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Math Help - inverse functions T.T please help !

  1. #1
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    inverse functions T.T please help !

    find the two functions which are the inverse of the functions
    1/(1+x^2), x is equal to and > 0 AND 1/(1+x^2), x is equal to and <0

    show that the following pairs of functions are inverses by showing that
    f(g(x)) = g(f(x)) = x
    f(x) = sqrt(16-x^2) -4<x<0 (x also equals -4 and 0)
    g(x) = - sqrt(16-x^2), 0<x<4 (x also equals 0 and 4)

    thank you so much for any help given !!!
    Last edited by CaptainBlack; October 30th 2009 at 01:21 PM.
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  2. #2
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    Inverse Functions

    Hello iiharthero
    Quote Originally Posted by iiharthero View Post
    find the two functions which are the inverse of the functions
    1/(1+x^2), x is equal to and > 0 AND 1/(1+x^2), x is equal to and <0

    show that the following pairs of functions are inverses by showing that
    f(g(x)) = g(f(x)) = x
    f(x) = sqrt(16-x^2) -4<x<0 (x also equals -4 and 0)
    g(x) = - sqrt(16-x^2), 0<x<4 (x also equals 0 and 4)

    thank you so much for any help given !!!
    1a) Let y=f(x) = \frac{1}{1+x^2}, x\ge0

    Then 1+x^2 = \frac{1}{y}

    \Rightarrow x^2 = \frac{1}{y}-1 = \frac{1-y}{y}

    \Rightarrow x = \sqrt{\frac{1-y}{y}}, since x \ge 0

    So the inverse function, replacing y by x, is f^{-1}(x)=\sqrt{\frac{1-x}{x}}

    1b) In the same way if g(x) = \frac{1}{1+x^2}, x \le 0 we get
    g^{-1}(x) =-\sqrt{\frac{1-x}{x}}
    2) f(x) = \sqrt{16-x^2}, -4 \le x \le 0 and g(x) = -\sqrt{16-x^2}, 0 \le x \le 4

    Note first of all that -4\le g(x) \le 0, so g(x) is a valid input into f, for 0\le x\le 4.

    Then for 0\le x\le 4
    f(g(x))= \sqrt{16-(-\sqrt{16-x^2})^2}
    =\sqrt{16-(16-x^2)}

    = \sqrt{x^2}

    =x
    Similarly 0\le f(x) \le 4, so f(x) is a valid input into g .

    Then for -4\le x \le 0,
    g(f(x)) = - \sqrt{16-(\sqrt{16-x^2})^2}
    =-\sqrt{16-(16-x^2)}
    = -\sqrt{x^2}
    =x, noting that for -4\le x \le 0, the negative square root is implicit in the value of x.
    Grandad
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