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Math Help - Integration question

  1. #1
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    Angry Integration question

    Question: The function f with the domain: x is greater than 1/2 is defined by f(x)=4x(2x-1)^7. Given that f(1)=4, find the value of d/dx{f^-1(x)} when x=4.

    I would try and find the inverse function of f(x), but im not sure how exactly this is related to integration in my textbook chapter. Can anyone shed some light as to which direction i proceed in finding d/dx{f^-1(x)}? Unsure how to start this one. Thanks
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  2. #2
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    Hello !
    Quote Originally Posted by fire.garden View Post
    Question: The function f with the domain: x is greater than 1/2 is defined by f(x)=4x(2x-1)^7. Given that f(1)=4, find the value of d/dx{f^-1(x)} when x=4.

    I would try and find the inverse function of f(x), but im not sure how exactly this is related to integration in my textbook chapter. Can anyone shed some light as to which direction i proceed in finding d/dx{f^-1(x)}? Unsure how to start this one. Thanks
    Let's do it in a general case first.

    Say you have a function f. Its inverse function f^{-1} is such that f(f^{-1}(x))=x
    Differentiate implicitly (use chain rule for the left-hand side of the equation) :

    \frac{d}{dx}(f^{-1}(x)) \cdot f'(f^{-1}(x))=1

    Hence [f^{-1}]'(x)=\frac{d}{dx}(f^{-1}(x))=\frac{1}{f'(f^{-1}(x))}

    Let x=4 : \underbrace{[f^{-1}]'(4)}_{\text{the value you're looking for}}=\frac{1}{f'(f^{-1}(4))}

    Now, how do you calculate f^{-1}(4) ?
    By solving 4=4x(2x-1)^7 or by using the information you're given : f(1)=4
    Hence f^{-1}(4)=1 and :

    \implies [f^{-1}]'(4)=\frac{1}{f'(1)}

    Can you take it from here ?
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  3. #3
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    Thanks! i think i can do it now
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