
by the factor 3.1; then shift the result up 19
units.
b. ( ) 3.1 19f x x= +
(48) 3.1 48 19 40.5f = + ≈
The model describes the actual data very well.
c. 2 1
2 1
( ) ( )f x f x
x x
−
−
( ) ( )
(10) (0)
10 0
3.1 10 19 3.1 0 19
10 0
28.8031 19
10
1.0
f f−
=
−
+ − +
=
−
−
=
≈
1.0 inches per month
d. 2 1
2 1
( ) ( )f x f x
x x
−
−
( ) ( )
(60) (50)
60 50
3.1 60 19 3.1 50 19
60 50
43.0125 40.9203
10
0.2
f f−
=
−
+ − +
=
−
−
=
≈
This rate of change is lower than the rate of
change in part (c). The relative leveling off of
the curve shows this difference.
135. a.
b.
136. a.
b.
137. makes sense
138. makes sense
Sample explanation: The reprogram should be
( 1).y f t= +
Sample explanation: The reprogram should be
( 1).y f t= −
A sample change is: The graph of g is a translation
of f three units to the left and three units upward.
A sample change is: The graph of f is a reflection of
the graph of y x= in the x-axis, while the graph of
g is a reflection of the graph of y x= in the y-axis.
A sample change is: The stretch will be 5 units and
the downward shift will be 10 units.

144. true
145. 2
( ) ( 4)g x x= − +
146. ( ) – – 5 1g x x= +
147. ( ) 2 2g x x= − − +
148. 21
( ) 16 – 1
4
g x x= − −
149. (–a, b)
150. (a, 2b)
151. (a + 3, b)
152. (a, b – 3)
153. Let x = the width of the rectangle.
Let x + 13 = the length of the rectangle.
2 2
2( 13) 2 82
2 26 2 82
4 26 82
4 56
56
4
14
13 27
l w P
x x
x x
x
x
x
x
x
+ =
+ + =
+ + =
+ =
=
=
=
+ =
The dimensions of the rectangle are 14 yards by 27
yards.
154. 10 4x x+ − =
( ) ( )
2 2
2
2
10 4
10 4
10 8 16
0 7 6
0 ( 6)( 1)
x x
x x
x x x
x x
x x
+ = +
+ = +
+ = + +
= + +
= + +
6 0 or 1 0
6 1
x x
x x
+ = + =
= − = −
–6 does not check and must be rejected.
The solution set is { }1 .−
155. 2
(3 7 )(5 2 ) 15 6 35 14
15 6 35 14( 1)
15 6 35 14
29 29
i i i i i
i i
i i
i
− + = + − −
= + − − −
= + − +
= −
156. 2 2 2
3 2 2
3 2 2
3 2
(2 1)( 2) 2 ( 2) 1( 2)
2 2 4 2
2 2 4 2
2 5 2
x x x x x x x x
x x x x x
x x x x x
x x x
− + − = + − − + −
= + − − − +
= + − − − +
= + − +
157. ( ) ( )
2 2
2
2
2
( ) 2 ( ) 6 3 4 2(3 4) 6
9 24 16 6 8 6
9 24 6 16 8 6
9 30 30
f x f x x x
x x x
x x x
x x
− + = − − − +
= − + − + +
= − − + + +
= − +
158.
2 2 2
3 3 31
x x
x xx
x x
= =
−− −
Section 2.6
1. a. The function 2
( ) 3 17f x x x= + − contains
neither division nor an even root. The domain of
f is the set of all real numbers or ( ),−∞ ∞ .
b. The denominator equals zero when x = 7 or x =
–7. These values must be excluded from the
domain.
domain of g = ( ) ( ) ( ), 7 7,7 7, .−∞ − − ∞ 
c. Since ( ) 9 27h x x= − contains an even root; the
quantity under the radical must be greater than or
equal to 0.
9 27 0
9 27
3
x
x
x
− ≥
≥
≥
Thus, the domain of h is { 3}x x ≥ , or the
interval [ )3, .∞

Section 2.6 Combinations of Functions; Composite Functions
d. Since the denominator of ( )j x contains an even
root; the quantity under the radical must be greater
than or equal to 0. But that quantity must also not
be 0 (because we cannot have division by 0).
Thus, 24 3x− must be strictly greater than 0.
24 3 0
3 24
8
x
x
x
− >
− > −
<
Thus, the domain of j is { 8}x x < , or the
interval ( ,8).−∞
2. a.
( )2
2
2
( )( ) ( ) ( )
5 1
5 1
6
f g x f x g x
x x
x x
x x
+ = +
= − + −
= − + −
= − + −
domain: ( , )−∞ ∞
b.
( )2
2
2
( )( ) ( ) ( )
5 1
5 1
4
f g x f x g x
x x
x x
x x
− = −
= − − −
= − − +
= − + −
domain: ( , )−∞ ∞
c. ( )( )
( ) ( )
2
2 2
3 2
3 2
( )( ) 5 1
1 5 1
5 5
5 5
fg x x x
x x x
x x x
x x x
= − −
= − − −
= − − +
= − − +
domain: ( , )−∞ ∞
d.
2
( )
( )
( )
5
, 1
1
f f x
x
g g x
x
x
x
 
= 
 
−
= ≠ ±
−
domain: ( , 1) ( 1,1) (1, )−∞ − − ∞ 
3. a. ( )( ) ( ) ( )
3 1
f g x f x g x
x x
+ = +
= − + +
b. domain of f: 3 0
3
[3, )
x
x
− ≥
≥
∞
domain of g: 1 0
1
[ 1, )
x
x
+ ≥
≥ −
− ∞
The domain of f + g is the set of all real
numbers that are common to the domain of f
and the domain of g. Thus, the domain of f + g
is [3, ∞).
4. a. ( )( )
( ) ( )
2 2
2 2
2
( 2.6 49 3994) ( 0.6 7 2412)
2.6 49 3994 0.6 7 2412
3.2 56 6406
B D x
B x D x
x x x x
x x x x
x x
+
= +
= − + + + − + +
= − + + − + +
= − + +
b. ( )( )
( )( )
2
2
3.2 56 6406
5 3.2(3) 56(3) 6406
6545.2
B D x x x
B D
+ = − + +
+ = − + +
=
The number of births and deaths in the U.S. in
2003 was 6545.2 thousand.
c. ( )( )B D x+ overestimates the actual number of
births and deaths in 2003 by 7.2 thousand.
5. a. ( ) ( )( ) ( )f g x f g x=
( )2
2
2
5 2 1 6
10 5 5 6
10 5 1
x x
x x
x x
= − − +
= − − +
= − +
b. ( ) ( )( ) ( )g f x g f x=
( ) ( )
2
2
2
2
2 5 6 5 6 1
2(25 60 36) 5 6 1
50 120 72 5 6 1
50 115 65
x x
x x x
x x x
x x
= + − + −
= + + − − −
= + + − − −
= + +
c. ( ) 2
( ) 10 5 1f g x x x= − +
( ) 2
( 1) 10( 1) 5( 1) 1
10 5 1
16
f g − = − − − +
= + +
=

6. a.
4 4
( )( )
1 1 22
x
f g x
x
x
= =
++

b. domain:
1
0,
2
x x x
 
≠ ≠ − 
 
or ( )
1 1
, ,0 0,
2 2
   
−∞ − − ∞   
   
 
7. ( ) 2
where ( ) ; ( ) 5h x f g f x x g x x= = = +

1. zero
2. negative
3. ( ) ( )f x g x+
4. ( ) ( )f x g x−
5. ( ) ( )f x g x⋅
6.
( )
( )
f x
g x
; ( )g x
7. ( , )−∞ ∞
8. (2, )∞
9. (0,3) ; (3, )∞
10. composition; ( )( )f g x
11. f; ( )g x
12. composition; ( )( )g f x
13. g; ( )f x
14. false
15. false
16. 2
Exercise Set 2.6
1. The function contains neither division nor an even
root. The domain ( ),= −∞ ∞
3. The denominator equals zero when 4.x = This value
must be excluded from the domain.
domain: ( ) ( ),4 4, .−∞ ∞
4. The denominator equals zero when 5.x = − This
value must be excluded from the domain.
domain: ( ) ( ), 5 5, .−∞ − − ∞
7. The values that make the denominator equal zero
domain: ( ) ( ) ( ), 3 3,5 5,−∞ − − ∞ 
8. The values that make the denominator equal zero
domain: ( ) ( ) ( ), 4 4,3 3,−∞ − − ∞ 
9. The values that make the denominators equal zero
domain: ( ) ( ) ( ), 7 7,9 9,−∞ − − ∞ 
10. The values that make the denominators equal zero
domain: ( ) ( ) ( ), 8 8,10 10,−∞ − − ∞ 
11. The first denominator cannot equal zero. The values
that make the second denominator equal zero must be
excluded from the domain.
domain: ( ) ( ) ( ), 1 1,1 1,−∞ − − ∞ 
12. The first denominator cannot equal zero. The values
that make the second denominator equal zero must be
excluded from the domain.
domain: ( ) ( ) ( ), 2 2,2 2,−∞ − − ∞ 
13. Exclude x for 0x = .
Exclude x for
3
1 0
x
− = .
( )
3
1 0
3
1 0
3 0
3
3
x
x x
x
x
x
x
− =
 
− = 
 
− =
− = −
=
domain: ( ) ( ) ( ),0 0,3 3,−∞ ∞ 

14. Exclude x for 0x = .
Exclude x for
4
1 0
x
− = .
( )
4
1 0
4
1 0
4 0
4
4
x
x x
x
x
x
x
− =
 
− = 
 
− =
− = −
=
domain: ( ) ( ) ( ),0 0,4 4,−∞ ∞ 
15. Exclude x for 1 0x − = .
1 0
1
x
x
− =
=
Exclude x for
4
2 0
1x
− =
−
.
( ) ( )( )
( )
4
2 0
1
4
1 2 1 0
1
4 2 1 0
4 2 2 0
2 6 0
2 6
3
x
x x
x
x
x
x
x
x
− =
−
 
− − = − − 
− − =
− + =
− + =
− = −
=
domain: ( ) ( ) ( ),1 1,3 3,−∞ ∞ 
16. Exclude x for 2 0x − = .
2 0
2
x
x
− =
=
Exclude x for
4
3 0
2x
− =
−
.
( ) ( )( )
( )
4
3 0
2
4
2 3 2 0
2
4 3 2 0
4 3 6 0
3 10 0
3 10
10
3
x
x x
x
x
x
x
x
x
− =
−
 
− − = − − 
− − =
− + =
− + =
− = −
=
domain: ( )
10 10
,2 2, ,
3 3
   
−∞ ∞   
   
 
17. The expression under the radical must not be
negative.
3 0
3
x
x
− ≥
≥
domain: [ )3,∞
negative.
2 0
2
x
x
+ ≥
≥ −
domain: [ )2,− ∞
19. The expression under the radical must be positive.
3 0
3
x
x
− >
>
domain: ( )3,∞
20. The expression under the radical must be positive.
2 0
2
x
x
+ >
> −
domain: ( )2,− ∞
negative.
5 35 0
5 35
7
x
x
x
+ ≥
≥ −
≥ −
domain: [ )7,− ∞
negative.
7 70 0
7 70
10
x
x
x
− ≥
≥
≥
domain: [ )10,∞
negative.
24 2 0
2 24
2 24
2 2
12
x
x
x
x
− ≥
− ≥ −
− −
≤
− −
≤
domain: ( ],12−∞

negative.
84 6 0
6 84
6 84
6 6
14
x
x
x
x
− ≥
− ≥ −
− −
≤
− −
≤
domain: ( ],14−∞
25. The expressions under the radicals must not be
negative.
2 0
2
x
x
− ≥
≥
and
3 0
3
x
x
+ ≥
≥ −
To make both inequalities true, 2x ≥ .
domain: [ )2,∞
26. The expressions under the radicals must not be
negative.
3 0
3
x
x
− ≥
≥
and
4 0
4
x
x
+ ≥
≥ −
To make both inequalities true, 3x ≥ .
domain: [ )3,∞
negative.
2 0
2
x
x
− ≥
≥
The denominator equals zero when 5.x =
domain: [ ) ( )2,5 5, .∞
negative.
3 0
3
x
x
− ≥
≥
The denominator equals zero when 6.x =
domain: [ ) ( )3,6 6, .∞
29. Find the values that make the denominator equal zero
and must be excluded from the domain.
( ) ( )
( )( )
3 2
2
2
5 4 20
5 4 5
5 4
( 5)( 2)( 2)
x x x
x x x
x x
x x x
− − +
= − − −
= − −
= − + −
–2, 2, and 5 must be excluded.
domain: ( ) ( ) ( ) ( ), 2 2,2 2,5 5,−∞ − − ∞  
30. Find the values that make the denominator equal zero
and must be excluded from the domain.
( ) ( )
( )( )
3 2
2
2
2 9 18
2 9 2
2 9
( 2)( 3)( 3)
x x x
x x x
x x
x x x
− − +
= − − −
= − −
= − + −
–3, 2, and 3 must be excluded.
domain: ( ) ( ) ( ) ( ), 3 3,2 2,3 3,−∞ − − ∞  
31. (f + g)(x) = 3x + 2
domain: ( , )−∞ ∞
(f – g)(x) = f(x) – g(x)
= (2x + 3) – (x – 1)
= x + 4
domain: ( , )−∞ ∞
2
( )( ) ( ) ( )
(2 3) ( 1)
2 3
fg x f x g x
x x
x x
= ⋅
= + ⋅ −
= + −
domain: ( , )−∞ ∞
( ) 2 3
( )
( ) 1
f f x x
x
g g x x
  +
= = 
− 
domain: ( ) ( ),1 1,−∞ ∞
32. (f + g)(x) = 4x – 2
domain: (–∞, ∞)
(f – g)(x) = (3x – 4) – (x + 2) = 2x – 6
(fg)(x) = (3x – 4)(x + 2) = 3x2 + 2x – 8
3 4
( )
2
f x
x
g x
  −
= 
+ 
domain: ( ) ( ), 2 2,−∞ − − ∞
33. 2
( )( ) 3 5f g x x x+ = + −
domain: ( , )−∞ ∞
2
( )( ) 3 5f g x x x− = − + −
domain: ( , )−∞ ∞
2 3 2
( )( ) ( 5)(3 ) 3 15fg x x x x x= − = −
domain: ( , )−∞ ∞
2
5
( )
3
f x
x
g x
  −
= 
 
domain: ( ) ( ),0 0,−∞ ∞

34. 2
( )( ) 5 – 6f g x x x+ = +
2
( )( ) –5 – 6f g x x x− = +
2 3 2
( )( ) ( – 6)(5 ) 5 – 30fg x x x x x= =
2
6
( )
5
f x
x
g x
  −
= 
 
domain: ( ) ( ),0 0,−∞ ∞
35. 2
( )( ) 2 – 2f g x x+ =
2
( – )( ) 2 – 2 – 4f g x x x=
2
3 2
( )( ) (2 – – 3)( 1)
2 – 4 – 3
fg x x x x
x x x
= +
= +
2
2 – – 3
( )
1
(2 – 3)( 1)
2 – 3
( 1)
f x x
x
g x
x x
x
x
 
