# `g(x) = root(3)(x), [-8,8]` Find the absolute extrema of the function on the closed interval.

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### 2 Answers

Given: `f(x)=root(3)(x), [-8, 8]`

Find the critical values for x by setting the derivative equal to zero and solving for the x value(s).

`f(x)=x^(1/3)`

`f'(x)=(1/3)x^(-2/3)=0 `

` f'(x)=1/(3x^(2/3))=0`

` 1=0 `

1=0 is not a true statement. Therefore the critical x value(s) does not exist.

Plug in the critical value(s), if one exists, and the endpoints of the interval into f(x).

`f(x)=root(3)(x)`

`f(8)=root(3)(8)=2`

`f(-8)=root(3)(-8)=-2`

Examine the f(x) values to determine the **absolute extrema**.

The **absolute minimum** value is the point **(-8 -2).**

The **absolute maximum** value is the point **(8, 2).**

You need to find out the absolute extrema of the given function, hence, you need to differentiate the function with respect to x, such that:

`g'(x) = (root(3)(x))'`

`g'(x) = (1/3)x^(1/3 - 1) g'(x) = (1/3)x^(-2/3)`

You need to solve for x the equation g'(x) = 0, such that:

`(1/3)x^(-2/3) = 0 => 1/(3root(3)(x^2)) = 0`

It can be noticed that there exists no values of x to verify the equation.

**Hence, there exists no absolute extrema of the given function, over the interval [-8,8].**