# Find the exact location of all the relative and absolute extrema of the function. g(x) = 4x^{3} -...

## Question:

Find the exact location of all the relative and absolute extrema of the function.

{eq}g(x) = 4x^{3} - 48x \quad {/eq} with domain {eq}~[-4,4] {/eq}

## Finding Minima & Maxima:

The function that is well defined in some domain can have the extremes at the endpoints of the domain too along with at the critical points of the function. The first derivative is zero at the extreme points of the function.

## Answer and Explanation:

In order to get the exact location of all the relative and absolute extrema of the function.

{eq}g(x) = 4x^{3} - 48x \quad {/eq} with domain {eq}~[-4,4] {/eq}, we have to find all the critical points using the first derivative test, where the first derivative is zero, as follows:

{eq}\Rightarrow g'(x)=0\\ \Rightarrow \frac{d}{dx}\left(4x^3-48x\right)=0\\ \Rightarrow 12x^2-48=0\\ \Rightarrow x=2,\:x=-2 {/eq}

So these are the two critical points.

Now we will find the value of the function at these critical points and the endpoints, as follows;

{eq}\Rightarrow g(-2) = 4(-2)^{3} - 48(-2)=64~~~ \\ \Rightarrow g(2) = 4(2)^{3} - 48(2)=-64 \\ \Rightarrow g(-4) = 4(-4)^{3} - 48(-4)=-64 \\ \Rightarrow g(4) = 4(4)^{3} - 48(4)=64 \\ {/eq}

Hence from above we can say that at x=-2 and x=4, we have the both relative extreme and absolute extreme.

Also, at x=2 and x=-4, it is both relative extreme and absolute extreme.

#### Learn more about this topic:

Finding Minima & Maxima: Problems & Explanation

from General Studies Math: Help & Review

Chapter 5 / Lesson 2
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