absolute maximum
posted by julie .
absolute maximum of (x^24)/(x^2+4)
the critical numbers i got was x=0,2.59, and 3.09
the absolute minimum is f(0)=1
what is the absolute maximum:
is it both f(4)=(f(4))=.6?
or can there only be one

The absolute maximum must occur where the deriviative is zero. But you also have to use othe tests to show that it not a minimum or only a relative maximum.
The derivative is [(x^2+4)(2x)  (x^24)(2x)]/(x^2+4)^2
=16 x/(x^2+4)^2
That only equals zero when x = 0. That happens to be the absolute minimum. There is no absolute maximum. The function approaches +1 as x goes to plus or minus infinity.