= 
+ 
+
= =
+
domain: ( ) ( ), 1 1,−∞ − − ∞
36. 2
( )( ) 6 – 2f g x x+ =
2
( )( ) 6 2f g x x x− = −
2 3 2
( )( ) (6 1)( 1) 6 7 1fg x x x x x x= − − − = − +
2
6 1
( )
1
f x x
x
g x
  − −
= 
− 
domain: ( ) ( ),1 1,−∞ ∞
37. 2 2
( )( ) (3 ) ( 2 15)f g x x x x+ = − + + −
2 12x= −
2 2
2
( )( ) (3 ) ( 2 15)
2 2 18
f g x x x x
x x
− = − − + −
= − − +
2 2
4 3 2
( )( ) (3 )( 2 15)
2 18 6 45
fg x x x x
x x x x
= − + −
= − − + + −
2
2
3
( )
2 15
f x
x
g x x
  −
= 
+ − 
domain: ( ) ( ) ( ), 5 5,3 3,−∞ − − ∞ 
38. 2 2
( )( ) (5 ) ( 4 12)f g x x x x+ = − + + −
4 7x= −
2 2
2
( )( ) (5 ) ( 4 12)
2 4 17
f g x x x x
x x
− = − − + −
= − − +
2 2
4 3 2
( )( ) (5 )( 4 12)
4 17 20 60
fg x x x x
x x x x
= − + −
= − − + + −
2
2
5
( )
4 12
f x
x
g x x
  −
= 
+ − 
domain: ( ) ( ) ( ), 6 6,2 2,−∞ − − ∞ 
39. ( )( ) 4f g x x x+ = + −
domain: [0, )∞
( )( ) 4f g x x x− = − +
domain: [0, )∞
( )( ) ( 4)fg x x x= −
domain: [0, )∞
( )
4
f x
x
g x
 
= 
− 
domain: [ ) ( )0,4 4,∞
40. ( )( ) 5f g x x x+ = + −
domain: [0, )∞
( )( ) 5f g x x x− = − +
domain: [0, )∞
( )( ) ( 5)fg x x x= −
domain: [0, )∞
( )
5
f x
x
g x
 
= 
− 
domain: [ ) ( )0,5 5,∞

41.
1 1 2 2 2
( )( ) 2 2
x
f g x
x x x x
+
+ = + + = + =
domain: ( ) ( ),0 0,−∞ ∞
1 1
( )( ) 2 2f g x
x x
− = + − =
domain: ( ) ( ),0 0,−∞ ∞
2 2
1 1 2 1 2 1
( )( ) 2
x
fg x
x x x x x
+ 
= + ⋅ = + = 
 
domain: ( ) ( ),0 0,−∞ ∞
1
1
2 1
( ) 2 2 1x
x
f
x x x
g x
+   
= = + ⋅ = +   
  
domain: ( ) ( ),0 0,−∞ ∞
42.
1 1
( )( ) 6 6f g x
x x
+ = − + =
domain: ( ) ( ),0 0,−∞ ∞
1 1 2 6 2
( )( ) 6 – 6
x
f g x
x x x x
−
− = − = − =
domain: ( ) ( ),0 0,−∞ ∞
2 2
1 1 6 1 6 1
( )( ) 6
x
fg x
x x x x x
− 
= − ⋅ = − = 
 
domain: ( ) ( ),0 0,−∞ ∞
1
1
6 1
( ) 6 6 1x
x
f
x x x
g x
−   
= = − ⋅ = −   
  
domain: ( ) ( ),0 0,−∞ ∞
43. ( )( ) ( ) ( )f g x f x g x+ = +
2 2
2
5 1 4 2
9 9
9 1
9
x x
x x
x
x
+ −
= +
− −
−
=
−
domain: ( ) ( ) ( ), 3 3,3 3,−∞ − − ∞ 
2 2
2
( )( ) ( ) ( )
5 1 4 2
9 9
3
9
1
3
f g x f x g x
x x
x x
x
x
x
− = −
+ −
= −
− −
+
=
−
=
−
domain: ( ) ( ) ( ), 3 3,3 3,−∞ − − ∞ 
( )
2 2
22
( )( ) ( ) ( )
5 1 4 2
9 9
(5 1)(4 2)
9
fg x f x g x
x x
x x
x x
x
= ⋅
+ −
= ⋅
− −
+ −
=
−
domain: ( ) ( ) ( ), 3 3,3 3,−∞ − − ∞ 
2
2
2
2
5 1
9( )
4 2
9
5 1 9
4 29
5 1
4 2
x
f xx
xg
x
x x
xx
x
x
+
  −=  − 
−
+ −
= ⋅
−−
+
=
−
The domain must exclude –3, 3, and any values that
make 4 2 0.x − =
4 2 0
4 2
1
2
x
x
x
− =
=
=
domain: ( ) ( ) ( ) ( )1 1
2 2
, 3 3, ,3 3,−∞ − − ∞  

44. ( )( ) ( ) ( )f g x f x g x+ = +
2 2
2
3 1 2 4
25 25
5 3
25
x x
x x
x
x
+ −
= +
− −
−
=
−
domain: ( ) ( ) ( ), 5 5,5 5,−∞ − − ∞ 
2 2
2
( )( ) ( ) ( )
3 1 2 4
25 25
5
25
1
5
f g x f x g x
x x
x x
x
x
x
− = −
+ −
= −
− −
+
=
−
=
−
domain: ( ) ( ) ( ), 5 5,5 5,−∞ − − ∞ 
( )
2 2
22
( )( ) ( ) ( )
3 1 2 4
25 25
(3 1)(2 4)
25
fg x f x g x
x x
x x
x x
x
= ⋅
+ −
= ⋅
− −
+ −
=
−
domain: ( ) ( ) ( ), 5 5,5 5,−∞ − − ∞ 
2
2
2
2
3 1
25( )
2 4
25
3 1 25
2 425
3 1
2 4
x
f xx
xg
x
x x
xx
x
x
+
  −=  − 
−
+ −
= ⋅
−−
+
=
−
make 2 4 0.x − =
2 4 0
2 4
2
x
x
x
− =
=
=
domain: ( ) ( ) ( ) ( ), 5 5,2 2,5 5,−∞ − − ∞  
45. ( )( ) ( ) ( )f g x f x g x+ = +
2
2
8 6
2 3
8 ( 3) 6( 2)
( 2)( 3) ( 2)( 3)
8 24 6 12
( 2)( 3) ( 2)( 3)
8 30 12
( 2)( 3)
x
x x
x x x
x x x x
x x x
x x x x
x x
x x
= +
− +
+ −
= +
− + − +
+ −
= +
− + − +
+ −
=
− +
domain: ( ) ( ) ( ), 3 3,2 2,−∞ − − ∞ 
( )( ) ( ) ( )f g x f x g x+ = −
2
2
8 6
2 3
8 ( 3) 6( 2)
( 2)( 3) ( 2)( 3)
8 24 6 12
( 2)( 3) ( 2)( 3)
8 18 12
( 2)( 3)
x
x x
x x x
x x x x
x x x
x x x x
x x
x x
= −
− +
+ −
= −
− + − +
+ −
= −
− + − +
+ +
=
− +
domain: ( ) ( ) ( ), 3 3,2 2,−∞ − − ∞ 
( )( ) ( ) ( )
8 6
2 3
48
( 2)( 3)
fg x f x g x
x
x x
x
x x
= ⋅
= ⋅
− +
=
− +
domain: ( ) ( ) ( ), 3 3,2 2,−∞ − − ∞ 
8
2( )
6
3
8 3
2 6
4 ( 3)
3( 2)
x
f xx
g
x
x x
x
x x
x
  −= 
 
+
+
= ⋅
−
+
=
−
make 3( 2) 0.x − =
3( 2) 0
3 6 0
3 6
2
x
x
x
x
− =
− =
=
=
domain: ( ) ( ) ( ), 3 3,2 2,−∞ − − ∞ 

46. ( )( ) ( ) ( )f g x f x g x+ = +
2
2
9 7
4 8
9 ( 8) 7( 4)
( 4)( 8) ( 4)( 8)
9 72 7 28
( 4)( 8) ( 4)( 8)
9 79 28
( 4)( 8)
x
x x
x x x
x x x x
x x x
x x x x
x x
x x
= +
− +
+ −
= +
− + − +
+ −
= +
− + − +
+ −
=
− +
domain: ( ) ( ) ( ), 8 8,4 4,−∞ − − ∞ 
( )( ) ( ) ( )f g x f x g x+ = −
2
2
9 7
4 8
9 ( 8) 7( 4)
( 4)( 8) ( 4)( 8)
9 72 7 28
( 4)( 8) ( 4)( 8)
9 65 28
( 4)( 8)
x
x x
x x x
x x x x
x x x
x x x x
x x
x x
= −
− +
+ −
= −
− + − +
+ −
= −
− + − +
+ +
=
− +
domain: ( ) ( ) ( ), 8 8,4 4,−∞ − − ∞ 
( )( ) ( ) ( )
9 7
4 8
63
( 4)( 8)
fg x f x g x
x
x x
x
x x
= ⋅
= ⋅
− +
=
− +
domain: ( ) ( ) ( ), 8 8,4 4,−∞ − − ∞ 
9
4( )
7
8
9 8
4 7
9 ( 8)
7( 4)
x
f xx
g
x
x x
x
x x
x
  −= 
 
+
+
= ⋅
−
+
=
−
make 7( 4) 0.x − =
7( 4) 0
7 28 0
7 28
4
x
x
x
x
− =
− =
=
=
domain: ( ) ( ) ( ), 8 8,4 4,−∞ − − ∞ 
47. ( )( ) 4 1f g x x x+ = + + −
domain: [1, )∞
( )( ) 4 1f g x x x− = + − −
domain: [1, )∞
2
( )( ) 4 1 3 4fg x x x x x= + ⋅ − = + −
domain: [1, )∞
4
( )
1
f x
x
g x
  +
= 
− 
domain: (1, )∞
48. ( )( ) 6 3f g x x x+ = + + −
domain: [3, ∞)
( )( ) 6 3f g x x x− = + − −
domain: [3, ∞)
2
( )( ) 6 3 3 18fg x x x x x= + ⋅ − = + −
domain: [3, ∞)
6
( )
3
f x
x
g x
  +
= 
− 
domain: (3, ∞)
49. ( )( ) 2 2f g x x x+ = − + −
domain: {2}
( )( ) 2 2f g x x x− = − − −
domain: {2}
2
( )( ) 2 2 4 4fg x x x x x= − ⋅ − = − + −
domain: {2}
2
( )
2
f x
x
g x
  −
= 
− 
domain: ∅
50. ( )( ) 5 5f g x x x+ = − + −
domain: {5}
( )( ) 5 5f g x x x− = − − −
domain: {5}
2
( )( ) 5 5 10 25fg x x x x x= − ⋅ − = − + −
domain: {5}
5
( )
5
f x
x
g x
  −
= 
− 
domain: ∅

51. f(x) = 2x; g(x) = x + 7
a. ( )( ) 2( 7) 2 14f g x x x= + = +
b. ( )( ) 2 7g f x x= +
c. ( )(2) 2(2) 14 18f g = + =
52. f(x) = 3x; g(x) = x – 5
a. ( )( ) 3( 5) 3 15f g x x x= − = −
b. ( )( ) 3 – 5g f x x=
c. ( )(2) 3(2) 15 9f g = − = −
53. f(x) = x + 4; g(x) = 2x + 1
a. ( )( ) (2 1) 4 2 5f g x x x= + + = +
b. ( )( ) 2( 4) 1 2 9g f x x x= + + = +
c. ( )(2) 2(2) 5 9f g = + =
54. f(x) = 5x + 2 ; g(x) = 3x – 4
a. ( )( ) 5(3 4) 2 15 18f g x x x= − + = −
b. ( )( ) 3(5 2) 4 15 2g f x x x= + − = +
c. ( )(2) 15(2) 18 12f g = − =
55. f(x) = 4x – 3; 2
( ) 5 2g x x= −
a. 2
2
( )( ) 4(5 2) 3
20 11
f g x x
x
= − −
= −

b. 2
2
2
( )( ) 5(4 3) 2
5(16 24 9) 2
80 120 43
g f x x
x x
x x
= − −
= − + −
= − +

c. 2
( )(2) 20(2) 11 69f g = − =
56. 2
( ) 7 1; ( ) 2 – 9f x x g x x= + =
a. 2 2
( )( ) 7(2 9) 1 14 62f g x x x= − + = −
b. 2
2
2
( )( ) 2(7 1) 9
2(49 14 1) 9
98 28 7
g f x x
x x
x x
= + −
= + + −
= + −

c. 2
( )(2) 14(2) 62 6f g = − = −
57. 2 2
( ) 2; ( ) 2f x x g x x= + = −
a. 2 2
4 2
4 2
( )( ) ( 2) 2
4 4 2
4 6
f g x x
x x
x x
= − +
= − + +
= − +

b. 2 2
4 2
4 2
( )( ) ( 2) 2
4 4 2
4 2
g f x x
x x
x x
= + −
= + + −
= + +

c. 4 2
( )(2) 2 4(2) 6 6f g = − + =
58. 2 2
( ) 1; ( ) 3f x x g x x= + = −
a. 2 2
4 2
4 2
( )( ) ( 3) 1
6 9 1
6 10
f g x x
x x
x x
= − +
= − + +
= − +

b. 2 2
4 2
4 2
( )( ) ( 1) 3
2 1 3
2 2
g f x x
x x
x x
= + −
= + + −
= + −

c. 4 2
( )(2) 2 6(2) 10 2f g = − + =
59. 2
( ) 4 ; ( ) 2 5f x x g x x x= − = + +
a. ( )2
2
2
( )( ) 4 2 5
4 2 5
2 1
f g x x x
x x
x x
= − + +
= − − −
= − − −

b. ( ) ( )
2
2
2
2
( )( ) 2 4 4 5
2(16 8 ) 4 5
32 16 2 4 5
2 17 41
g f x x x
x x x
x x x
x x
= − + − +
= − + + − +
= − + + − +
= − +

c. 2
( )(2) 2(2) 2 1 11f g = − − − = −

60. 2
( ) 5 2; ( ) 4 1f x x g x x x= − = − + −
a. ( )2
2
2
( )( ) 5 4 1 2
5 20 5 2
5 20 7
f g x x x
x x
x x
= − + − −
= − + − −
= − + −

b. ( ) ( )
2
2
2
2
( )( ) 5 2 4 5 2 1
(25 20 4) 20 8 1
25 20 4 20 8 1
25 40 13
g f x x x
x x x
x x x
x x
= − − + − −
= − − + + − −
= − + − + − −
= − + −

c. 2
( )(2) 5(2) 20(2) 7 13f g = − + − =
61. ( ) ;f x x= g(x) = x – 1
a. ( )( ) 1f g x x= −
b. ( )( ) 1g f x x= −
c. ( )(2) 2 1 1 1f g = − = =
62. ( ) ; ( ) 2f x x g x x= = +
a. ( )( ) 2f g x x= +
b. ( )( ) 2g f x x= +
c. ( )(2) 2 2 4 2f g = + = =
63. f(x) = 2x – 3;
3
( )
2
x
g x
+
=
a.
3
( )( ) 2 3
2
3 3
x
f g x
x
x
+ 
= − 
 
