Answer
$$\eqalign{
& \left( { - 2, - 1} \right),{\text{ Absolute minimum }} \cr
& \left( { - 1,7} \right),{\text{ Absolute maximum }} \cr
& \left( {1, - 1} \right){\text{, Absolute minimum }} \cr
& \left( {2,7} \right),{\text{ Absolute maximum }} \cr} $$
Work Step by Step
$$\eqalign{
& g\left( x \right) = 2{x^3} - 6x + 3{\text{ with domain }}\left[ { - 2,2} \right] \cr
& {\text{Differentiate}} \cr
& g'\left( x \right) = \frac{d}{{dx}}\left[ {2{x^3} - 6x + 3} \right] \cr
& g'\left( x \right) = 6{x^2} - 6 \cr
& {\text{Find the stationary points}}{\text{, set }}g'\left( x \right) = 0 \cr
& g'\left( x \right) = 6{x^2} - 6 \cr
& 6{x^2} - 6 = 0 \cr
& 6{x^2} = 6 \cr
& x = \pm 1 \cr
& {\text{There are no singular points}}{\text{, the derivative is defined for every }}x \cr
& {\text{The domain of the function is }}\left[ { - 2,2} \right],{\text{ }} \cr
& {\text{The endpoints are: }}x = - 2,{\text{ and }}x = 2 \cr
& g\left( { - 2} \right) = 2{\left( { - 2} \right)^3} - 6\left( { - 2} \right) + 3 = - 1,{\text{ }}\left( { - 2, - 1} \right),{\text{ }}\left( {{\text{endpoint}}} \right) \cr
& g\left( { - 1} \right) = 2{\left( { - 1} \right)^3} - 6\left( { - 1} \right) + 3 = 7,{\text{ }}\left( { - 1,7} \right){\text{, }}\left( {{\text{stationary point}}} \right) \cr
& g\left( 1 \right) = 2{\left( 1 \right)^3} - 6\left( 1 \right) + 3 = - 1,{\text{ }}\left( {1, - 1} \right){\text{, }}\left( {{\text{stationary point}}} \right) \cr
& g\left( 2 \right) = 2{\left( 2 \right)^3} - 6\left( 2 \right) + 3 = 7,{\text{ }}\left( {2,7} \right){\text{, }}\left( {{\text{endpoint}}} \right) \cr
& {\text{Thus}},{\text{ we have found the following extrema:}} \cr
& \left( { - 2, - 1} \right),{\text{ Absolute minimum }}\left( {{\text{endpoint}}} \right) \cr
& \left( { - 1,7} \right),{\text{ Absolute maximum }}\left( {{\text{Stationary point}}} \right) \cr
& \left( {1, - 1} \right){\text{, Absolute minimum }}\left( {{\text{Stationary point}}} \right) \cr
& \left( {2,7} \right),{\text{ Absolute maximum }}\left( {{\text{endpoint}}} \right) \cr} $$
