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Math Help - [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 f(x)=\int_0^x 1+sin(sin(t))dt and I must find [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 f is 1+sin(sin(x)). But looking back, if I replace this in the upper limit of the integral of f(x), I doubt it will give me the identity. I also know that 1+sin(sin(x)) is f'(x). Hmm... can you help me a bit?
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  2. #2
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    f(f^{-1} (x)) = x take derivative (f^{-1}(x))' f'(f^{-1} (x)) = 1\implies (f^{-1} (0))' f'(f^{-1} (0)) = 1.

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

    Thus, (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 f^{-1}(0)=0? Is it because f^{-1}(f(x))=x, replacing x by 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 [f^{-1}(x)]'. Because, as I said, it's maybe equal to the derivative of f. Note that, maybe by a coincidence we have that f'(0)=1.
    EDIT : Nevermind my P.D. I just realized it's totally false and worthless.
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    Because 1 + \sin \left( {\sin (x)} \right) \ge 0 (i.e. non-negative bounded and continuous) we know that \int\limits_0^a {1 + \sin \left( {\sin (x)} \right)}  = 0\quad  \Leftrightarrow \quad a = 0.
    Thus only 0 is mapped into 0,
    \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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  5. #5
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    Quote Originally Posted by arbolis View Post
    Why does it imply that f^{-1}(0)=0?
    Because it is an inverse function. If f(a) = b then (a,b)\in f and so (b,a)\in f^{-1} (assuming the function is invertible) and so 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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