Suppose that a function $f\colon[0,1]\rightarrow \mathbb{R}$ is continuous, $f\left(0\right)=0$ and $f\left(1\right)>0$. Prove that there is a number $x_0$ where $0\leq x_0<1$ such that $f\left(x_0\right)=0$ and $f\left(x\right)>0$ $\forall x$ where $x_0<x\leq 1$.
My attempt:
Argue by contradiction.
$x_0$ where $0\leq x_0<1$ such that $f\left(x_0\right)=0$ $\exists$ $x_n$ where $f\left(x_n\right)\leq 0$.
However, I have no been able to go any further. Am I on the right track? Please do not provide a complete solution.