Chapter 5.1, Problem 36E

To determine

To find: thecritical points and local extreme values and critical points that are not stationary points.

Answer to Problem 36E

Critical points	derivative	extremum	value
$x=0$	Undefined	Local min	$0$
$x=2$	$0$	Local min	$0$
$x=-2$	$0$	Local min	$0$

Explanation of Solution

Given information:

The function is $y=x^{\frac{2}{3}}\left(x^2-4\right)$

Calculation: to find the critical points and local extreme points, first derivate the function,

  $\begin{array}{l}y=x^{\frac{2}{3}}\left(x^2-4\right)\\ y=x^{\frac{8}{3}}-4x^{\frac{2}{3}}\text{ [simplify]}\\ y\text{'}=\frac{8}{3}{x}^{\frac{5}{3}}-\frac{8}{3}{x}^{\frac{-1}{3}}\text{ [derivative the function]}\end{array}$

To find the critical points , equal the derivative of function to zero,

  $\begin{array}{l}\frac{8}{3}{x}^{\frac{5}{3}}-\frac{8}{3}{x}^{\frac{-1}{3}}=0\\ x=2,0,-2\end{array}$

Now, substitute the value of critical points in the function and find the value of function so the critical points and extreme values can be following as:

Critical points	derivative	extremum	value
$x=0$	Undefined	Local min	$0$
$x=2$	$0$	Local min	$0$
$x=-2$	$0$	Local min	$0$

And $\left(0,0\right)$ is not a stationary point.