Calculus early trans 6e solution ron larson bruce edwards
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C H A P T E R 1
Preparation for Calculus
Section 1.1
Graphs and Models.................................................................................2
Section 1.2
Linear Models and Rates of Change....................................................11
Section 1.3
Functions and Their Graphs.................................................................22
Section 1.4
Fitting Models to Data..........................................................................34
Section 1.5
Inverse Functions..................................................................................37
Section 1.6
Exponential and Logarithmic Functions .............................................54
Review Exercises ..........................................................................................................63
Problem Solving ...........................................................................................................73
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C H A P T E R 1
Preparation for Calculus
Section 1.1 Graphs and Models
3x
2
1. y
7. y
3
x-intercept: (2, 0)
x2
4
x
3
y
5
2
0
2
3
4
0
5
y-intercept: (0, 3)
Matches graph (b).
0
y
2. y
x2
9
6
(0, 4)
3, 0 , 3, 0
x-intercepts:
y-intercept: (0, 3)
2
( 2, 0)
(2, 0)
x
6
Matches graph (d).
4
4
( 3, 5)
3. y
6
x-intercepts:
3, 0 ,
3, 0
8. y
y-intercept: (0, 3)
Matches graph (a).
4. y
x
3
(3, 5)
4
x2
3
6
2
x
x-intercepts: 0, 0 ,
2
x
3
x
0
1
2
3
4
5
6
y
9
4
1
0
1
4
9
1, 0 , 1, 0
y
y-intercept: (0, 0)
10
(0, 9)
(6, 9)
8
Matches graph (c).
6
1x
2
5. y
(1, 4)
(2, 1)
4
2
2
(5, 4)
(4, 1)
x
4
x
y
2
0
1
0
2
4
2
3
4
6
4
2
2
9. y
4
2
x
6
(3, 0)
2
y
5
x
6
4
3
2
1
0
1
2
3
(4, 4)
4
3
y
(2, 3)
2
1
0
1
(0, 2)
( 2, 1)
y
x
4
2
2
( 4, 0)
4
6
2
( 5, 3)
6. y
5
4
( 4, 2) 2
2x
( 1, 1)
( 3, 1)
1
x
0
1
2
5
2
5
3
1
0
3
(1, 3)
(0, 2)
x
4
6
4
2
( 2, 0)
2
7
y
1
3
y
8
( 1, 7)
(0, 5)
4
2
(1, 3)
(2, 1)
x
6
4
(3, 1)
2
2
4
2
5,0
2
(4, 3)
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Section 1.1
10. y
x
x
1
14. y
3
y
2
2
1
1
0
0
1
1
2
3
0
1
2
1
x
2
x
6
4
3
y
1
4
1
2
1
y
5
4
3
2
3
(3, 2)
2
( 2, 1)
( 1, 1)
(2, 1)
1
2
1
( 1, 0)
2
1
( 6, 14 )
( 4, 12 )
( 3,
2
1
2
1
4
1
1 2 3
2
3
4
5
1)
6
x
x
Undef.
0
x
3
(1, 0)
(0, 1)
2
11. y
1
(0, 12 )
(2, 14 )
x
3
2
y
4
( 3, 2)
3
Graphs and Models
0
1
4
9
15. y
16
5
x
5
y
6
4
5
2
3
( 4.00, 3)
(2, 1.73)
y
6
6
2
3
x
4
4
2
8
12
16
(9, 3)
(16, 2)
4
(4, 4)
(1, 5)
6
(0, 6)
(a)
2, y
2, 1.73
(b)
x, 3
4, 3
y
3
5
2
5
3
1.73
4
8
12. y
16. y
2
x
x5
5x
6
x
2
y
1
0
0
1
2
2
7
14
2
3
4
( 0.5, 2.47)
9
9
(1, 4)
y
6
5
4
3
(b)
(7, 3)
( 1, 1)
2
(a)
(14, 4)
(2, 2)
(0, 2 )
17. y
x
( 2, 0)
5
10
15
x
3
2
1
0
1
2
3
y
1
3
2
3
Undef.
3
3
2
1
(1, 3)
3
3
2
1
2
3
20
5
2x
x
5;
2
18. y
4x2
4
1,
4
5
2x
5; 0, 5
5
5,
2
0
3
40
2
x-intercept: 0
4 x2
3
2
1
1
4 and x,
5
(3, 1)
x
1.65,
x-intercept: 0
2, 2
2
3
2x
4
y-intercept: y
y
( 3, 1)
x,
0.5, 2.47
y-intercept: y
20
3
x
13. y
0.5, y
4x
3
3; 0, 3
3
None. y cannot equal 0.
1
3
2
2, 2
( 1, 3)
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4
Chapter 1
x2
19. y
Preparation for Calculus
x
2
24. y
y-intercept: y
0
y
2
0
2; 0,
x-intercepts: 0
x2
0
x
x
2, 1;
20. y
2
x
3
2
2
x
x2
3x
3x
1
2
02
y-intercept: y
y
1
2, 0 , 1, 0
y-intercept: y
3
0
x-intercepts: 0
y
0; 0, 0
x3
x-intercepts: 0
0
40
x
4x
0
x x
x
0,
25. x 2 y
2 x
2
2; 0, r0 ,
2, 0
x2
x2
3x
3x
1
x x
3
3x
1
2
2
0, 3; 0, 0 ,
4y
02
4y
0
y
21. y
x 16
x
2
x-intercept: x 2 0
y-intercept: y
0 16
x2
0
x
x 4
x
0, 4,
x
x2
1
y-intercept: y
y
x-intercept: 0
x
23. y
2
5x
4
26. y
x2
2x
y-intercept: y
4; 0, 0 , 4, 0 ,
4, 0
02
x2
1
y -intercept: y
x-intercept: 0
2
5x
x
1
4x
x
3x
2
1
x2
1
3
2
x
4;
x
2
0; 0, 0
1
1
3
3
r
x
3
;
3
0, 2
1
1
2
x
Note: x
0
x
1
2;
x2
2x
2
x
1
0
1
0
1; 0, 1
2x
1
1; 1, 0
2
50
02
20
0
1; 0, 1
x
40
1
y
x-intercept:
1
0; 0, 0
x
x
1
0
x2
0; 0, 0
x 16
x-intercepts: 0
22. y
0
2
3, 0
0
y-intercept: 02 y
r
1…”
0; 0, 0
4x
2
2
3 0
2
2 x
30
3
,
3
Đ
0Ô
Ô
'
ã
3 3 is an extraneous solution.
