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