y= x^3 - 3x^2 + 6
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)