Help me find the absolute extrema on the closed function [3,5]
$\displaystyle h(t)=\frac{t}{t-2}$
So far I've got $\displaystyle h'(t)=\frac{-2}{(t-2)^2}$
Where do I go from here?
you did good taking the derivative
now, you want to check the end points, that is, find h(3) and h(5). then you find the critical points (by setting h'(t) = 0, and solving for t), then you find the value of the function at the critical points. whichever value is the greatest is the absolute maximum, whichever is least is the absolute minimum
Note: points that make the function undefined are also critical points, so t = 2 would be a critical point here. however, that is not in the interval we are considering, so we forget about that...