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Multivariable Calculus 9th Edition by James Stewart solution manual

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resulting graph 1 unit upward.
(d) To obtain the graph of  = ( − 2) − 2, we shift the graph of  = () 2 units to the right (for the “−2” inside the
parentheses), and then shift the resulting graph 2 units downward.
(e) To obtain the graph of  = −(), we reflect the graph of  = () about the ­axis.
(f) To obtain the graph of  =  −1 (), we reflect the graph of  = () about the line  =  (assuming  is one–to­one).
10. (a) To obtain the graph of  = ( − 8), we shift the
graph of  = () right 8 units.
(b) To obtain the graph of  = −(), we reflect the
graph of  = () about the ­axis.
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
(c) To obtain the graph of  = 2 − (), we reflect the
graph of  = () about the ­axis, and then shift the
resulting graph 2 units upward.
(d) To obtain the graph of  =
1
2 () − 1, we shrink the
graph of  = () by a factor of 2, and then shift the
resulting graph 1 unit downward.
(e) To obtain the graph of  =  −1 (), we reflect the
graph of  = () about the line  = .
(f) To obtain the graph of  =  −1 ( + 3), we reflect the
graph of  = () about the line  =  [see part (e)],
and then shift the resulting graph left 3 units.
11. () =  3 + 2. Start with the graph of  =  3 and
shift 2 units upward.
12 . () = ( − 3) 2 . Start with the graph of  =  2 and
shift 3 units to the right.
13.  =
√  + 2. Start with the graph of  = √  and shift
2 units to the left.
14 .  = ln( + 5). Start with the graph of  = ln and
shift 5 units to the left.
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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15. () = 1 + cos2. Start with the graph of  = cos, compress horizontally by a factor of 2, and then shift 1 unit upward.
16. () = −  + 2. Start with the graph of  =   , reflect about the x­axis, and then shift 2 units upward.
17. () = 1 + 05  . Start with the graph of  = 05  =
 1
2
 
and shift 1 unit upward.
18. () =

− if   0
  − 1 if  ≥ 0
On (−∞0), graph  = − (the line with slope −1 and ­intercept 0)
with open endpoint (00).
On [0∞), graph  =   − 1 (the graph of  =   shifted 1 unit downward)
with closed endpoint (00).
19. (a) () = 2 5 − 3 2 + 2 ⇒ (−) = 2(−) 5 − 3(−) 2 + 2 = −2 5 − 3 2 + 2. Since (−) 6= () and
(−) 6= −(),  is neither even nor odd.
(b) () =  3 −  7 ⇒ (−) = (−) 3 − (−) 7 = − 3 +  7 = −( 3 −  7 ) = −(), so  is odd.
(c) () =  −
2
⇒ (−) =  −(−)
2
=  −
2
= (), so  is even.
(d) () = 1 + sin ⇒ (−) = 1 + sin(−) = 1 − sin. Now (−) 6= () and (−) 6= −(), so  is neither
even nor odd.
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
(e) () = 1 − cos2 ⇒ (−) = 1 − cos[2(−)] = 1 − cos(−2) = 1 − cos2 = (), so  is even.
(f) () = ( + 1) 2 =  2 + 2 + 1. Now (−) = (−) 2 + 2(−) + 1 =  2 − 2 + 1. Since (−) 6= () and
(−) 6= −(),  is neither even nor odd.
20. For the line segment from (−22) to (−10), the slope is
0 − 2
−1 + 2
= −2, and an equation is  − 0 = −2( + 1) or,
equivalently,  = −2 − 2. The circle has equation  2 +  2 = 1; the top half has equation  =
√ 1 −  2
(we have solved for
positive ). Thus, () =

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