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

Section 4.5 Summary of Curve Sketching 319
y
y=x´
s22, `d and concave downward on s2`, 22d. The inflection point is
s22, 22e 22 d < s22, 20.27d.
H. We use this information to sketch the curve in Figure 8. n
FIGURE 8
_2
1
_1
(_1, _1/e)
x
ExamplE 4 Sketch the graph of f sxd −
cos x
2 1 sin x .
A. The domain is R.
B. The y-intercept is f s0d − 1 2 . The x-intercepts occur when cos x − 0, that is,
x − sy2d 1 n, where n is an integer.
C. f is neither even nor odd, but f sx 1 2d − f sxd for all x and so f is periodic and
has period 2. Thus, in what follows, we need to consider only 0 < x < 2 and
then extend the curve by translation in part H.
D. Asymptotes: None
E. f 9sxd −
s2 1 sin xds2sin xd 2 cos x scos xd
s2 1 sin xd 2 − 2 2 sin x 1 1
s2 1 sin xd 2
The denominator is always positive, so f 9sxd . 0 when 2 sin x 1 1 , 0 &?
sin x , 2 1 2 &? 7y6 , x , 11y6. So f is increasing on s7y6, 11y6d and
decreasing on s0, 7y6d and s11y6, 2d.
F. From part E and the First Derivative Test, we see that the local minimum value is
f s7y6d − 21ys3 and the local maximum value is f s11y6d − 1ys3 .
G. If we use the Quotient Rule again and simplify, we get
2 cos x s1 2 sin xd
f 0sxd − 2
s2 1 sin xd 3
Because s2 1 sin xd 3 . 0 and 1 2 sin x > 0 for all x, we know that f 0sxd . 0
when cos x , 0, that is, y2 , x , 3y2. So f is concave upward on sy2, 3y2d
and concave downward on s0, y2d and s3y2, 2d. The inflection points are
sy2, 0d and s3y2, 0d.
H. The graph of the function restricted to 0 < x < 2 is shown in Figure 9. Then we
extend it, using periodicity, to the complete graph in Figure 10.
y
1
2
11π
6
1
œ„3
” , ’
y
1
2
π
2
π
3π
2
2π
x
_π
π
2π
3π
x
” 7π 1
-
6 œ„3
, ’
FIGURE 9
FIGURE 10
n
ExamplE 5 Sketch the graph of y − lns4 2 x 2 d.
A. The domain is
hx | 4 2 x 2 . 0j − hx | x 2 , 4j − hx | | x |
, 2j − s22, 2d
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