Answer
$$\eqalign{
& \left( { - 4, - 16} \right),{\text{ Absolute minimum }} \cr
& \left( { - 2,16} \right),{\text{ Absolute maximum }} \cr
& \left( {2, - 16} \right){\text{, Absolute minimum }} \cr
& \left( {4, - 16} \right),{\text{ Absolute maximum }} \cr} $$
Work Step by Step
$$\eqalign{
& g\left( x \right) = {x^3} - 12x{\text{ with domain }}\left[ { - 4,4} \right] \cr
& {\text{Differentiate}} \cr
& g'\left( x \right) = \frac{d}{{dx}}\left[ {{x^3} - 12x} \right] \cr
& g'\left( x \right) = 3{x^2} - 12 \cr
& {\text{Find the stationary points}}{\text{, set }}g'\left( x \right) = 0 \cr
& g'\left( x \right) = 3{x^2} - 12 \cr
& 3{x^2} - 12 = 0 \cr
& {x^2} = 4 \cr
& x = \pm 2 \cr
& {\text{There are no singular points}}{\text{, the derivative is defined for every }}x \cr
& {\text{The domain of the function is }}\left[ { - 4,4} \right],{\text{ }} \cr
& {\text{The endpoints are: }}x = - 4,{\text{ and }}x = 4 \cr
& g\left( { - 4} \right) = {\left( { - 4} \right)^3} - 12\left( { - 4} \right) = - 16,{\text{ }}\left( { - 4, - 16} \right),{\text{ }}\left( {{\text{endpoint}}} \right) \cr
& g\left( { - 2} \right) = {\left( { - 2} \right)^3} - 12\left( { - 2} \right) = 16,{\text{ }}\left( { - 2,16} \right){\text{, }}\left( {{\text{stationary point}}} \right) \cr
& g\left( 2 \right) = {\left( 2 \right)^3} - 12\left( 2 \right) = - 16,{\text{ }}\left( {2, - 16} \right){\text{, }}\left( {{\text{stationary point}}} \right) \cr
& g\left( 4 \right) = {\left( 4 \right)^3} - 12\left( 4 \right) = 16,{\text{ }}\left( {4,16} \right){\text{, }}\left( {{\text{endpoint}}} \right) \cr
& {\text{Thus}},{\text{ we have found the following extrema:}} \cr
& \left( { - 4, - 16} \right),{\text{ Absolute minimum }}\left( {{\text{endpoint}}} \right) \cr
& \left( { - 2,16} \right),{\text{ Absolute maximum }}\left( {{\text{Stationary point}}} \right) \cr
& \left( {2, - 16} \right){\text{, Absolute minimum }}\left( {{\text{Stationary point}}} \right) \cr
& \left( {4, - 16} \right),{\text{ Absolute maximum }}\left( {{\text{endpoint}}} \right) \cr} $$
