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Thread: Interesting Question

  1. #1
    MHF Contributor
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    Interesting Question

    This is from a Mu Alpha Theta Competition I took today.

    19. Let f and g be functions such that $\displaystyle f(x)^2=g^{-1}(4x-1)$. What is $\displaystyle f(x)f'(x)g'(f(x)^2)$?

    A) Not enough information
    B)2
    C) $\displaystyle \frac{2}{4x-1}$
    D)8x-2


    I didn't really know how to approach this, so a little push would be great.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Jameson
    This is from a Mu Alpha Theta Competition I took today.

    19. Let f and g be functions such that $\displaystyle f(x)^2=g^{-1}(4x-1)$. What is $\displaystyle f(x)f'(x)g'(f(x)^2)$?

    A) Not enough information
    B)2
    C) $\displaystyle \frac{2}{4x-1}$
    D)8x-2


    I didn't really know how to approach this, so a little push would be great.
    $\displaystyle
    f(x)^2=g^{-1}(4x-1)
    $

    Rewrite as:

    $\displaystyle
    g([f(x)]^2)=4x-1
    $

    Now differentiate wrt x.

    RonL
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  3. #3
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    Quote Originally Posted by CaptainBlack
    $\displaystyle
    f(x)^2=g^{-1}(4x-1)
    $

    Rewrite as:

    $\displaystyle
    g([f(x)]^2)=4x-1
    $

    Now differentiate wrt x.

    RonL
    How do you know it is differenciable
    Last edited by ThePerfectHacker; Mar 4th 2006 at 03:14 PM.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by ThePerfectHacker
    How do you know it is differenciable
    1. It is a linear form in $\displaystyle x$ and so differentiable wrt $\displaystyle x$.

    2. If $\displaystyle f$ and $\displaystyle g$ are differentiable then this is differentiable. Asking the
    question implies we may assume that $\displaystyle f$ and $\displaystyle g$ are differentiable, just
    as it allows us to assume that $\displaystyle '$ denotes the derivative.


    RonL
    Last edited by CaptainBlack; Mar 5th 2006 at 09:39 AM.
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  5. #5
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    I was making a remark that in analysis you always need to show diffrenciability before taking the derivative, I just found that funny.
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  6. #6
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    Quote Originally Posted by CaptainBlack
    $\displaystyle
    f(x)^2=g^{-1}(4x-1)
    $

    Rewrite as:

    $\displaystyle
    g([f(x)]^2)=4x-1
    $

    Now differentiate wrt x.

    RonL
    So I get that $\displaystyle g'[f(x)^2]*2f(x)*f'(x)=4$, so my answer is 2.
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  7. #7
    Grand Panjandrum
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    Quote Originally Posted by Jameson
    So I get that $\displaystyle g'[f(x)^2]*2f(x)*f'(x)=4$, so my answer is 2.
    That's what I get

    RonL
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