Answer
$$\eqalign{
& \left( { - 1, - 4} \right),{\text{ Absolute minimum }} \cr
& \left( {0,0} \right),{\text{ Relative maximum }} \cr
& \left( {2, - 4} \right),{\text{ Absolute minimum }} \cr} $$
Work Step by Step
$$\eqalign{
& h\left( t \right) = {t^3} - 3{t^2}{\text{ with domain }}\left[ { - 1, + \infty } \right) \cr
& {\text{Differentiate}} \cr
& h'\left( t \right) = \frac{d}{{dt}}\left[ {{t^3} - 3{t^2}} \right] \cr
& h'\left( t \right) = 3{t^2} - 6t \cr
& {\text{Find the stationary points}}{\text{, set }}h'\left( t \right) = 0 \cr
& h'\left( t \right) = 3{t^2} - 6t \cr
& 3{t^2} - 6t = 0 \cr
& 3t\left( {t - 2} \right) = 0 \cr
& t = 0,{\text{ }}t = 2 \cr
& {\text{There are no singular points}}{\text{, the derivative is defined for every }}t \cr
& {\text{The domain of the function is }}\left[ { - 1, + \infty } \right) \cr
& {\text{The endpoint is: }}t = - 1 \cr
& h\left( { - 1} \right) = {\left( { - 1} \right)^3} - 3{\left( { - 1} \right)^2} = - 4,{\text{ }}\left( { - 1, - 4} \right),{\text{ }}\left( {{\text{endpoint}}} \right) \cr
& h\left( 0 \right) = {\left( 0 \right)^3} - 3{\left( 0 \right)^2} = ,{\text{ }}\left( {0,0} \right){\text{, }}\left( {{\text{stationary point}}} \right) \cr
& h\left( 2 \right) = {\left( 2 \right)^3} - 3{\left( 2 \right)^2} = - 4,{\text{ }}\left( {2, - 4} \right){\text{, }}\left( {{\text{stationary point}}} \right) \cr
& {\text{Thus}},{\text{ we have found the following extrema:}} \cr
& \left( { - 1, - 4} \right),{\text{ Absolute minimum }}\left( {{\text{endpoint}}} \right) \cr
& \left( {0,0} \right),{\text{ Relative maximum }}\left( {{\text{Stationary point}}} \right) \cr
& \left( {2, - 4} \right),{\text{ Absolute minimum }}\left( {{\text{Stationary point}}} \right) \cr} $$
