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

First find the critical x value(s) of the function. To find the critical x value(s), set the derivative equal to zero and solve for the x value(s).

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

`3x^2-3x=0`

`3x(x-1)=0`

`x=0,x=1`

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

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

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

First find the critical x value(s) of the function. To find the critical x value(s), set the derivative equal to zero and solve for the x value(s).

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

`3x^2-3x=0`

`3x(x-1)=0`

`x=0,x=1`

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

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

`f(-1)=(-1)^3-(3/2)(-1)^2=-5/2`

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

`f(1)=(1)^3-(3/2)(1)^2=-1/2`

`f(2)=(2)^3-(3/2)(2)^2=2`

Examine the f(x) values.

The absolute maximum is at the point (2, 2).

The absolute minimum is at the point `(-1,-5/2).`

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