Show that for any $0Suppose $f:[0,1]\to\mathbb{R}$ is continuous and $\min_{x\in[0,1]}f(x)=f(0)=f(1)$. Show that for any  $y\in(0,1)$, there is $c\in(0,1)$ s.t. $f(c)=f(y+c)$.
Intuitively I can see what is going on by drawing a graph, but I am having difficulty writing the proof down formally using mathematical languages. Can some one help me?

Hint:
Since f is continuous in $[0,1]$ and f(0)=f(1), hence there exists at least one x$\in~(0,1)$ such that $$f\prime(x)=0 \text { (Rolle's theorem)}$$
If you have drawn the graph you will notice that the curve will be symmetric about x
