• Support PF! Buy your school textbooks, materials and every day products Here!

Simple Calculus Questions Having to Do With Inverses

  • Thread starter jsewell94
  • Start date
  • #1
23
0

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!
 
Last edited by a moderator:

Answers and Replies

  • #2
22,097
3,277
Please show an attempt or this will be deleted.
 
  • #3
Curious3141
Homework Helper
2,843
86
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
23
0
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
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,224
947
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
Homework Helper
2,843
86
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
23
0
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
Science Advisor
4,790
131
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
Science Advisor
1,089
10
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.
 

Related Threads for: Simple Calculus Questions Having to Do With Inverses

  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
4
Views
1K
Replies
1
Views
799
  • Last Post
Replies
3
Views
1K
Replies
1
Views
1K
  • Last Post
Replies
10
Views
1K
Replies
3
Views
2K
Top