# Real Analysis: Use Intermediate value Thm to prove the functions

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?

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,
• Where you wrote "...where $\;f(c)=0\;$..." it must be "...where $\;g(c)=0\;$...now substitute for the definition of $\;g\;$ ! – DonAntonio Nov 24 '13 at 18:15