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A difficult problem regarding inverses and derivatives

  • Thread starter nietzsche
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  • #1
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Homework Statement



Is there a continuous function f:R->R such that f'(f(x)) = x ?

Homework Equations





The Attempt at a Solution



I've been working on this problem for quite some time now. I can see that, supposing there is such a function f, then [tex]f^{-1}(x) = f'(x)[/tex]. So we are looking for a function whose derivative is its inverse.

I'm trying to see what other information I can gather from the givens, but I'm pretty much clueless as to where to go from here. Any suggestions?
 

Answers and Replies

  • #2
Dick
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Think about the properties that a continuous function has to have to have an inverse. Draw some graphs. It has to hit every horizontal line exactly once, right? Can you make a statement about f(x) involving the words 'increasing' and 'decreasing'?
 
  • #3
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hmm...

f(x) has to be one-to-one in order for it to have an inverse function, so it must be either increasing for all x or decreasing for all x.

if we suppose f(x) is increasing, then f'(x) will be positive for all x. but f inverse will have negative values, so f'(x) can never equal f inverse. a similar argument applies for decreasing functions.

how's that? i'm not sure how to make it more rigorous... that always seems to be my problem.
 
  • #4
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according to all the graphs i drew, i figured out that the inverse of an increasing function is also an increasing function, and the inverse of a decreasing function is also a decreasing function. i'm not sure if i'm looking at all the cases though. is that a valid assumption?
 
  • #5
Dick
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hmm...

f(x) has to be one-to-one in order for it to have an inverse function, so it must be either increasing for all x or decreasing for all x.

if we suppose f(x) is increasing, then f'(x) will be positive for all x. but f inverse will have negative values, so f'(x) can never equal f inverse. a similar argument applies for decreasing functions.

how's that? i'm not sure how to make it more rigorous... that always seems to be my problem.
That's what I was thinking. If f(x) is increasing then f'(x)>=0 for all x, but f'(f(-1))=(-1), that's a contradiction. As you say, similar argument for decreasing. That might be all they are really expecting. If you want totally rigorous, then you have to prove two things, if you haven't already proved them. i) if f(x) is invertible then f(x) is either increasing or decreasing and ii) if f(x) is increasing e.g. then f'(x)>=0. If you want to continue, any ideas on proving either one?
 
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  • #6
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That's what I was thinking. If f(x) is increasing then f'(x)>=0 for all x, but f'(f(-1))=(-1), that's a contradiction. As you say, similar argument for decreasing. That might be all they are really expecting. If you want totally rigorous, then you have to prove two things, if you haven't already proved them. i) if f(x) is invertible then f(x) is either increasing or decreasing and ii) if f(x) is increasing e.g. then f'(x)>=0. If you want to continue, any ideas on proving either one?
thanks for the help. with regards to the proofs for i and ii, i'm going to have to look over it a bit more, but i get the main idea now.
 
  • #7
lanedance
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you should probably demonstrate for the case when f is not invertible as well, should be easy enough if you assume
[tex] f(x_1) = f(x_2) [/tex] with
[tex] x_1 \neq x_2 [/tex]
 

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