# Thread: Absolute extrema

1. ## Absolute extrema

Help me find the absolute extrema on the closed function [3,5]

$h(t)=\frac{t}{t-2}$

So far I've got $h'(t)=\frac{-2}{(t-2)^2}$

Where do I go from here?

2. Originally Posted by XIII13Thirteen
Help me find the absolute extrema on the closed function [3,5]

$h(t)=\frac{t}{t-2}$

So far I've got $h'(t)=\frac{-2}{(t-2)^2}$

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...

3. So ...

$t=\frac{2}{\sqrt-2}$? Am I doing something wrong setting the equation to 0? If this is true is there no crictical points in the interval?

4. Originally Posted by XIII13Thirteen
So ...

$t=\frac{2}{\sqrt-2}$? Am I doing something wrong setting the equation to 0? If this is true is there no crictical points in the interval?
yes, you did something wrong...i'm not even sure what
the derivative is never zero (since the numerator is never zero) so just check the end points

5. There's nothing wrong. This function doesn't have a real extremum. If the function is continous, just check the end points. But if it isn't continous, check the break off points too.