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Thread: [SOLVED] Inverse function of an integral

  1. #1
    MHF Contributor arbolis's Avatar
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    [SOLVED] Inverse function of an integral

    I have $\displaystyle f(x)=\int_0^x 1+sin(sin(t))dt$ and I must find $\displaystyle [f^{-1}(0)]'$. I'm asking you to help me finding the inverse function of f, I'll try to do the derivative.
    One of my guesses would be that as I have an integral of a function, the inverse operation is the derivative of the integral. In other words I would say that maybe the inverse function of $\displaystyle f$ is $\displaystyle 1+sin(sin(x))$. But looking back, if I replace this in the upper limit of the integral of $\displaystyle f(x)$, I doubt it will give me the identity. I also know that $\displaystyle 1+sin(sin(x))$ is $\displaystyle f'(x)$. Hmm... can you help me a bit?
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    $\displaystyle f(f^{-1} (x)) = x$ take derivative $\displaystyle (f^{-1}(x))' f'(f^{-1} (x)) = 1\implies (f^{-1} (0))' f'(f^{-1} (0)) = 1$.

    Now $\displaystyle f(0) = 0 \implies f^{-1} (0) = 0$.
    And $\displaystyle f'(x) = 1 + \sin (\sin x)$ so $\displaystyle f'(f^{-1}(0)) = f'(0) = 1 + \sin (\sin 0) = 1$.

    Thus, $\displaystyle (f^{-1}(0))' \cdot 1 = 1 $ and that is the answer.
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  3. #3
    MHF Contributor arbolis's Avatar
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    take derivative .

    Now .
    And so .

    Thus, and that is the answer.
    Nicely done! I wouldn't think about it... I understand everything except one thing :
    Now .
    . Why does it imply that $\displaystyle f^{-1}(0)=0$? Is it because $\displaystyle f^{-1}(f(x))=x$, replacing $\displaystyle x$ by $\displaystyle 0$ we would get it? If yes, then I understood everything, and that's a beautiful answer.
    P.D.: Just curious if it's possible to get the general formula $\displaystyle [f^{-1}(x)]'$. Because, as I said, it's maybe equal to the derivative of $\displaystyle f$. Note that, maybe by a coincidence we have that $\displaystyle f'(0)=1$.
    EDIT : Nevermind my P.D. I just realized it's totally false and worthless.
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    Because $\displaystyle 1 + \sin \left( {\sin (x)} \right) \ge 0$ (i.e. non-negative bounded and continuous) we know that $\displaystyle \int\limits_0^a {1 + \sin \left( {\sin (x)} \right)} = 0\quad \Leftrightarrow \quad a = 0$.
    Thus only 0 is mapped into 0,
    $\displaystyle \begin{array}{l} f(x) = \int\limits_0^x {1 + \sin \left( {\sin (t)} \right)} = 0\quad \Leftrightarrow \quad x = 0 \\ f^{ - 1} (0) = 0 \\ \end{array}$.
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    Quote Originally Posted by arbolis View Post
    Why does it imply that $\displaystyle f^{-1}(0)=0$?
    Because it is an inverse function. If $\displaystyle f(a) = b$ then $\displaystyle (a,b)\in f$ and so $\displaystyle (b,a)\in f^{-1}$ (assuming the function is invertible) and so $\displaystyle b = f^{-1} (a)$. By the way, I never proved the function is invertible. I just assumed that. That is something you need to prove yourself.
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  6. #6
    MHF Contributor arbolis's Avatar
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    By the way, I never proved the function is invertible. I just assumed that. That is something you need to prove yourself.
    f is strictly increasing (But note that f'(x) is equal to 0 in some separate points, but I think it means that f is strictly increasing and has critical points other than extrema, but I'm not sure.) and continuous. Does that make f one-to-one, or have I to work harder due to the fact that f'(x) can be equal to 0?
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