True or false: If $f(x)$ is continuous on $[0, 2]$ and $f(0)=f(2)$, then there exists a number $c\in [0, 1]$ such that $f(c) = f(c + 1)$.

Define a new function by $g(x)=f(x+1)-f(x)$. Notice that $$g(0) = f(1) - f(0)$$ $$g(1) = f(2) - f(1) = f(0) - f(1)$$ and therefore $g(0) = -g(1)$.

Now use the Intermediate Value Theorem.

Hint: Consider $\phi(x) = f(x+1)-f(x)$ on $[0,1]$.

Note that $\phi(1) = - \phi(0)$.