Answer
$$\eqalign{
& {\text{Relative maximum at }}\left( {0,3} \right) \cr
& {\text{Relative minimum at }}\left( {2, - 1} \right) \cr} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {x^3} - 3{x^2} + 3 \cr
& {\text{*Calculate the first derivative}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {{x^3} - 3{x^2} + 3} \right] \cr
& f'\left( x \right) = 3{x^2} - 6x \cr
& {\text{Set }}f'\left( x \right) = 0 \cr
& 3{x^2} - 6x = 0 \cr
& 3x\left( {x - 2} \right) = 0 \cr
& x = 0,{\text{ }}x = 2 \cr
& \cr
& *{\text{Calculate the second derivative}} \cr
& f''\left( x \right) = \frac{d}{{dx}}\left[ {f'\left( x \right)} \right] \cr
& f''\left( x \right) = \frac{d}{{dx}}\left[ {3{x^2} - 6x} \right] \cr
& f'\left( x \right) = 6x - 6 \cr
& \cr
& {\text{Evaluate the second derivative at }}x = 0{\text{ and }}x = 2 \cr
& *f''\left( 0 \right) = 6\left( 0 \right) - 6 < 0 \cr
& {\text{Then by the second derivative test }}\left( {{\text{Theorem 3}}{\text{.9}}} \right) \cr
& f\left( x \right){\text{ has a relative maximum at }}\left( {0,f\left( 0 \right)} \right) \cr
& f\left( 0 \right) = {\left( 0 \right)^3} - 3{\left( 0 \right)^2} + 3 = 3 \cr
& {\text{Relative maximum at }}\left( {0,3} \right) \cr
& *f''\left( 2 \right) = 6\left( 2 \right) - 6 > 0 \cr
& {\text{Then by the second derivative test }}\left( {{\text{Theorem 3}}{\text{.9}}} \right) \cr
& f\left( x \right){\text{ has a relative minimum at }}\left( {2,f\left( 2 \right)} \right) \cr
& f\left( 2 \right) = {\left( 2 \right)^3} - 3{\left( 2 \right)^2} + 3 = - 1 \cr
& {\text{Relative minimum at }}\left( {2, - 1} \right) \cr} $$
