Asked by julie
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
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
Answers
Answered by
drwls
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.
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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