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

Asked on by enotes

### Textbook Question

Chapter 3, 3.1 - Problem 24 - Calculus of a Single Variable (10th Edition, Ron Larson).
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mathace | (Level 3) Assistant Educator

Posted on

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).

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

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].

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