I hope I am understanding the concept correctly. I'm just checking my work here...

$\displaystyle f(x)=2+3x^2-x^3$

$\displaystyle f'(x)=-3x^2+6x$ =$\displaystyle 3x(x-2)$?

So 0 and 2 are the critical numbers?

If this is the case how does the second derivative give me the maximum and minimum?