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Math Help - Inverse function, in terms with another inverse function

  1. #1
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    Inverse function, in terms with another inverse function

    f is a one-to-one function with inverse f^{-1}.
    p(x) = \frac{1}{1+f(x)}
    Caculate the inverse of the function in terms of f^{-1}

    So I tried swapping p(x) and f(x).
    f^{-1}(x) = \frac{1}{1+p^{-1}(x)}

    \frac{1}{f^{-1}(x)} = 1 + p^{-1}(x)

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

    The answer in the back of the book however, says this:
    p^{-1}(x) = f^{-1}\left(\frac{1}{x} - 1\right)

    How do I get to that answer?
    Last edited by Chris L T521; September 22nd 2009 at 08:35 PM. Reason: reformatted LaTeX
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  2. #2
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    Quote Originally Posted by BlackBlaze View Post
    f is a one-to-one function with inverse f^{-1}.
    p(x) = \frac{1}{1+f(x)}
    Caculate the inverse of the function in terms of f^{-1}

    So I tried swapping p(x) and f(x).
    f^{-1}(x) = \frac{1}{1+p^{-1}(x)}

    \frac{1}{f^{-1}(x)} = 1 + p^{-1}(x)

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

    The answer in the back of the book however, says this:
    p^{-1}(x) = f^{-1}\left(\frac{1}{x} - 1\right)

    How do I get to that answer?
    It helps to apply the procedure correctly.

    Let y = \frac{1}{1 + f(x)}. Swap y and x:

    x = \frac{1}{1 + f(y)}.

    Your job is to make y the subject:

    \frac{1}{x} = 1 + f(y) \Rightarrow \frac{1}{x} - 1 = f(y)

    and it should be clear what the last step is.
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  3. #3
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    Right at the second step, I saw what a gross error I made. Don't I feel silly.

    Thanks for the speedy response, as well as the formatting.
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