27. Symmetric with respect to the y-axis because
y
x
2
x2
6
6.
4, 0
x2
28. y
x
No symmetry with respect to either axis or the origin.
29. Symmetric with respect to the x-axis because
y
2
y2
x3
8 x.
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Section 1.1
30. Symmetric with respect to the origin because
y
3
x
y
x
y
3
x
40. y
x
3
x
2x
3
y
2
3
0
2x
3
x.
1
0
1
1, y -intercept
2x
3
1
3,
2
Intercepts: 0, 1 ,
31. Symmetric with respect to the origin because
x y
xy
4.
1
3
,
2
x
y
2
Symmetry: none
3
,0
2
y
xy 2
(0, 1)
)
x
32. Symmetric with respect to the x-axis because
x
x-intercept
0
(
2
5
Graphs and Models
1
1
2
1
10.
2
33. y
4
x
3
No symmetry with respect to either axis or the origin.
34. Symmetric with respect to the origin because
x
y
4
x
xy
2
x
9
0
9
2
y
9
3, x -interceptsr
x
y
3, 0
9
2
x
(0, 9)
x2
9
6
4
2
1
( 3, 0)
(3, 0)
x
4
2
2
4
6
2
2x2
x
x 2x
1
2
x2
1
is symmetric with respect to the y-axis
x
because y
x
x3
37. y
x2
Symmetry: y-axis
42. y
36. y
9, y -intercept
.
1
x2
x
x
2
6
x
y
y
10
x
2
x2
Intercepts: 0, 9 , 3, 0 ,
0.
35. Symmetric with respect to the origin because
y
9
0
0
x2
4
41. y
2
2
x2
x
1
2
.
1
y
020
0
x 2x
1
0, y -intercept
1
x
Intercepts: 0, 0 ,
x is symmetric with respect to the y-axis
0,
1,
2
1,
2
x-intercepts
y
0
5
Symmetry: none
4
3
because y
x
3
x3
x
x3
x
x.
2
(
38. y
x
1
,
2
0
)
1
(0, 0)
x
3 is symmetric with respect to the x-axis
3
2
1
1
2
3
because
y
x
3
y
x
3.
39. y
2
3x
y
2
30
0
2
3x
Intercepts: 0, 2 ,
43. y
x3
2
y
3
2
3
2
0
2,
3
2
x
Intercepts:
2, y -intercept
3x
0
2
,
3
x
x-intercept
2, y -intercept
x3
3
2
3
x
2, x-intercept
2, 0 , 0, 2
y
Symmetry: none
5
4
y
0
3
(0, 2)
Symmetry: none
2
(0, 2)
3
(
2, 0)
1
x
1
1
3
2
1
2
3
1
2
,0
3
x
2
3
1
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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6
Chapter 1
Preparation for Calculus
44. y
x3
4x
y
03
40
x x
y3
47. x
y3
0, y -intercept
x3
4x
0
x x2
4
0
2 x
2
0
x
y
0, y -intercept
0, x-intercept
Intercept: (0, 0)
x
3
y
y3
x
Symmetry: origin
0, 2, x-intercepts
r
x
0
y
Intercepts: 0, 0 , 2, 0 ,
y
4
3
3
x
2, 0
4
x3
x
x3
4x
4x
2
(0, 0)
x
Symmetry: origin
4
3
2
1
y
1
2
3
4
2
3
3
4
( 2, 0)
(0, 0)
3
(2, 0)
1
1
y2
48. x
x
4
3
1
y
2
y
3
45. y
x
y
x
x
5
0 0
5
5
0
x
0, y -intercept
x
Intercepts: 0, 0 ,
0, 5, x -intercepts
5, 0
2
4
2 y
0
2
0
y
2,
r y -intercepts
x
2
4
Intercepts: 0, 2 , 0,
2,
x
y
2
0
y2
4
4, x-intercept
4, 0
4
y
y
Symmetry: x-axis
3
Symmetry: none
3
2
( 5, 0)
(0, 0)
(0, 2)
x
4
3
2
1
1
( 4, 0)
2
x
5
2
1
1
3
(0, 2)
4
3
46. y
y
5
25
x2
25
2
0
25
49. y
8
x
5, y -intercept
25
x2
0
y
8
0
Undefined
25
x2
0
x 5
x
0
8
x
0
No solution
5,
r x-intercept
x
Intercepts: 0, 5 , 5, 0 ,
y
25
x
2
5, 0
25
no x-intercept
Intercepts: none
y
x2
no y -intercept
8
x
8
x
y
Symmetry: origin
y
Symmetry: y-axis
8
y
6
7
6
( 5, 0)
4
3
2
1
4
(0, 5)
2
x
2
2
4
6
8
(5, 0)
x
4 3 2 1
1 2 3 4 5
2
3
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Section 1.1
10
x2 1
50. y
53. y 2
y
10
y
02
x
9
xr
0
1
No solution
no x-intercepts
y
Intercept: (0, 10)
xr
9
0
x
9
0
12
10
y
2
x
10
x2 1
1
10
x
(0, 10)
9
r 0
y
10
x2
9
2
y
10, y -intercept
1
x
r9
9
2
y
3, ry -intercepts
9, x-intercept
Intercepts: 0, 3 , 0, 3 ,
Symmetry: y-axis
7
Graphs and Models
x
y2
9
9, 0
x
9
2
x
6
4
2
2
4
Symmetry: x-axis
6
y
51. y
6
x
y
6
0
6
x
0
6
x
6
4
6, y -intercept
( 9, 0)
2
(0, 3)
2
(0, 3)
x
10
4
2
2
4
6
6,
r x-intercepts
x
Intercepts: 0, 6 ,
y
6
6
x
6, 0 , 6, 0
6
x
54. x 2
4 y2
Symmetry: y-axis
4
r
y
x2
(0, 6)
6
2
(6, 0)
8
4
2
4
2
2
6
8
1, ry -intercepts
2,
r x-intercepts
2, 0 , 2, 0 , 0, 1 , 0, 1
Intercepts:
4
4
r
2
2
4
x
x
02
4
2
x
4
( 6, 0)
2
40
r
2
4
y
8
x2
4
y
6
8
x
52. y
6
x
y
6
0
2
4
y
2
x2
4
4 y2
4
Symmetry: origin and both axes
6
y
6, y -intercept
3
6
x
0
6
x
0
6
2
(0, 1)
(2, 0)
( 2, 0)
x, x -intercept
3
x
1
1
3
(0, 1)
2
Intercepts: (0, 6), (6, 0)
3
Symmetry: none
y
8
(0, 6)
4
2
(6, 0)
x
2
4
6
8
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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8
Chapter 1
55. x
Preparation for Calculus
3y2
6
3y2
6
x
6
r
y
30
x
0
r 2, y -intercepts
3
y
8
4x
y
7
y
4x
x
4x
15
5x
3
x
7
6, x-intercept
2
y
2 , 0,
3y2
x
6
58. 3x
2y
4
4x
2y
10
4
4x
6
3x
2
5.
