(adsbygoogle = window.adsbygoogle || []).push({}); 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

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

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