y= x^3 - 3x^2 + 2

First we need to calculate the critical values of y, to do so, we need to find the first derivative's zeros.

Let us differentiate:

y' = 3x^2 - 6x

3x^2 - 6x = 0

Factor 3x:

==> 3x(x-2) = 0

The, critical values are...

## Unlock

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

y= x^3 - 3x^2 + 2

First we need to calculate the critical values of y, to do so, we need to find the first derivative's zeros.

Let us differentiate:

y' = 3x^2 - 6x

3x^2 - 6x = 0

Factor 3x:

==> 3x(x-2) = 0

The, critical values are x= {0, 2}

when x= 0==> y= 2

when x= 2 ==> y= -2

Then extreme values are: (0, 2) and (2, -2)