3
2)
4
2
4x
y
10
2
10
2
3x
4
1
3x
y
2
y
( 0,
7
Point of intersection: (3, 5)
Symmetry: x-axis
2
x
The corresponding y-value is y
6
Intercepts: 6, 0 , 0,
3
8
8
x
6
r
x
x
y
3
y
2
x
57.
4
4x
7x
14
x
2
(6, 0)
10
x
1
1
2
3
( 0,
2
6
7
The corresponding y-value is y
2)
Point of intersection:
3
1.
2, 1
4
56. 3x
4 y2
8
4 y2
3x
8
y
r
3x
4
3x
40
4
x2
x
y
4
y
4
x
2
4
x
x
2
x
x
2 x
x
0
r 2
2
1
2, 1
The corresponding y-values are y
no y -intercepts
y
5 for x
60.
3x
8
4 y2
8
x
3
y2
y
x
1
3
x
x
3
x
x2
0
2
Symmetry: x-axis
y
x
6
x
2 for x
2 and
1.
Points of intersection: 2, 2 ,
0
2
x
0
8
, x-intercept
3
8,
3
y
6
8
x
3x
y
8
3x
Intercept:
6
6
3
0
2
4
no solution
2
y
2
r
y
59. x 2
1
y2
3
1, 5
x
2
2x
x
1 or x
1
2
x
1 x
2
2
4
2
The corresponding y-values are y
( 83, 0)
x
2
2
6
8
10
and y
1 for x
2 for x
1
2.
2
4
Points of intersection:
1,
2 , 2, 1
6
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Section 1.1
61. x 2
x
5
5
y2
5
y2
y
1
y
x
2
x
x
2
2
x
x2
5
x
64. y
1
y
x4
2 x2
1
2
x
Graphs and Models
9
1
2
1
y = x4
2x
2x 2 + 1
2
1
(0, 1)
0
2x
x
2
2x
4
1 or x
2 x
1 x
3
2
The corresponding y-values are y
1 for x
and y
2
2 for x
( 1, 0)
1
2
1,
25
y2
y
15
y
25
x2
3x
25
x2
9x2
3x
90 x
15
90 x
0
x2
9x
20
0
x
5 x
4
65. y
200
4 or x
y
x3
x
2x2
2
4
1
6
x2
4x
3 for x
4
x+6
(3,
y=
x2
3)
( 2, 2)
7
2
4x
5, 0
2, 2 ,
Points of intersection:
x
Analytically,
1
x
1
y = x3
1 x
1, 0, 1.
2
4, 3 ,
x
3x
x x
1
2
4
y=
5.
Points of intersection:
x
y
5
0 for x
x
2
0
The corresponding y-values are y
63. y
x
2x2
225
10 x
and y
0
4
x
0
x
x4
2
15
2
x2
x2
25
1, 0 , 0, 1 , 1, 0
Points of intersection:
2 , 2, 1
Analytically, 1
y2
3x
x2
y=1
2.
Points of intersection:
62. x 2
3
(1, 0)
2x2 + x
x
1
2
5x
3 x
x
3,
3
x2
6
6
6
2
x
2
3, 1.732
4x
4x
0
0
(2, 1)
4
6
(0, 1)
3,
x
2.
( 1, 5)
66. y
8
y = x2 + 3x
1
Points of intersection:
Analytically, x
3
2x
x
3
x x
y
2x
6
1, 5 , 0, 1 , 2, 1
2
x
x
2
2 x
1
x
2x
0
1
0
2
3x
6
x
7
1
y=6
(1, 5)
x
(3, 3)
4
8
1
x
3
1, 0, 2.
y = ⏐2x
3⏐+ 6
Points of intersection: (3, 3), (1, 5)
Analytically,
2x
2x
3
6
6
2x
3
x
3
x or 2 x
3
x
3 or
x
x
x
1.
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|
10
Chapter 1
Preparation for Calculus
67. (a) Using a graphing utility, you obtain
y
0.005t 2 0.27t 2.7.
(b)
71. y
kx3
(a) 1, 4 :
k1
1
k
0
k 0
3
k
16
2, 1 :
(b)
(c)
0
(c) For 2020, t
y
0, 0 :
(d)
30
0
40.
1, 1 :
72. y 2
0.005 40
2
0.27 40
1
(b)
(c)
4k
k
1
4
4
0, 0 :
2
k can be any real number.
3
k
k
1
4k 2
16
8k
k
2
0
2
4k 0
k can be any real number.
20
30
(d)
The model is a good fit for the data.
3, 3 :
3
30.
0.24 30
1
8
k
4k 1
1
2, 4 :
330
y
3
k 1
12
(a) 1, 1 :
68. (a) Using a graphing utility, you obtain
y
0.24t 2 12.6t 40.
