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Math Help - inverse of a function

  1. #1
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    inverse of a function

    Let f(x)=\frac{x}{1+|x|}.
    1. Show f(x) is differentiable.
    2. Find the inverse of f(x).
    3. Show f^{-1}(x) is also differentiable.

    I did (1). I have trouble finding the inverse of f(x) so that it is differentiable. Some help please.
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  2. #2
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    Hi

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

    f^{-1}\left(f(x)\right) = \frac{x}{1-|\frac{x}{1+|x|}|} = \frac{x}{1-\frac{|x|}{1+|x|}} = x
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  3. #3
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    Quote Originally Posted by running-gag View Post
    Hi

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

    f^{-1}\left(f(x)\right) = \frac{x}{1-|\frac{x}{1+|x|}|} = \frac{x}{1-\frac{|x|}{1+|x|}} = x
    But f^{-1}(x) = \frac{x}{1-|x|} is not differentiable at x=1 and x=-1
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  4. #4
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    Quote Originally Posted by dori1123 View Post
    Let f(x)=\frac{x}{1+|x|}.
    1. Show f(x) is differentiable.
    2. Find the inverse of f(x).
    3. Show f^{-1}(x) is also differentiable.

    I did (1). I have trouble finding the inverse of f(x) so that it is differentiable. Some help please.
    The first thing I'd do is draw a graph of y = f(x).

    Note that f(x) = \frac{x}{1+x} = \frac{-1}{x + 1} + 1 if x > 0 and f(x) = \frac{x}{1 - x} = \frac{-1}{x - 1} - 1 if x \leq 0.

    Clearly the domain of y = f(x) is all real numbers and the range is (-1, 1).

    Use the above to inform your thinking on the inverse function and its differentiability. In particular, what's the domain of y = f^{-1}(x) \, .... ?
    Last edited by mr fantastic; February 20th 2009 at 12:50 PM.
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  5. #5
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    The domain for y = f^{-1}(x) is (-1, 1), so f^{-1}(x) is differentiable since f^{-1}(x) is differentiable on (-1,1), right?
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  6. #6
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    Quote Originally Posted by dori1123 View Post
    The domain for y = f^{-1}(x) is (-1, 1), so f^{-1}(x) is differentiable since f^{-1}(x) is differentiable on (-1,1), right?
    Correct. I hope you checked for differentiability at the point (0, 0) ....
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