Simple Calculus Questions Having to Do With Inverses

  • Thread starter jsewell94
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  • #1
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Homework Statement


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]

The Attempt at a Solution


http://i.imgur.com/X5Q9A.jpg?1?7662
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!
 
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Answers and Replies

  • #2
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Please show an attempt or this will be deleted.
 
  • #3
Curious3141
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2. Stuff to know..
(f-1)'(x) = 1/f(f-1(x))
What you posted here makes no sense. Use Latex to format these expressions so they're more legible.
 
  • #4
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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. :)
 
  • #5
SammyS
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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. :)
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.
 
  • #6
Curious3141
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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. :)
Yes, it looks good.

But can you assume [itex](f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}[/itex] without proof?
 
  • #7
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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?
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]
 
  • #8
chiro
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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?
 
  • #9
Bacle2
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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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