# Find the absolute maximum value and the absolute minimum value for the following function: f(x)=x...

## Question:

Find the absolute maximum value and the absolute minimum value for the following function:

{eq}f(x)=x \sqrt {16-x^2} \ \text {on} \ [-4,0] {/eq}

## Absolute Extreme Values

The absolute extreme values of a function {eq}\displaystyle y=f(x), x\in[a,b] {/eq} are the values of the function evaluated at the local extreme points or the end points of the domain,

where the function attain the maximum or minimum values in the domain.

To find the absolute extrema, we will find the critical points, *c* from the given domain

where {eq}\displaystyle f'(c)= 0, \text{ or } f'(c) - \text{ does not exist} {/eq}

and evaluate the function at the critical points and at the end-points to obtain the maximum and minimum values.

## Answer and Explanation:

To find the absolute extrema (maximum and minimum) of the function {eq}\displaystyle f(x) = x \sqrt {16-x^2}, \text{ on } [-4,0] {/eq} we will first determine the critical points and then evaluate the function at the critical points and the end-points of the domain.

Obtaining the derivative function,

{eq}\displaystyle f'(x)=\frac{d}{dx}\left( x\sqrt{16-x^2}\right)=\sqrt{16-x^2}-\frac{x^2}{\sqrt{16-x^2}}=\frac{2(8-x^2)}{\sqrt{16-x^2}}, {/eq}

the critical points are given by

{eq}\displaystyle \begin{align} & f'(x)=0 \text{ or } f'(x)-\text{ does not exist}\\ &\frac{2(8-x^2)}{\sqrt{16-x^2}}=0 \text{ or } \frac{2(8-x^2)}{\sqrt{16-x^2}} -\text{ does not exist} \\ &8-x^2=0 \text{ or } \sqrt{16-x^2}=0 \\ &x=-2\sqrt{2},2\sqrt{2} \text{ or } x=-4,4, \text{ but } 2\sqrt{2} \notin [-4,0] \text{ and } 4\notin [-4,0] \\ \implies &\text{ the critical points are }x=-4, x=-2\sqrt{2}. \end{align} {/eq}

To find the absolute minimum and maximum, we will evaluate the function at the critical points and at the end points,

{eq}\displaystyle f(-4)=0, f(-2\sqrt{2})=-2\sqrt{2}\cdot 2\sqrt{2}=-8, f(0)=0\implies \\ \displaystyle \boxed{\text{ the absolute maximum value is } f(-4)=f(0)=0 \text{ and the absolute minimum value is }f(-2\sqrt{2})=-8}. {/eq}

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from Math 104: Calculus

Chapter 9 / Lesson 3