Answer
Maximum is $(6, 2/3)$ & minimum is $(-1, -1/2)$
Work Step by Step
Step-1: Put the extreme values of $t$ in $h$,
$$h(-1)=\frac{-1}{-1+3}=-\frac{1}{2}$$
$$h(6) = \frac{6}{6+3}=\frac{2}{3}$$
Step-2: Thus, maximum is $(6, 2/3)$ & minimum is $(-1, -1/2)$
