Multivariable Calculus 9th Edition by James Stewart solution manual

5. True. See the Vertical Line Test.
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CHAPTER 1 FUNCTIONSAND MODELS
6. False. Let () =  2 and () = 2. Then ( ◦ )() = (()) = (2) = (2) 2 = 4 2 and
( ◦ )() = (()) = ( 2 ) = 2 2 . So  ◦  6=  ◦ .
7. False. Let () =  3 . Then  is one­to­one and  −1 () =
3
√ . But 1() = 1 3 , which is not equal to  −1 ().
8. True. We can divide by   since   6= 0 for every .
9. True. The function ln is an increasing function on (0∞).
10. False. Let  = . Then (ln) 6 = (ln) 6 = 1 6 = 1, but 6ln = 6ln = 6 · 1 = 6 6= 1 = (ln) 6 . What is true, however,
is that ln( 6 ) = 6ln for   0.
11. False. Let  =  2 and  = . Then
ln
ln
=
ln 2
ln
=
2ln
ln
= 2 and ln


= ln
 2

= ln = 1, so in general the statement
is false. What is true, however, is that ln


= ln − ln.
12. False. It is true that tan
3
4
= −1, but since the range of tan −1 is
 − 
2 

2
 , we must have tan −1 (−1) = − 
4 .
13. False. For example, tan −1 20 is defined; sin −1 20 and cos −1 20 are not.
14. False. For example, if  = −3, then
 (−3) 2
=
√ 9 = 3, not −3.
1. (a) When  = 2,  ≈ 27. Thus, (2) ≈ 27.
(b) () = 3 ⇒  ≈ 23, 56
(c) The domain of  is −6 ≤  ≤ 6, or [−66].
(d) The range of  is −4 ≤  ≤ 4, or [−44].
(e)  is increasing on [−44], that is, on −4 ≤  ≤ 4.
(f)  is not one­to­one because it fails the Horizontal Line Test.
(g)  is odd because its graph is symmetric about the origin.
2. (a) When  = 2,  = 3. Thus, (2) = 3.
(b)  is one­to­one because it passes the Horizontal Line Test.
(c) When  = 2,  ≈ 02. So  −1 (2) ≈ 02.
(d) The range of  is [−135], which is the same as the domain of  −1 .
(e) We reflect the graph of  through the line  =  to obtain the graph of  −1 .
3. () =  2 − 2 + 3, so ( + ) = ( + ) 2 − 2( + ) + 3 =  2 + 2 +  2 − 2 − 2 + 3, and
( + ) − ()

=
( 2 + 2 +  2 − 2 − 2 + 3) − ( 2 − 2 + 3)

=
(2 +  − 2)

= 2 +  − 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.
CHAPTER 1 REVIEW
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4. There will be some yield with no fertilizer, increasing yields with increasing
fertilizer use, a leveling­off of yields at some point, and disaster with too
much fertilizer use.
5. () = 2(3 − 1). Domain: 3 − 1 6= 0 ⇒ 3 6= 1 ⇒  6=
1
3 .
 =
 −∞
1
3

∪
 1
3 ∞

Range: all reals except 0 ( = 0 is the horizontal asymptote for .)
 = (−∞0) ∪ (0∞)
6. () =
√ 16 −  4 .
Domain: 16 −  4 ≥ 0 ⇒  4 ≤ 16 ⇒ || ≤
4
√ 16
⇒ || ≤ 2.  = [−22]
Range:  ≥ 0 and  ≤
√ 16
⇒ 0 ≤  ≤ 4.
 = [04]
7. () = ln( + 6). Domain:  + 6  0 ⇒   −6.  = (−6∞)
Range:  + 6  0, so ln( + 6) takes on all real numbers and, hence, the range is .
 = (−∞∞)
8.  = () = 3 + cos2. Domain: .  = (−∞∞)
Range: −1 ≤ cos2 ≤ 1 ⇒ 2 ≤ 3 + cos2 ≤ 4 ⇒ 2 ≤  ≤ 4.
 = [24]
9. (a) To obtain the graph of  = () + 5, we shift the graph of  = () 5 units upward.
(b) To obtain the graph of  = ( + 5), we shift the graph of  = () 5 units to the left.
(c) To obtain the graph of  = 1 + 2(), we stretch the graph of  = () vertically by a factor of 2, and then shift the