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: 38


$f(x)$ increasing on $\left( \dfrac{16}{25} ,\infty\right )$ and decreasing on $\left(0, \dfrac{16}{25}\right )$. $f(x)$ has a local minimum at $x= \dfrac{16}{25}$.

Work Step by Step

Given $$y=x^{5 / 2}-x^{2} \quad(x>0)$$ Since $$f'(x) = \frac{5}{2}x^{3/2}-2 x = x ((5/2)x^{1/2}-2)$$ Then $f'(x)=0$ for $x= 0 $ and $x= \dfrac{16}{25}$. Since $x=0$ is not in the interval, choose $x=0.5 $ and $x=1$. Then \begin{align*} f'(0.5)& <0\\ f'(1)&>0 \end{align*} Thus $f(x)$ is increasing on $\left( \dfrac{16}{25} ,\infty\right )$ and decreasing on $\left(0, \dfrac{16}{25}\right )$. Hence, $f(x)$ has a local minimum at $x= \dfrac{16}{25}$.
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