(c) For 2020, t
8k
4kx
The GDP in 2020 will be $21.5 trillion.
5
3
2
4
2.7
21.5
(b)
4
2
12.6 30
40
2
4k 3
9
12k
k
9
12
3
4
554
The number of cellular phone subscribers in 2020
will be 554 million.
C
69.
2.04 x
R
5600
3.29 x
5600
3.29 x
5600
1.25 x
x
5600
1.25
73. Answers may vary. Sample answer:
y
x
x
4, x
4 x
3 x
8 has intercepts at
3, and x
8.
74. Answers may vary. Sample answer:
2.04 x
y
3
2
x
3,
2
x
4480
x
5
2
4 x
5.
2
4, and x
x
75. (a) If (x, y) is on the graph, then so is
To break even, 4480 units must be sold.
70. y
is
0.37
y by x-axis symmetry. So, the graph is
symmetric with respect to the origin. The converse is
not true. For example, y
x3 has origin symmetry
but is not symmetric with respect to either the x-axis
or the y-axis.
400
0
x,
x, y by y-axis
x, y is on the graph, then so
symmetry. Because
10,770
x2
has intercepts at
(b) Assume that the graph has x-axis and origin
symmetry. If (x, y) is on the graph, so is x,
100
0
If the diameter is doubled, the resistance is changed by
approximately a factor of 14. For instance,
y 20 | 26.555 and y 40 | 6.36125.
x-axis symmetry. Because x,
then so is
x,
y
y by
y is on the graph,
x, y by origin
symmetry. Therefore, the graph is symmetric with
respect to the y-axis. The argument is similar for
y-axis and origin symmetry.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Section 1.2
x3
x:
y
03
0
0 ; 0, 0
x-intercepts: 0
x3
x
x x2
76. (a) Intercepts for y
y -intercept:
0, 0 , 1, 0
x2
2:
y -intercept:
0
2
x-intercepts: 0
x
2
x x
1 x
1;
1, 0
Intercepts for y
y
1
11
Linear Models and Rates of Change
2 ; 0, 2
2
None. y cannot equal 0.
(b) Symmetry with respect to the origin for y
y
x
3
x
x
3
y
x
2
x
x3
x
x2
2
x
2
0
2
x
1
0
x
2
(c)
x
3
x
x
2 x
2
x because
x2
2 because
x.
Symmetry with respect to the y-axis for y
2
x3
2.
2
y
6
Point of intersection : (2, 6)
Note: The polynomial x 2
1 has no real roots.
x
77. False. x-axis symmetry means that if
4,
5 is on the
4, 5 is also on the graph. For example,
graph, then
4, 5 is not on the graph of x
f
r b2
2a
4ac Đ
, 0Ô .
Ô
'
29, whereas
b Đ
, 0Ô .
2a '
80. True. The x-intercept is
4, 5 is on the graph.
78. True. f 4
y2
b
79. True. The x-intercepts are
„
•
‚
4.
Section 1.2 Linear Models and Rates of Change
1. m
2
2. m
0
7 1
2 1
6. m
6
3
2
y
3. m
1
4. m
12
( 2, 7)
7
6
5
3
2
5. m
4
3
5
2
6
2
1
3
x
4
3
2
1
1
3
4
y
3
2
(5, 2)
1
x
1
1
2
3
5
6
7
2
3
4
(3, 4)
5
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12
Chapter 1
1
4
7. m
Preparation for Calculus
6
4
y
11.
5
, undefined.
0
m= 2
m is undefined.
The line is vertical.
3
2
m=
m=1
8
y
6
(3, 4)
4
7
2
(4, 6)
6
x
5
6
4
2
2
4
4
8 10
3
2
y
12.
(4, 1)
1
x
2
1
1
2
3
5
m= 3
m=
6
6
( 2, 5)
5
5
8. m
5
3
m=0
4
0
2
0
m=3
x
6
The line is horizontal.
2
2
4
2
y
1
13. Because the slope is 0, the line is horizontal and its
2. Therefore, three additional points are
equation is y
x
1
1
1
2
3
4
5
6
0, 2 , 1, 2 , 5, 2 .
2
3
4
14. Because the slope is undefined, the line is vertical and its
equation is x
4. Therefore, three additional points
(3, 5) (5, 5)
6
4, 0 ,
are
2
3
1
2
9. m
1
6
1
2
2
1 ã
4
3Đ
Ô
4'
4, 1 ,
4, 2 .
15. The equation of this line is
y
y
7
3x
1
y
3x
10.
Therefore, three additional points are (0, 10), (2, 4), and
(3, 1).
3
2
(
1 2
,
2 3
)
(
3 1
,
4 6
)
16. The equation of this line is
x
3
2
1
2
y
3
1
2
3
2
2 x
2
y
2x
2.
Therefore, three additional points are
3,
4,
1, 0 ,
and (0, 2).
3Đ
ã 1
Ô
4'
4
7Đ
ã5
Ô
8'
4
10. m
'
Đ
Ô
ã
1
Đ 3ã
Ô 8
'
8
3
17.
y
3x
4
3
4y
3x
12
0
3x
4y
12
y
y
5
3
4
(0, 3)
2
1
2
( 78 , 34 )
1
x
2
1
1
1
x
( 54 , 14 )
4
3
2
1
1
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Section 1.2
18. The slope is undefined so the line is vertical.
x
x
21. y
2
3x
y
2
3x
9
y
3x
11
0
3x
y
5
5
Linear Models and Rates of Change
0
y
1
13
3
11
y
x
4
3
2
1
1
3
1
2
2
1
x
3
2
1
4
1
1
2
3
5
6
(3, 2)
2
5
4
3
4
5
y
2x
3
3y
2x
0
2x
19.
y
22.
