Prove that there is a number $x_0 \in (0,1]$ such that $f(x_0)=0$ and $f(x)>0$ for $0\le x < x_0$. Suppose that the function $f:[0,1]\rightarrow \mathbb{R}$ is continuous, $f(0)>0$, and $f(1)=0$. Prove that there is a number $x_0 \in (0,1]$ such that $f(x_0)=0$ and $f(x)>0$ for $0\le x < x_0$; that is, there is a smallest point in the interval $[0,1]$ at which the function $f$ attains the value $0$. 

Since $f$ is continuous on a closed an bounded interval, $[0,1]$, then by the extreme value theorem, $f$ attains both a minimum and a maximum value. The minimum value will obviously be $0$, but I'm not really sure where to go after stating this. Any suggestions?
 A: The existence of a root is apparent in the region $(0,1]$ as 1 is a root.  So, now that we know the roots exist, we can take the least root in the set $S:=\{x_i\mid f(x_i)=0\}$ and call that element $x_0$, since the set of roots of a continuous function is closed.  Since, for all $x\in(0,x_0)$, $f(x)\not=0$, $f(0)=1$, $f(x_0)=0$, and $f$ is continuous, all values between $x_0$ and $0$ must be greater than 0.
A: For a proof by contradiction, assume:
$$\forall x_0 \in [0,1] \,| \, f(x_0) = 0, \exists x_1 \in [0,1]\,|\, f(x_1) = 0, x_1 < x_0$$
Then, since we are given  $f$ is continuous on $[0,1]$ and $f(0) > 0$, it should be relatively simple to show a contradiction exists.
A: If there is no smallest root,
then, for any $c > 0$,
there is a root $r$
such that
$0 < r < c$.
Use this to prove that
$f$ is not continuous at zero.
A: Starting from the hypothesis that $f:[0,1]\rightarrow \mathbb{R}$ is a continuous function (on a compact set $[0,1]$) satisfying $f(0)>0$ and $f(1)=0$ one has in particular that there exist $c,d \in [0,1]$ such that
$\displaystyle f(c)=\inf_{x\in [0,1]} f(x)$ and $\displaystyle f(d)=\sup_{x \in [0,1]}f(x)$.
In particular, one has the flag of inequalities 
$$f(c)\leq f(1)=0<f(0)\leq f(d).$$

The proof that $f(x_0)=0$ for some $0\leq x_0<1$ is a direct consequence for the intermediate value theorem for continuous functions. Indeed, if $f(c)=0$ then $x_0$ is a root of the equation $f(x)$ on the interval $[0,x_0]$. Otherwise, one has $f(c)<0<f(0)$ together with the continuity condition assures the existence of $x_0$ such that $f(x_0)=0$ fulfils on the open interval $(0,c)$.

The proof that the inequality $f(x)>0$ holds on the interval $0\leq x<x_0$ is a natural consequence of the continuity of the function on the point $0$.
In concrete, for every $\varepsilon>0$ it is possible to find a $\delta>0$ such that for every $x \in (-\delta,\delta)\cap [0,x_0)$ there holds
$$ -\varepsilon<f(x)-f(0)<\varepsilon.$$
In particular, for the choice $\varepsilon=f(0)$ we then have that $f(x)>0$ in a neighborhood $(0,\delta)$ of the interval $[0,x_0)$ ($\delta\leq x_0$). This proves that $f(x)>0$ on $(0,x_0)$. Thereby $f(x)>0$ on $[0,x_0)$, as desired.
