Given: `y=3x^(2/3)-2x,[-1,1]`

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

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

`2/x^(1/3)=2`

`x^(1/3)=1`

`x=1`

Plug in the critical x value(s) and the endpoints of the closed interval into the y(x) function.

`y(x)=3x^(2/3)-2x`

`y(-1)=5`

`y(1)=1`

Examine the y(x) values to determine...

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Given: `y=3x^(2/3)-2x,[-1,1]`

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

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

`2/x^(1/3)=2`

`x^(1/3)=1`

`x=1`

Plug in the critical x value(s) and the endpoints of the closed interval into the y(x) function.

`y(x)=3x^(2/3)-2x`

`y(-1)=5`

`y(1)=1`

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

The **absolute maximum** is the point (-1, 5).

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

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:

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

You need to solve for x the equation y' = 0

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

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

Factoring out 2 yields:

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

`1 - root(3)x = 0 => root(3)x = 1 => x = 1^3 => x = 1 in [-1,1]`

**Hence, evaluating the absolute extrema of the given function, over the interval [-1,1], yields that it reaches it's extrema at x = 1.**