`h(t) = t/(t + 3), [-1,6]` Find the absolute extrema of the function on the closed interval.
Find the critical values for t by setting the derivative equal to zero and solving for the t value(s).
A critical value will not exists for this function.
Plug in the critical t value(s) and the endpoints of the closed interval into the h(t) function.
Examine the h(t) values to determine the absolute extrema.
The absolute maximum is the point `(6,2/3).`
The absolute minimum is the point `(-1,-1/2)`