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Homework Help: Simple Calculus Questions Having to Do With Inverses

  1. Aug 22, 2012 #1
    1. The problem statement, all variables and given/known data
    Suppose [itex]f^{-1}(x)[/itex] is the inverse function of a differentiable function [itex]f[/itex] Let [itex]G(x) = \frac{1}{f^{-1}(x)}[/itex] If [itex]f(3) = 2 [/itex] and [itex]f'(3) = 1/9[/itex] , find [itex]G'(2) [/itex].

    2. Stuff to know..
    [itex](f^{-1}(x))' = \frac{1}{f'(f^{-1}(x))} [/itex]

    3. The attempt at a solution
    I just have no idea if I'm doing this right or not..or if I've made a stupid calculation error. Thanks for the help!
    Last edited by a moderator: Aug 22, 2012
  2. jcsd
  3. Aug 22, 2012 #2
    Please show an attempt or this will be deleted.
  4. Aug 22, 2012 #3


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    What you posted here makes no sense. Use Latex to format these expressions so they're more legible.
  5. Aug 22, 2012 #4
    Okay, sorry about the sucky post. I was in a rush to get it up, and I didn't realize the image was screwed up. I also didn't take the time to format it neatly. Hope this is more readable. :)
  6. Aug 22, 2012 #5


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    Yes that makes sense now. I see that you also fixed one or more typos.

    Your method & answer look correct.

    BTW: You may be warned about an overly large image by one of the moderators.
  7. Aug 22, 2012 #6


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    Yes, it looks good.

    But can you assume [itex](f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}[/itex] without proof?
  8. Aug 22, 2012 #7
    Yeah, you can assume this. We went over it in class, and I managed to prove it (maybe?).

    According to the chain rule

    [itex](f(g(x))' = f'(g(x))g'(x) [/itex]

    So, consider: [itex] f(f^{-1}(x)) = x [/itex]

    According to the chain rule,[itex] f'(f^{-1}(x))(f^{-1})'(x) = 1 [/itex]

    So, [itex](f^{-1})'(x) = \frac{1}{f'(f^{-1})} [/itex]
  9. Aug 22, 2012 #8


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    You can prove it if you want. Assuming an inverse exists, then y = f(x) which implies x = f-1(y) = f(f-1(x)) = f-1(f(x)) (by the properties of the inverse). Now can you differentiate these with respect to x using the differentiability rules to get it in the form of the above?
  10. Aug 23, 2012 #9


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    A kind-of-obvious point for the OP: you may want to also address the issue of the

    possibility of your denominator being zero when you answer your question.
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