Suppose that a function $f\colon[0,1]\rightarrow \mathbb{R}$ is continuous, $f\left(0\right)=0$

Question

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.

Answer 1

Let $x_0 = \sup \{x \in [0,1]: f(x) = 0\}$. You need to show two things. First, that $f(x_0) = 0$, and second that $f(x) > 0$ for all $x > x_0$. The intermediate value theorem may be helpful for the latter part.

Answer 2

Let $f : I \rightarrow \mathbb{R}$ is a continuous function where $I \subseteq \mathbb{R}$ is an interval. If for some $x \in I, f(x) > 0$ then there exists a nbd $V$ of $x$ such that for each $y \in V \cap I, f(y) > 0.$