1
− 1
+
1
− 2 =
+ − 1 − 2( − 1)
( − 1)
=
2 − 1 − 2 2 + 2
2 −
= − 2
2
− 4 + 1
2 −
, { | 6= 01}
(b) ( − )() =
1
− 1
−
1
− 2
=
− ( − 1) + 2( − 1)
( − 1)
=
1 + 2 2 − 2
2 −
=
2 2 − 2 + 1
2 −
, { | 6= 01}
(c) ()() =
1
− 1
1
− 2
=
1
2 −
−
2
− 1
=
1 − 2
2 − , { | 6= 01}
(d)
() =
1
− 1
1
− 2
=
1
− 1
1 − 2
=
1
− 1
·
1 − 2
=
( − 1)(1 − 2)
= −
( − 1)(2 − 1)
= −
2 2 − 3 + 1 ,
| 6= 0
1
2 1
[Note the additional domain restriction () 6= 0 ⇒ 6=
1
2 .]
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.
38
¤
CHAPTER 1 FUNCTIONSAND MODELS
35. () = 3 + 5 and () =
3
√
. The domain of each function is (−∞∞).
(a) ( ◦ )() = (()) =
3
√
=
3
√
3
+ 5 = + 5. The domain is (−∞∞).
(b) ( ◦ )() = (()) = ( 3 + 5) =
3
3 + 5. The domain is (−∞∞).
(c) ( ◦ )() = (()) = ( 3 + 5) = ( 3 + 5) 3 + 5. The domain is (−∞∞).
(d) ( ◦ )() = (()) =
3
√
=
3
3
√
=
9
√
. The domain is (−∞∞).
36. () = 1 and () = 2 + 1. The domain of is (−∞0) ∪ (0∞). The domain of is (−∞∞).
(a) ( ◦ )() = (()) = (2 + 1) =
1
2 + 1 . The domain is
{ | 2 + 1 6= 0} =
| 6= − 1
2
=
−∞− 1
2
∪
− 1
2 ∞
.
(b) ( ◦ )() = (()) =
1
= 2
1
+ 1 =
2
+ 1. We must have 6= 0, so the domain is (−∞0) ∪ (0∞).
(c) ( ◦ )() = (()) =
1
=
1
1
= . Since requires 6= 0, the domain is (−∞0) ∪ (0∞).
(d) ( ◦ )() = (()) = (2 + 1) = 2(2 + 1) + 1 = 4 + 3. The domain is (−∞∞).
37. () =
1
√ and () = + 1. The domain of is (0∞). The domain of is (−∞∞).
(a) ( ◦ )() = (()) = ( + 1) =
1
√ + 1 . We must have + 1 0, or −1, so the domain is (−1∞).
(b) ( ◦ )() = (()) =
1
√
=
1
√ + 1. We must have 0, so the domain is (0∞).
(c) ( ◦ )() = (()) =
1
√
=
1
1 √
=
1
1
√
=
√
=
4
√ . We must have 0, so the domain
is (0∞).
(d) ( ◦ )() = (()) = ( + 1) = ( + 1) + 1 = + 2. The domain is (−∞∞).
38. () =
+ 1
and () = 2 − 1. The domain of is (−∞−1) ∪ (−1∞). The domain of is (−∞∞).
(a) ( ◦ )() = (()) = (2 − 1) =
2 − 1
(2 − 1) + 1
=
2 − 1
2
. We must have 2 6= 0 ⇔ 6= 0. Thus, the domain
is (−∞0) ∪ (0∞).