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