Multivariable Calculus 9th Edition by James Stewart solution manual

 −2 − 2
if −2 ≤  ≤ −1
√ 1 −  2
if −1   ≤ 1
.
21. () = ln,  = (0∞); () =  2 − 9,  = .
(a) ( ◦ )() = (()) = ( 2 − 9) = ln( 2 − 9).
Domain:  2 − 9  0 ⇒  2  9 ⇒ ||  3 ⇒  ∈ (−∞−3) ∪ (3∞)
(b) ( ◦ )() = (()) = (ln) = (ln) 2 − 9. Domain:   0, or (0∞)
(c) ( ◦ )() = (()) = (ln) = lnln. Domain: ln  0 ⇒    0 = 1, or (1∞)
(d) ( ◦ )() = (()) = ( 2 − 9) = ( 2 − 9) 2 − 9. Domain:  ∈ , or (−∞∞)
22. Let () =  +
√ , () = √ , and () = 1. Then ( ◦  ◦ )() = 1
  + √  = ().
23. More than one model appears to be plausible. Your choice of model depends
on whether you think medical advances will keep increasing life expectancy, or
if there is bound to be a natural leveling­off of life expectancy. A linear model,
 = 02441 − 4133960, gives us an estimate of 821 years for the
year 2030.
24. (a) Let  denote the number of toaster ovens produced in one week and
 the associated cost. Using the points (10009000) and
(150012,000), we get an equation of a line:
 − 9000 =
12,000 − 9000
1500 − 1000
( − 1000) ⇒
 = 6( − 1000) + 9000 ⇒  = 6 + 3000.
(b) The slope of 6 means that each additional toaster oven produced adds $6 to the weekly production cost.
(c) The ­intercept of 3000 represents the overhead cost—the cost incurred without producing anything.
25. The value of  for which () = 2+4  equals 6 will be  −1 (6). To solve 2+4  = 6, we either observe that letting  = 1
gives us equality, or we graph  1 = 2 + 4  and  2 = 6 to find the intersection at  = 1. Since (1) = 6,  −1 (6) = 1.
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CHAPTER 1 REVIEW
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26. We write  =
2 + 3
1 − 5
and solve for : (1 − 5) = 2 + 3 ⇒  − 5 = 2 + 3 ⇒  − 3 = 2 + 5 ⇒
 − 3 = (2 + 5) ⇒  =
 − 3
2 + 5
. Interchanging  and  gives  =
 − 3
2 + 5 , so 
−1 () =
 − 3
2 + 5 .
27. (a) ln √  + 1 = ln + ln
√  + 1
[Law 1]
= ln + ln( + 1) 12 = ln +
1
2
ln( + 1) [Law 3]
(b) log 2

 2 + 1
 − 1
= log 2
  2
+ 1
 − 1
 12
=
1
2
log 2
  2
+ 1
 − 1

[Law 3]
=
1
2
 log
2 (
2
+ 1) − log 2 ( − 1)  [Law 2]
=
1
2
log 2 ( 2 + 1) −
1
2
log 2 ( − 1)
28. (a)
1
2
ln − 2ln( 2 + 1) = ln 12 − ln( 2 + 1) 2 = ln
√ 
( 2 + 1) 2
(b) ln( − 3) + ln( + 3) − 2ln( 2 − 9) = ln[( − 3)( + 3)] − ln( 2 − 9) 2
= ln
( − 3)( + 3)
( 2 − 9) 2
= ln
 2 − 9
( 2 − 9) 2
= ln
1
 2 − 9
29. (a)  2 ln 5 =  ln 5
2
= 5 2 = 25
(b) log 6 4 + log 6 54 = log 6 (4 · 54) = log 6 216 = log 6 6 3 = 3
(c) Let  = arcsin
4
5 , so sin =
4
5 . Draw a right triangle with angle  as shown
in the figure. By the Pythagorean Theorem, the adjacent side has length 3,
and tan

arcsin
4
5

= tan =
opp
adj
=
4
3 .
30. (a) ln
1
 3
= ln −3 = −3
(b) sin(tan −1 1) = sin

4
=
√ 2
2
(c) 10 −3 log 4 = 10 log 4
−3
= 4 −3 =
1
4 3
=
1
64
31.  2 = 3 ⇒ ln( 2 ) = ln3 ⇒ 2 = ln3 ⇒  =
1
2
ln3 ≈ 0549
32. ln 2 = 5 ⇒  ln 
2
=  5 ⇒  2 =  5 ⇒  = ± √  5 ≈ ±12182
33.  

= 10 ⇒ ln

 
 
= ln10 ⇒   = ln10 ⇒ ln  = ln(ln10) ⇒  = ln(ln10) ≈ 0834