Interval around a root of a function I have a question that may seems stupid and obvious, but for me it's not. The question is the following:
Edit: I've explained bad my question
Suppose we have $f: \Bbb{R} \to \mathbb{R}$ a non constant continuous differentiable function on $\Bbb{R}$. Suppose also that we have $x_0 \in \Bbb{R}$ and that $f(x_0) = 0$. Is it true that $f$ is equal to $0$ only in $x_0$? 
This means: is it possible to find a non constant function that for some point $x_0$ is equal to $0$ and that is equal to $0$ also for the point in $[x_0 - \epsilon, x_0 + \epsilon]$?
2nd edit:
It seems everyone is thinking about the function $x \sin(\frac1x)$ but this function, tell me if I'm wrong, is equal to zero only for singular points. There doesn't exist an interval $[a,b]$ such that $\forall x \in [a,b] x\sin(\frac1x) = 0$

The function $g(x)=\begin{cases} e^{-1/x^2} & x<0\\0 & x\geq 0\end{cases}$ is differentiable (in fact, it is $C^{\infty}$). By properly shifting, you can get a differentiable $g$ such that $g(x)=0$ for all $x$ in some interval $[a,b]$. 
To compute explicitly, we can do the following:
$$f(x)=\begin{cases} e^{-1/x^2} & x<0 \\ 0 & x\in[0,1] \\ e^{-1/(x-1)^2} & x>1 \end{cases}$$ 
The function $f$ is differentiable ($C^{\infty}$, I'll leave it to you) and satisfies $f(x)=0$ on the interval $[0,1]$.  