= + −
=

b.
(2 3) 3 2
( )( )
2 2
x x
g f x x
− +
= = =
c. ( )(2) 2f g =
64.
3
( ) 6 3; ( )
6
x
f x x g x
+
= − =
a.
3
( )( ) 6 3 3 3
6
x
f g x x x
+ 
= − = + − = 
 

b.
6 3 3 6
( )( )
6 6
x x
g f x x
− +
= = =
c. ( )(2) 2f g =
65.
1 1
( ) ; ( )f x g x
x x
= =
a. 1
1
( )( )
x
f g x x= =
b. 1
1
( )( )
x
g f x x= =
c. ( )(2) 2f g =
66.
2 2
( ) ; ( )f x g x
x x
= =
a. 2
2
( )( )
x
f g x x= =
b. 2
2
( )( )
x
g f x x= =
c. ( )(2) 2f g =
67. a.
1 2
( )( ) , 0
1
3
f g x f x
x
x
 
= = ≠ 
  +

( )
2( )
1
3
2
1 3
x
x
x
x
x
=
 
+ 
 
=
+
b. We must exclude 0 because it is excluded from
g.
We must exclude
1
3
− because it causes the
denominator of f g to be 0.
domain: ( )
1 1
, ,0 0, .
3 3
   
−∞ − − ∞   
   
 

68. a.
1 5 5
( )
1 1 44
x
f g x f
x x
x
 
= = = 
+  +

b. We must exclude 0 because it is excluded from g.
We must exclude
1
4
domain: ( )
1 1
, ,0 0, .
4 4
   
−∞ − − ∞   
   
 
69. a.
4
4
( )( )
4
1
xf g x f
x
x
 
= = 
  +

( )
4
( )
4
1
4
, 4
4
x
x
x
x
x
x
 
 
 =
 
+ 
 
= ≠ −
+
g.
We must exclude 4− because it causes the
domain: ( ) ( ) ( ), 4 4,0 0, .−∞ − − ∞ 
70. a. ( )
6
6 6
6 6 55
xf g x f
x x
x
 
= = = 
+  +

g.
We must exclude
6
5
domain: ( )
6 6
, ,0 0, .
5 5
   
−∞ − − ∞   
   
 
71. a. ( ) ( )2 2f g x f x x= − = −
b. The expression under the radical in f g must
not be negative.
2 0
2
x
x
− ≥
≥
domain: [ )2, .∞
72. a. ( ) ( )3 3f g x f x x= − = −
b. The expression under the radical in f g must
not be negative.
3 0
3
x
x
− ≥
≥
domain: [ )3, .∞
73. a. ( )( ) ( 1 )f g x f x= −
( )
2
1 4
1 4
5
x
x
x
= − +
= − +
= −
b. The domain of f g must exclude any values
that are excluded from g.
1 0
1
1
x
x
x
− ≥
− ≥ −
≤
domain: (−•, 1]., 1].
74. a. ( )( ) ( 2 )f g x f x= −
( )
2
2 1
2 1
3
x
x
x
= − +
= − +
= −
b. The domain of f g must exclude any values
that are excluded from g.
2 0
2
2
x
x
x
− ≥
− ≥ −
≤
domain: (−•, 2]., 2].
75. ( )4
( ) 3 1f x x g x x= = −
76. ( ) ( )3
; 2 5f x x g x x= = −
77. ( ) ( ) 23
9f x x g x x= = −
78. ( ) ( ) 2
; 5 3f x x g x x= = +
79. f(x) = |x| g(x) = 2x – 5
80. f (x) = |x|; g(x) = 3x – 4
81.
1
( ) ( ) 2 3f x g x x
x
= = −
82. ( ) ( )
1
; 4 5f x g x x
x
= = +

83. ( )( ) ( ) ( )3 3 3 4 1 5f g f g+ − = − + − = + =
84. ( )( ) ( ) ( )2 2 2 2 3 1g f g f− − = − − − = − = −
85. ( )( ) ( ) ( ) ( )( )2 2 2 1 1 1fg f g= = − = −
86. ( )
( )
( )
3 0
3 0
3 3
gg
f f
 
= = = 
− 
87. The domain of f g+ is [ ]4,3− .
88. The domain of
f
g
is ( )4,3− .
89. The graph of f g+
90. The graph of f g−
91. ( )( ) ( ) ( )1 ( 1) 3 1f g f g f− = − = − =
92. ( )( ) ( ) ( )1 (1) 5 3f g f g f= = − =
93. ( )( ) ( ) ( )0 (0) 2 6g f g f g= = = −
94. ( )( ) ( ) ( )1 ( 1) 1 5g f g f g− = − = = −
95. ( )( ) 7f g x =
( )2
2
2
2
2
2 3 8 5 7
2 6 16 5 7
2 6 11 7
2 6 4 0
3 2 0
( 1)( 2) 0
x x
x x
x x
x x
x x
x x
− + − =
− + − =
− + =
− + =
− + =
− − =
1 0 or 2 0
1 2
x x
x x
− = − =
= =
96. ( )( ) 5f g x = −
( )2
2
2
2
2
1 2 3 1 5
1 6 2 2 5
6 2 3 5
6 2 8 0
3 4 0
(3 4)( 1) 0
x x
x x
x x
x x
x x
x x
− + − = −
− − + = −
− − + = −
− − + =
+ − =
+ − =
3 4 0 or 1 0
3 4 1
4
3
x x
x x
x
+ = − =
= − =
= −
97. a. ( )( ) ( ) ( )
(1.48 115.1) (1.44 120.9)
2.92 236
M F x M x F x
x x
x
+ = +
= + + +
= +
b. ( )( ) 2.92 236
( )(25) 2.92(25) 236
309
M F x x
M F
+ = +
+ = +
=
The total U.S. population in 2010 was 309
million.
c. It is the same.
98. a. ( )( ) ( ) ( )
(1.44 120.9) (1.48 115.1)
0.04 5.8
F M x F x M x
x x
x
− = −
= + − +
= − +
b. ( )( ) 0.04 5.8
( )(25) 0.04(25) 5.8
4.8
F M x x
F M
− = − +
− = − +
=
In 2010 there were 4.8 million more women
than men.
c. The result in part (b) underestimates the actual
difference by 0.2 million.

99. ( )(20,000)
65(20,000) (600,000 45(20,000))
200,000
R C−
= − +
= −
The company lost $200,000 since costs exceeded
revenues.
(R – C)(30,000)
= 65(30,000) – (600,000 + 45(30,000))
= 0
The company broke even.
(R – C)(40,000)
= 65(40,000) – (600,000 + 45(40,000))
= 200,000
The company gained $200,000 since revenues
exceeded costs.
100. a. The slope for f is -0.44 This is the decrease in
profits for the first store for each year after
2012.
b. The slope of g is 0.51 This is the increase in
profits for the second store for each year after
2012.
c. f + g = -.044x + 13.62 + 0.51x + 11.14
= 0.07x + 24.76
The slope for f + g is 0.07 This is the profit for
the two stores combined for each year after
2012.
101. a. f gives the price of the computer after a $400
discount. g gives the price of the computer after
a 25% discount.
b. ( )( ) 0.75 400f g x x= −
This models the price of a computer after first a
25% discount and then a $400 discount.
c. ( )( ) 0.75( 400)g f x x= −
This models the price of a computer after first a
$400 discount and then a 25% discount.
d. The function f g models the greater discount,
since the 25% discount is taken on the regular
price first.
102. a. f gives the cost of a pair of jeans for which a $5
rebate is offered.
g gives the cost of a pair of jeans that has been
discounted 40%.
b. ( )( ) 0.6 5f g x x= −
The cost of a pair of jeans is 60% of the regular
price minus a $5 rebate.
c. ( )( ) ( )0.6 5g f x x= −
= 0.6x – 3
The cost of a pair of jeans is 60% of the regular
price minus a $3 rebate.
d. f g because of a $5 rebate.
108. When your trace reaches x = 0, the y value disappears
because the function is not defined at x = 0.
109.
( )( ) 2f g x x= −
The domain of g is [ )0,∞ .
The expression under the radical in f g must not
be negative.
2 0
2
2
4
x
x
x
x
− ≥
− ≥ −
≤
≤
domain: [ ]0,4
110. makes sense
111. makes sense
Sample explanation: It is common that f g and
g f are not the same.

Sample explanation: The diagram illustrates
( ) 2
( ) 4.g f x x= +
A sample change is: ( )( ) ( )
( )
2
2
2
2
2
4
4 4
4 4
8
f g x f x
x
x
x
= −
= − −
= − −
= −

A sample change is:
( ) ( )
( ) ( )( ) ( )
( )( ) ( )( ) ( ) ( )
2 ; 3
( ) 3 2(3 ) 6
( ) 3 2 6
f x x g x x
f g x f g x f x x x
g f x g f x g f x x x
= =
= = = =
= = = =


A sample change is:
( ) ( )( ) ( )( ) 4 4 7 5f g f g f= = =
117. true
118. ( )( ) ( )( )f g x f g x= − 
( ( )) ( ( )) since is even
( ( )) ( ( )) so is even
f g x f g x g
f g x f g x f g
= −
= 
120.
1 3
1
5 2 4
x x x− +
− = −
( ) ( )
1 3
20 20 1
5 2 4
4 1 10 3 20 5
4 4 10 30 20 5
6 34 20 5
6 5 20 34
1 54
54
x x x
x x x
x x x
x x
x x
x
x
− +   
− = −   
   
− − + = −
− − − = −
− − = −
− + = +
− =
= −
The solution set is {-54}.
121. Let x = the number of bridge crossings at which the
costs of the two plans are the same.
 Discount PassNo Pass
6 30 4
6 4 30
2 30
15
x x
x x
x
x
= +
− =
=
=

The two plans cost the same for 15 bridge crossings.
The monthly cost is ( )$6 15 $90.=
122.
( )
Ax By Cy D
By Cy D Ax
y B C D Ax
D Ax
y
B C
+ = +
− = −
− = −
−
=
−
123. {(4, 2),(1, 1),(1,1),(4,2)}− −
The element 1 in the domain corresponds to two
elements in the range.
Thus, the relation is not a function.
124.
5
4
5
( ) 4
5 4
4 5
( 4) 5
5
4
x
y
y x y
y
xy y
xy y
y x
y
x
= +
 
= + 
 
= +
− =
− =
=
−
125. 2
2
2
1
1
1
1
1
x y
x y
x y
x y
y x
= −
+ =
+ =
+ =
= +

Section 2.7 Inverse Functions
Section 2.7
1. ( )
7
( ) 4 7
4
7 7
x
f g x
x
x
+ 
= − 
 
= + −
=
( )
(4 7) 7
( )
4
4 7 7
4
4
4
x
g f x
x
x
x
− +
=
− +
=
=
=
( ) ( )( ) ( )f g x g f x x= =
2. ( ) 2 7f x x= +
Replace ( )f x with y:
2 7y x= +
Interchange x and y:
2 7x y= +
Solve for y:
2 7
7 2
7
2
x y
x y
x
y
= +
− =
−
=
Replace y with 1
( )f x−
:
1 7
( )
2
x
f x− −
=
3. 3
( ) 4 1f x x= −
3
4 1y x= −
3
4 1x y= −
Solve for y:
3
3
3
3
4 1
1 4
1
4
1
4
x y
x y
x
y
x
y
= −
+ =
+
=
+
=
Replace y with 1
( )f x−
:
1 3
1
( )
4
x
f x− +
=
Alternative form for answer:
3
1 3
3
33 3
3 3 3
3
1 1
( )
4 4
1 2 2 2
4 2 8
2 2
2
x x
f x
x x
x
− + +
= =
+ +
= ⋅ =
+
=
4.
1
( ) , 5
5
x
f x x
x
+
= ≠
−
1
5
x
y
x
+
=
−
1
5
y
x
y
+
=
−
Solve for y:
( )
1
5
5 1
5 1
5 1
( 1) 5 1
5 1
1
y
x
y
x y y
xy x y
xy y x
y x x
x
y
x
+
=
−
− = +
− = +
− = +
− = +
+
=
−
Replace y with 1
( )f x−
:
1 5 1
( )
1
x
f x
x
− +
=
−
5. The graphs of (b) and (c) pass the horizontal line test
and thus have an inverse.
6. Find points of 1
f −
.
( )f x 1
( )f x−
( 2, 2)− − ( 2, 2)− −
( 1,0)− (0, 1)−
(1,2) (2,1)

7. 2
( ) 1f x x= +
2
1y x= +
2
1x y= +
Solve for y:
2
2
1
1
1
x y
x y
x y
= +
− =
− =
Replace y with 1
( )f x−
:
1
( ) 1f x x−
= −
1. inverse
2. x; x
3. horizontal; one-to-one
4. y x=
Exercise Set 2.7
1. ( ) 4 ; ( )
4
( ( )) 4
4
4
( ( ))
4
x
f x x g x
x
f g x x
x
g f x x
= =
 
= = 
 
= =
f and g are inverses.
2.
( )
( )
( ) 6 ; ( )
6
( ) 6
6
6
( )
6
x
f x x g x
x
f g x x
x
g f x x
= =
 
