Multivariable Calculus 9th Edition by James Stewart solution manual

Thus, even if  1 does not equal  2 , ( 1 ) may equal ( 2 ), so  is not 1­1.
16.  is not 1­1 because eventually we all stop growing and therefore, there are two times at which we have the same height.
17. (a) Since  is 1­1, (6) = 17 ⇔  −1 (17) = 6.
(b) Since  is 1­1,  −1 (3) = 2 ⇔ (2) = 3.
18. First, we must determine  such that () = 3. By inspection, we see that if  = 1, then (1) = 3. Since  is 1­1 ( is an
increasing function), it has an inverse, and  −1 (3) = 1. If  is a 1­1 function, then ( −1 ()) = , so ( −1 (2)) = 2.
19. First, we must determine  such that () = 4. By inspection, we see that if  = 0, then () = 4. Since  is 1­1 ( is an
increasing function), it has an inverse, and  −1 (4) = 0.
20. (a)  is 1­1 because it passes the Horizontal Line Test.
(b) Domain of  = [−33] = Range of  −1 . Range of  = [−13] = Domain of  −1 .
(c) Since (0) = 2,  −1 (2) = 0.
(d) Since (−17) ≈ 0,  −1 (0) ≈ −17.
21. We solve  =
5
9 ( − 32) for :
9
5  =  − 32
⇒  =
9
5  + 32. This gives us a formula for the inverse function, that
is, the Fahrenheit temperature  as a function of the Celsius temperature .  ≥ −45967 ⇒
9
5  + 32 ≥ −45967
⇒
9
5  ≥ −49167
⇒  ≥ −27315, the domain of the inverse function.
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 1.5 INVERSEFUNCTIONSAND LOGARITHMS
¤
51
22.  =
 0
 1 −  2  2 ⇒ 1 −
 2
 2
=
 2
0
 2
⇒
 2
 2
= 1 −
 2
0
 2
⇒  2 =  2

1 −
 2
0
 2

⇒  = 

1 −
 2
0
 2
.
This formula gives us the speed  of the particle in terms of its mass , that is,  =  −1 ().
23. First note that () = 1 −  2 ,  ≥ 0, is one­to­one. We first write  = 1 −  2 ,  ≥ 0, and solve for :
 2 = 1 −  ⇒  =
√ 1 −  (since  ≥ 0). Interchanging  and  gives  = √ 1 − , so the inverse function is
 −1 () =
√ 1 − .
24. Completing the square, we have () =  2 − 2 = ( 2 − 2 + 1) − 1 = ( − 1) 2 − 1 and, with the restriction  ≥ 1,
 is one­to­one. We write  = ( − 1) 2 − 1,  ≥ 1, and solve for :  − 1 =
√  + 1 (since  ≥ 1
⇔  − 1 ≥ 0),
so  = 1 +
√  + 1. Interchanging  and  gives  = 1 + √  + 1, so  −1 () = 1 + √  + 1.
25. First write  = () = 2 +
√  + 1 and note that  ≥ 2. Solve for :  − 2 = √  + 1
⇒ ( − 2) 2 =  + 1 ⇒
 = ( − 2) 2 − 1( ≥ 2). Interchanging  and  gives  = ( − 2) 2 − 1 so  −1 () = ( − 2) 2 − 1 with domain  ≥ 2.
26. We write  = () =
6 − 3
5 + 7
and solve for : (5 + 7) = 6 − 3 ⇒ 5 + 7 = 6 − 3 ⇒
5 + 3 = 6 − 7 ⇒ (5 + 3) = 6 − 7 ⇒  =
6 − 7
5 + 3 . Interchanging  and  gives  =
6 − 7
5 + 3 ,
so  −1 () =
6 − 7
5 + 3 .
27. We solve  =  1− for : ln = ln 1− ⇒ ln = 1 −  ⇒  = 1 − ln. Interchanging  and  gives the inverse
function  = 1 − ln.
28. We solve  = 3ln( − 2) for : 3 = ln( − 2) ⇒  3 =  − 2 ⇒  = 2 +  3 . Interchanging  and  gives the
inverse function  = 2 +  3 .
29. We solve  =

2 +
3
√   5
for :
5
  = 2 +
3
√ 
⇒
3
√  =
5
  − 2
⇒  =

5
  − 2  3 . Interchanging  and 
gives the inverse function  =

5
√  − 2  3 .