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Math Help - 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 f has an inverse and f(2)=-4, f'(2)=\frac{3}{4}. If g=\frac{1}{f^-1}, what is g'(-4)?

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

    Thanks!
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  2. #2
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    Hi

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

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

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

    Therefore

    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

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

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

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

    Therefore

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

    I think that will clear things up for me.

    Thanks!
    Well the derivative of f^{-1} is \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 (fog)' = f'og \times g'

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

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

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

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