= = 
 
= =
3. f(x) = 3x + 8;
8
( )
3
x
g x
−
=
8
( ( )) 3 8 8 8
3
(3 8) 8 3
( ( ))
3 3
x
f g x x x
x x
g f x x
− 
= + = − + = 
 
+ −
= = =
4.
( )
( )
9
( ) 4 9; ( )
4
9
( ) 4 9 9 9
4
(4 9) 9 4
( )
4 4
x
f x x g x
x
f g x x x
x x
g f x x
−
= + =
− 
= + = − + = 
 
+ −
= = =
5. f(x) = 5x – 9;
5
( )
9
x
g x
+
=
5
( ( )) 5 9
9
5 25
9
9
5 56
9
5 9 5 5 4
( ( ))
9 9
x
f g x
x
x
x x
g f x
+ 
= − 
 
+
= −
−
=
− + −
= =
f and g are not inverses.
6.
( )
( )
3
( ) 3 7; ( )
7
3 3 9 3 40
( ) 3 7 7
7 7 7
3 7 3 3 4
( )
7 7
x
f x x g x
x x x
f g x
x x
g f x
+
= − =
+ + − 
= − = − = 
 
− + −
= =
f and g are not inverses.
7.
3 3
3 3
( ) ; ( ) 4
4
3 3
( ( ))
4 4x x
f x g x
x x
f g x x
= = +
−
= = =
+ −
3
4
3
( ( )) 4
4
3 4
3
4 4
x
g f x
x
x
x
−
= +
− 
= ⋅ + 
 
= − +
=

8.
( )
( )
2
2
5
2 2
( ) ; ( ) 5
5
2 2
( ( ))
5 5 2
2 5
( ) 5 2 5 5 5
2
x
x
f x g x
x x
x
f g x x
x
g f x x x
−
= = +
−
= = =
+ −
− 
= + = + = − + = 
 
9. ( ) ; ( )
( ( )) ( )
( ( )) ( )
f x x g x x
f g x x x
g f x x x
= − = −
= − − =
= − − =
10.
( )
( ) ( )
33
3 33 3
3
3
( ) 4; ( ) 4
( ) 4 4
( ) 4 4 4 4
f x x g x x
f g x x x x
g f x x x x
= − = +
= + − = =
= − + = − + =
11. a. f(x) = x + 3
y = x + 3
x = y + 3
y = x – 3
1
( ) 3f x x−
= −
b. 1
1
( ( )) 3 3
( ( )) 3 3
f f x x x
f f x x x
−
−
= − + =
= + − =
12. a.
1
( ) 5
5
5
5
( ) 5
f x x
y x
x y
y x
f x x−
= +
= +
= +
= −
= −
b. ( )
( )
1
1
( ) 5 5
( ) 5 5
f f x x x
f f x x x
−
−
= − + =
= + − =
13. a.
1
( ) 2
2
2
2
( )
2
f x x
y x
x y
x
y
x
f x−
=
=
=
=
=
b. 1
1
( ( )) 2
2
2
( ( ))
2
x
f f x x
x
f f x x
−
−
 
= = 
 
= =
14. a.
1
( ) 4
4
4
4
( )
4
f x x
y x
x y
x
y
x
f x−
=
=
=
=
=
b. ( )
( )
1
1
( ) 4
4
4
( )
4
x
f f x x
x
f f x x
−
−
 
= = 
 
= =
15. a.
1
( ) 2 3
2 3
2 3
3 2
3
2
3
( )
2
f x x
y x
x y
x y
x
y
x
f x−
= +
= +
= +
− =
−
=
−
=
b. 1
1
3
( ( )) 2 3
2
3 3
2 3 3 2
( ( ))
2 2
x
f f x
x
x
x x
f f x x
−
−
− 
= + 
 
= − +
=
+ −
= = =
16. a. ( ) 3 1
3 1
3 1
1 3
f x x
y x
x y
x y
= −
= −
= −
+ =
1
1
3
1
( )
3
x
y
x
f x−
+
=
+
=
b. ( )
( )
1
1
1
( ) 3 1 1 1
3
3 1 1 3
( )
3 3
x
f f x x x
x x
f f x x
−
−
+ 
= − = + − = 
 
− +
= = =

3
3
3
3
3
1 3
( ) 2
2
2
2
2
( ) 2
. . f x x
y x
x y
x y
y x
f x x−
= +
= +
= +
− =
= −
= −
17 a
b. ( )
3
1 3
( ( )) 2 2
2 2
f f x x
x
x
−
= − +
= − +
=
3 31 3 3
( ( )) 2 2f f x x x x−
= + − = =
18. a. 3
3
3
3
3
1 3
( ) 1
1
1
1
1
( ) 1
f x x
y x
x y
x y
y x
f x x−
= −
= −
= −
+ =
= +
= +
b. ( )
3
1 3
3 31 3 3
( ( )) 1 1
1 1
( ( )) 1 1
f f x x
x
x
f f x x x x
−
−
= + −
= + −
=
= − + = =
19. a. 3
3
3
3
3
1 3
( ) ( 2)
( 2)
( 2)
2
2
( ) 2
f x x
y x
x y
x y
y x
f x x−
= +
= +
= +
= +
= −
= −
b. ( ) ( )
3 3
1 3 3
1 33
( ( )) 2 2
( ( )) ( 2) 2
2 2
f f x x x x
f f x x
x
x
−
−
= − + = =
= + −
= + −
=
20. a. 3
3
3
3
3
( ) ( 1)
( 1)
( 1)
1
1
f x x
y x
x y
x y
y x
= −
= −
= −
= −
= +
b. ( ) ( ) ( )
( )
3 3
1 3 3
1 33
( ) 1 1
( ) ( 1 1 1 1
f f x x x x
f f x x x x
−
−
= + − = =
= − + = − + =
21. a.
1
1
( )
1
1
1
1
1
( )
f x
x
y
x
x
y
xy
y
x
f x
x
−
=
=
=
=
=
=
b. 1
1
1
( ( ))
1
1
( ( ))
1
f f x x
x
f f x x
x
−
−
= =
= =
22. a.
1
2
( )
2
2
2
2
2
( )
f x
x
y
x
x
y
xy
y
x
f x
x
−
=
=
=
=
=
=
b.
21( ( )) 2
2 2
x
f f x x
x
− = = ⋅ =
( )
21( ) 2
2 2
x
f f x x
x
− = = ⋅ =

23. a.
2
1 2
( )
( ) , 0
f x x
y x
x y
y x
f x x x−
=
=
=
=
= ≥
b. 1 2
1 2
( ( )) for 0.
( ( )) ( )
f f x x x x x
f f x x x
−
−
= = = ≥
= =
24. a. 3
3
3
3
1 3
( )
( )
f x x
y x
x y
y x
f x x−
=
=
=
=
=
b. ( )
( ) ( )
31 3
3
1 3
( )
( )
f f x x x
f f x x x
−
−
= =
= =
25. a.
4
( )
2
x
f x
x
+
=
−
( )
1
4
2
4
2
2 4
2 4
1 2 4
2 4
1
2 4
( ) , 1
1
x
y
x
y
x
y
xy x y
xy y x
y x x
x
y
x
x
f x x
x
−
+
=
−
+
=
−
− = +
− = +
− = +
+
=
−
+
= ≠
−
b. ( )
( )
( )
1
2 4
4
1( )
2 4
2
1
2 4 4 1
2 4 2 1
6
6
x
xf f x
x
x
x x
x x
x
x
−
+
+
−=
+
−
−
+ + −
=
+ − −
=
=
( )
( )
( )
1
4
2 4
2
( )
4
1
2
2 8 4 2
4 2
6
6
x
x
f f x
x
x
x x
x x
x
x
−
+ 
+ − =
+
−
−
+ + −
=
+ − −
=
=
26. a.
5
( )
6
x
f x
x
+
=
−
( )
1
5
6
5
6
6 5
6 5
1 6 5
6 5
1
6 5
( ) , 1
1
x
y
x
y
x
y
xy x y
xy y x
y x x
x
y
x
x
f x x
x
−
+
=
−
+
=
−
− = +
− = +
− = +
+
=
−
+
= ≠
−

b. ( )
( )
( )
1
6 5
5
1( )
6 5
6
1
6 5 5 1
6 5 6 1
11
11
x
xf f x
x
x
x x
x x
x
x
−
+
+
−=
+
−
−
+ + −
=
+ − −
=
=
( )
( )
( )
1
5
6 5
6
( )
5
1
6
6 30 5 6
5 6
11
11
x
x
f f x
x
x
x x
x x
x
x
−
+ 
+ − =
+
−
−
+ + −
=
+ − −
=
=
27. a. ( )
2 1
3
2 1
3
2 1
3
x
f x
x
x
y
x
y
x
y
+
=
−
+
=
−
+
=
−
x(y – 3) = 2y + 1
xy – 3x = 2y + 1
xy – 2y = 3x + 1
y(x – 2) = 3x + 1
3 1
2
x
y
x
+
=
−
( )1 3 1
2
x
f x
x
− +
=
−
b.
( )–1
3 12 1
2( ( ))
3 1 3
2
x
xf f x
x
x
+ +
−=
+ −
−
( )
( )
2 3 1 2 6 2 2
3 1 3 2 3 1 3 6
x x x x
x x x x
+ + − + + −
= =
+ − − + − +
7
7
x
x= =
( )
( )
( )
1
2 13 1
3( ( ))
2 1 2
3
3 2 1 3
2 1 2 3
6 3 3 7
2 1 2 6 7
x
xf f x
x
x
x x
x x
x x x
x
x x
−
+ +
−=
+ −
−
+ + −
=
+ − −
+ + −
= = =
+ − +
28. a. ( )
2 3
1
x
f x
x
−
=
+
2 3
1
x
y
x
−
=
+
2 3
1
y
x
y
−
=
+
xy + x = 2y – 3
y(x – 2) = –x – 3
3
2
x
y
x
− −
=
−
( )1 3
2
x
f x
x
− − −
=
−
, 2x ≠
b. ( )( )
( )1
32 3
2
3 1
2
x
xf f x
x
x
−
− − −
−=
− − +
−
2 6 3 6 5
3 2 5
x x x
x
x x
− − − + −
= = =
− − + − −
( )1
2 3 3
1( ( ))
2 3 2
1
x
xf f x
x
x
−
−− −
+=
− −
+
2 3 3 3 5
2 3 2 2 5
x x x
x
x x
− + − − −
= = =
− − − −
29. The function fails the horizontal line test, so it does
not have an inverse function.
30. The function passes the horizontal line test, so it does
have an inverse function.

35.
36.
37.
38.
39. a. ( ) 2 1f x x= −
1
2 1
2 1
1 2
1
2
1
( )
2
y x
x y
x y
x
y
x
f x−
= −
= −
+ =
+
=
+
=
b.
c. domain of f : ( ),−∞ ∞
range of f : ( ),−∞ ∞
domain of 1
f −
: ( ),−∞ ∞
range of 1
f −
: ( ),−∞ ∞
40. a. ( ) 2 3f x x= −
1
2 3
2 3
3 2
3
2
3
( )
2
y x
x y
x y
x
y
x
f x−
= −
= −
+ =
+
=
+
=
b.
range of f : ( ),−∞ ∞
domain of 1
f −
: ( ),−∞ ∞
range of 1
f −
: ( ),−∞ ∞

41. a. 2
( ) 4f x x= −
2
2
2
1
4
4
4
4
( ) 4
y x
x y
x y
x y
f x x−
= −
= −
+ =
+ =
= +
b.
c. domain of f : [ )0,∞
range of f : [ )4,− ∞
domain of 1
f −
: [ )4,− ∞
range of 1
f −
: [ )0,∞
42. a. 2
( ) 1f x x= −
2
2
2
1
1
1
1
1
( ) 1
y x
x y
x y
x y
f x x−
= −
= −
+ =
− + =
= − +
b.
c. domain of f : ( ],0−∞
range of f : [ )1,− ∞
domain of 1
f −
: [ )1,− ∞
range of 1
f −
: ( ],0−∞
43. a. ( )
2
( ) 1f x x= −
( )
( )
2
2
1
1
1
1
1
( ) 1
y x
x y
x y
x y
f x x−
= −
= −
− = −
− + =
= −
b.
c. domain of f : ( ],1−∞
range of f : [ )0,∞
domain of 1
f −
: [ )0,∞
range of 1
f −
: ( ],1−∞
44. a. ( )
2
( ) 1f x x= −
( )
( )
2
2
1
1
1
1
1
( ) 1
y x
x y
x y
x y
f x x−
= −
= −
= −
+ =
= +
b.
domain of 1
f −
: [ )0,∞
range of 1
f −
: [ )1,∞
45. a. 3
( ) 1f x x= −
3
3
3
3
1 3
1
1
1
1
( ) 1
y x
x y
x y
x y
f x x−
= −
= −
+ =
+ =
= +
b.

range of f : ( ),−∞ ∞
domain of 1
f −
: ( ),−∞ ∞
range of 1
f −
: ( ),−∞ ∞
46. a. 3
( ) 1f x x= +
3
3
3
3
1 3
1
1
1
1
( ) 1
y x
x y
x y
x y
f x x−
= +
= +
− =
− =
= −
b.
range of f : ( ),−∞ ∞
domain of 1
f −
: ( ),−∞ ∞
range of 1
f −
: ( ),−∞ ∞
47. a. 3
( ) ( 2)f x x= +
3
3
3
3
1 3
( 2)
( 2)
2
2
( ) 2
y x
x y
x y
x y
f x x−
= +
= +
= +
− =
= −
b.
range of f : ( ),−∞ ∞
domain of 1
f −
: ( ),−∞ ∞
range of 1
f −
: ( ),−∞ ∞
48. a. 3
( ) ( 2)f x x= −
3
3
3
3
1 3
( 2)
( 2)
2
2
( ) 2
y x
x y
x y
x y
f x x−
= −
= −
= −
+ =
= +
b.
range of f : ( ),−∞ ∞
domain of 1
f −
: ( ),−∞ ∞
range of 1
f −
: ( ),−∞ ∞
49. a. ( ) 1f x x= −
2
2
1 2
1
1
1
1
( ) 1
y x
x y
x y
x y
f x x−
= −
= −
= −
+ =
= +
b.
domain of 1
f −
: [ )0,∞
range of 1
f −
: [ )1,∞
50. a. ( ) 2f x x= +
2
1 2
2
2
2
( 2)
( ) ( 2)
y x
x y
x y
x y
f x x−
= +
= +
− =
− =
= −

b.
domain of 1
f −
: [ )2,∞
range of 1
f −
: [ )0,∞
51. a. 3
( ) 1f x x= +
3
3
3
3
1 3
1
1
1
( 1)
( ) ( 1)
y x
x y
x y
x y
f x x−
= +
= +
− =
− =
= −
b.
range of f : ( ),−∞ ∞
domain of 1
f −
: ( ),−∞ ∞
range of 1
f −
: ( ),−∞ ∞
52. a. 3
( ) 1f x x= −
3
3
3
3
1 3
1
1
1
1
( ) 1
y x
x y
x y
x y
f x x−
= −
= −
= −
+ =
= +
b.
range of f : ( ),−∞ ∞
domain of 1
f −
: ( ),−∞ ∞
range of 1
f −
: ( ),−∞ ∞
53. ( ) ( )(1) 1 5f g f= =
54. ( ) ( )(4) 2 1f g f= = −
55. ( )( ) ( ) ( )1 ( 1) 1 1g f g f g− = − = =
56. ( )( ) ( ) ( )0 (0) 4 2g f g f g= = =
57. ( ) ( )1 1
(10) 1 2f g f− −
= − = , since ( )2 1f = − .
58. ( ) ( )1 1
(1) 1 1f g f− −
= = − , since ( )1 1f − = .
59. ( )( ) ( )
( )
( ) ( )
0 (0)
4 0 1
1 2 1 5 7
f g f g
f
f
=
= ⋅ −
= − = − − = −

