Calculus: Early Transcendentals 8th Edition

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

Chapter 4 - Section 4.5 - Summary of Curve Sketching - 4.5 Exercises - Page 322: 43

Answer

A. The domain is $(-\infty, \infty)$ B. The y-intercept is $\frac{1}{2}$ There are no x-intercepts. C. The function is not an even function or an odd function. D. $\lim\limits_{x \to -\infty} \frac{1}{1+e^{-x}} = 0$ $y = 0$ is a horizontal asymptote. $\lim\limits_{x \to \infty} \frac{1}{1+e^{-x}} = 1$ $y = 1$ is a horizontal asymptote. E. The function is increasing on the interval $(-\infty, \infty)$ F. There is no local maximum or local minimum. G. The graph is concave down on the interval $(0, \infty)$ The graph is concave up on the interval $(-\infty, 0)$ The point of inflection is $(0, \frac{1}{2})$ H. We can see a sketch of the curve below.
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Work Step by Step

$y = \frac{1}{1+e^{-x}}$ A. The function is defined for all real numbers. The domain is $(-\infty, \infty)$ B. When $x=0$, then $y = \frac{1}{1+e^{-0}} = \frac{1}{1+1} = \frac{1}{2}$ The y-intercept is $\frac{1}{2}$ When $y = 0$: $\frac{1}{1+e^{-x}} = 0$ There are no values of $x$ such that $y = 0$ There are no x-intercepts. C. The function is not an even function or an odd function. D. $\lim\limits_{x \to -\infty} \frac{1}{1+e^{-x}} = 0$ $y = 0$ is a horizontal asymptote. $\lim\limits_{x \to \infty} \frac{1}{1+e^{-x}} = 1$ $y = 1$ is a horizontal asymptote. E. We can find the values of $x$ such that $y' = 0$: $y' = \frac{-(-e^{-x})}{(1+e^{-x})^2} = \frac{e^{-x}}{(1+e^{-x})^2} = 0$ There are no values of $x$ such that $y' = 0$ For all values of $x,~~~$ $y' \gt 0$ The function is increasing on the interval $(-\infty, \infty)$ F. There is no local maximum or local minimum. G. We can find the values of $x$ such that $y'' = 0$: $y'' = \frac{(-e^{-x})(1+e^{-x})^2-(e^{-x})(2)(1+e^{-x})(-e^{-x})}{(1+e^{-x})^4}$ $y'' = \frac{(-e^{-x})(1+e^{-x})-(e^{-x})(2)(-e^{-x})}{(1+e^{-x})^3}$ $y'' = \frac{e^{-2x}-e^{-x}}{(1+e^{-x})^3} = 0$ $e^{-2x}-e^{-x} = 0$ $e^{-2x} = e^{-x}$ $e^{-x} = 1$ $\frac{1}{e^x} = 1$ $e^x = 1$ $x = 0$ When $x \gt 0$, then $y'' \lt 0$ The graph is concave down on the interval $(0, \infty)$ When $x \lt 0$, then $y'' \gt 0$ The graph is concave up on the interval $(-\infty, 0)$ When $x = 0$, then $y = \frac{1}{1+e^{-0}} = \frac{1}{1+1} = \frac{1}{2}$ The point of inflection is $(0, \frac{1}{2})$ H. We can see a sketch of the curve below.
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