Calculus 8th Edition

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

Chapter 6 - Inverse Functions - 6.3* The Natural Exponential Function - 6.3* Exercises - Page 453: 70


Absolute maximum value is 1.6487 and the absolute minimum value is -0.7358.

Work Step by Step

$f(x) =xe^{\frac{x}{2}}$ Apply product rule of differentiation. $f'(x) =e^{\frac{x}{2}}(1+\frac{x}{2})$ Now put, $f'(x) =0$ $e^{\frac{x}{2}}(1+\frac{x}{2})=0$ Since, $e^{\frac{x}{2}}\ne0$ Thus, $(1+\frac{x}{2})=0$ This implies $x=-2$ As we are given $f(x) =xe^{\frac{x}{2}}$ at intervals [-3,1] $f(-3)=-3e^{\frac{-3}{2}}=-0.6694$ $f(-2)=-2e^{\frac{-2}{2}}=-0.7358$ $f(1)=1e^{\frac{1}{2}}=1.6487$ Therefore, the absolute maximum value at x =-3,-2,1 for $f(x) =xe^{\frac{x}{2}}$ is 1.6487 and the absolute minimum value is -0.7358.
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