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Math Help - inverse function problem

  1. #1
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    inverse function problem

    I have a function

    f(x) = 4x^2 - x^3

    If the domain of this function is restricted to 0 < x < 8/3 then f-1(x) can be defined - ie inverse function.

    I am told to find the gradient of f^-1(x) at the point ( \frac{7}{8}, \frac{1}{2})

    My first thought was to work out the inverse function. But this looks hard? Is this what I should do? Or is there another way? I know that an inverse function is a reflection of function on y=x - can I somehow use this to calculate? If so how?

    Angus
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: inverse function problem

    Quote Originally Posted by angypangy View Post
    I am told to find the gradient of f^-1(x) at the point ( \frac{7}{8}, \frac{1}{2})
    (f^{-1})'(7/8)=\frac{1}{f'(1/2)}=\ldots

    Edited: Corrected.
    Last edited by FernandoRevilla; December 16th 2011 at 04:16 AM.
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  3. #3
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    Re: inverse function problem

    Quote Originally Posted by FernandoRevilla View Post
    (f^{-1})'(1/2)=\frac{1}{f'(7/8)}=\ldots
    I have already calculated dy/dx of f(x) which is 8x - 3x^2. If I use the y value of f(x) when x=7/8, then I get:

    8(1/2) - 3(1/2)^2 = 4 - 3/4 = 13/4.

    The reciprocal of this is 4/13 - which is correct answer.
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Re: inverse function problem

    Quote Originally Posted by angypangy View Post
    I have already calculated dy/dx of f(x) which is 8x - 3x^2. If I use the y value of f(x) when x=7/8, then I get:

    8(1/2) - 3(1/2)^2 = 4 - 3/4 = 13/4.

    The reciprocal of this is 4/13 - which is correct answer.
    You are right, I didn't check that f(1/2)=7/8 . I supposed (without computing) f(7/8)=1/2 .
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