46. Let () = cos and () = 2 . Then ( ◦ )() = (()) = (cos) = (cos) 2 = cos 2 = ().
47. Let () =
3
√ and () =
1 + . Then ( ◦ )() = (()) =
3
√
=
3
√
1 +
3
√ = ().
c ° 2021 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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CHAPTER 1 FUNCTIONSAND MODELS
48. Let () =
1 +
and () =
3
√ . Then ( ◦ )() = (()) =
1 +
=
3
1 +
= ().
49. Let () = 2 and () = sec tan. Then ( ◦ )() = (()) = ( 2 ) = sec( 2 ) tan( 2 ) = ().
50. Let () =
√ and () = √ 1 + . Then ( ◦ )() = (()) = √
=
1 +
√ = ().
51. Let () =
√ , () = − 1, and () = √ . Then
( ◦ ◦ )() = ((())) =
√
=
√
− 1
=
√
− 1 = ().
52. Let () = ||, () = 2 + , and () =
8
√ . Then
( ◦ ◦ )() = ((())) = ((||)) = (2 + ||) =
8
2 + || = ().
53. Let () = cos, () = sin, and () = 2 . Then
( ◦ ◦ )() = ((())) = ((cos)) = (sin(cos)) = [sin(cos)] 2 = sin 2 (cos) = ().
54. Let () = tan, () =
√ + 1, and () = cos. Then
( ◦ ◦ )() = ((())) = ((tan)) = √ tan + 1 = cos √ tan + 1 = ().
55. (a) ((3)) = (4) = 6. (b) ((2)) = (1) = 5.
(c) ( ◦ )(5) = ((5)) = (3) = 5. (d) ( ◦ )(5) = ((5)) = (2) = 3.
56. (a) (((2))) = ((3)) = (4) = 1. (b) ( ◦ ◦ )(1) = (((1))) = ((3)) = (5) = 2.
(c) ( ◦ ◦ )(1) = (((1))) = ((5)) = (2) = 1. (d) ( ◦ ◦ )(3) = (((3))) = ((4)) = (6) = 2.
57. (a) (2) = 5, because the point (25) is on the graph of . Thus, ((2)) = (5) = 4, because the point (54) is on the
graph of .
(b) ((0)) = (0) = 3
(c) ( ◦ )(0) = ((0)) = (3) = 0
(d) ( ◦ )(6) = ((6)) = (6). This value is not defined, because there is no point on the graph of that has
coordinate 6.
(e) ( ◦ )(−2) = ((−2)) = (1) = 4
(f) ( ◦ )(4) = ((4)) = (2) = −2
58. To find a particular value of (()), say for = 0, we note from the graph that (0) ≈ 28 and (28) ≈ −05. Thus,
((0)) ≈ (28) ≈ −05. The other values listed in the table were obtained in a similar fashion.
() (())
−5 −02 −4
−4 12 −33
−3 22 −17
−2 28 −05
−1 3 −02
() (())
0 28 −05
1 22 −17
2 12 −33
3 −02 −4
4 −19 −22
5 −41 19
c ° 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 1.3 NEW FUNCTIONSFROM OLD FUNCTIONS
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41
59. (a) Using the relationship distance = rate · time with the radius as the distance, we have () = 60.
(b) = 2 ⇒ ( ◦ )() = (()) = (60) 2 = 3600 2 . This formula gives us the extent of the rippled area
(in cm 2 ) at any time .
60. (a) The radius of the balloon is increasing at a rate of 2 cms, so () = (2 cms)( s) = 2 (in cm).
(b) Using =
4
3
3 , we get ( ◦ )() = (()) = (2) = 4
3 (2)
3
=
32
3
3 .
The result, =
32
3
3 , gives the volume of the balloon (in cm 3 ) as a function of time (in s).
61. (a) From the figure, we have a right triangle with legs 6 and , and hypotenuse .
By the Pythagorean Theorem, 2 + 6 2 = 2 ⇒ = () =