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

Section 4.3 How Derivatives Affect the Shape of a Graph 299
Try reproducing the graph in Fig ure 12
with a graphing calculator or computer.
Some machines produce the complete
graph, some produce only the portion
to the right of the y-axis, and some
produce only the portion between
x − 0 and x − 6. For an explanation
and cure, see Example 7 in “Graphing
Calculators and Computers” at
www.stewartcalculus.com. An equivalent
expression that gives the correct
graph is
y − sx 2 d 1y3 ?
y
4
3
2
0
FIGURE 12
6 2 x
| 6 2 x | | 6 2 x | 1y3
(4, 2%?#)
1 2 3 4 5 7
y=x@?#(6-x)!?#
TEC In Module 4.3 you can practice
using information about f 9, f 0, and
asymptotes to determine the shape of
the graph of f.
x
Interval 4 2 x x 1y3 s6 2 xd 2y3 f 9sxd f
x , 0 1 2 1 2 decreasing on s2`, 0d
0 , x , 4 1 1 1 1 increasing on s0, 4d
4 , x , 6 2 1 1 2 decreasing on s4, 6d
x . 6 2 1 1 2 decreasing on s6, `d
To find the local extreme values we use the First Derivative Test. Since f 9 changes
from negative to positive at 0, f s0d − 0 is a local minimum. Since f 9 changes from positive
to negative at 4, f s4d − 2 5y3 is a local maximum. The sign of f 9 does not change
at 6, so there is no minimum or maximum there. (The Second Derivative Test could be
used at 4 but not at 0 or 6 since f 0 does not exist at either of these numbers.)
Looking at the expression for f 0sxd and noting that x 4y3 > 0 for all x, we have
f 0sxd , 0 for x , 0 and for 0 , x , 6 and f 0sxd . 0 for x . 6. So f is concave
downward on s2`, 0d and s0, 6d and concave upward on s6, `d, and the only inflection
point is s6, 0d. The graph is sketched in Figure 12. Note that the curve has vertical
l ` as x l 0 and as x l 6.
n
tangents at s0, 0d and s6, 0d because | f 9sxd |
ExamplE 8 Use the first and second derivatives of f sxd − e 1yx , together with asymptotes,
to sketch its graph.
SOLUtion Notice that the domain of f is hx | x ± 0j, so we check for vertical
asymptotes by computing the left and right limits as x l 0. As x l 0 1 , we know that
t − 1yx l `, so
lim e 1yx − lim e t − `
xl 0 1 tl `
and this shows that x − 0 is a vertical asymptote. As x l 0 2 , we have
t − 1yx l 2`, so
lim e 1yx − lim e t − 0
xl 0 2 tl2`
As x l 6`, we have 1yx l 0 and so
lim e 1yx − e 0 − 1
xl 6`
This shows that y − 1 is a horizontal asymptote (both to the left and right).
Now let’s compute the derivative. The Chain Rule gives
f 9sxd − 2 e 1yx
Since e 1yx . 0 and x 2 . 0 for all x ± 0, we have f 9sxd , 0 for all x ± 0. Thus f is
decreasing on s2`, 0d and on s0, `d. There is no critical number, so the function has no
local maximum or minimum. The second derivative is
x 2
f 0sxd − 2 x 2 e 1yx s21yx 2 d 2 e 1yx s2xd
x 4
− e 1yx s2x 1 1d
x 4
Since e 1yx . 0 and x 4 . 0, we have f 0sxd . 0 when x . 2 1 2 sx ± 0d and f 0sxd , 0
when x , 2 1 2 . So the curve is concave downward on s2`, 21 2 d and concave upward on
s2 1 2 , 0d and on s0, `d. The inflection point is s21 2 , e22 d.
To sketch the graph of f we first draw the horizontal asymptote y − 1 (as a dashed
line), together with the parts of the curve near the asymptotes in a preliminary sketch
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