Multivariable Calculus 9th Edition by James Stewart solution manual

2
 15 .
(b) From the graph, we estimate that the BAC is 008 gdL when
 ≈ 12 hours.
33. From the graph, it appears that  is an odd function ( is undefined for  = 0).
To prove this, we must show that (−) = −().
(−) =
1 −  1(−)
1 +  1(−)
=
1 −  (−1)
1 +  (−1)
=
1 −
1
 1
1 +
1
 1
·
 1
 1
=
 1 − 1
 1 + 1
= − 1 − 
1
1 +  1
= −()
so  is an odd function.
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SECTION 1.5 INVERSEFUNCTIONSAND LOGARITHMS
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34. We’ll start with  = −1 and graph () =
1
1 +  
for  = 01, 1, and 5.
From the graph, we see that there is a horizontal asymptote  = 0 as  → −∞
and a horizontal asymptote  = 1 as  → ∞. If  = 1, the y­intercept is
 0
1
2
 .
As  gets smaller (close to 0), the graph of  moves left. As  gets larger, the graph
of  moves right.
As  changes from −1 to 0, the graph of  is stretched horizontally. As 
changes through large negative values, the graph of  is compressed horizontally.
(This takes care of negatives values of .)
If  is positive, the graph of  is reflected through the y­axis.
Last, if  = 0, the graph of  is the horizontal line  = 1(1 + ).
35. We graph the function () =

2 (

+  − ) for  = 1, 2, and 5. Because
(0) = , the y­intercept is , so the y­intercept moves upward as  increases.
Notice that the graph also widens, becoming flatter near the y­axis as  increases.
1.5 Inverse Functions and Logarithms
1. (a) See Definition 1.
(b) It must pass the Horizontal Line Test.
2. (a)  −1 () =  ⇔ () =  for any  in . The domain of  −1 is  and the range of  −1 is .
(b) See the steps in Box 5.
(c) Reflect the graph of  about the line  = .
3.  is not one­to­one because 2 6= 6, but (2) = 20 = (6).
4.  is one­to­one because it never takes on the same value twice.
5. We could draw a horizontal line that intersects the graph in more than one point. Thus, by the Horizontal Line Test, the
function is not one­to­one.
6. No horizontal line intersects the graph more than once. Thus, by the Horizontal Line Test, the function is one­to­one.
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.
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CHAPTER 1 FUNCTIONSAND MODELS
7. No horizontal line intersects the graph more than once. Thus, by the Horizontal Line Test, the function is one­to­one.
8. We could draw a horizontal line that intersects the graph in more than one point. Thus, by the Horizontal Line Test, the
function is not one­to­one.
9. The graph of () = 2 − 3 is a line with slope 2. It passes the Horizontal Line Test, so  is one­to­one.
Algebraic solution: If  1 6=  2 , then 2 1 6= 2 2 ⇒ 2 1 − 3 6= 2 2 − 3 ⇒ ( 1 ) 6= ( 2 ), so  is one­to­one.
10. The graph of () =  4 − 16 is symmetric with respect to the ­axis. Pick any ­values equidistant from 0 to find two equal
function values. For example, (−1) = −15 and (1) = −15, so  is not one­to­one.
11. No horizontal line intersects the graph of () =  3 + 4 more than once. Thus, by the Horizontal Line Test, the function is
one­to­one.
Algebraic solution: If  1 6=  2 , then  3
1 6= 
3
2
⇒  3
1 + 4 6= 
3
2 + 4
⇒ ( 1 ) 6= ( 2 ), so  is one­to­one.
12. The graph of () =
3
√  passes the Horizontal Line Test, so  is one­to­one.
13. () = 1 − sin. (0) = 1 and () = 1, so  is not one­to­one.
14. The graph of () =  4 − 1 passes the Horizontal Line Test when  is restricted to the interval [0,10], so  is one­to­one.
15. A football will attain every height  up to its maximum height twice: once on the way up, and again on the way down.