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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 $0le x < x_0$; 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?

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.

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 $S:={x_imid f(x_i)=0}$ and call that element $x_0$, since the set of roots of a continuous function is closed. Since, for all $xin(0,x_0)$, $f(x)not=0$, $f(0)=1$, $f(x_0)=0$, and $f$ is continuous, all values between $x_0$ and $0$ must be greater than 0.

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.

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
$displaystyle f(c)=inf_{xin [0,1]} f(x)$ 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 $0leq x_0<1$ 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 $0leq x<x_0$ 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 $[0,x_0)$ ($deltaleq x_0$). This proves that $f(x)>0$ on $(0,x_0)$. Thereby $f(x)>0$ on $[0,x_0)$, as desired.