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Thread: If g=1/f^-1, what is g^-1(-4)? Please help!

  1. #1
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    Exclamation If g=1/f^-1, what is g^-1(-4)? Please help!

    Please help. I think I'm close but I am just not getting the answer.
    The problem reads:
    Suppose the $\displaystyle f$ has an inverse and $\displaystyle f(2)=-4$, $\displaystyle f'(2)=\frac{3}{4}$. If $\displaystyle g=\frac{1}{f^-1}$, what is $\displaystyle g'(-4)$?

    I have
    $\displaystyle f^-1(-4)=2, f(2)=(-4)$
    $\displaystyle g(-4)=\frac{1}{2}$
    $\displaystyle g'(-4)=\frac{1}{f'(\frac{1}{2})}$

    Thanks!
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  2. #2
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    Hi

    $\displaystyle g=\frac{1}{f^{-1}} = \frac{1}{u}$ where $\displaystyle u = f^{-1}$

    $\displaystyle g'=-\frac{u'}{u^2}$

    $\displaystyle u' = \frac{1}{f'of^{-1}}$

    Therefore

    $\displaystyle g'=-\frac{1}{f'of^{-1} \times \left(f^{-1}\right)^2}$
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  3. #3
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    question on 3rd line

    Quote Originally Posted by running-gag View Post
    Hi

    $\displaystyle g=\frac{1}{f^{-1}} = \frac{1}{u}$ where $\displaystyle u = f^{-1}$

    $\displaystyle g'=-\frac{u'}{u^2}$

    $\displaystyle u' = \frac{1}{f'of^{-1}}$

    Therefore

    $\displaystyle g'=-\frac{1}{f'of^{-1} \times \left(f^{-1}\right)^2}$
    Can you explain how you got the 3rd line...
    $\displaystyle u' = \frac{1}{f'of^{-1}}$?

    I think that will clear things up for me.

    Thanks!
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  4. #4
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    Quote Originally Posted by yvonnehr View Post
    Can you explain how you got the 3rd line...
    $\displaystyle u' = \frac{1}{f'of^{-1}}$?

    I think that will clear things up for me.

    Thanks!
    Well the derivative of $\displaystyle f^{-1}$ is $\displaystyle \frac{1}{f'of^{-1}}$

    Haven't you already seen this relation ?
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  5. #5
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    thank!

    No I don't remember seeing that. I'm going to look for it in my book and see what else I missed in that section.

    Now I can try to make sense of your entire response. Thanks again.
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  6. #6
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    It comes from the formula $\displaystyle (fog)' = f'og \times g'$

    applied to $\displaystyle (fof^{-1})' = f'of^{-1} \times \left(f^{-1}\right)'$

    Due to the fact that $\displaystyle fof^{-1} = Id$, $\displaystyle (fof^{-1})' = 1$

    Therefore $\displaystyle f'of^{-1} \times \left(f^{-1}\right)' = 1$

    which leads to $\displaystyle \left(f^{-1}\right)' = \frac{1}{f'of^{-1}}$
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