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Math Help - Inverse Function

  1. #1
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    Self Inverse



    f(x) = \frac{2}{x }  - x so i have done  f(f(x)) -x = 0 so \frac{2}{\frac{2}{ x}  -x}   - (\frac{2}{ x}  -x) - x = 0 How does that equal \frac{4{x}^{ 2} -4}{x(2-{x}^{ 2} ) } ???
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  2. #2
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    Shouldn't it be f(f^{-1}(x)) - x =0?

    it may be easier to write
    y=\frac{2}{x} - x
    yx=2 - x^2

    Which is a quadratic, so
    x=\frac{-y \pm \sqrt{y^2 + 8}}{2}

    and hence the inverse is
    f^{-1}(x)=\frac{-x \pm \sqrt{x^2 + 8}}{2}
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  3. #3
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    Quote Originally Posted by SpringFan25 View Post
    Shouldn't it be f(f^{-1}(x)) - x =0?
    who knows what the name is called all i know is i have to do the formula i've put up on there.
    Last edited by mr fantastic; May 22nd 2011 at 12:35 PM.
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  4. #4
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    (not sure if you saw my previous post before i edited it to include the answer)

    using your formula you should get the same answer as i posted:

    f(f^{-1}(x)) -x =0
    \frac{2}{f^{-1}(x)} - f^{-1}(x) -x =0

    multiple everything by f^-1(x):
    2 - \left[f^{-1}(x)\right]^2 -xf^{-1}(x) =0

    this is a quadratic in f^{-1}(x). the coefficients (a,b,c) are (-1,-x,2)

    Solve with quadratic formula

    f^{-1}(x) = \frac{x \pm \sqrt{x^2 + 8}}{-2} = \frac{-x \pm \sqrt{x^2 + 8}}{2}
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