Multivariable Calculus 9th Edition by James Stewart solution manual

30. We solve  =
1 −  −
1 +  −
for : (1 +  − ) = 1 −  − ⇒  +  − = 1 −  − ⇒  − +  − = 1 −  ⇒
 − (1 + ) = 1 −  ⇒  − =
1 − 
1 + 
⇒ − = ln 1 − 
1 + 
⇒  = −ln 1 − 
1 + 
or, equivalently,
 = ln
 1 − 
1 + 
 −1
= ln
1 + 
1 − 
. Interchanging  and  gives the inverse function  = ln 1 + 
1 −  .
31.  = () =
√ 4 + 3 ( ≥ 0)
⇒  2 = 4 + 3 ⇒  =
 2 − 3
4
.
Interchange  and :  =
 2 − 3
4
. So  −1 () =
 2 − 3
4
( ≥ 0). From
the graph, we see that  and  −1 are reflections about the line  = .
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CHAPTER 1 FUNCTIONSAND MODELS
32.  = () = 1 +  − ⇒  − =  − 1 ⇒ − = ln( − 1) ⇒
 = −ln( − 1). Interchange  and :  = −ln( − 1).
So  −1 () = −ln( − 1). From the graph, we see that  and  −1 are
reflections about the line  = .
33. Reflect the graph of  about the line  = . The points (−1−2), (1−1),
(22), and (33) on  are reflected to (−2−1), (−11), (22), and (33)
on  −1 .
34. Reflect the graph of  about the line  = .
35. (a)  = () =
√ 1 −  2
(0 ≤  ≤ 1 and note that  ≥ 0) ⇒
 2 = 1 −  2 ⇒  2 = 1 −  2 ⇒  =
 1 −  2 . So
 −1 () =
√ 1 −  2 , 0 ≤  ≤ 1.
We see that  −1 and  are the same
function.
(b) The graph of  is the portion of the circle  2 +  2 = 1 with 0 ≤  ≤ 1 and
0 ≤  ≤ 1 (quarter­circle in the first quadrant). The graph of  is symmetric
with respect to the line  = , so its reflection about  =  is itself, that is,
 −1 = .
36. (a)  = () =
3
√ 1 −  3
⇒  3 = 1 −  3 ⇒  3 = 1 −  3 ⇒
 =
3
 1 −  3 . So  −1
() =
3
√ 1 −  3 . We see that  and  −1
are the
same function.
(b) The graph of  is symmetric with respect to the line  = , so its reflection
about  =  is itself, that is,  −1 = .
37. (a) It is defined as the inverse of the exponential function with base , that is, log   =  ⇔   = .
(b) (0∞) (c)  (d) See Figure 11.
38. (a) The natural logarithm is the logarithm with base , denoted ln.
(b) The common logarithm is the logarithm with base 10, denoted log.
(c) See Figure 13.
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SECTION 1.5 INVERSEFUNCTIONSAND LOGARITHMS
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39. (a) log 3 81 = log 3 3 4 = 4 (b) log 3 
1
81

= log 3 3 −4 = −4 (c) log 9 3 = log 9 9 12 =
1
2
40. (a) ln
1
 2
= ln −2 = −2 (b) ln
√  = ln 12
=
1
2
(c) ln

ln 
50 
= ln( 50 ) = 50
41. (a) log 2 30 − log 2 15 = log 2
 30
15

= log 2 2 = 1
(b) log 3 10 − log 3 5 − log 3 18 = log 3
 10
5

− log 3 18 = log 3 2 − log 3 18 = log 3

2
18

= log 3
 1
9

= log 3 3 −2 = −2
(c) 2log 5 100 − 4log 5 50 = log 5 100 2 − log 5 50 4 = log 5
 100 2
50 4

= log 5

10 4
5 4 · 10 4

= log 5 5 −4 = −4
42. (a)  3 ln 2 =  ln 2
3
= 2 3 = 8 (b)  −2 ln 5 =  ln 5
−2
= 5 −2 =
1
25
(c)  ln(ln 
3 )
=  ln(3) = 3
43. (a) log 10
  2  3  
= log 10  2 + log 10  3 + log 10  [Law 1]
= 2log 10  + 3log 10  + log 10  [Law 3]
(b) ln