e x- and y-intercepts are both mmetry: None. x →∞0 cause both 6x and ex become large as x→ ∞o, we have lim_ 6xe* = ∞o. As x - have an indeterminate product that requires the use of l'Hospital's Rule. im 6xe = lim →-00 x→→∞ e-x 6x H lim -6e* X4-00 x) = 6xe* + 6e* = lim =x-00 us the x-axis is a horizontal asymptote. 6 ce ex is always positive, we see that f'(x) > 0 when x + 1 > 0, and f'(x) < 0 cand decreasing on • (-00₁ |
e x- and y-intercepts are both mmetry: None. x →∞0 cause both 6x and ex become large as x→ ∞o, we have lim_ 6xe* = ∞o. As x - have an indeterminate product that requires the use of l'Hospital's Rule. im 6xe = lim →-00 x→→∞ e-x 6x H lim -6e* X4-00 x) = 6xe* + 6e* = lim =x-00 us the x-axis is a horizontal asymptote. 6 ce ex is always positive, we see that f'(x) > 0 when x + 1 > 0, and f'(x) < 0 cand decreasing on • (-00₁ |
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
f(x) = 6xex
Transcribed Image Text:A.
B. The x- and y-intercepts are both
C.
the graph of f(x) = 6xe*.
The domain of f is R.
E.
D. Because both 6x and e* become large as x→ ∞, we have lim 6xe* = ∞o. As x→-co, however, e* →
x → 00
we have an indeterminate product that requires the use of l'Hospital's Rule.
F.
Symmetry: None.
lim 6xe*
X→-00
lim
=
X→-00
=
x →-00
lim -6e* =
f'(x) = 6xe* + 6e* =
6x
Thus the x-axis is a horizontal asymptote.
Because f'(-1) =
absolute) minimum.
H
lim
= X→-00
G. F"(x) = (6 + 6x)e* + 6e* =
Since f"(x) > 0 if x >
Since e* is always positive, we see that f'(x) > 0 when x + 1 > 0, and f'(x) < 0 when x + 1 < 0. So f is increasing on
∞) and decreasing on -00 2,
6
and f' changes from negative to positive at x =
and f"(x) < 0 if x <
and concave downward on the interval
, f(-1) = -6e
I
f is concave upward on the interval
SO
The inflection point is (x, y) =
is a local (and
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