3y
y
3x
3
5
4
5y
20
5y
14
x
2
3x
6
0
4
y
3
5
2
4
( 2, 4)
(0, 0)
x
1
2
3
4
2
1
1
x
3
20.
y
y
4
4
0
y
2
1
1
23. (a) Slope
’y
’x
2
1
3
(b)
5
x
10 ft
(0, 4)
3
30 ft
2
By the Pythagorean Theorem,
1
x2
x
3
2
1
1
2
x
y
Population (in millions)
24. (a)
(b) The slopes are:
310
305
300
295
290
t
4
5
6
7
8
9
Year (4 j 2004)
302
102
10 10
295.8
5
298.6
6
301.6
7
304.4
8
307.0
9
1000
31.623 feet.
293.0
4
295.8
5
298.6
6
301.6
7
304.4
8
|
2.8
2.8
3.0
2.8
2.6
The population increased least rapidly from 2008 to 2009.
(c) Average rate of change from 2004 to 2009:
307.0
9
293.0
4
14
5
2.8 million per yr
(d)
For 2020, t
20 and y
Equivalently, y “ 11 2.8
16 2.8
307.0
293.0
337.8.‹
|337.8 million.
|
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www.pdfgrip.com
14
Chapter 1
25. y
4x
Preparation for Calculus
3
33. y
The slope is m
2x
1
y
4 and the y-intercept is 0, 3 .
3
x
26.
y
1
y
x
1
1
1 and the y-intercept is (0, 1).
The slope is m
x
2
27. x
5y
1
1
1
20
1x
5
y
2
4
1
5
Therefore, the slope is m
and the y-intercept is
1x
3
34. y
1
y
(0, 4).
2
28. 6 x
5y
15
y
6x
5
1
x
3
3
2
3
1
(0, 1)
Therefore, the slope is m
6
5
2
and the y-intercept is
3
0, 3 .
29. x
4
4
35. y
The line is vertical. Therefore, the slope is undefined and
there is no y-intercept.
30. y
2
3
2
y
3x
2
x
1
1
2
y
1
4
The line is horizontal. Therefore, the slope is m
the y-intercept is 0, 1 .
0 and
3
2
1
x
4
31. y
3
2
1
3
2
3
4
2
3
y
4
2
1
x
3
2
1
1
2
3
4
36. y
5
2
1
3x
y
3x
4
13
4
y
5
6
16
12
32. x
4
y
x
16
8
12
4
8
4
3
8
2
1
x
1
2
3
37. 2 x
y
3
0
y
2x
5
1
2
3
y
1
x
2
2
1
3
1
2
3
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
www.pdfgrip.com
Section 1.2
38. x
2y
6
0
y
8
6
43. m
1x
2
15
Linear Models and Rates of Change
3
3
6
5
, undefined
0
The line is horizontal.
y
x
6
6
0
4
x
2
y
x
10
8
6
2
(6, 8)
8
6
4
4
6
(6, 3)
2
x
39.
0
8 0
2
4 0
2 x 0
y
2x
m
y
0
2
y
4
8
2
(4, 8)
8
6
4
2x
2
2
3
44. m
2
y
(0, 0)
x
4
2
2
4
2
y
6
y
0
2
1
0
2
2
0
y
7
1
40. m
2
2
9
3
1
y
3
x
1
8
1
2
3
4
1
y
y
41. m
3x
2
2
3x
y
3x
4
0
3x
y
8
2
0
5
2
1x
y
2
x
3
4
4
2
4
(3, 2)
4
6
4
45. m
y
9
8
7
6
5
4
3
2
1
5
40
3
(2, 8)
y
3
4
y
(5, 0)
1 2 3 4
6 7 8 9
0
y
1
7
46. m
6
( 3, 6)
3
4
5
1
0
11
x
2
11
x
2
22 x
(1, 2)
2
x
3
2
1
y
4
( 12 , 72 )
1
2
2
x
3
4
4y
4
3
2
Đ
Ô
ã
1
'
Đ 3ã
Ô 8
'
2
3
4
y
y
5Đ
Ô
4'
12 y
3
32 x
40
37
1
8
3
8
x
3
12 y
1
3
3
ã
2
3
32 x
( 0, 34 )
1
1
4
1
4
11
2
0
3Đ
ã 1
Ô
4'
4
7Đ
ã5
Ô
8'
4
3
0
11
4
1
2
3
2
1
7
2
1
2
x
1
4
4
y
y
x
6
( 2, 2)
0
6 2
3 1
42. m
x
40
(1, 2)
3
8
x
3
8
x
3
0
y
3y
4
2
8
3
y
8x
6
2
1
0
( 78 , 34 )
x
2
1
1
1
( 54 , 14 )
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16
Chapter 1
47.
x
x
3
3
0
Preparation for Calculus
52.
y
x
a
3
a
y
a
4
a
1
a
a
2
1
(3, 0)
1
2
x
4
1
1
1
1
1
2
53.
b
a
48. m
b
x
a
y
b
x
a
x
a
y
b
y
b
1
x
2a
9
2a
9
b
(a, 0)
x
54.
3x
50.
3x
x
2
3
3x
2
3x
y
2
y
2
5
2
2y
5
2y
y
a
a
2
3
0
x
1
a
a
4
3
1
4
3
x
y
3y
4
0
1
1
2
y
4
3
2
0
2
a
1
5
x
a
1
y
2
y
1
5
2
3
1
1
2y
3x
y
a
2
a
3
a
a
2a
y
x
49.
1
5
2
x
5
x
y
3
6
1
5
2
(0, b)
1
0
y
1
1
x
2
x
2
2y
y
a
2
a
4
2a
5
a
y
x
51.
a
1
a
x
y
x
4
3
0
55. The given line is vertical.
1
1
(a) x
7, or x
7
0
(b) y
2, or y
2
0
56. The given line is horizontal.
3
x
x
y
3
(a) y
y
3
0
(b) x
0
1, or x
1
0
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Section 1.2
57. x
y
2
y
x
m
1
(a)
x
61. 5 x
3y
2
5x
3
5
3
m
y
5
1x
y
5
x
y
3
0
2
7
8
y
(a)
2
24 y
5
1x
y
5
x
x
y
7
y
7
y
x
m
1
(a)
7
62. 3x
y
2
1x
y
2
x
x
y
1
(b) y
2
1x
y
2
x
3
0
x
y
59. 4 x
2y
m
2x
4
0
2x
y
(b)
24 x
7
4
y
1
2y
2
x
2y
4
0
7x
8
x
3
4
18
0
3
4
20
8
4
3x
4
3
3x
12
5
4y
x
0
4
3
5
x
4
5
4x
3
16
3
15
4x
16
0
4x
3y
31
63. The slope is 250.
2
2
V
1850 when t
V
250 t
2
250t
8
7
x
4
7
4
24 x
5
3y
3
x
3
4
4y
2
y
2.
