James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

A86 Appendix I Answers to Odd-Numbered Exercises
37. A. f22, 2g B. y-int s3 ; x-int 24y3, 2y3, 2y3,
5y3 C. Period 2 D. None
E. Inc on s22, 211y6d, s25y6, y6d, s7y6, 2d;
dec on s211y6, 25y6d, sy6, 7y6d
F. Loc max f s211y6d − f sy6d − 2; loc min
f s25y6d − f s7y6d − 22
G. CU on s24y3, 2y3d,
s2y3, 5y3d;
CD on s22, 24y3d,
s2y3, 2y3d, s5y3, 2d;
IPs s24y3, 0d, s2y3, 0d,
s2y3, 0d, s5y3, 0d
H. See graph at right.
4π
3
_2π
y
11π
” 6 ,2’
2
π
” 6 ,2’
π 2π 5π
3 3 3
0 2π x
5π
7π
” 6 ,_2’ _2 ” 6
,_2’
39. A. All reals except s2n 1 1d (n an integer)
B. y-int 0; Stewart x-int / 2n Calculus: C. ET About 5 th Editio the origin, n period 2
Solution Ar t
D. VA x − s2n 1 1d E. Inc on ss2n 2 1d, s2n 1 1dd
5et040539
F. None 5.20.02 G. CU on s2n, s2n 1 1dd;
CD on ss2n 2 1d, 2nd; IPs s2n, 0d
H. x=_3π x=_π x=π x=3π
y
Stewart / Calculus: ET 5 th Editio n
Solution Ar t
5et040547
49. A.
5.20.02
All x in s2n, s2n 1 1dd (n an integer)
B. x-int y2 1 2n C. Period 2 D. VA x − n
E. Inc on s2n, y2 1 2nd; dec on sy2 1 2n, s2n 1 1dd
F. Loc max f sy2 1 2nd − 0 G. CD on s2n, s2n 1 1dd
H.
y
_4π _3π _2π _π π 2π 3π 4π
0
51. A. s2`, 0d ø s0, `d
B. None C. None
D. VA x − 0
E. Inc on s2`, 21d, s0, `d;
dec on s21, 0d
F. Loc max f s21d − 2e
G. CU on s0, `d; CD on s2`, 0d
H. See graph at right.
x
y
(_1, _e)
(0, 0)
x
_2π 0 2π
41. A. R B. y-int y4
C. None
D. HA y − 0, y − y2
E. Inc on s2`, `d F. None
G. CU on s2`, 0d;
CD on s0, `d; IP s0, y4d
H. See graph at right.
x
y
0
(0, π/4)
y=π/2
y=0
43. A. R B. y-int 1 2 C. None y
y 1
D. HA y − 0, y − 1
E. Inc on R F. None
G. CU on s2`, 0d;
CD on s0, `d; IP s0, 1 2 d
H. See graph at right.
0
x
45. A. s0, `d B. None y
C. None D. VA x − 0
4
E. Inc on s1, `d; dec on s0, 1d
F. Loc min f s1d − 1
3
G. CU on s0, 2d; CD on s2, `d; 2
IP (2, 1 2 1 ln 2)
1
H. See graph at right.
(1, 1)
1
”2, +ln 2’
2
0 x
1 2 3 4
x
53. A. R B. y-int 1 y
C. None D. HA y − e 6y2
E. Inc on R F. None
G. CU on (2`, 1 2 ); CD on ( 1 2 , `);
IP ( 1 2 , e arctans1y2d )
H. See graph at right.
55.
m
(0, m¸) √=c
0 √
ms80354-1
6et 4.5.53
8.12.06
4
3
2
1
y=e
π/2
1 arctan(1/2)
” 2, e ’
y=e _π/2
_2 0 2 4 6 8 x
57. (a) When t − sln adyk (b) When t − sln adyk
(c) y
1
1/2
0 ln a
k
y=p(t)
t
47. A. R B. y-int 1 4 y
C. None
D. HA y − 0, y − 1
y=1
E. Dec on R F. None
G. CU on (ln 1 2 , `);
1 4
”ln , ’
CD on s2`, ln 1 2 9
2d; IP sln 1 2 , 4 9d
0 x
H. See graph at right.
59.
y
0
L/2
L x
61. y − x 2 1 63. y − 2x 2 3
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