60. ( )( ) ( )
( )
( ) ( )
0 (0)
2 0 5
5 4 5 1 21
g f g f
g
g
=
= ⋅ −
= − = − − = −

61. Let ( )1
1f x−
= . Then
( ) 1
2 5 1
2 6
3
f x
x
x
x
=
− =
=
=
Thus, ( )1
1 3f −
=
62. Let ( )1
7g x−
= . Then
( ) 7
4 1 7
4 8
2
g x
x
x
x
=
− =
=
=
Thus, ( )1
7 2g−
=

63. [ ]( ) ( )
( )
( )
( )
2
(1) 1 1 2
(4)
2 4 5
3
4 3 1 11
g f h g f
g f
g
g
 = + + 
=
= ⋅ −
=
= ⋅ − =
64. [ ]( ) ( )
( )
( )
( )
2
(1) 1 1 2
(4)
4 4 1
15
2 15 5 25
f g h f g
f g
f
f
 = + + 
=
= ⋅ −
=
= ⋅ − =
65. a. {(Zambia, 4.2), (Colombia, 4.5),
(Poland, 3.3), (Italy, 3.3),
(United States, 2.5)}
b. {(4.2, Zambia), (4.5 , Colombia),
(3.3 , Poland), (3.3, Italy),
(2.5, United States)}
f is not a one-to-one function because the
inverse of f is not a function.
66. a. {(Zambia,- 7.3), (Colombia, - 4.5),
(Poland, - 2.8), (Italy, - 2.8),
(United States, - 1.9)}
b. { (- 7.3, Zambia), (- 4.5, Colombia),
(- 2.8, Po land), (- 2.8, Italy),
(- 1.9, United States)}
g is not a one-to-one function because the
inverse of g is not a function.
67. a. It passes the horizontal line test and is one-to-
one.
b. f--1
(0.25) = 15 If there are 15 people in the
room, the probability that 2 of them have the
same birthday is 0.25.
f--1
(0.5) = 21 If there are 21 people in the room,
the probability that 2 of them have the same
birthday is 0.5.
f--1
(0.7) = 30 If there are 30 people in the room,
the probability that 2 of them have the same
birthday is 0.7.
68. a. This function fails the horizontal line test.
Thus, this function does not have an inverse.
b. The average happiness level is 3 at 12 noon and
at 7 p.m. These values can be represented as
(12,3) and (19,3) .
c. The graph does not represent a one-to-one
function. (12,3) and (19,3) are an example of
two x-values that correspond to the same y-
value.
69.
9 5
( ( )) ( 32) 32
5 9
32 32
f g x x
x
x
 
= − +  
= − +
=
5 9
( ( )) 32 32
9 5
32 32
g f x x
x
x
  
= + −  
  
= + −
=
76.
not one-to-one
77.
one-to-one

78.
one-to-one
79.
not one-to-one
80.
not one-to-one
81.
not one-to-one
82.
one-to-one
83.
not one-to-one
84.
f and g are inverses
85.
86.
87. makes sense
88. makes sense
89. makes sense
Sample explanation: The vertical line test is used to
determine if a relation is a function, but does not tell
us if a function is one-to-one.

A sample change is: The inverse is {(4,1), (7,2)}.
A sample change is: f(x) = 5 is a horizontal line, so
it does not pass the horizontal line test.
A sample change is: 1
( ) .
3
x
f x−
=
94. true
95. ( )( ) 3( 5) 3 15.f g x x x= + = +
3 15
3 15
15
3
y x
x y
x
y
= +
= +
−
=
( )
1 15
( )
3
x
f g x
− −
=
1
( ) 5
5
5
5
( ) 5
g x x
y x
x y
y x
g x x−
= +
= +
= +
= −
= −
1
( ) 3
3
3
3
( )
3
f x x
y x
x y
x
y
x
f x−
=
=
=
=
=
( )1 1 15
( ) 5
3 3
x x
g f x− − −
= − =
96.
1
3 2
( )
5 3
3 2
5 3
3 2
5 3
(5 3) 3 2
5 3 3 2
5 3 3 2
(5 3) 3 2
3 2
5 3
3 2
( )
5 3
x
f x
x
x
y
x
y
x
y
x y y
xy x y
xy y x
y x x
x
y
x
x
f x
x
−
−
=
−
−
=
−
−
=
−
− = −
− = −
− = −
− = −
−
=
−
−
=
−
Note: An alternative approach is to show that
( )( ) .f f x x=
97. No, there will be 2 times when the spacecraft is at the
same height, when it is going up and when it is
coming down.
98. 1
8 ( 1) 10f x−
+ − =
1
( 1) 2
(2) 1
6 1
7
7
f x
f x
x
x
x
−
− =
= −
= −
=
=

100. 2
2 5 1 0x x− + =
2
2
2
2
5 1
0
2 2
5 1
2 2
5 25 1 25
2 16 2 16
5 17
4 16
5 17
4 16
5 17
4 4
5 17
4 4
5 17
4
x x
x x
x x
x
x
x
x
x
− + =
− = −
− + = − +
 
− = 
 
− = ±
− = ±
= ±
±
=
The solution set is
5 17
4
 ± 
 
  
.
101.
( ) ( )
3/4
3/2
3/4
4 3 4 33/4
4/3
5 15 0
5 15
3
3
3
x
x
x
x
x
− =
=
=
=
=
The solution set is { }4/3
3 .
102. 3 2 1 21
2 1 7
x
x
− ≥
− ≥
2 1 7 2 1 7
or
2 6 2 8
2 6 2 8
2 2 2 2
3 4
x x
x x
x x
x x
− ≤ − − ≥
≤ − ≥
−
≤ ≥
≤ − ≥
The solution set is { }3 or 4x x x≤ − ≥
or ( ] [ ), 3 4, .−∞ − ∞
103. 2 2 2 2
2 1 2 1
2 2
( ) ( ) (1 7) ( 1 2)
( 6) ( 3)
36 9
45
3 5
x x y y− + − = − + − −
= − + −
= +
=
=
104.
105. 2
2
2
2
6 4 0
6 4
6 9 4 9
( 3) 13
3 13
3 13
y y
y y
y y
y
y
y
− − =
− =
− + = +
− =
− = ±
= ±
Solution set: { }3 13±

Section 2.8 Distance and Midpoint Formulas; Circles
Section 2.8
1. ( ) ( )
2 2
2 1 2 1d x x y y= − + −
( ) ( )
2 2
2 2
2 ( 1) 3 ( 3)
3 6
9 36
45
3 5
6.71
d = − − + − −
= +
= +
=
=
≈
2.
1 7 2 ( 3) 8 1 1
, , 4,
2 2 2 2 2
+ + − −     
= = −     
     
3.
2 2 2
2 2
0, 0, 4;
( 0) ( 0) 4
16
h k r
x y
x y
= = =
− + − =
+ =
4.
2 2 2
2 2
2 2
0, 6, 10;
( 0) [ ( 6)] 10
( 0) ( 6) 100
( 6) 100
h k r
x y
x y
x y
= = − =
− + − − =
− + + =
+ + =
5. a. 2 2
2 2 2
( 3) ( 1) 4
[ ( 3)] ( 1) 2
x y
x y
+ + − =
− − + − =
So in the standard form of the circle’s equation
2 2 2
( ) ( )x h y k r− + − = ,
we have 3, 1, 2.h k r= − = =
center: ( , ) ( 3, 1)h k = −
radius: r = 2
b.
c. domain: [ ]5, 1− −
range: [ ]1,3−
6. 2 2
4 4 1 0x y x y+ + − − =
( ) ( )
( ) ( )
2 2
2 2
2 2
2 2
2 2 2
4 4 1 0
4 4 0
4 4 4 4 1 4 4
( 2) ( 2) 9
[ ( )] ( 2) 3
x y x y
x x y y
x x y y
x y
x x y
+ + − − =
+ + − =
+ + + + + = + +
+ + − =
− − + − =
So in the standard form of the circle’s equation
2 2 2
( ) ( )x h y k r− + − = , we have
2, 2, 3h k r= − = = .
1. 2 2
2 1 2 1( ) ( )x x y y− + −
2. 1 2
2
x x+
; 1 2
2
y y+
3. circle; center; radius
4. 2 2 2
( ) ( )x h y k r− + − =
5. general
6. 4; 16
Exercise Set 2.8
1. 2 2
(14 2) (8 3)d = − + −
2 2
12 5
144 25
169
13
= +
= +
=
=

2. 2 2
(8 5) (5 1)d = − + −
2 2
3 4
9 16
25
5
= +
= +
=
=
3. ( ) ( )
2 2
6 4 3 ( 1)d = − − + − −
( ) ( )
2 2
10 4
100 16
116
2 29
10.77
= − +
= +
=
=
≈
4. ( ) ( )
2 2
1 2 5 ( 3)d = − − + − −
( ) ( )
2 2
3 8
9 64
73
8.54
= − +
= +
=
≈
5. 2 2
( 3 0) (4 0)d = − − + −
2 2
3 4
9 16
25
5
= +
= +
=
=
6. ( )
22
(3 0) 4 0d = − + − −
( )
22
3 4
9 16
25
5
= + −
= +
=
=
7. 2 2
[3 ( 2)] [ 4 ( 6)]d = − − + − − −
2 2
5 2
25 4
29
5.39
= +
= +
=
≈
8. 2 2
[2 ( 4)] [ 3 ( 1)]d = − − + − − −
( )
22
6 2
36 4
40
2 10
6.32
= + −
= +
=
=
≈
9. 2 2
(4 0) [1 ( 3)]d = − + − −
2 2
4 4
16 16
32
4 2
5.66
= +
= +
=
=
≈
10. ( ) ( )
2 2
2 2
2
4 0 [3 2 ]
4 [3 2]
16 5
16 25
41
6.40
d = − + − −
= + +
= +
= +
=
≈
11.
2 2
2 2
( .5 3.5) (6.2 8.2)
( 4) ( 2)
16 4
20
2 5
4.47
d = − − + −
= − + −
= +
=
=
≈
12. ( )
( ) ( )
22
2 2
(1.6 2.6) 5.7 1.3
1 7
1 49
50
5 2
7.07
d = − + − −
= − + −
= +
=
=
≈

13.
2 2
2 2
( 5 0) [0 ( 3)]
( 5) ( 3)
5 3
8
2 2
2.83
d = − + − −
= +
= +
=
=
≈
14. ( ) ( )
( )
22
2 2
7 0 0 2
7 2
7 2
9
3
d  = − + − −
 
 = + − 
= +
=
=
15.
2 2
2 2
( 3 3 3) (4 5 5)
( 4 3) (3 5)
16(3) 9(5)
48 45
93
9.64
d = − − + −
= − +
= +
= +
=
≈
16. ( ) ( )
( ) ( )
2 2
2 2
3 2 3 5 6 6
3 3 4 6
9 3 16 6
27 96
123
11.09
d = − − + −
= − +
= ⋅ + ⋅
= +
=
≈
17.
2 2
2 2
1 7 6 1
3 3 5 5
( 2) 1
4 1
5
2.24
d
   
= − + −   
   
= − +
= +
=
≈
18.
2 2
2 2
2 2
3 1 6 1
4 4 7 7
3 1 6 1
4 4 7 7
1 1
2
1.41
d
      
= − − + − −      
      
   
= + + +   
   
= +
=
≈
19.
6 2 8 4 8 12
, , (4,6)
2 2 2 2
+ +   
= =   
   
20.
10 2 4 6 12 10
, , (6,5)
2 2 2 2
+ +   
= =   
   
21.
2 ( 6) 8 ( 2)
,
2 2
8 10
, ( 4, 5)
2 2
− + − − + − 
 
 
− − 
= = − − 
 
22.
( ) ( )4 1 7 3 5 10
, ,
2 2 2 2
5
, 5
2
− + − − + −  − − 
=   
  
− 
= − 
 
23.
3 6 4 ( 8)
,
2 2
3 12 3
, , 6
2 2 2
− + − + − 
 
 
−   
= = −   
   
24.
( )2 8 1 6 10 5 5
, 5,
2 2 2 2 2
− + − − + −   
= = −     
    
25.
( )
7 5 3 11
2 2 2 2
,
2 2
12 8
6 42 2, , 3, 2
2 2 2 2
 −    
+ − + −    
    
 
 
 
− − 
  − 
= = − = − −   
   
 

26. 2 2 7 4 4 3
5 5 15 15 5 15, ,
2 2 2 2
4 1 3 1 2 1
, ,
5 2 15 2 5 10
      − + − + − −           =  
        
   
= − ⋅ ⋅ = −   
   
27.
( )
8 ( 6) 3 5 7 5
,
2 2
2 10 5
, 1,5 5
2 2
 + − +
  
 
 
