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Homework Help: For a continuous function on [0,2] , f(0)=f(2)

  1. Jul 13, 2010 #1
    1. The problem statement, all variables and given/known data

    Suppose f is a continuous function on [0,2] with f(0) = f(2). Show that there is an x in [0,1] where f(x) = f(x+1).

    2. Relevant equations

    By the Intermediate Value Theorem, we know that any values between sup{f(x)} and inf {f(x)} over x in [0,2] will be repeated, since the function returns to where it began and must retrace, as it were, its steps (or perhaps it is a constant function).


    3. The attempt at a solution

    What I have so far is:

    Let A = {f(x) : x in [0,1]} and let B={f(x) : x in [1,2]} , then (A [tex]\cap[/tex] B) contains all the values of f(x) that are repeated in these intervals. So f[tex]^{-1}[/tex](A[tex]\cap[/tex] B) contains all the values of x in [0,2] for which f(x) repeats itself. It remains to be shown that there is x1 and x2 in f[tex]^{-1}[/tex](A [tex]\cap[/tex]B) for which x2 = x1 + 1, and I am unsure of how to do this, as I am unsure of my entire approach.

    Any suggestions would be greatly appreciated.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 13, 2010 #2

    Office_Shredder

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    You might want to look at the function f(x)-f(x+1) on the interval [0,1].

    See if you can prove it has a zero using the IVT
     
  4. Jul 13, 2010 #3
    Wow, thanks a lot! Got it.
     
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