Answer
$$\eqalign{
& {\text{max}} = \frac{1}{2}{\text{ at }}x = 5 \cr
& {\text{There is no minimum value}}{\text{.}} \cr} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{{x - 2}}{{x + 1}};{\text{ }}\left( { - 1,5} \right] \cr
& {\text{Differentiate}} \cr
& f'\left( x \right) = \frac{{\left( {x + 1} \right)\left( 1 \right) - \left( {x - 2} \right)\left( 1 \right)}}{{{{\left( {x + 1} \right)}^2}}} \cr
& f'\left( x \right) = \frac{{x + 1 - x + 2}}{{{{\left( {x + 1} \right)}^2}}} \cr
& f'\left( x \right) = \frac{3}{{{{\left( {x + 1} \right)}^2}}} \cr
& f'\left( x \right) = 0 \cr
& {\text{There are no real values at which }}f'\left( x \right) = 0 \cr
& {\text{Evaluate }}f{\text{ at the end points }}a{\text{ and }}b \cr
& f\left( { - 1} \right) = {\text{undefined}} \cr
& f\left( 5 \right) = \frac{1}{2} \cr
& {\text{The largest value of }}f{\text{ in the interval }}\left( { - 1,5} \right]{\text{ is }}\frac{1}{2},{\text{ so}} \cr
& {\text{there is a max}} = \frac{1}{2}{\text{ at }}x = 5 \cr
& {\text{There is no minimum value}}{\text{.}} \cr} $$
