Answer
$$\eqalign{
& \left( { - 1,5} \right),{\text{ Relative maximum}} \cr
& \left( {\frac{1}{2}, - \frac{1}{{16}}} \right){\text{, Absolute minimum}} \cr} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = 3{x^4} - 2{x^3}{\text{ with domain }}\left[ { - 1, + \infty } \right) \cr
& {\text{Differentiate}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {3{x^4} - 2{x^3}} \right] \cr
& f'\left( x \right) = 12{x^3} - 6{x^2} \cr
& {\text{Find the stationary points}}{\text{, set }}f'\left( x \right) = 0 \cr
& 12{x^3} - 6{x^2} = 0 \cr
& 6{x^2}\left( {2x - 1} \right) = 0 \cr
& x = 0,{\text{ }}x = \frac{1}{2} \cr
& {\text{There are no singular points}}{\text{, the derivative is defined for every }}x \cr
& {\text{The domain of the function is }}\left[ { - 1, + \infty } \right) \cr
& {\text{The endpoint is: }}x = - 1 \cr
& f\left( { - 1} \right) = 3{\left( { - 1} \right)^4} - 2{\left( { - 1} \right)^3} = 5,{\text{ }}\left( { - 1,5} \right),{\text{ }}\left( {{\text{endpoint}}} \right) \cr
& f\left( 0 \right) = 3{\left( 0 \right)^4} - 2{\left( 0 \right)^3} = 0,{\text{ }}\left( {0,0} \right){\text{, }}\left( {{\text{stationary point}}} \right) \cr
& f\left( {\frac{1}{2}} \right) = 3{\left( {\frac{1}{2}} \right)^4} - 2{\left( {\frac{1}{2}} \right)^3} = - \frac{1}{{16}},{\text{ }}\left( {\frac{1}{2}, - \frac{1}{{16}}} \right){\text{, }}\left( {{\text{stationary point}}} \right) \cr
& {\text{Thus}},{\text{ we have found the following extrema:}} \cr
& \left( { - 1,5} \right),{\text{ Relative maximum}} \cr
& \left( {\frac{1}{2}, - \frac{1}{{16}}} \right){\text{, Absolute minimum}} \cr} $$
