1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A difficult problem regarding inverses and derivatives

  1. Oct 18, 2009 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant equations



    3. 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?
     
  2. jcsd
  3. Oct 18, 2009 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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'?
     
  4. Oct 18, 2009 #3
    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.
     
  5. Oct 18, 2009 #4
    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?
     
  6. Oct 18, 2009 #5

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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?
     
    Last edited: Oct 18, 2009
  7. Oct 18, 2009 #6
    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.
     
  8. Oct 19, 2009 #7

    lanedance

    User Avatar
    Homework Helper

    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]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: A difficult problem regarding inverses and derivatives
  1. Derivatives of inverses (Replies: 10)

  2. Derivative of Inverse (Replies: 7)

Loading...