Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 4 - Section 4.3 - How Derivatives Affect the Shape of a Graph - 4.3 Exercises - Page 302: 53

Answer

(a) $y = 0$ is a horizontal asymptote. (b) $f$ is decreasing on the interval $(0, \infty)$ $f$ is increasing on the interval $(-\infty,0)$ (c) The local maximum is $(0,1)$ (d) The graph is concave down on this interval: $(-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$ The graph is concave up on this interval: $(-\infty,-\frac{\sqrt{2}}{2})\cup (\frac{\sqrt{2}}{2}, \infty)$ The points of inflection are $(-\frac{\sqrt{2}}{2}, \frac{1}{\sqrt{e}})$ and $(\frac{\sqrt{2}}{2}, \frac{1}{\sqrt{e}})$ (e) We can see a sketch of the graph below.

Work Step by Step

(a) $f(x) = e^{-x^2}$ $f(x)$ is defined for all values of $x$ There are no vertical asymptotes. $\lim\limits_{x \to -\infty}f(x) = 0$ $\lim\limits_{x \to \infty}f(x) = 0$ $y = 0$ is a horizontal asymptote. (b) We can find the points where $f'(x) = 0$: $f'(x) = -2x~e^{-x^2} = 0$ $x = 0$ When $x \gt 0$ then $f'(x) \lt 0$ $f$ is decreasing on the interval $(0, \infty)$ When $x \lt 0$ then $f'(x) \gt 0$ $f$ is increasing on the interval $(-\infty,0)$ (c) $f(0) = e^{-(0)^2} = 1$ The local maximum is $(0,1)$ There is no local minimum. (d) We can find the points where $f''(x) = 0$: $f''(x) = -2~e^{-x^2}+(-2x)(-2x)e^{-x^2}$ $f''(x) = -2~e^{-x^2}+4x^2e^{-x^2}$ $f''(x) = (4x^2-2)e^{-x^2} = 0$ $4x^2-2 = 0$ $x^2 = \frac{1}{2}$ $x = \pm \frac{\sqrt{2}}{2}$ The graph is concave down when $f''(x) \lt 0$ The graph is concave down on this interval: $(-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$ The graph is concave up when $f''(x) \gt 0$ The graph is concave up on this interval: $(-\infty,-\frac{\sqrt{2}}{2})\cup (\frac{\sqrt{2}}{2}, \infty)$ $f(-\frac{\sqrt{2}}{2}) = e^{-(-\frac{\sqrt{2}}{2})^2} = \frac{1}{\sqrt{e}}$ $f(\frac{\sqrt{2}}{2}) = e^{-(\frac{\sqrt{2}}{2})^2} = \frac{1}{\sqrt{e}}$ The points of inflection are $(-\frac{\sqrt{2}}{2}, \frac{1}{\sqrt{e}})$ and $(\frac{\sqrt{2}}{2}, \frac{1}{\sqrt{e}})$ (e) We can see a sketch of the graph below.
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