Prove that if $f$ is a continuous function on the interval $[a, b]$ with $f(a)<0<f(b)$ then $f(c)=0$ for some $c \in(a, b)$.

Let $g$ be a continuous function on $[0,1]$ satisfying $g(0)=g(1)$. By considering the function $f(x)=g\left(x+\frac{1}{2}\right)-g(x)$ on $\left[0, \frac{1}{2}\right]$, show that $g\left(c+\frac{1}{2}\right)=g(c)$ for some $c \in\left[0, \frac{1}{2}\right]$. Show, more generally, that for any positive integer $n$ there exists a point $c_{n} \in\left[0, \frac{n-1}{n}\right]$ for which $g\left(c_{n}+\frac{1}{n}\right)=g\left(c_{n}\right)$.