Calculus 8th Edition

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

Chapter 3 - Applications of Differentiation - 3.1 Maximum and Minimum Values - 3.1 Execises - Page 212: 39


Critical numbers are $x=0, \frac87, 4$

Work Step by Step

Given, $F(x)=x^{4/5}(x-4)^2$. Differentiating $F$ with respect to $x$ using product and chain rule, we get, $$\frac{dF}{dx}=\frac{4}{5}x^{-\frac{1}{5}}(x-4)^2+2x^{4/5}(x-4)$$ At critical points, the derivative of the function is either equal to $0$ or does not exist. Note, $\frac{dF}{dx}$ does not exist at $x=0$. Thus, $0$ is a critical number of this function. At, $\frac{dF}{dx}=0$. $$\frac{\frac{4}{5}(x-4)^2}{x^{1/5}}+2x^{4/5}(x-4)=0$$ $$\frac{4}{5}(x-4)^2+2x(x-4)=0$$ $$4(x-4)^2+10x(x-4)=0\implies 4(x^2+16-8x)+10x^2-40x=0$$ $$4x^2+64-32x+10x^2-40x=0$$ $$14x^2-72x+64=0$$ Dividing by $2$, $$7x^2-36x+32=0$$On solving the quadratic equation, we get, $$x=\frac{8}7, 4$$
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