Answer
$$\eqalign{
& \left( {0,3} \right),{\text{ Relative maximum }} \cr
& \left( {\frac{1}{2}{\text{,}}\frac{5}{2}} \right),{\text{ Absolute minimum }} \cr
& \left( {3,15} \right){\text{, Absolute maximum }} \cr} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = 2{x^2} - 2x + 3{\text{ with domain }}\left[ {0,3} \right] \cr
& {\text{Differentiate}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {2{x^2} - 2x + 3} \right] \cr
& f'\left( x \right) = 4x - 2 \cr
& {\text{Find the stationary points}}{\text{, set }}f'\left( x \right) = 0 \cr
& f'\left( x \right) = 4x - 2 \cr
& 4x - 2 = 0 \cr
& 4x = 2 \cr
& x = \frac{1}{2} \cr
& {\text{There are no singular points}}{\text{, the derivative is defined for every }}x \cr
& {\text{The domain of the function is }}\left[ {0,3} \right],{\text{ }} \cr
& {\text{The endpoints are: }}x = 0,{\text{ and }}x = 3 \cr
& f\left( 0 \right) = 2{\left( 0 \right)^2} - 2\left( 0 \right) + 3 = 3,{\text{ }}\left( {0,3} \right),{\text{ }}\left( {{\text{endpoint}}} \right) \cr
& f\left( {\frac{1}{2}} \right) = 2{\left( {\frac{1}{2}} \right)^2} - 2\left( {\frac{1}{2}} \right) + 3 = \frac{5}{2},{\text{ }}\left( {\frac{1}{2}{\text{,}}\frac{5}{2}} \right){\text{, }}\left( {{\text{stationary point}}} \right) \cr
& f\left( 3 \right) = 2{\left( 3 \right)^2} - 2\left( 3 \right) + 3 = 15,{\text{ }}\left( {3,15} \right){\text{, }}\left( {{\text{endpoint}}} \right) \cr
& {\text{Thus}},{\text{ we have found the following extrema:}} \cr
& \left( {0,3} \right),{\text{ Relative maximum }}\left( {{\text{endpoint}}} \right) \cr
& \left( {\frac{1}{2}{\text{,}}\frac{5}{2}} \right),{\text{ Absolute minimum }}\left( {{\text{Stationary point}}} \right) \cr
& \left( {3,15} \right){\text{, Absolute maximum }}\left( {{\text{endpoint}}} \right) \cr} $$
