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January 31, 2015

January 31, 2015

Posted by **julie** on Sunday, October 28, 2007 at 8:44pm.

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

- absolute maximum -
**drwls**, Monday, October 29, 2007 at 12:04amThe 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^2-4)(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.

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