Calculus: Early Transcendentals 8th Edition

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

Chapter 4 - Section 4.1 - Maximum and Minimum Values - 4.1 Exercises: 43

Answer

$x=0$ and $x=\frac{2}{3}$ are critical numbers of $f$.

Work Step by Step

How to find the critical numbers of a function $f$ according to the definition 1) Find all numbers of $c$ satisfying either $f'(c)=0$ or $f'(c)$ does not exist. 2) See that all the $c$ we get lie in the domain of $f$ $(D_f)$ or not. - If $c$ lies in $D_f$, $c$ is a critical number of $f$. - If $c$ does not lie in $D_f$, $c$ is not a critical number of $f$. $$f(x)=x^2e^{-3x}$$ $D_f=R$ *First, we find $f'(x)$ $$f'(x)=2xe^{-3x}+x^2e^{-3x}(-3x)'$$ $$f'(x)=2xe^{-3x}+x^2e^{-3x}(-3)$$ $$f'(x)=2xe^{-3x}-3x^2e^{-3x}$$ $$f'(x)=xe^{-3x}(2-3x).$$ We find $f'(x)=0$, $$f'(x)=0$$ $$x=0\hspace{0.5cm}or\hspace{0.5cm}e^{-3x}=0\hspace{0.5cm}or\hspace{0.5cm}(2-3x)=0$$ - For $e^{-3x}=0,$ this equation has no satisfying values of $x$ since $e^{-3x}\gt0$ for all $x$. - For $2-3x=0$ or $x=\frac{2}{3,}$ this value belongs to $D_f$, so it is a critical number of $f$ - For $x=0,$ this value belongs to $D_f$, so it is a critical number of $f$. We conclude that $x=0$ and $x=\frac{2}{3}$ are critical numbers of $f$.
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