1850
1350
64. The slope is 4.50.
2
V
156 when t
V
4.5 t
2
4.5t
147
2.
156
65. The slope is 1600.
1
2
7
x
4
5Đ
Ô
6'
24 y
1
2
12
7
x
4
42 x
35
24
35
24 y
23
y
1
2
4
x
7
5Đ
Ô
6'
42 y
21
24 x
20
42 y
41
0
y
42 x
y
(a)
1
2
y
(a)
7
y
1
m
53
3x
4
(b) y
y
y
40 y
3x
5
1
4y
35
y
(a) y
3
5
40 y
m
3
2
7
8
9
7
3x
2x
4y
24 y
4y
3
3
2
2 x
60. 7 x
30
40 x
0
y
x
4y
3
3
(b)
40 x
y
0
3
4
x
0
(b)
24 x
58. x
5
3
21
2
2
17
0
y
y
(b)
Linear Models and Rates of Change
„
•
‚
V
V
17,200 when t
1600 t
1600t
2
2.
17,200
20,400
66. The slope is 5600.
0
V
ã
V
245,000 when t
5600 t
5600t
2
2.
245,000
256,200
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18
Chapter 1
67. m1
m2
Preparation for Calculus
1
2
0
2
0
1
2
71. Equations of altitudes:
a b
y
x a
c
x
b
a b
y
x a
c
Solving simultaneously, the point of intersection is
•
a 2 b 2Đ
b,
Ô.
c '
1
1
2
3
m1 z m2
The points are not collinear.
6 4
7 0
11 4
m2
5 0
m1 z m2
10
7
7
5
68. m1
y
(b, c)
The points are not collinear.
(a, 0)
69. Equations of perpendicular bisectors:
y
c
2
a
y
c
2
a
b
a
x
c
b
b
x
c
x
bĐ
Ô
2 '
aĐ
Ô
2 '
( a, 0)
ã
ã
Setting the right-hand sides of the two equations equal
and solving for x yields x
0.
b,
a2
b2
2c
a2
b 2Đ
Ô is:
c '
m1
c 2Đ
Ô.
'
ã
b 2 c
b
b3
a2
Letting x
0 in either equation gives the point of
intersection:
0,
b cĐ
, Ô to
3 3'
72. The slope of the line segment from
3a 2
3b 2
3c
”
…
3a 2
2b 3
This point lies on the third perpendicular bisector,
x
0.
The slope of the line segment from
y
0,
a2
b2
2c
a
( b 2 a , 2c )
(
a
b, c
2
2
m2
)
x
3a
(a, 0)
( a, 0)
b
2
c
0
2
70. Equations of medians:
y
y
y
y
c2
cĐ
Ô to
3'
ã
3b
2
2c
2
b3
3c 2
2c 2
6c
3a 2
3b 2
2bc
m2
Therefore, the points are collinear.
c
x
b
c
3a
b
x
c
x
3a b
( b 2 a , 2c )
a
(b, c)
( a 2 b , 2c )
a
x
( a, 0)
(0, 0) (a, 0)
Solving simultaneously, the point of intersection is
b cĐ
, Ô.
3 3'
ã
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•
‚
”
…
c3
b3
m1
b
,
3
3b 2
2bc
c 2Đ
Ô is:
'
(b, c)
2
ã
ã
c3
c2
c2
Section 1.2
73. ax
by
(a) The line is parallel to the x-axis if a
b z 0.
0 and
(b) The line is parallel to the y-axis if b
a z 0.
0 and
(c) Answers will vary. Sample answer: a
b 8.
8y
4
y
1
8
5x
4
(d) The slope must be
5.
2
5x
8
2y
4
y
1
2
5x
and b
3.
5
2
(e) a
5x
2
3y
4
5x
6y
8
Lines d and f appear parallel.
(d) Lines b and f appear perpendicular.
Lines b and d appear perpendicular.
5 and
75. Find the equation of the line through the points (0, 32)
and (100, 212).
180
100
9 C
5
9C
5
m
F
32
F
5 and
9
5
0
32
or
C
5x
2
4
2
For x
W1
0.07 s
2000
New job offer: W2
0.05s
2300
1
9
5F
160
5F 9C 160
0
72 , C q 22.2 . |
For F
76. C
77. (a) Current job:
(b)
(b) Lines a and b have negative slopes.
(c) Lines c and e appear parallel.
1
2
Answers will vary. Sample answer: a
b
2.
5x
19
74. (a) Lines c, d, e and f have positive slopes.
4
5x
Linear Models and Rates of Change
0.51x
q
200
137, C
0.51 137
200
$269.87.
3500
(15,000, 3050)
0
1500
20,000
Using a graphing utility, the point of intersection is (15,000, 3050).
Analytically, W1
0.07 s
W2
2000
0.05s
0.02 s
300
s
So, W1
W2
2300
15,000
0.07 15,000
2000
3050.
When sales exceed $15,000, the current job pays more.
(c) No, if you can sell $20,000 worth of goods, then W1 ! W2 .
(Note: W1
3400 and W2
78. (a) Depreciation per year:
875
5
3300 when s
20,000.)
1000
$175
y
875
175 x
xd
where 0
5.
d
0
6
0
(b) y
(c)
875
175 2
200
875
175 x
675
$525
175 x
x | 3.86 years
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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20
Chapter 1
Preparation for Calculus
79. (a) Two points are (50, 780) and (47, 825).
The slope is
825 780
45
m
15.
