Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.2 Exponential Functions and Their Derivatives - 6.2 Exercises: 69

Answer

The absolute maximum value at (2, 1.213) and the absolute minimum value at (-1, -0.882).

Work Step by Step

$f(x) =xe^{\frac{-x^{2}}{8}}$ Apply product rule of differentiation. $f'(x) =e^{\frac{-x^{2}}{8}}(1-\frac{x^{2}}{4})$ Now put, $f'(x) =0$ $e^{\frac{-x^{2}}{8}}(1-\frac{x^{2}}{4})=0$ Since, $e^{\frac{-x^{2}}{8}}\ne0$ Thus, $(1-\frac{x^{2}}{4})=0$ This implies $x=2$ Now at $x=-1, 2, 4$, we have $f(-1)=-1 e^{\frac{-1}{8}}=-0.882$ $f(2) =2e^{\frac{-2^{2}}{8}}=\frac{2}{\sqrt e}=1.213$ $f(4)=4e^{-2}= 0.54134$ Therefore, the absolute maximum value at (2, 1.213) and the absolute minimum value at (-1, -0.882).
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