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?
Best Answer
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.