(b) ( ◦ )() = (()) = 2
+ 1
− 1 =
2
+ 1
− 1 =
2 − 1( + 1)
+ 1
=
− 1
+ 1 . We must have + 1 6= 0
⇔
6= −1. Thus, the domain is (−∞−1) ∪ (−1∞).
(c) ( ◦ )() = (()) =
+ 1
+ 1
+ 1
=
+ 1
+ 1
+ 1
·
+ 1
+ 1
=
+ ( + 1)
=
2 + 1 . We must have both + 1 6= 0
and 2 + 1 6= 0, so the domain excludes both −1 and − 1
2 . Thus, the domain is (−∞−1) ∪
−1− 1
2
∪
− 1
2 ∞
.
(d) ( ◦ )() = (()) = (2 − 1) = 2(2 − 1) − 1 = 4 − 3. The domain is (−∞∞).
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SECTION 1.3 NEW FUNCTIONSFROM OLD FUNCTIONS
¤
39
39. () =
2
and () = sin. The domain of is (−∞0) ∪ (0∞). The domain of is (−∞∞).
(a) ( ◦ )() = (()) = (sin) =
2
sin
= 2csc. We must have sin 6= 0, so the domain is
{ | 6= , an integer}.
(b) ( ◦ )() = (()) =
2
= sin
2
. We must have 6= 0, so the domain is (−∞0) ∪ (0∞).
(c) ( ◦ )() = (()) =
2
=
2
2
= . Since requires 6= 0, the domain is (−∞0) ∪ (0∞).
(d) ( ◦ )() = (()) = (sin) = sin(sin). The domain is (−∞∞).
40. () =
√ 5 − and () = √ − 1. The domain of is (−∞5] and the domain of is [1∞).
(a) ( ◦ )() = (()) = √ − 1
=
5 −
√ − 1. We must have − 1 ≥ 0
⇔ ≥ 1 and
5 −
√ − 1 ≥ 0
⇔
√ − 1 ≤ 5
⇔ 0 ≤ − 1 ≤ 25 ⇔ 1 ≤ ≤ 26. Thus, the domain is [126].
(b) ( ◦ )() = (()) = √ 5 −
=
√
5 − − 1. We must have 5 − ≥ 0 ⇔ ≤ 5 and
√ 5 − − 1 ≥ 0
⇔
√ 5 − ≥ 1
⇔ 5 − ≥ 1 ⇔ ≤ 4. Intersecting the restrictions on gives a domain
of (−∞4].
(c) ( ◦ )() = (()) = √ 5 −
=
5 −
√ 5 − . We must have 5 − ≥ 0
⇔ ≤ 5 and
5 −
√ 5 − ≥ 0
⇔
√ 5 − ≤ 5
⇔ 0 ≤ 5 − ≤ 25 ⇔ −5 ≤ − ≤ 20 ⇔ −20 ≤ ≤ 5.
Intersecting the restrictions on gives a domain of [−205].
(d) ( ◦ )() = (()) = √ − 1
=
√
− 1 − 1. We must have − 1 ≥ 0 ⇔ ≥ 1 and
√ − 1 − 1 ≥ 0
⇔
√ − 1 ≥ 1
⇔ − 1 ≥ 1 ⇔ ≥ 2. Intersecting the restrictons on gives a domain
of [2∞).
41. ( ◦ ◦ )() = ((())) = (( 2 )) = (sin( 2 )) = 3sin( 2 ) − 2
42. ( ◦ ◦ )() = ((())) =
√
=
2
√
=
2
√
− 4
43. ( ◦ ◦ )() = ((())) = (( 3 + 2)) = [( 3 + 2) 2 ] = ( 6 + 4 3 + 4)
=
( 6
+ 4 3 + 4) − 3 =
√ 6
+ 4 3 + 1
44. ( ◦ ◦ )() = ((())) =
3
√
=
3
√
3
√ − 1
= tan
3
√
3
√ − 1
45. Let () = 2 + 2 and () = 4 . Then ( ◦ )() = (()) = (2 + 2 ) = (2 + 2 ) 4 = ().