47 50
3
p
780
p
15 x
50
15 x
750
82. The tangent line is perpendicular to the line joining the
point 4, 3 and the center of the circle, (1, 1).
y
4
780
15 x
1530
6
(1, 1)
2
2
4
2
1
1530
15
x
(b)
2
x
or
p
(4, 3)
6
50
Slope of the line joining (1, 1) and 4, 3 is
1
1
0
4
.
3
1600
Tangent line:
0
If p
855, then x
(c) If p
795, then x
80. (a) y
18.91
x
(b)
3
4
45 units.
1
1530
15
y
0
y
2
49 units
795
y
3.97 x
quiz score, y
3
x 4
4
3
x 6
4
3x 4 y
3
24
test score
83. x
100
1 2
d
0
1 1
2
1
5
2
0
2
2
1
5 2
2
20
0
(c) If x
17, y
18.91
3.97 17
84. 4 x
86.4.
3y
10
42
d
0
33
4
(d) The slope shows the average increase in exam score
for each unit increase in quiz score.
(e) The points would shift vertically upward 4 units.
The new regression line would have a y-intercept
22.91 3.97 x.
4 greater than before: y
85. A point on the line x y
from the point (0, 1) to x
d
81. The tangent line is perpendicular to the line joining the
point (5, 12) and the center (0, 0).
10
11
12
5
2
10
7
5
2
3
1 is (0, 1). The distance
y 5 0 is
1
12
5
4
2
2
2 2.
y
86. A point on the line 3x
(5, 12)
4
(0, 0) 8
4
x
d
16
3 1
4 1
2
3
8
4
10
2
1, 1 . The
1 is
1, 1 to 3x
distance from the point
8
8
4y
3
4
5
4y
10
10
0 is
9
.
5
16
Slope of the line joining (5, 12) and (0, 0) is
12
.
5
The equation of the tangent line is
5
y 12
x 5
12
5
169
y
x
12
12
5 x 12 y 169 0.
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Section 1.2
87. If A
d
0, then By
d
0, then Ax
C
By1
A
2
C
B
2
.
0 is the vertical line x
Ax1ã C
A
CĐ
Ô
A'
x1
Ax1
Ax1
By1
A
2
C
B
2
C A. The distance to x1 , y1 is
.
(Note that A and B cannot both be zero.) The slope of the line Ax
The equation of the line through x1 , y1 perpendicular to Ax
Ay
Ay1
B
x x1
A
Bx Bx1
Bx1
Ay1
Bx
y
y1
21
C B. The distance to x1 , y1 is
0 is the horizontal line y
By1ã C
B
CĐ
Ô
B'
y1
If B
C
Linear Models and Rates of Change
By
By
C
0 is
A B.
0 is:
C
Ay
The point of intersection of these two lines is:
Ax
By
Bx
C
Ay
Bx1
Ay1
A2 x
B2 x
A
ABy
ABy
2
2
B x1
2
B x
x
Ax
By
Bx
C
Ay
Bx1
Ay1
A
2
B 2 x1 ABy1
,
A2 B 2
2
2
AC
B x1
AC
2
ABy1
By adding equations 1 and 2
B x1 ABy1
A
B2
3
BC
ABx1
2
B y
2
A y1
BC
BC
y
AC
ABy1
2
B2 y
A2 y
ABx
ABx
1
AC
4
2
ABx1
A y1 By adding equations 3 and 4
ABx1 A2 y1
A2 B 2
ABx1 A2 y1Đ
Ô point of intersection
A2 B 2
'
BC
The distance between x1 , y1 and this point gives you the distance between x1 , y1 and the line Ax
d
AC
B 2 x1 ABy1
2
A
B2
AC
ABy1 A2 x1“
«
2
A
B2
‹
“ AC
88. y
d
mx
1y
C
m3
By1
A
2
B
2
BC
4
2
ABx1 A2 y1
2
A
B2
ABx1 B 2 y1“
«
2
A
B2
‹
Ax1 ” By1
«
2
A
B2
‹
“
y«1
‹
2
2
”
»
…
A2
B2 C
»
… A2
0.
”
»
…
”
»
…
2
C
”
»
…
Ax1
B
2
By1
2
2
Ax1 By1 C
»
… A2 B 2
0
1 1
m
The distance is 0 when m
BC
BC
2
mx
2
2
By1 Ax
”1 “
«
2
2
A
B
‹
4
Ax1
“
x«
1
‹
By
4
1
2
3m
m
3
2
1. In this case, the line y
1
x
4 contains the point (3, 1).
8
9
9
( 1, 0)
4
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•
‚
22
Chapter 1
Preparation for Calculus
89. For simplicity, let the vertices of the rhombus be (0, 0),
(a, 0), (b, c), and a b, c , as shown in the figure.
c
The slopes of the diagonals are then m1
c
m2
b
equal, a
b
and
y2*
x2*
. Because the sides of the rhombus are
a
y1*
x1*
y2
x2
y1
x1
y
2
b
2
2
c , and you have
c
m1m2
a
91. Consider the figure below in which the four points are
collinear. Because the triangles are similar, the result
immediately follows.
a
c2
c
b
b
b2
a
c2
c2
a2
(x 2 , y2 )
(x *2 , y*2 )
(x1, y1 )
1.
(x *1, y*1 )
Therefore, the diagonals are perpendicular.
y
x
(b, c)
(a + b, c)
1 m2 , then m1m2
92. If m1
1. Let L3 be a line with
slope m3 that is perpendicular to L1. Then m1m3
1.
x
(0, 0)
(a , 0)
So, m2
L 2 and L 3 are parallel. Therefore,
m3
L 2 and L1 are also perpendicular.
90. For simplicity, let the vertices of the quadrilateral be
(0, 0), (a, 0), (b, c), and (d, e), as shown in the figure.
The midpoints of the sides are
a
,
2
b§ d c e ã
dĐ e
Đ a bã c
0Ô ,
, , ¤
,
, ‚and ¤, .
