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Let $f \colon [0,2] \to \Bbb R$ be a continuous function such that $f(0) = f(2)$. Use the intermediate value theorem to prove that there exist numbers $x, y \in [0, 2]$ such that $f (x) = f (y)$ and $|x − y| = 1$.

Hint: Introduce the auxiliary function $g \colon [0, 1] \to \Bbb R$ defined by $g(x) = f (x + 1) − f (x)$.

I still do not know how to prove it. Could anyone help?

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The given function $\;g\;$ is continous in $\;[0,1]\;$ and

$$\begin{cases}g(0)=f(1)-f(0)\\{}\\g(1)=f(2)-f(1)\end{cases}\implies g(0)g(1)<0\;\;\text{, unless}\;f(1)=f(0)\;\;\text{(why?)}$$

and so by the IVM there exists $\;c\in (0,1)\;$ s.t. $\;g(c)=0\;\ldots$ ...end the exercise now,

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  • $\begingroup$ I see this relationship: g(0) is positive < 0 < g(1) is negative or g(0) is negative < 0 < g(1) is positive. The. by IVP, there exist c in (0, 1) where f(c) = 0. Correct me if I am wrong . But how does this help us prove f(x) = f(y) and |x-y| = 1 $\endgroup$ – afsdf dfsaf Nov 24 '13 at 18:03
  • $\begingroup$ Where you wrote "...where $\;f(c)=0\;$..." it must be "...where $\;g(c)=0\;$...now substitute for the definition of $\;g\;$ ! $\endgroup$ – DonAntonio Nov 24 '13 at 18:15
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    $\begingroup$ just figure out...f(c+1)-f(c) = g(c) = 0. Then, |x-y| = |c+1-c| = 1? right $\endgroup$ – afsdf dfsaf Nov 24 '13 at 18:56
  • $\begingroup$ Yes @afsdfdfsaf . Now don't forget to upvote answers that helped you and, eventually, accept the answer that is the best for you. $\endgroup$ – DonAntonio Nov 24 '13 at 19:49
  • $\begingroup$ Sure will do...it is not necessarily to show the case f(0) = f(1) right? $\endgroup$ – afsdf dfsaf Nov 24 '13 at 19:55

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