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Math Help - Find intervals of f increasing/decreasing

  1. #1
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    Find intervals of f increasing/decreasing

    1. The problem statement, all variables and given/known data

    (a) Find the intervals on which f is increasing or decreasing.
    (b) Find the local maximum and minimum values of f.
    (c) Find the intervals of concavity and the inflection points.

    x2/(x2+3)

    2. The attempt at a solution

    a) I find that
    f'(x)=6x/(x2+3)2

    I am stuck because (x2+3)2 has no solutions and I don't know how to define x in order to proceed further.
    If anybody can help me out understanding this, I will be very thankful.
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  2. #2
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    Quote Originally Posted by phillyolly View Post
    1. The problem statement, all variables and given/known data

    (a) Find the intervals on which f is increasing or decreasing.
    (b) Find the local maximum and minimum values of f.
    (c) Find the intervals of concavity and the inflection points.

    x2/(x2+3)

    2. The attempt at a solution

    a) I find that
    f'(x)=6x/(x2+3)2

    I am stuck because (x2+3)2 has no solutions and I don't know how to define x in order to proceed further.
    If anybody can help me out understanding this, I will be very thankful.
    \displaystyle f(x)=\frac{x^2}{x^2+3}

    \displaystyle f'(x)=\frac{(x^2+3)(2x) - (x^2)(2x)}{(x^2+3)^2}

    \displaystyle =\frac{6x}{(x^2+3)^2}

    like you said.

    Set \displaystyle f'(x) = 0

    So \displaystyle 6x = 0 \implies x=0

    and so on..
    Last edited by mr fantastic; July 9th 2010 at 07:37 PM.
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  3. #3
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    Hi!

    How can I solve (x^2+3)^2?
    I cannot put x^2+3=0 because x^2=-3 does not exist.
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  4. #4
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    Quote Originally Posted by phillyolly View Post
    Hi!

    How can I solve (x^2+3)^2?
    I cannot put x^2+3=0 because x^2=-3 does not exist.
    Why are you trying to solve \displaystyle(x^2+3)^2=0? Suppose we have two real numbers \displaystyle a and \displaystyle b, b\ne0, then \displaystyle \frac{a}{b}=0 \iff a=0.
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