Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 4 - Applications of the Derivative - 4.3 The Mean Value Theorem and Monotonicity - Exercises - Page 189: 37


$f(x)$ is increasing on $\left( \left( \dfrac{3}{2}\right)^{2/5},\infty\right )$ and decreasing on $\left( -\infty, \left(\dfrac{3}{2}\right)^{2/5}\right )$. $f(x)$ has local a minimum at $x=\left( \dfrac{3}{2}\right)^{2/5} $

Work Step by Step

Given $$y=x^{4}-4 x^{3 / 2} \quad(x>0)$$ Since $$f'(x) = 4x^{3}-6 x^{1 / 2} =2x^{1/2} (2x^{5/2}-3)$$ Then $f'(x)=0$ for $x= 0 $ and $x= \left( \dfrac{3}{2}\right)^{2/5}$. Since $x=0$ is not in the interval, choose $x=1 $ and $x=2$. Then: \begin{align*} f'(1)&= -2<0\\ f'(2)&>0 \end{align*} Then $f(x)$ is increasing on $\left( \left( \dfrac{3}{2}\right)^{2/5},\infty\right )$ and decreasing on $\left( -\infty, \left(\dfrac{3}{2}\right)^{2/5}\right )$. Hence, $f(x)$ has a local minimum at $x=\left( \dfrac{3}{2}\right)^{2/5} $
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