## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 4 - Applications of the Derivative - 4.4 The Shape of a Graph - Exercises - Page 195: 29

#### Answer

$f(\frac{2}{9})$ is a local minimum. $f''(0)$ is not defined and the second derivative test can not be applied at the point $0$.

#### Work Step by Step

$f'(x)=9x^{1/2}-2x^{-1/2}$ is differentiable everywhere except at $0$. Therefore, $0$ is a critical point of $y$. The other critical point is the solution of $9x^{1/2}-2x^{-1/2}=0$. $9x^{1/2}-2x^{-1/2}=0\implies9x^{1/2}=2x^{-1/2}$ $\implies \frac{2}{9}=x$. The critical points are $\frac{2}{9}$ and $0$. Now, $f''(0)=\frac{9}{2}(0)^{-1/2}+(0)^{-3/2}$ is not defined and the second derivative test can't be applied at the point $0$. $f''(\frac{2}{9})=\frac{9}{2}(\frac{2}{9})^{-1/2}+(\frac{2}{9})^{-3/2}=positive$ According to the second derivative test, if $f'(\frac{2}{9})=0$ and $f''(\frac{2}{9})$ is positive, then $f(\frac{2}{9})$ is a local minimum.

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