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I am not sure if the IVT should be applied here. I am try to do this problem and am stuck an how proceed:

Suppose $f: [-1,1] \rightarrow \mathbb{R}$ is continuous and satisfies $f(-1)=f(1)$. Prove that there exists $y \in [0,1]$ such that $f(y)=f(y-1)$.

So far, I have considered a new function $g(x)=f(x)-f(x-1)$, $x \in [0,1]$, but am stuck after this.

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    $\begingroup$ evaluate $g$ at $0$ and $1$ $\endgroup$ – Aram Nov 17 '14 at 1:01
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Using the fact that $f(1)=f(-1)$, we have
$g(0)=f(0)-f(-1)=f(0)-f(1)$
$g(1)=f(1)-f(0)=-[f(0)-f(1)]$

So, $g(0)$ and $g(1)$ have opposite signs. By continuity of $g$, it must have a root in $[0,1]$, so there exists $y \in [0,1]$ such that $g(y)=0$, i.e. $f(y)=f(y-1)$.

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We have $g(x)=f(x)-f(x-1)$, $g$ is continuous because its a sum of continuous functions, now if $$g(0)=f(0)-f(-1) = 0$$ we're done, otherwise if $$g(0) > 0$$ because $g(1) = g(-1)$ we also know that $$g(1) < 0$$ then by continuity and the IVT there is some $c \in [0,1]$ such that $$g(c) = 0$$ and we're done, the other case ($g(0) < 0$ has the same proof)

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