# `y = 3x^(2/3) - 2x, [-1,1]` Find the absolute extrema of the function on the closed interval. 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).

Approved by eNotes Editorial Team 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.

Approved by eNotes Editorial Team