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

Section 4.5 Summary of Curve Sketching 321
gives
f sxd − x 3
x 2 1 1 − x 2
x
x 2 1 1
This equation suggests that y − x is a candidate for a slant asymptote. In fact,
1
f sxd 2 x − 2
x
x 2 1 1 − 2 x
1 1 1 l 0 as x l 6`
x 2
So the line y − x is a slant asymptote.
E. f 9sxd − sx 2 1 1ds3x 2 d 2 x 3 ? 2x
sx 2 1 1d 2
− x 2 sx 2 1 3d
sx 2 1 1d 2
Since f 9sxd . 0 for all x (except 0), f is increasing on s2`, `d.
F. Although f 9s0d − 0, f 9 does not change sign at 0, so there is no local maximum or
minimum.
G. f 0sxd − sx 2 1 1d 2 s4x 3 1 6xd 2 sx 4 1 3x 2 d ? 2sx 2 1 1d2x
sx 2 1 1d 4
− 2xs3 2 x 2 d
sx 2 1 1d 3
y
y= ˛
≈+1
Since f 0sxd − 0 when x − 0 or x − 6s3 , we set up the following chart:
3œ„3
”_œ„3, _ ’
4
0
3œ„3
”œ„3, ’
4
x
inflection
points
Interval x 3 2 x 2 sx 2 1 1d 3 f 99sxd f
x , 2s3 2 2 1 1 CU on (2`, 2s3 )
2s3 , x , 0 2 1 1 2 CD on (2s3 , 0)
0 , x , s3 1 1 1 1 CU on (0, s3 )
x . s3 1 2 1 2 CD on (s3 , `)
y=x
FIGURE 13
The points of inflection are (2s3 , 2 3 4 s3 ), s0, 0d, and (s3 , 3 4 s3 ).
H. The graph of f is sketched in Figure 13. n
1–54 Use the guidelines of this section to sketch the curve.
11. y − x 2 x 2
2 2 3x 1 x 12. y − 1 1 1 2 x 1 1 x 2
1. y − x 3 1 3x 2 2. y − 2 1 3x 2 2 x 3
3. y − x 4 2 4x 4. y − x 4 2 8x 2 1 8
13. y − x
14. y − 1
x 2 2 4
x 2 2 4
5. y − xsx 2 4d 3 6. y − x 5 2 5x
7. y − 1 5 x 5 2 8 3 x 15. y − x 2
sx 2 1d2
16. y − 3 1 16x 8. y − s4 2 x 2 d 5
x 2 1 3
x 2 1 1
9. y − x
10. y − x 2 1 5x
17. y − x 2 1
18. y − x
x 2 1
25 2 x 2 x 2 x 3 2 1
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