Calculus of a Single Variable, Chapter 3, 3.4, Section 3.4, Problem 33

First derivative to locate the x coordinates of critical points.
f'(x) = 3*x^2-3*2*x+0 = 3x^2-6x
At critical points, f'(x) = 0,
3x^2-6x = 0
x^2-2x=0
x(x-2) =0
This gives, x = 0 or x =2
Therefore, there are two critical points at x = 0 and x = 2.
Second derivative test.
f''(x) = 6x -6
At x = 0, f"(x) = 6*0-6 = -6 lt0 . Therefore, at x =0, there is a local maximum.
at x = 0, f(x) = 0^3-3*(0)^2+3 = 3 The local maximum is (0,3)
at x=2, f''(x) = 6*2-6 = +6 gt3 Therefore, at x = 2, there is a local minimum
at x = 2, f'(x) = 2^3-3*(2)^2+3 = 8-12+3 = -1 The local minimum is (2,-1)