Chapter 5.1, Problem 37E

To determine

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

Answer to Problem 37E

Critical points	derivative	extremum	value
$x=-2$	Undefined	Local max	$0$
$x=-\sqrt{2}$	$0$	minimum	$-2$
$x=\sqrt{2}$	$0$	maximum	$2$
$x=2$	undefined	Local min	$0$

Explanation of Solution

Given information:

The function is $y=x\sqrt{4-x^2}$

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

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

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

  $\begin{array}{l}\frac{4-2x^2}{\sqrt{4-x^2}}=0\\ x=\sqrt{2},-\sqrt{2}\end{array}$

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
$x=-2$	Undefined	Local max	$0$
$x=-\sqrt{2}$	$0$	minimum	$-2$
$x=\sqrt{2}$	$0$	maximum	$2$
$x=2$	undefined	Local min	$0$

And $\left(-2,0\right)\text{ and }\left(2,0\right)$ is not a stationary point.