Suppose f is continous on [0,1] and f(0)=f(1). Let n be a natural number. Prove that there is some number x, such that f(x)=f(x+1/n).
The hints says to consider g(x)=f(x)-f(x+1/n)
The Attempt at a Solution
I've tried to consider the function g(x), but I haven't gotten anything useful from it. When I've tried various values of n, like 1/2, 1/3...I've noticed that there are repeating terms and I can manipulate the terms a bit to get like g(0)=-g(1/2) for n=1/2 and the like, but I am not sure where to go with this