Chapter 5.3, Problem 5E

To determine

The local extreme values of the function, and identify any absolute extrema and support your answer graphically.

Answer to Problem 5E

Local maxima are $\left(-\sqrt{8},0\right)$ and $\left(2,4\right)$ and Local minima are $\left(-2,-4\right)$ and $\left(\sqrt{8},\text{ 0}\right)$ . Absolute maxima is 4 and absolute minima is $-4$ .

Explanation of Solution

Given information :

The given function is $y=x\sqrt{8-x^2}$ .

Calculation:

For the continuous function $f\left(x\right)$ .

At the critical point $c$ :

1. If $f$ changes sign from positive to negative at $c$ ( $f\text{'}>c$ for $x<c$ and $f\text{'}<0$ for $x>c$ ), then $f$ has a local maximum value at $c$ .
2. if $f$ changes sign from negative to positive at $c$ ( $f\text{'}<0$ for $x<c$ and $f\text{'}>c$ for $x>c$ ), then $f$ has a local minimum value at $c$ .
3. if $f$ does not changes its sign at $c$ ( $f\text{'}$ has same sign on both sides of $c$ ), then $f$ has no local extreme value at $c$ .

Now let $f\left(x\right)=x\sqrt{8-x^2}$ .

Find the first derivative of $f\left(x\right)$ :

  $\begin{array}{l}f\text{'}\left(x\right)=\sqrt{8-x^2}+\frac{x\left(-2x\right)}{2\sqrt{8-x^2}}\\ f\text{'}\left(x\right)=\sqrt{8-x^2}-\frac{x^2}{\sqrt{8-x^2}}\\ f\text{'}\left(x\right)=\frac{8-x^2-x^2}{\sqrt{8-x^2}}\\ f\text{'}\left(x\right)=\frac{8-2x^2}{\sqrt{8-x^2}}\end{array}$

Find the critical value by setting $f\text{'}\left(x\right)=0$

  $\begin{array}{l}\frac{8-2x^2}{\sqrt{8-x^2}}=0\\ 8-2x^2=0\\ 8=2x^2\\ x^2=\frac{8}{2}\\ x^2=4\\ x=\pm 2\end{array}$

Also, domain of the function $f\left(x\right)=x\sqrt{8-x^2}$ is

  $\begin{array}{l}8-x^2\ge 0\\ 8\ge x^2\\ -\sqrt{8}\le x\le \sqrt{8}\end{array}$

The critical point partition the x-axis into three intervals as shown below:

  

From the first derivative test the sign of $f\text{'}$ is negative for $-\sqrt{8}<x<-2$ , positive for $-2<x<2$ and again negative for $2<x<\sqrt{8}$ .

Therefore, there is local maximum is at $x=-\sqrt{8}$ and at $x=2$ and local minimum is at $x=-2$ and at $x=\sqrt{8}$ .

Now, the local maximum values are:

  $\begin{array}{l}f\left(-\sqrt{8}\right)=-\sqrt{8}\sqrt{8-{\left(-\sqrt{8}\right)}^2}\\ f\left(-\sqrt{8}\right)=-\sqrt{8}\sqrt{8-8}\\ f\left(-\sqrt{8}\right)=0\end{array}$

And

  $\begin{array}{l}f\left(2\right)=2\sqrt{8-{\left(2\right)}^2}\\ f\left(2\right)=2\sqrt{8-4}\\ f\left(2\right)=2\sqrt{4}\\ f\left(2\right)=2\end{array}$

Now, the local minimum values are:

  $\begin{array}{l}f\left(\sqrt{8}\right)=\sqrt{8}\sqrt{8-{\left(\sqrt{8}\right)}^2}\\ f\left(\sqrt{8}\right)=\sqrt{8}\sqrt{8-8}\\ f\left(\sqrt{8}\right)=0\end{array}$

And

  $\begin{array}{l}f\left(-2\right)=-2\sqrt{8-{\left(-2\right)}^2}\\ f\left(-2\right)=-2\sqrt{8-4}\\ f\left(-2\right)=-2\sqrt{4}\\ f\left(-2\right)=-4\end{array}$

Now, the highest value is the absolute maximum and the lowest value is the absolute minimum.

The graph of the function to support the answer is drawn below:

  

Hence,

Local maxima are $\left(-\sqrt{8},0\right)$ and $\left(2,4\right)$ and Local minima are $\left(-2,-4\right)$ and $\left(\sqrt{8},\text{ 0}\right)$ . Absolute maxima is 4 and absolute minima is $-4$ .