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Thread: inverse of a function

  1. #1
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    inverse of a function

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

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

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

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

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

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

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

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

    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 $\displaystyle y = f^{-1}(x) \, .... ?$
    Last edited by mr fantastic; Feb 20th 2009 at 12:50 PM.
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  5. #5
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    The domain for $\displaystyle y = f^{-1}(x)$ is $\displaystyle (-1, 1)$, so $\displaystyle f^{-1}(x)$ is differentiable since $\displaystyle f^{-1}(x)$ is differentiable on $\displaystyle (-1,1)$, right?
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  6. #6
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    Quote Originally Posted by dori1123 View Post
    The domain for $\displaystyle y = f^{-1}(x)$ is $\displaystyle (-1, 1)$, so $\displaystyle f^{-1}(x)$ is differentiable since $\displaystyle f^{-1}(x)$ is differentiable on $\displaystyle (-1,1)$, right?
    Correct. I hope you checked for differentiability at the point (0, 0) ....
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