Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 4: Applications of Derivatives - Section 4.1 - Extreme Values of Functions - Exercises 4.1 - Page 192: 47


$x=0$ and $x=4$

Work Step by Step

An interior point of the domain is a critical point of $f$ if $f'$ is zero or undefined. --- $y=x^{2}-32x^{1/2}$ $y'=2x-32\displaystyle \cdot\frac{1}{2}x^{-1/2}=2x-\frac{16}{\sqrt{x}}$ $x=0$ is a critical point as $y'$ is undefined there Checking where the derivative is zero, $2x-\displaystyle \frac{16}{\sqrt{x}}=0$ $\displaystyle \frac{2x\sqrt{x}-16}{\sqrt{x}}=0\quad\Rightarrow\quad 2x\sqrt{x}-16=0$ $\Rightarrow\quad 2x^{3/2}=16$ $\Rightarrow\quad x^{3/2}=8$ $\Rightarrow\quad x=8^{2/3}=[(2^{3})^{1/3}]^{2}=2^{2}=4$ $x=4$ is a critical point
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