Continuous function of one variable Let $f(x)$ continuous function on $R$ wich can be in different signs. Prove, that there is exists an arithmetic progression $a, b, c   (a&lt;b&lt;c)$,  such that $f(a)+f(b)+f(c)=0$.
Try defining the following function on $R \times R^+$

$ g(x,h) = f(x-h) + f(x) + f(x+h) $

Argue that if $f$ can take positive and negative values, then so can $g$.  Then, use the intermediate value theorem.
HINT:

1) Think about the intermediate value theorem. 

2) Think about some $x$ and some $y$ with $f(x)\gt0$ and $f(y)\lt0$.