Answer
$${\text{Relative maximum at }}\left( {3, - 5} \right)$$
Work Step by Step
$$\eqalign{
& h\left( t \right) = t - 4\sqrt {t + 1} \cr
& {\text{Domain: }}\left[ { - 1,\infty } \right) \cr
& {\text{*Calculate the first derivative}} \cr
& h'\left( t \right) = \frac{d}{{dt}}\left[ {t - 4\sqrt {t + 1} } \right] \cr
& h'\left( t \right) = 1 - 4\left( {\frac{1}{{2\sqrt {t + 1} }}} \right) \cr
& h'\left( t \right) = 1 - 4\left( {\frac{1}{{2\sqrt {t + 1} }}} \right) \cr
& h'\left( t \right) = 1 - \frac{2}{{\sqrt {t + 1} }} \cr
& {\text{Set }}h'\left( t \right) = 0 \cr
& 1 - \frac{2}{{\sqrt {t + 1} }} = 0 \cr
& \frac{2}{{\sqrt {t + 1} }} = 1 \cr
& \sqrt {t + 1} = 2 \cr
& t + 1 = 4 \cr
& t = 3 \cr
& *{\text{Calculate the second derivative}} \cr
& h''\left( t \right) = \frac{d}{{dt}}\left( {1 - 2{{\left( {t + 1} \right)}^{1/2}}} \right) \cr
& h''\left( t \right) = - {\left( {t + 1} \right)^{ - 3/2}} \cr
& {\text{Evaluate the second derivative at }}t = 3 \cr
& h''\left( 3 \right) = - {\left( {3 + 1} \right)^{ - 3/2}} \cr
& h''\left( 3 \right) = - \frac{1}{8} < 0 \cr
& {\text{Then by the second derivative test }}\left( {{\text{Theorem 3}}{\text{.9}}} \right) \cr
& h\left( t \right){\text{ has a relative maximum at }}\left( {0,h\left( 3 \right)} \right) \cr
& h\left( 3 \right) = 3 - 4\sqrt {3 + 1} = - 5 \cr
& {\text{Relative maximum at }}\left( {3, - 5} \right) \cr} $$
