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

318 Chapter 4 Applications of Differentiation
B. The x- and y-intercepts are both 0.
C. Symmetry: None
D. Since
lim
x l `
x 2
sx 1 1 − `
y
there is no horizontal asymptote. Since sx 1 1 l 0 as x l 21 1 and f sxd is
always positive, we have
x 2
lim
x l21 1 sx 1 1 − `
and so the line x − 21 is a vertical asymptote.
E. f 9sxd − sx 1 1 s2xd 2 x 2 ? 1y(2sx 1 1)
x 1 1
−
3x 2 1 4x
2sx 1 1d 3y2
xs3x 1 4d
−
2sx 1 1d 3y2
We see that f 9sxd − 0 when x − 0 (notice that 2 4 3 is not in the domain of f ), so the
only critical number is 0. Since f 9sxd , 0 when 21 , x , 0 and f 9sxd . 0 when
x . 0, f is decreasing on s21, 0d and increasing on s0, `d.
F. Since f 9s0d − 0 and f 9 changes from negative to positive at 0, f s0d − 0 is a local
(and absolute) minimum by the First Derivative Test.
x=_1
FIGURE 7
0
y= ≈
œ„„„„ x+1
x
G. f 0sxd − 2sx 1 1d3y2 s6x 1 4d 2 s3x 2 1 4xd3sx 1 1d 1y2
− 3x 2 1 8x 1 8
4sx 1 1d 3 4sx 1 1d 5y2
Note that the denominator is always positive. The numerator is the quadratic
3x 2 1 8x 1 8, which is always positive because its discriminant is
b 2 2 4ac − 232, which is negative, and the coefficient of x 2 is positive. Thus
f 0sxd . 0 for all x in the domain of f, which means that f is concave upward on
s21, `d and there is no point of inflection.
H. The curve is sketched in Figure 7. n
ExamplE 3 Sketch the graph of f sxd − xe x .
A. The domain is R.
B. The x- and y-intercepts are both 0.
C. Symmetry: None
D. Because both x and e x become large as x l `, we have lim x l ` xe x − `. As
x l 2`, however, e x l 0 and so we have an indeterminate product that requires
the use of l’Hospital’s Rule:
lim xe x − lim
x l2` x l2`
x
e 2x
Thus the x-axis is a horizontal asymptote.
− lim 1
x l2` 2e 2x
E. f 9sxd − xe x 1 e x − sx 1 1de x
− lim
x l2`s2e x d − 0
Since e x is always positive, we see that f 9sxd . 0 when x 1 1 . 0, and f 9sxd , 0
when x 1 1 , 0. So f is increasing on s21, `d and decreasing on s2`, 21d.
F. Because f 9s21d − 0 and f 9 changes from negative to positive at x − 21,
f s21d − 2e 21 < 20.37 is a local (and absolute) minimum.
G. f 0sxd − sx 1 1de x 1 e x − sx 1 2de x
Since f 0sxd . 0 if x . 22 and f 0sxd , 0 if x , 22, f is concave upward on
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