= =  
 
28.
( )
7 3 3 3 6 ( 2) 10 3 8
, ,
2 2 2 2
5 3, 4
   + − + − −
=      
   
= −
29. 18 2 4 4
,
2 2
3 2 2 0 4 2
, ,0 (2 2,0)
2 2 2
 + − +
  
 
   +
= = =      
   
30.
( )
50 2 6 6 5 2 2 0
, ,
2 2 2 2
6 2
,0 3 2,0
2
   + − + +
=      
   
 
= =  
 
31. 2 2 2
2 2
( 0) ( 0) 7
49
x y
x y
− + − =
+ =
32. 2 2 2
( 0) ( 0) 8x y− + − =
2 2
64x y+ =
33. ( ) ( )
( ) ( )
2 2 2
2 2
3 2 5
3 2 25
x y
x y
− + − =
− + − =
34. ( ) [ ]
22 2
2 ( 1) 4x y− + − − =
( ) ( )
2 2
2 1 16x y− + + =
35. [ ] ( )
( ) ( )
2 2 2
2 2
( 1) 4 2
1 4 4
x y
x y
− − + − =
+ + − =
36. [ ] ( )
2 2 2
( 3) 5 3x y− − + − =
( ) ( )
2 2
3 5 9x y+ + − =
37. [ ] [ ] ( )
( ) ( )
22 2
2 2
( 3) ( 1) 3
3 1 3
x y
x y
− − + − − =
+ + + =
38. [ ] [ ] ( )
22 2
( 5) ( 3) 5x y− − + − − =
( ) ( )
2 2
5 3 5x y+ + + =
39. [ ] ( )
( ) ( )
2 2 2
2 2
( 4) 0 10
4 0 100
x y
x y
− − + − =
+ + − =
40. [ ] ( )
2 2 2
( 2) 0 6x y− − + − =
( )
2 2
2 36x y+ + =
41. 2 2
2 2 2
16
( 0) ( 0)
0, 0, 4;
x y
x y y
h k r
+ =
− + − =
= = =
center = (0, 0); radius = 4
domain: [ ]4,4−
range: [ ]4,4−
42. 2 2
49x y+ =
2 2 2
( 0) ( 0) 7
0, 0, 7;
x y
h k r
− + − =
= = =
domain: [ ]7,7−
range: [ ]7,7−

43. ( ) ( )
( ) ( )
2 2
2 2 2
3 1 36
3 1 6
3, 1, 6;
x y
x y
h k r
− + − =
− + − =
= = =
domain: [ ]3,9−
range: [ ]5,7−
44. ( ) ( )
2 2
2 3 16x y− + − =
2
2 2
( 2) ( 3) 4
2, 3, 4;
x y
h k r
− + − =
= = =
domain: [ ]2,6−
range: [ ]1,7−
45. 2 2
2 2 2
( 3) ( 2) 4
[ ( 3)] ( 2) 2
3, 2, 2
x y
x y
h k r
+ + − =
− − + − =
= − = =
center = (–3, 2); radius = 2
domain: [ ]5, 1− −
range: [ ]0,4
46. ( ) ( )
2 2
1 4 25x y+ + − =
[ ]
2 2 2
( 1) ( 4) 5
1, 4, 5;
x y
h k r
− − + − =
= − = =
domain: [ ]6,4−
range: [ ]1,9−
47. 2 2
2 2 2
( 2) ( 2) 4
[ ( 2)] [ ( 2)] 2
2, 2, 2
x y
x y
h k r
+ + + =
− − + − − =
= − = − =
center = (–2, –2); radius = 2
domain: [ ]4,0−
range: [ ]4,0−
48. ( ) ( )
2 2
4 5 36x y+ + + =
[ ] [ ]
2 2 2
( 4) ( 5) 6
4, 5, 6;
x y
h k r
− − + − − =
= − = − =
domain: [ ]10,2−
range: [ ]11,1−

49. ( )
22
1 1x y+ − =
0, 1, 1;h k r= = =
domain: [ ]1,1−
range: [ ]0,2
50. ( )
22
2 4x y+ − =
0, 2, 2;h k r= = =
center = (0,2); radius = 2
domain: [ ]2,2−
range: [ ]0,4
51. ( )
2 2
1 25x y+ + =
1, 0, 5;h k r= − = =
center = (–1,0); radius = 5
domain: [ ]6,4−
range: [ ]5,5−
52. ( )
2 2
2 16x y+ + =
2, 0, 4;h k r= − = =
center = (–2,0); radius = 4
domain: [ ]6,2−
range: [ ]4,4−
53.
( ) ( )
( ) ( )
( ) ( )
[ ] [ ]
2 2
2 2
2 2
2 2
2 2 2
6 2 6 0
6 2 6
6 9 2 1 9 1 6
3 1 4
( 3) 9 ( 1) 2
x y x y
x x y y
x x y y
x y
x
+ + + + =
+ + + = −
+ + + + + = + −
+ + + =
− − + − − =
54. 2 2
8 4 16 0x y x y+ + + + =
( ) ( )2 2
8 4 16x x y y+ + + = −
( ) ( )2 2
8 16 4 4 20 16x x y y+ + + + + = −
( ) ( )
2 2
4 2 4x y+ + + =
[ ] [ ]
2 2 2
( 4) ( 2) 2x y− − + − − =

55.
( ) ( )
( ) ( )
( ) ( )
2 2
2 2
2 2
2 2
2 2 2
10 6 30 0
10 6 30
10 25 6 9 25 9 30
5 3 64
( 5) ( 3) 8
x y x y
x x y y
x x y y
x y
x y
+ − − − =
− + − =
− + + − + = + +
− + − =
− + − =
56. 2 2
4 12 9 0x y x y+ − − − =
( ) ( )2 2
4 12 9x x y y− + − =
( ) ( )2 2
4 4 12 36 4 36 9x x y y− + + − + = + +
( ) ( )
2 2
2 6 49x y− + − =
2 2 2
( 2) ( 6) 7x y− + − =
57.
( ) ( )
( ) ( )
( ) ( )
[ ]
2 2
2 2
2 2
2 2
2 2 2
8 2 8 0
8 2 8
8 16 2 1 16 1 8
4 1 25
( 4) ( 1) 5
x y x y
x x y y
x x y y
x y
x y
+ + − − =
+ + − =
+ + + − + = + +
+ + − =
− − + − =
58. 2 2
12 6 4 0x y x y+ + − − =
( ) ( )2 2
12 6 4x x y y+ + − =
( ) ( )2 2
12 36 6 9 36 9 4x x y y+ + + − + = + +
[ ] ( )
2 2 2
( 6) 3 7x y− − + − =
59.
( )
( ) ( )
( ) ( )
( ) ( )
2 2
2 2
22
2 2
2 2 2
2 15 0
2 15
2 1 0 1 0 15
1 0 16
1 0 4
x x y
x x y
x x y
x y
x y
− + − =
− + =
− + + − = + +
− + − =
− + − =
60. 2 2
6 7 0x y y+ − − =
( )2 2
6 7x y y+ − =
( ) ( )2 2
0 6 9 0 9 7x y y− = − + = + +
( ) ( )
2 2
0 3 16x y− + − =
2 2 2
( 0) ( 3) 4x y− + − =

61. 2 2
2 1 0x y x y+ − + + =
( )
2 2
2 2
2
2
2 1
1 1
2 1 1 1
4 4
1 1
1
2 4
x x y y
x x y y
x y
− + + = −
− + + + + = − + +
 
− + + = 
 
center =
1
, 1
2
 
− 
 
; radius =
1
2
62. 2 2 1
0
2
x y x y+ + + − =
2 2
2 2
2 2
1
2
1 1 1 1 1
4 4 2 4 4
1 1
1
2 2
x x y y
x x y y
x y
+ + + =
+ + + + + = + +
   
− + − =   
   
center =
1 1
,
2 2
 
 
 
; radius = 1
63. 2 2
3 2 1 0x y x y+ + − − =
( )
2 2
2 2
2
2
3 2 1
9 9
3 2 1 1 1
4 4
3 17
1
2 4
x x y y
x x y y
x y
+ + − =
+ + + − + = + +
 
+ + − = 
 
center =
3
,1
2
 
− 
 
; radius =
17
2
64. 2 2 9
3 5 0
4
x y x y+ + + + =
2 2
2 2
2 2
9
3 5
4
9 25 9 9 25
3 5
4 4 4 4 4
3 5 25
2 2 4
x x y y
x x y y
x y
+ + + = −
+ + + + + = − + +
   
+ + + =   
   
center =
3 5
,
2 2
 
− − 
 
; radius =
5
2
65. a. Since the line segment passes through the
center, the center is the midpoint of the
segment.
( )
1 2 1 2
,
2 2
3 7 9 11 10 20
, ,
2 2 2 2
5,10
x x y y
M
+ + 
=  
 
+ +   
= =   
   
=
The center is ( )5,10 .

b. The radius is the distance from the center to
one of the points on the circle. Using the
point ( )3,9 , we get:
( ) ( )
2 2
2 2
5 3 10 9
2 1 4 1
5
d = − + −
= + = +
=
The radius is 5 units.
c.
( ) ( ) ( )
( ) ( )
22 2
2 2
5 10 5
5 10 5
x y
x y
− + − =
− + − =
66. a. Since the line segment passes through the
center, the center is the midpoint of the
segment.
( )
1 2 1 2
,
2 2
3 5 6 4 8 10
, ,
2 2 2 2
4,5
x x y y
M
+ + 
=  
 
+ +   
= =   
   
=
The center is ( )4,5 .
b. The radius is the distance from the center to
one of the points on the circle. Using the
point ( )3,6 , we get:
( ) ( )
( )
2 2
22
4 3 5 6
1 1 1 1
2
d = − + −
= + − = +
=
The radius is 2 units.
c.
( ) ( ) ( )
( ) ( )
22 2
2 2
4 5 2
4 5 2
x y
x y
− + − =
− + − =
67.
Intersection points: ( )0, 4− and ( )4,0
Check ( )0, 4− :
( )
22
0 4 16
16 16 true
+ − =
=
( )0 4 4
4 4 true
− − =
=
Check ( )4,0 :
2 2
4 0 16
16 16 true
+ =
=
4 0 4
4 4 true
− =
=
The solution set is ( ) ( ){ }0, 4 , 4,0− .
68.
Check ( )0, 3− :
( )
22
0 3 9
9 9 true
+ − =
=
( )0 3 3
3 3 true
− − =
=
Check ( )3,0 :
2 2
3 0 9
9 9 true
+ =
=
3 0 3
3 3 true
− =
=

69.
Intersection points: ( )0, 3− and ( )2, 1−
Check ( )0, 3− :
( ) ( )
( )
2 2
2 2
0 2 3 3 9
2 0 4
4 4
true
− + − + =
− + =
=
3 0 3
3 3 true
− = −
− = −
Check ( )2, 1− :
( ) ( )
2 2
2 2
2 2 1 3 4
0 2 4
4 4
true
− + − + =
+ =
=
1 2 3
1 1 true
− = −
− = −
The solution set is ( ) ( ){ }0, 3 , 2, 1− − .
70.
Check ( )0, 1− :
( ) ( )
( )
2 2
2 2
0 3 1 1 9
3 0 9
9 9
true
− + − + =
− + =
=
1 0 1
1 1 true
− = −
− = −
Check ( )3,2 :
( ) ( )
2 2
2 2
3 3 2 1 9
0 3 9
9 9
true
− + + =
+ =
=
2 3 1
2 2 true
= −
=
71. 2 2
(8495 4422) (8720 1241) 0.1
72,524,770 0.1
2693
d
d
d
= − + − ⋅
= ⋅
≈
The distance between Boston and San Francisco is
about 2693 miles.
72. 2 2
(8936 8448) (3542 2625) 0.1
1,079,033 0.1
328
d
d
d
= − + − ⋅
= ⋅
≈
The distance between New Orleans and Houston is
about 328 miles.
73. If we place L.A. at the origin, then we want the
equation of a circle with center at ( )2.4, 2.7− − and
radius 30.
( )( ) ( )( )
( ) ( )
2 2 2
2 2
2.4 2.7 30
2.4 2.7 900
x y
x y
− − + − − =
+ + + =
74. C(0, 68 + 14) = (0, 82)
2 2 2
2 2
( 0) ( 82) 68
( 82) 4624
x y
x y
− + − =
+ − =
83.
84.

85.
86. makes sense
87. makes sense
Sample explanation: Since 2
4r = − this is not the
equation of a circle.
89. makes sense
A sample change is: The equation would be
2 2
256.x y+ =
A sample change is: The center is at (3, –5).
A sample change is: This is not an equation for a
circle.
A sample change is: Since 2
36r = − this is not the
equation of a circle.
94. The distance for A to B:
( )2 2
2 2
(3 1) [3 1 ]
2 2
4 4
8
2 2
AB d d= − + + − +
= +
= +
=
=
The distance from B to C:
( )
( )
2 2
22
(6 3) [3 6 ]
3 3
9 9
18
3 2
BC d d= − + + − +
= + −
= +
=
=
The distance for A to C:
2 2
2 2
(6 1) [6 (1 )]
5 5
25 25
50
5 2
AC d d= − + + − +
= +
= +
=
=
2 2 3 2 5 2
5 2 5 2
AB BC AC+ =
+ =
=
95. a. is distance from ( , ) to midpoint
1 1 2
d x x
( )
2 2
1 2 1 2
1 1 1
2 2
1 2 1 1 2 1
1
2 2
2 1 2 1
1
2 2 2
2 1 2 1 2 2 1 1
1
2 2
1 2 1 2 1 2 2 1 1
2 2
1 2 1 2 1 2 2 1 1
2 2
2 2
2 2
2 2
2 2
4 4
1
2 2
4
1
2 2
2
x x y y
d x y
x x x y y y
d
x x y y
d
x x x x y y y y
d
d x x x x y y y y
d x x x x y y y y
+ +   
= − + −   
   
+ − + −   
= +   
   
− −   
= +   
   
− + − +
= +
= − + + − +
= − + + − +
( )
1
2 2 2
2 2
2 1 2
2 2 2
2 2
1 2 2 1 2 2
2
2 2
1 2 1 2
2
2 2 2 2
1 1 2 2 1 2 1 2
2
2 2 2
2 1 1 2 2 1 2
is distance from midpoint to ,
2 2
2 2
2 2
2 2
2 2
4 4
1
2 2
4
d x y
x x y y
d x y
x x x y y y
d
x x y y
d
x x x x y y y y
d
d x x x x y y y
+ +   
= − + −   
   
