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: 39

Answer

$x=0$, $x=4$ and $x=\frac{8}{7}$ are critical numbers of $F$.

Work Step by Step

How to find the critical numbers of a function $f$ according to 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 $f$. $$F(x)=x^{4/5}(x-4)^2$$ $$F(x)=\sqrt[5]{x^4}(x-4)^2$$ $D_F=R$ 1) Now, we find $F'(x)$ $$F'(x)=\frac{4}{5}x^{-1/5}(x-4)^2+x^{4/5}2(x-4)$$ $$F'(x)=\frac{4}{5}x^{-1/5}(x-4)^2+2x^{4/5}(x-4)$$ $$F'(x)=\frac{4}{5}\sqrt[5]{\frac{1}{x}}(x-4)^2+2x^{4/5}(x-4)$$ We find $F'(x)=0$, $$F'(x)=0$$ $$\frac{4}{5}x^{-1/5}(x-4)^2+2x^{4/5}(x-4)=0$$ $$\frac{4(x-4)^2}{5x^{1/5}}+2x^{4/5}(x-4)=0$$ $$(x-4)[\frac{4(x-4)}{5x^{1/5}}+2x^{4/5}]=0$$ $$(x-4)[\frac{4(x-4)+10x}{5x^{1/5}}]=0$$ $$(x-4)[\frac{14x-16}{5x^{1/5}}]=0$$ $$x=4\hspace{0.5cm}or\hspace{0.5cm}x=\frac{8}{7}$$ Also, looking at $$F'(x)=\frac{4}{5}\sqrt[5]{\frac{1}{x}}(x-4)^2+2x^{4/5}(x-4)$$ reveals that for $x=0$, F'(x) would not exist. 2) Examine whether values of $x$ lie in $D_F$ or not. We see that all of $x=0$, $x=4$ and $x=\frac{8}{7}\in R$, so they all lie in $D_F$. We conclude that $x=0$, $x=4$ and $x=\frac{8}{7}$ are critical numbers of $F$.
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