Functional equaliton on continuous function $f:[0,1]\to \mathbb{R}$

Question

Let $f:[0,1]\to \mathbb{R}$ be continuous, and suppose that $f(0)=f(1).$ Show that there is a value $x\in [0, 1998/1999]$ satisfying $f(x)=f(x+1/1999)$.

Answer 1

Hint: examine the sign of the function$$g(x)=f(x+1/1999)-f(x)$$ on that domain and use the intermediate value theorem.