+ − + −   
= +   
   
− −   
= +   
   
− + − +
= +
= − + + −( )2
1 2
2 2 2 2
2 1 1 2 2 1 2 1 2
1 2
1
2 2
2
y
d x x x x y y y y
d d
+
= − + + − +
=

b. ( ) ( )3 1 1 2 2is the distance from , tod x y x y
2 2
3 2 1 2 1
2 2 2 2
3 2 1 2 1 2 2 1 1
1 2 3
( ) ( )
2 2
1 1
because
2 2
d x x y y
d x x x x y y y y
d d d a a a
= − + −
= − + + − +
+ = + =
96. Both circles have center (2, –3). The smaller circle
has radius 5 and the larger circle has radius 6. The
smaller circle is inside of the larger circle. The area
between them is given by
( ) ( )
2 2
6 5π π− 36 25π π= −
11
34.56squareunits.
π=
≈
97. The circle is centered at (0,0). The slope of the radius
with endpoints (0,0) and (3,–4) is
4 0 4
.
3 0 3
m
− −
= − = −
−
The line perpendicular to the
radius has slope
3
.
4
The tangent line has slope
3
4
and
passes through (3,–4), so its equation is:
3
4 ( 3).
4
y x+ = −
98. ( )7 2 5 7 9x x− + = −
7 14 5 7 9
7 9 7 9
9 9
x x
x x
− + = −
− = −
− = −
The original equation is equivalent to the
statement 9 9,− = − which is true for every value
of x.
The equation is an identity, and all real numbers
are solutions. The solution set
{ is a real number}.x x
99.
4 7 4 7 5 2
5 2 5 2 5 2
i i i
i i i
+ + +
= ⋅
− − +
2
2
20 8 35 14
25 10 10 4
34 8 35
25 4
34 27
29
27 34
29 29
i i i
i i i
i
i
i
+ + +
=
+ − −
− +
=
+
+
=
= +
100. 9 4 1 15
8 4 16
2 4
x
x
x
− ≤ − <
− ≤ <
− ≤ <
The solution set is { }2 4 or [ 2, 4).x x− ≤ < −
101. 2
2
2
0 2( 3) 8
2( 3) 8
( 3) 4
3 4
3 2
1, 5
x
x
x
x
x
x
= − − +
− =
− =
− = ±
= ±
=
102. 2
2
2 1 0
2 1 0
x x
x x
− − + =
+ − =
2
2
4
2
( 2) ( 2) 4(1)( 1)
2(1)
2 8
2
2 2 2
2
1 2
b b ac
x
a
x
− ± −
=
− − ± − − −
=
±
=
±
=
= ±
The solution set is {1 2}.±
103. The graph of g is the graph of f shifted 1 unit up and
3 units to the left.

Chapter 2 Review Exercises
1. function
domain: {2, 3, 5}
range: {7}
2. function
domain: {1, 2, 13}
range: {10, 500, π}
3. not a function
domain: {12, 14}
range: {13, 15, 19}
4. 2 8
2 8
x y
y x
+ =
= − +
5. 2
2
3 14
3 14
x y
y x
+ =
= − +
6. 2
2
2 6
2 6
2 6
x y
y x
y x
+ =
= − +
= ± − +
Since more than one value of y can be obtained from
some values of x, y is not a function of x.
7. f(x) = 5 – 7x
a. f(4) = 5 – 7(4) = –23
b. ( 3) 5 7( 3)
5 7 21
7 16
f x x
x
x
+ = − +
= − −
= − −
c. f(–x) = 5 – 7(–x) = 5 + 7x
8. 2
( ) 3 5 2g x x x= − +
a. 2
(0) 3(0) 5(0) 2 2g = − + =
b. 2
( 2) 3( 2) 5( 2) 2
12 10 2
24
g − = − − − +
= + +
=
c. 2
2
2
( 1) 3( 1) 5( 1) 2
3( 2 1) 5 5 2
3 11 10
g x x x
x x x
x x
− = − − − +
= − + − + +
= − +
d. 2
2
( ) 3( ) 5( ) 2
3 5 2
g x x x
x x
− = − − − +
= + +
9. a. (13) 13 4 9 3g = − = =
b. g(0) = 4 – 0 = 4
c. g(–3) = 4 – (–3) = 7
10. a.
2
( 2) 1 3
( 2) 1
2 1 3
f
− −
− = = = −
− − −
b. f(1) = 12
c.
2
2 1 3
(2) 3
2 1 1
f
−
= = =
−
11. The vertical line test shows that this is not the graph
of a function.
12. The vertical line test shows that this is the graph of a
function.
function.
of a function.
of a function.
function.
17. a. domain: [–3, 5)
b. range: [–5, 0]
e. increasing: ( 2, 0) or (3, 5)−
decreasing: ( 3, 2) or (0, 3)− −
f. f(–2) = –3 and f(3) = –5

18. a. domain: ( , )−∞ ∞
b. range: ( ],3−∞
d. y-intercept: 3
e. increasing: (–, 0)
decreasing: (0, )∞
f. f(–2) = 0 and f(6) = –3
19. a. domain: ( , )−∞ ∞
b. range: [–2, 2]
c. x-intercept: 0
d. y-intercept: 0
e. increasing: (–2, 2)
constant: ( , 2) or (2, )−∞ − ∞
f. f(–9) = –2 and f(14) = 2
20. a. 0, relative maximum −2
b. −2, 3, relative minimum −3, –5
21. a. 0, relative maximum 3
b. none
( )
2
2
2
8
8
8
y x
y x
y x
= +
= − +
= +
axis.
2
2
2
8
8
8
y x
y x
y x
= +
− = +
= − −
( )
2
2
2
2
8
8
8
2
y x
y x
y x
y x
= +
− = − +
− = +
= − −
( )
2 2
2 2
2 2
17
17
17
x y
x y
x y
+ =
− + =
+ =
axis.
( )
2 2
22
2 2
17
17
17
x y
x y
x y
+ =
+ − =
+ =
axis.
( ) ( )
2 2
2 2
2 2
17
17
17
x y
x y
x y
+ =
− + − =
+ =
origin.
( )
3 2
3 2
3 2
5
5
5
x y
x y
x y
− =
− − =
− − =
( )
3 2
23
3 2
5
5
5
x y
x y
x y
− =
− − =
− =
axis.

( ) ( )
3 2
3 2
3 2
5
5
5
x y
x y
x y
− =
− − − =
− − =
26. The graph is not symmetric with respect to the y-axis
or the origin. The function is neither even nor odd.
27. The graph is symmetric with respect to the y-axis.
The function is even.
28. 3
3
3
( ) 5
( ) ( ) 5( )
5
( )
f x x x
f x x x
x x
f x
= −
− = − − −
= − +
= −
The function is odd. The function is symmetric with
29. 4 2
4 2
4 2
( ) 2 1
( ) ( ) 2( ) 1
2 1
( )
f x x x
f x x x
x x
f x
= − +
− = − − − +
= − +
=
The function is even. The function is symmetric with
30. 2
2
2
( ) 2 1
( ) 2( ) 1 ( )
2 1
( )
f x x x
f x x x
x x
f x
= −
− = − − −
= − −
= −
The function is odd. The function is symmetric with
31. a.
b. range: {–3, 5}
32. a.
b. range: { }0y y ≤
33.
8( ) 11 (8 11)
8 8 11 8 11
8
8
8
x h x
h
x h x
h
h
+ − − −
+ − − +
=
=
=
34.
( )2 2
2( ) ( ) 10 2 10x h x h x x
h
− + + + + − − + +
( )
( )
2 2 2
2 2 2
2
2 2 10 2 10
2 4 2 10 2 10
4 2
4 2 1
4 2 1
x xh h x h x x
h
x xh h x h x x
h
xh h h
h
h x h
h
x h
− + + + + + + − −
=
− − − + + + + − −
=
− − +
=
− − +
=
− − +
35. a. Yes, the eagle’s height is a function of time
since the graph passes the vertical line test.
b. Decreasing: (3, 12)
The eagle descended.
c. Constant: (0, 3) or (12, 17)
The eagle’s height held steady during the first 3
seconds and the eagle was on the ground for 5
seconds.
d. Increasing: (17, 30)
The eagle was ascending.

36.
37.
1 2 1 1
;
5 3 2 2
m
− −
= = = −
−
falls
38.
4 ( 2) 2
1;
3 ( 1) 2
m
− − − −
= = =
− − − −
rises
39.
1 1
4 4 0
0;
6 ( 3) 9
m
−
= = =
− −
horizontal
40.
10 5 5
2 ( 2) 0
m
−
= =
− − −
undefined; vertical
41. point-slope form: y – 2 = –6(x + 3)
42.
2 6 4
2
1 1 2
m
− −
= = =
− − −
point-slope form: y – 6 = 2(x – 1)
or y – 2 = 2(x + 1)
slope-intercept form: y = 2x + 4
43. 3x + y – 9 = 0
y = –3x + 9
m = –3
point-slope form:
y + 7 = –3(x – 4)
y = –3x + 12 – 7
y = –3x + 5
44. perpendicular to
1
4
3
y x= +
m = –3
point-slope form:
y – 6 = –3(x + 3)
y = –3x – 9 + 6
y = –3x – 3
45. Write 6 4 0x y− − = in slope intercept form.
6 4 0
6 4
6 4
x y
y x
y x
− − =
− = − +
= −
The slope of the perpendicular line is 6, thus the
slope of the desired line is
1
.
6
m = −
( )
1 1
1
6
1
6
1
6
( )
( 1) ( 12)
1 ( 12)
1 2
6 6 12
6 18 0
y y m x x
y x
y x
y x
y x
x y
− = −
− − = − − −
+ = − +
+ = − −
+ = − −
+ + =
46. slope:
2
;
5
y-intercept: –1
47. slope: –4; y-intercept: 5
48. 2 3 6 0
3 2 6
2
2
3
x y
y x
y x
+ + =
= − −
= − −
slope:
2
;
3
− y-intercept: –2

49. 2 8 0
2 8
4
y
y
y
− =
=
=
slope: 0; y-intercept: 4
50. 2 5 10 0x y− − =
Find x-intercept:
2 5(0) 10 0
2 10 0
2 10
5
x
x
x
x
− − =
− =
=
=
Find y-intercept:
2(0) 5 10 0
5 10 0
5 10
2
y
y
y
y
− − =
− − =
− =
= −
51. 2 10 0x − =
2 10
5
x
x
=
=
52. a. First, find the slope using the points
(2,28.2) and (4,28.6).
28.6 28.2 0.4
0.2
4 2 2
m
−
= = =
−
Then use the slope and one of the points to write
the equation in point-slope form.
( )
( )
( )
1 1
28.2 0.2 2
or
28.6 0.2 4
y y m x x
y x
y x
− = −
− = −
− = −
b. Solve for y to obtain slope-intercept form.
( )28.2 0.2 2
28.2 0.2 0.4
0.2 27.8
( ) 0.2 27.8
y x
y x
y x
f x x
− = −
− = −
= +
= +
c. ( ) 0.2 27.8
(7) 0.2(12) 27.8
30.2
f x x
f
= +
= +
=
The linear function predicts men’s average age
of first marriage will be 30.2 years in 2020.
53. a.
27 21 6
0.2
2010 1980 30
m
−
= = =
−
b. For the period shown, the number of the
percentage of liberal college freshman increased
each year by approximately 0.2. The rate of
change was 0.2% per year.
54.
( )2 2
2 1
2 1
[9 4 9 ] [4 4 5]( ) ( )
10
9 5
f x f x
x x
− − − ⋅−
= =
− −
55.
56.

57.
58.
59.
60.
61.
62.
63.
64.
65.
66.
67.
68.

69.
70.
71.
72.
73.
74.
75.
76. domain: ( , )−∞ ∞
77. The denominator is zero when x = 7. The domain is
( ) ( ),7 7,−∞ ∞ .
78. The expressions under each radical must not be
negative.
8 – 2x ≥ 0
–2x ≥ –8
x ≤ 4
domain: ( , 4].−∞
79. The denominator is zero when x = –7 or
x = 3.
domain: ( ) ( ) ( ), 7 7,3 3,−∞ − − ∞ 
negative. The denominator is zero when x = 5.
x – 2 ≥ 0
x ≥ 2
domain: [ ) ( )2,5 5,∞
negative.
1 0 and 5 0
1 5
x x
x x
− ≥ + ≥
≥ ≥ −
domain: [ )1,∞
82. f(x) = 3x – 1; g(x) = x – 5
(f + g)(x) = 4x – 6
domain: ( , )−∞ ∞
(f – g)(x) = (3x – 1) – (x – 5) = 2x + 4
domain: ( , )−∞ ∞
2
( )( ) (3 1)( 5) 3 16 5fg x x x x x= − − = − +
domain: ( , )−∞ ∞
3 1
( )
5
f x
x
g x
  −
= 
− 
domain: ( ) ( ),5 5,−∞ ∞

83. 2 2
( ) 1; ( ) 1f x x x g x x= + + = −
2
( )( ) 2f g x x x+ = +
domain: ( , )−∞ ∞
2 2
( )( ) ( 1) ( 1) 2f g x x x x x− = + + − − = +
domain: ( , )−∞ ∞
2 2
4 3
2
2
( )( ) ( 1)( 1)
1
1
( )
1
fg x x x x
x x x
f x x
x
g x
= + + −
= + − −
  + +
= 
− 
domain: ( ) ( ) ( ), 1 1,1 1,−∞ − − ∞ 
84. ( ) 7; ( ) 2
( )( ) 7 2
f x x g x x
f g x x x
= + = −
+ = + + −
domain: [2, )∞
( )( ) 7 2f g x x x− = + − −
domain: [2, )∞
2
( )( ) 7 2
5 14
fg x x x
x x
= + ⋅ −
= + −
domain: [2, )∞
7
( )
2
f x
x
g x
  +
= 
− 
domain: (2, )∞
85. 2
( ) 3; ( ) 4 1f x x g x x= + = −
a. 2
2
( )( ) (4 1) 3
16 8 4
f g x x
x x
= − +
= − +

b. 2
2
( )( ) 4( 3) 1
4 11
g f x x
x
= + −
= +

c. 2
( )(3) 16(3) 8(3) 4 124f g = − + =
86. ( ) ;f x x= g(x) = x + 1
a. ( )( ) 1f g x x= +
b. ( )( ) 1g f x x= +
c. ( )(3) 3 1 4 2f g = + = =
87. a.
( )( )
1
11 11
1
1 1 1 2
2 2
f g x f
x
x
xxx
x
x
x x
 
=  
 
 
++   + = = =
− − − 
 

b. 0 1 2 0
1
2
x x
x
≠ − ≠
≠
( )
1 1
,0 0, ,
2 2
   
−∞ ∞   
   
 
88. a. ( )( ) ( 3) 3 1 2f g x f x x x= + = + − = +
b. 2 0
2
x
x
+ ≥
≥ −
[ 2, )− ∞
89. 4 2
( ) ( ) 2 1f x x g x x x= = + −
90. ( ) 3
( ) 7 4f x x g x x= = +
91.
3 1 5
( ) ; ( ) 2
5 2 3
f x x g x x= + = −
3 5 1
( ( )) 2
5 3 2
6 1
5 2
7
10
f g x x
x
x
 
= − + 
 
= − +
= −
5 3 1
( ( )) 2
3 5 2
5
2
6
7
6
g f x x
x
x
 
= + − 
 
= + −
= −
f and g are not inverses of each other.

