Answer
$${\text{Relative maximum at }}\left( {3,9} \right)$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = 6x - {x^2} \cr
& {\text{*Calculate the first derivative}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {6x - {x^2}} \right] \cr
& f'\left( x \right) = 6 - 2x \cr
& {\text{Set }}f'\left( x \right) = 0 \cr
& 6 - 2x = 0 \cr
& x = 3 \cr
& \cr
& *{\text{Calculate the second derivative}} \cr
& f''\left( x \right) = \frac{d}{{dx}}\left[ {f'\left( x \right)} \right] \cr
& f''\left( x \right) = \frac{d}{{dx}}\left[ {6 - 2x} \right] \cr
& f'\left( x \right) = - 2 \cr
& \cr
& {\text{Evaluate the second derivative at }}x = 3 \cr
& f''\left( 3 \right) = - 2 \cr
& f''\left( 3 \right) < 0,{\text{ Then by the second derivative test }}\left( {{\text{Theorem 3}}{\text{.9}}} \right) \cr
& f\left( x \right){\text{ has a relative maximum at }}\left( {3,f\left( 3 \right)} \right) \cr
& f\left( 3 \right) = 6\left( 3 \right) - {\left( 3 \right)^2} = 9 \cr
& {\text{Relative maximum at }}\left( {3,9} \right) \cr} $$
