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

Section 4.5 Summary of Curve Sketching 317
If additional accuracy is desired near any point, you can compute the value of the
derivative there. The tangent indicates the direction in which the curve proceeds.
ExamplE 1 Use the guidelines to sketch the curve y − 2x 2
x 2 2 1 .
A. The domain is
hx | x 2 2 1 ± 0j − hx |
x ± 61j − s2`, 21d ø s21, 1d ø s1, `d
B. The x- and y-intercepts are both 0.
C. Since f s2xd − f sxd, the function f is even. The curve is symmetric about the y-axis.
D. lim
x l6`
2x 2
x 2 2 1 − lim
x l6`
2
1 2 1yx 2 − 2
Therefore the line y − 2 is a horizontal asymptote.
Since the denominator is 0 when x − 61, we compute the following limits:
y=2
x=_1
y
0
FIGURE 5
Preliminary sketch
x=1
We have shown the curve approaching
its horizontal asymptote from above
in Figure 5. This is confirmed by the
intervals of increase and decrease.
y=2
x=_1
x=1
FIGURE 6
Finished sketch of y − 2x 2
x 2 2 1
y
0
x
x
2x 2
lim
x l1 1 x 2 2 1 − `
2x 2
lim
x l21 1 x 2 2 1 − 2`
2x 2
lim
x l1 2 x 2 2 1 − 2`
lim 2x 2
x l21 2 x 2 2 1 − `
Therefore the lines x − 1 and x − 21 are vertical asymptotes. This information
about limits and asymptotes enables us to draw the preliminary sketch in Figure 5,
showing the parts of the curve near the asymptotes.
E. f 9sxd − sx 2 2 1ds4xd 2 2x 2 ? 2x
sx 2 2 1d 2 − 24x
sx 2 2 1d 2
Since f 9sxd . 0 when x , 0 sx ± 21d and f 9sxd , 0 when x . 0 sx ± 1d, f is
increasing on s2`, 21d and s21, 0d and decreasing on s0, 1d and s1, `d.
F. The only critical number is x − 0. Since f 9 changes from positive to negative at 0,
f s0d − 0 is a local maximum by the First Derivative Test.
G. f 0sxd − sx 2 2 1d 2 s24d 1 4x ? 2sx 2 2 1d2x
− 12x 2 1 4
sx 2 2 1d 4 sx 2 2 1d 3
Since 12x 2 1 4 . 0 for all x, we have
and f 0sxd , 0 &? | x |
f 0sxd . 0 &? x 2 2 1 . 0 &? | x | . 1
, 1. Thus the curve is concave upward on the intervals
s2`, 21d and s1, `d and concave downward on s21, 1d. It has no point of inflection
since 1 and 21 are not in the domain of f.
H. Using the information in E–G, we finish the sketch in Figure 6. n
ExamplE 2 Sketch the graph of f sxd − x 2
sx 1 1 .
A. Domain − hx | x 1 1 . 0j − hx |
x . 21j − s21, `d
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