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Composition of Functions, Derivatives, Hard Problem

  1. Sep 29, 2009 #1
    1. The problem statement, all variables and given/known data

    Does there exist a continuous function f: R -> R such that [tex] f'(f(x)) = x [/tex]?

    2. Relevant equations

    3. The attempt at a solution

    I was trying to find an example, but wasn't able to, and if I had to take a guess whether such a function exists or not, I'd say no. Here is my "proof" (quotation marks because I think the proof is flawed):

    Assume [tex] f'(f(x)) = x [/tex]. Then integration gives us [tex] f(f(x)) = \frac{x^2}{2} [/tex]. Now, [tex] (f(f(x))' = f'(f(x))*f'(x) = xf'(x) [/tex]. But since [tex] f(f(x)) = \frac{x^2}{2} [/tex] then [tex] (f(f(x))' = (x^2/2)' = x [/tex]. So [tex] xf'(x) = x [/tex], implying f'(x) = 1. But this contradicts the assumption that f'(f(x)) = x.

    I think the flaw is where I did the integration. I don't think I integrated properly.
  2. jcsd
  3. Sep 29, 2009 #2


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    Homework Helper

    try differentiating as a check
    [tex]\frac{d}{dx}f(f(x)) = f'(f(x))*f'(x) [/tex]
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