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Thread: Inverse function, in terms with another inverse function

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

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

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

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

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

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

    How do I get to that answer?
    Last edited by Chris L T521; Sep 22nd 2009 at 07:35 PM. Reason: reformatted LaTeX
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  2. #2
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    Quote Originally Posted by BlackBlaze View Post
    $\displaystyle f$ is a one-to-one function with inverse $\displaystyle f^{-1}$.
    $\displaystyle p(x) = \frac{1}{1+f(x)}$
    Caculate the inverse of the function in terms of $\displaystyle f^{-1}$

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

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

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

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

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

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

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

    Your job is to make y the subject:

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

    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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