Given: `f(x)=2x^3-6x,[0, 3]`

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

`f'(x)=6x^2-6=0`

`6(x^2-1)=0`

`6(x+1)(x-1)=0`

`x=-1, x=1`

The critical values for x are x=1 and x=-1. Plug in the critical value(s) and the endpoints of the interval into f(x). Because x=-1 is not in the interval [0, 3], it is not necessary to plug in the x=-1

`f(x)=2x^3-6x`

`f(0)=2(0)^3-6(0)=0`

`f(1)=2(1)^3-6(1)=-4`

`f(3)=2(3)^2-6(3)=36`

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

The **absolute minimum** value is the point **(1,
-4**).

The a**bsolute maximum** value is the point **(3,
36)**.

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

`f'(x) = (2x^3 - 6x)'`

`f'(x) = 6x^2 - 6`

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

`6x^2 - 6 = 0`

6`(x^2 - 1) = 0 => x^2 - 1 = 0 => (x - 1)(x + 1) = 0 => x = 1` and `x = -1`

You need to notice that only `x = 1 in [0,3]`

**Hence, evaluating the absolute extrema of the function, in interval
[0,3], yields x= 1.**

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