Answer
$$\eqalign{
& {\text{max}} = 20{\text{ at }}x = - 2 \cr
& {\text{min}} = - 7{\text{ at }}x = 1 \cr} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = 2{x^3} + 3{x^2} - 12x;\,\,\left[ { - 3,2} \right] \cr
& f\left( x \right){\text{ is }}\,{\text{continuous }}\,{\text{and}}\,{\text{differentiable everywhere}}{\text{, the }} \cr
& {\text{absolute}}\,{\text{ extrema }}\,{\text{must }}\,{\text{occur}}\,{\text{ }}\,{\text{either }}\,{\text{at }}\,{\text{endpoints }}\,{\text{of }} \cr
& {\text{the interval or at solutions to the equation }}f'\left( x \right) = 0{\text{ in}} \cr
& {\text{the open interval }}\left( { - 3,2} \right){\text{.}} \cr
& \cr
& {\text{Differentiating }}f\left( x \right) \cr
& f'\left( x \right) = 6{x^2} + 6x - 12 \cr
& {\text{setting }}f'\left( x \right) = 0 \cr
& 6{x^2} + 6x - 12 = 0 \cr
& {x^2} + x - 2 = 0 \cr
& \left( {x + 2} \right)\left( {x - 1} \right) = 0 \cr
& x = - 2,\,\,x = 1 \cr
& {\text{Thus}}{\text{, there are stationary points at }}x = - 2{\text{ and }}x = 1 \cr
& {\text{Evaluating }}f\left( x \right){\text{ at the endpoints}}{\text{, }}x = - 3,{\text{ }}x = 2{\text{ and the}} \cr
& {\text{stationary points}}{\text{.}} \cr
& f\left( { - 3} \right) = 2{\left( { - 3} \right)^3} + 3{\left( { - 3} \right)^2} - 12\left( { - 3} \right) = 9 \cr
& f\left( { - 2} \right) = 2{\left( { - 2} \right)^3} + 3{\left( { - 2} \right)^2} - 12\left( { - 2} \right) = 20 \cr
& f\left( 1 \right) = 2{\left( 1 \right)^3} + 3{\left( 1 \right)^2} - 12\left( 1 \right) = - 7 \cr
& f\left( 2 \right) = 2{\left( 2 \right)^3} + 3{\left( 2 \right)^2} - 12\left( 2 \right) = 4 \cr
& {\text{Then we can conclude that the absolute minimum of }}f\left( x \right){\text{ on}} \cr
& {\text{the interval }}\left[ { - 3,2} \right]{\text{ is }} - 7{\text{ at }}x = 1.{\text{ And the absolute maximum is}} \cr
& {\text{20 at }}x = - 2. \cr
& \cr
& {\text{max}} = 20{\text{ at }}x = - 2 \cr
& {\text{min}} = - 7{\text{ at }}x = 1 \cr} $$