92.
2
( ) 2 5 ; ( )
5
x
f x x g x
−
= − =
2
( ( )) 2 5
5
2 (2 )
2 (2 5 ) 5
( ( ))
5 5
x
f g x
x
x
x x
g f x x
− 
= −  
 
= − −
=
− −
= = =
f and g are inverses of each other.
93. a. ( ) 4 3f x x= −
1
4 3
4 3
3
4
3
( )
4
y x
x y
x
y
x
f x−
= −
= −
+
=
+
=
b. 1 3
( ( )) 4 3
4
x
f f x− + 
= − 
 
3 3x
x
= + −
=
1 (4 3) 3 4
( ( ))
4 4
x x
f f x x− − +
= = =
94. a. 3
( ) 8 1f x x= +
3
3
3
3
3
3
3
1
8 1
8 1
1 8
1
8
1
8
1
2
1
( )
2
y x
x y
x y
x
y
x
y
x
y
x
f x−
= +
= +
− =
−
=
−
=
−
=
−
=
b. ( )
3
3
1 1
( ) 8 1
2
x
f f x−
 −
= +  
 
1
8 1
8
1 1
x
x
x
− 
= + 
 
= − +
=
( )
( )33
1
33
8 1 1
( )
2
8
2
2
2
x
f f x
x
x
x
−
+ −
=
=
=
=
95. a.
7
( )
2
x
f x
x
−
=
+
( )
1
7
2
7
2
2 7
2 7
1 2 7
2 7
1
2 7
( ) , 1
1
x
y
x
y
x
y
xy x y
xy y x
y x x
x
y
x
x
f x x
x
−
−
=
+
−
=
+
+ = −
− = − −
− = − −
− −
=
−
− −
= ≠
−

b. ( )
( )
( )
1
2 7
7
1( )
2 7
2
1
2 7 7 1
2 7 2 1
9
9
x
xf f x
x
x
x x
x x
x
x
−
− −
−
−=
− −
+
−
− − − −
=
− − + −
−
=
−
=
( )
( )
( )
1
7
2 7
2
( )
7
1
2
2 14 7 2
7 2
9
9
x
x
f f x
x
x
x x
x x
x
x
−
− 
− − + =
−
−
+
− + − +
=
− − +
−
=
−
=
96. The inverse function exists.
97. The inverse function does not exist since it does not
pass the horizontal line test.
98. The inverse function exists.
99. The inverse function does not exist since it does not
pass the horizontal line test.
100.
101. 2
( ) 1f x x= −
2
2
2
1
1
1
1
1
( ) 1
y x
x y
y x
y x
f x x−
= −
= −
= −
= −
= −
102. ( ) 1f x x= +
2
1 2
1
1
1
( 1)
( ) ( 1) , 1
y x
x y
x y
x y
f x x x−
= +
= +
− =
− =
= − ≥
103. 2 2
2 2
[3 ( 2)] [9 ( 3)]
5 12
25 144
169
13
d = − − + − −
= +
= +
=
=
104. ( )
22
2 2
[ 2 ( 4)] 5 3
2 2
4 4
8
2 2
2.83
d = − − − + −
= +
= +
=
=
≈
105.
( )
( )
2 12 6 4 10 10
, , 5,5
2 2 2 2
+ − + − 
= = −   
  

Chapter 2 Test
106.
4 ( 15) 6 2 11 4 11
, , , 2
2 2 2 2 2
+ − − + − − −     
= = −     
     
107. 2 2 2
2 2
3
9
x y
x y
+ =
+ =
108. 2 2 2
2 2
( ( 2)) ( 4) 6
( 2) ( 4) 36
x y
x y
− − + − =
+ + − =
109. center: (0, 0); radius: 1
domain: [ ]1,1−
range: [ ]1,1−
110. center: (–2, 3); radius: 3
domain: [ ]5,1−
range: [ ]0,6
111. 2 2
2 2
2 2
2 2
4 2 4 0
4 2 4
4 4 2 1 4 4 1
( 2) ( 1) 9
x y x y
x x y y
x x y y
x y
+ − + − =
− + + =
− + + + + = + +
− + + =
center: (2, –1); radius: 3
domain: [ ]1,5−
range: [ ]4,2−
Chapter 2 Test
1. (b), (c), and (d) are not functions.
2. a. f(4) – f(–3) = 3 – (–2) = 5
b. domain: (–5, 6]
c. range: [–4, 5]
d. increasing: (–1, 2)
e. decreasing: ( 5, 1) or (2, 6)− −
f. 2, f(2) = 5
g. (–1, –4)
h. x-intercepts: –4, 1, and 5.
i. y-intercept: –3
3. a. –2, 2
b. –1, 1
c. 0
d. even; ( ) ( )f x f x− =
e. no; f fails the horizontal line test
f. (0)f is a relative minimum.
g.
h.

i.
j. 2 1
2 1
( ) ( ) 1 0 1
1 ( 2) 3
f x f x
x x
− − −
= = −
− − −
4.
domain: ( ),−∞ ∞
range: ( ),−∞ ∞
5.
domain: [ ]2,2−
range: [ ]2,2−
6.
domain: ( ),−∞ ∞
range: {4}
7.
domain: ( ),−∞ ∞
range: ( ),−∞ ∞
8.
domain: [ ]5,1−
range: [ ]2,4−
9.
domain: ( ),−∞ ∞
range: { }1,2−
10.
domain: [ ]6,2−
range: [ ]1,7−
11.
domain of f: ( ),−∞ ∞
range of f: [ )0,∞
domain of g: ( ),−∞ ∞
range of g: [ )2,− ∞

Chapter 2 Test
12.
range of g: ( ],4−∞
13.
range of f: ( ),−∞ ∞
domain of 1
f −
: ( ),−∞ ∞
range of 1
f −
: ( ),−∞ ∞
14.
domain of 1
f −
: ( ),−∞ ∞
range of 1
f −
: ( ),−∞ ∞
15.
domain of f: [ )0,∞
range of f: [ )1,− ∞
domain of 1
f −
: [ )1,− ∞
range of 1
f −
: [ )0,∞
16. 2
( ) 4f x x x= − −
2
2
2
( 1) ( 1) ( 1) 4
2 1 1 4
3 2
f x x x
x x x
x x
− = − − − −
= − + − + −
= − −
17.
( ) ( )f x h f x
h
+ −
( )
( )
2 2
2 2 2
2
( ) ( ) 4 4
2 4 4
2
2 1
2 1
x h x h x x
h
x xh h x h x x
h
xh h h
h
h x h
h
x h
+ − + − − − −
=
+ + − − − − + +
=
+ −
=
+ −
=
= + −
18. ( )2
( )( ) 2 6 4g f x x x x− = − − − −
2
2
2 6 4
3 2
x x x
x x
= − − + +
= − + −
19.
2
4
( )
2 6
f x x
x
g x
  − −
= 
− 
domain: ( ) ( ),3 3,−∞ ∞
20. ( )( )( ) ( )f g x f g x=
2
2
2
(2 6) (2 6) 4
4 24 36 2 6 4
4 26 38
x x
x x x
x x
= − − − −
= − + − + −
= − +
21. ( )( )( ) ( )g f x g f x=
( )2
2
2
2 4 6
2 2 8 6
2 2 14
x x
x x
x x
= − − −
= − − −
= − −
22. ( ) ( )2
( 1) 2 ( 1) ( 1) 4 6g f − = − − − − −
( )
( )
2 1 1 4 6
2 2 6
4 6
10
= + − −
= − −
= − −
= −

23. 2
( ) 4f x x x= − −
2
2
( ) ( ) ( ) 4
4
f x x x
x x
− = − − − −
= + −
f is neither even nor odd.
( )
2 3
2 3
2 3
7
7
7
x y
x y
x y
+ =
− + =
+ =
axis.
( )
2 3
32
2 3
7
7
7
x y
x y
x y
+ =
+ − =
− =
( ) ( )
2 3
2 3
2 3
7
7
7
x y
x y
x y
+ =
− + − =
− =
25.
8 1 9
3
1 2 3
m
− − −
= = =
− − −
point-slope form: y – 1 = 3(x – 2)
or y + 8 = 3(x + 1)
26.
1
5
4
y x= − + so m = 4
point-slope form: y – 6 = 4(x + 4)
27. Write 4 2 5 0x y+ − = in slope intercept form.
4 2 5 0
2 4 5
52
2
x y
y x
y x
+ − =
= − +
= − +
The slope of the parallel line is –2, thus the slope of
the desired line is 2.m = −
( )
1 1( )
( 10) 2 ( 7)
10 2( 7)
10 2 14
2 24 0
y y m x x
y x
y x
y x
x y
− = −
− − = − − −
+ = − +
+ = − −
+ + =
28. a. Find slope:
25.8 24.6 1.2
0.12
20 10 10
m
−
= = =
−
point-slope form:
( )
( )
1 1
24.6 0.12 10
y y m x x
y x
− = −
− = −
b. slope-intercept form:
( )24.6 0.12 10
24.6 0.12 1.2
0.12 23.4
( ) 0.12 23.4
y x
y x
y x
f x x
− = −
− = −
= +
= +
c. ( ) 0.12 23.4
0.12(40) 23.4
28.2
f x x= +
= +
=
According to the model, 28.2% of U.S.
households will be one-person households in
2020.
29.
2 2
3(10) 5 [3(6) 5]
10 6
205 103
4
192
4
48
− − −
−
−
=
=
=
30. g(–1) = 3 – (–1) = 4
(7) 7 3 4 2g = − = =
31. The denominator is zero when x = 1 or
x = –5.
domain: ( ) ( ) ( ), 5 5,1 1,−∞ − − ∞ 
negative.
5 0 and 1 0
5 1
x x
x x
+ ≥ − ≥
≥ − ≥
domain: [ )1,∞

Cumulative Review
33.
7 7
( )( )
2 2 44
x
f g x
x
x
= =
−−

0, 2 4 0
1
2
x x
x
≠ − ≠
≠
domain: ( )
1 1
,0 0, ,
2 2
   
−∞ ∞   
   
 
34. ( ) ( )7
2 3f x x g x x= = +
35. 2 2
2 1 2 1( ) ( )d x x y y= − + −
( )
( )
22
2 1 2 1
22
2 2
( )
(5 2) 2 ( 2)
3 4
9 16
25
5
d x x y y= − + −
= − + − −
= +
= +
=
=
1 2 1 2 2 5 2 2
, ,
2 2 2 2
7
,0
2
x x y y+ + + − +   
=   
  
 
=  
 
The length is 5 and the midpoint is
( )
7
,0 or 3.5,0
2
 
 
 
.
Cumulative Review Exercises (Chapters 1–2)
1. domain: [ )0,2
range: [ ]0,2
2. ( ) 1f x = at 1
2
and 3
2
.
3. relative maximum: 2
4.
5.
6.
2
2
( 3)( 4) 8
12 8
20 0
( 4)( 5) 0
x x
x x
x x
x x
+ − =
− − =
− − =
+ − =
x + 4 = 0 or x – 5 = 0
x = –4 or x = 5
7. 3(4 1) 4 6( 3)
12 3 4 6 18
18 25
25
18
x x
x x
x
x
− = − −
− = − +
=
=
8.
2 2
2
2
2
2
( ) ( 2)
4 4
0 5 4
0 ( 1)( 4)
x x
x x
x x
x x x
x x
x x
+ =
= −
= −
= − +
= − +
= − −
x – 1 = 0 or x – 4 = 0
x = 1 or x = 4
A check of the solutions shows that x = 1 is an
extraneous solution.
9. 2/ 3 1/ 3
6 0x x− − =
Let 1/ 3
.u x= Then 2 2/ 3
.u x=
2
6 0
( 2)( 3) 0
u u
u u
− − =
+ − =
1/3 1/3
3 3
–2 or 3
–2 or 3
(–2) or 3
–8 or 27
u u
x x
x x
x x
= =
= =
= =
= =

10. 3 2
2 4
x x
− ≤ +
4 3 4 2
2 4
2 12 8
20
x x
x x
x
   
− ≤ +   
   
− ≤ +
≤
The solution set is ( ,20].−∞
11.
domain: ( ),−∞ ∞
range: ( ),−∞ ∞
12.
domain: [ ]0,4
range: [ ]3,1−
13.
range of g: ( ),−∞ ∞
14.
domain of f: [ )3,∞
domain of 1
f −
: [ )2,∞
range of 1
f −
: [ )3,∞
15.
( ) ( )f x h f x
h
+ −
( ) ( )
( )
( )
2 2
2 2 2
2 2 2
2
4 ( ) 4
4 ( 2 ) 4
4 2 4
2
2
2
x h x
h
x xh h x
h
x xh h x
h
xh h
h
h x h
h
x h
− + − −
=
− + + − −
=
− − − − +
=
− −
=
− −
=
= − −
16. ( )( )( ) ( )f g x f g x=
( )
( )
2
2
2
2
2
( )( ) 5
0 4 5
0 4 ( 10 25)
0 4 10 25
0 10 21
0 10 21
0 ( 7)( 3)
f g x f x
x
x x
x x
x x
x x
x x
= +
= − +
= − + +
= − − −
= − − −
= + +
= + +

The value of ( )( )f g x will be 0 when 3x = − or
7.x = −
17.
1 1
,
4 3
y x= − + so m = 4.
point-slope form: y – 5 = 4(x + 2)
general form: 4 13 0x y− + =

Cumulative Review
18. 0.07 0.09(6000 ) 510
0.07 540 0.09 510
0.02 30
1500
6000 4500
x x
x x
x
x
x
+ − =
+ − =
− = −
=
− =
$1500 was invested at 7% and $4500 was
invested at 9%.
19. 200 0.05 .15
200 0.10
2000
x x
x
x
+ =
=
=
For $2000 in sales, the earnings will be the
same.
20. width = w
length = 2w + 2
2(2w + 2) + 2w = 22
4w + 4 + 2w = 22
6w = 18
w = 3
2w + 2 = 8
The garden is 3 feet by 8 feet.

College algebra 7th edition by blitzer solution manual

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College algebra 7th edition by blitzer solution manual