Answer
$$\eqalign{
& \left( { - 1,5} \right),{\text{ Absolute maximum }} \cr
& \left( {1, - 5} \right),{\text{ Absolute minimum}} \cr} $$
Work Step by Step
$$\eqalign{
& f\left( t \right) = - 2{t^3} - 3t{\text{ with domain }}\left[ { - 1,1} \right] \cr
& {\text{Differentiate}} \cr
& f'\left( t \right) = \frac{d}{{dt}}\left[ { - 2{t^3} - 3t} \right] \cr
& f'\left( t \right) = - 6{t^2} - 3 \cr
& {\text{Find the stationary points}}{\text{, set }}f'\left( t \right) = 0 \cr
& f'\left( t \right) = - 6{t^2} - 3 \cr
& - 6{t^2} - 3 = 0 \cr
& - 6{t^2} = 3 \cr
& {\text{No real solutions}}{\text{, there are no stationary points}} \cr
& {\text{There are no singular points}}{\text{, the derivative is defined for every }}t \cr
& {\text{The domain of the function is }}\left[ { - 1,1} \right],{\text{ }} \cr
& {\text{The endpoints are: }}t = - 1,{\text{ and }}t = 1 \cr
& f\left( { - 1} \right) = - 2{\left( { - 1} \right)^3} - 3\left( { - 1} \right) = 5,{\text{ }}\left( { - 1,5} \right),{\text{ }}\left( {{\text{endpoint}}} \right) \cr
& f\left( 1 \right) = - 2{\left( 1 \right)^3} - 3\left( 1 \right) = - 5,{\text{ }}\left( {1, - 5} \right){\text{, }}\left( {{\text{endpoint}}} \right) \cr
& {\text{Thus}},{\text{ we have found the following extrema:}} \cr
& \left( { - 1,5} \right),{\text{ Absolute maximum }}\left( {{\text{endpoint}}} \right) \cr
& \left( {1, - 5} \right),{\text{ Absolute minimum }}\left( {{\text{endpoint}}} \right) \cr} $$