2 „ 2 ' 2
2 „
'
'2 2
The slope of the opposite sides are equal:
c
2
a
0
b
a
2
e
e
2
d
2
2
a
2
2
0
c
e
2
d
2
b
2
c
c
2
a
d
c
b
ã
bx
c1
a
x
b
y
Đ
ay Ô c2
'
1
m1
c1
b
ãb
y ‚ x
„
a
a
b
m1
c2
a
m2
b
a
1 m1 is negative.
95. True. The slope must be positive.
d
2
„
by
94. False; if m1 is positive, then m2
e
2
b
ax
m2
e
b
93. True.
a
96. True. The general form Ax By
horizontal and vertical lines.
d
2
0 includes both
C
Therefore, the figure is a parallelogram.
y
(d, e)
(b 2 d ,
c
e
2
)
(b, c)
( d2 , 2e )
(a 2 b , 2c )
x
(0, 0)
( a2 , 0)
(a, 0)
Section 1.3 Functions and Their Graphs
1. (a) f 0
(b) f
70
3
7 3
(c) f b
(d) f x
4
7b
1
4
4
4
7 x
2. (a) f
25
7b
1
4
4
7x
11
4
4
(b) f 11
11
(c) f 4
4
(d) f x
x’
5
5
5
x
1
1
16
4
9
3
x
5
’
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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3. (a) g 0
(b) g
5
1
3
2
3
2
(c) g c
2
8.
9.
4
2t
t2
4
0
3
2
’x
c
2
t
4
t
4 t
x
3x
1
f 2
x
2
x
x3
f 1
f x
t
4x2
t2
2t
45
8
2
4
4
t3
8t 2
3
x3
x3
3 x 2 x
3x 2 x
3
x
1
x
1
x
2
x
0
1
Đ 2
Ô
' 2
(c) f
2 Đ
Ô
3'
2
ãS
sin
3
Đ3
Ô
2
'
(d) f
Đ
Ô
6'
1
x
1
3, x
1
x
2
3
ãS
sin
6
x3
3x 2
1
1
1
x x
x
x
x
1
1
S
ã
S
ã
S
x
0
, x
2
„
2
3x x
x ,
1 x
1
1
x x
1, x
1
x
11
x
1
x
11
’
’
6x
>0,
f
5
> 5,
16. h x
f
Domain: x
x3
Domain:
f f,
f f,
Range:
4
x2
Domain:
f f,
14. h x
Range:
f , 4@
17. f x
16
3
x
Range:
13. f x
z
z
Range: >0, f
Range:
’
1
1
0t
Domain:
f
x2
t x2
0
d
16
4, 4>
Range: 0, 4>
Note: y
> 3,
f , 0@
16
x2
0t
3
@
@
16
x 2 is a semicircle of radius 4.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
www.pdfgrip.com
ã
S
Domain: 6 x t 0
Domain: f f ,
S
ã
1Đ
Ô
'2
x
2
x
x
x2
z
Range: >0, f
12. g x
Ô
S
0
5
ãS
sin
4
x
'
1
1
x
2
3x
S
1
5Đ
Ô
4'
2
x
1
cos 2S
(b) f
16t
S Đ
ã
0
2
Đ
cos
Ô
'
1
2
sinS
23
1
S Đ Đ
2ã ã
ã
cos
2 Ô Ô cos
3
3
' '
Đ
Ô
3'
(c) f
1
15. g x
f f,
Domain:
5
x
1
x
x
1
11. f x
1
cos 0
ã S Đ
cos 2
Ô
4'
Đ
Ô
4'
(b) f
x
f 1
f x
4
4c
x
1
cos 2 0
0
5
2
3
x
f x
5. (a) f 0
6. (a) f
2
f x
Functions and Their Graphs
(d) f
9
4
4
1
x
10.
1
4
4
f x
2
t
2
5
5
5
c c
(d) g t
5
2
2
42 4
4. (a) g 4
2
5
5
(d) g t
(b) g
5
5
(c) g 2
7.
02
5
Section 1.3
24
Chapter 1
18. f x
Preparation for Calculus
3
x
sec
sin x
St
4
1S
2n
4
2
20. h t
z
4n
t
4nz
Domain: all t
z
2
2, n an integer
nS, n an integer
Domain: all t
0z
f ,0
Range:
0,
f
,0
3 z 0
x
3 z 0
f 0,
2 x
x
0 t
x
x d1
Domain: 0
xd 1
2
d>0, 1@
(a) f
24. f x
x
2
x
x
3x
3x
2
0
2 x
1
0
2 or xt
Domain: x
1
1
cos x
2,
f
2t 2
4
0
2, x t 0
1
2 1
1
1
(b) f 0
20
2
2
(c) f 2
22
2
6
2 t2
1
(d) f t 2
1
(Note: t 2
1 t 0 for all t.)
f,
2
f
>2,
, 1f
Range:
t
1
f
z
cos x z 1
Domain: all x z 2nS, n an integer
› x2
fi 2
fl2 x
2, x ! 1
2
2
30. f x
d
(a) f
2
cos x
0
2, 2
t
f , 1@ >
f 2,
Domain:
25. g x
2
2
1, x
Domain:
2
z
0
f,
› 2x
fi
fl2 x
29. f x
x t 0 and
4 z 0
Domain:
[Note: You can see that the range is all y z 1 by
graphing f.]
x t 0 and 1
4
Domain: all x z r 2
Range: all y z 1
1
f
3,
1
x2
x
x 2
x 4
Domain: all x z 4
x
S
3
f, 3
x2
f
22. f x
23. f x
x
28. g x
Domain: all x
n , n integer
2S
6
3
Domain:
3
x
21. f x
x
Domain: all x z
f f,
Range:
S
5
2n ,
1
27. f x
cot t
z
6
Domain: all x
1@ f >1,
f ,
Range:
12
1
z 0
2
1
sin x z
2
Range: >0, f
St
sin x
f f,
Domain:
19. f t
1
26. h x
2, x d 1
2
2
6
(b) f 0
02
(c) f 1
12
(d) f s 2
2
(Note: s 2
2 ! 1 for all s.)
Domain:
f,
2
2
2 s2
2
3
2
2
2
2s 4
8s 2
10
f
Range: >2, f
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S
