Find all relative extremas. Use the Second Derivative Test when applicable. f(x)=(x^3)-(3x^2)+3... What I did:f'(x)=(3x^2)-6x 0=(3x^2)-6x -> 6x=3x^2 -> 6x=x(3x)6=3x -> x=2, 0, f''(x)=6x-60=6x-6 -> 6=6x -> x=1 After getting the critical numbers, how do I get the relative extremas again? Through the sign chart I think, but I'm a bit confused how to set that sign chart up. Answers: Relative minimum: (0, 3)Relative maximum: (2, -1)