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March 30, 2017

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absolute maximum of (x^2-4)/(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

  • absolute maximum - ,

    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^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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