Critical values are values where f'(x)=0 and the bounds of a function. Thus, let's solve for f'(x)!
f(x)=2x^3-3x^2+3x+8
f'(x)=6x^2-6x+3
Now let's set f'(x)=0
0=6x^2-6x+3
0=2x^2-2x+1
As it turns out, 2x^2-2x+1 isn't factorable!
This saves me some time because this means there are no critical numbers!