Answer
$$\eqalign{
& \left( { - 2, - 10} \right),{\text{ Absolute minimum }} \cr
& \left( {2,10} \right),{\text{ Absolute maximum}} \cr} $$
Work Step by Step
$$\eqalign{
& f\left( t \right) = {t^3} + t{\text{ with domain }}\left[ { - 2,2} \right] \cr
& {\text{Differentiate}} \cr
& f'\left( t \right) = \frac{d}{{dt}}\left[ {{t^3} + t} \right] \cr
& f'\left( t \right) = 3{t^2} + 1 \cr
& {\text{Find the stationary points}}{\text{, set }}f'\left( t \right) = 0 \cr
& f'\left( t \right) = 3{t^2} + 1 \cr
& 3{t^2} + 1 = 0 \cr
& 3{t^2} = - 1 \cr
& {\text{No real solutions}}{\text{, there are no stationary points}} \cr
& {\text{There are no singular points}}{\text{, the derivative is defined for every }}t \cr
& {\text{The domain of the function is }}\left[ { - 2,2} \right],{\text{ }} \cr
& {\text{The endpoints are: }}t = - 2,{\text{ and }}t = 2 \cr
& f\left( { - 2} \right) = {\left( { - 2} \right)^3} + \left( { - 2} \right) = - 10,{\text{ }}\left( { - 2, - 10} \right),{\text{ }}\left( {{\text{endpoint}}} \right) \cr
& f\left( 2 \right) = {\left( 2 \right)^3} + \left( 2 \right) = 10,{\text{ }}\left( {2,10} \right){\text{, }}\left( {{\text{endpoint}}} \right) \cr
& {\text{Thus}},{\text{ we have found the following extrema:}} \cr
& \left( { - 2, - 10} \right),{\text{ Absolute minimum }}\left( {{\text{endpoint}}} \right) \cr
& \left( {2,10} \right),{\text{ Absolute maximum }}\left( {{\text{endpoint}}} \right) \cr} $$
