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 $\forall x \in [a,b] x\sin(\frac1x) = 0$

It is not always true. Let $f(x)=x^2\sin(1/x)$ when $x\ne 0$, and let $f(0)=0$.
Note that $f(x)=0$ whenever $x=\frac{1}{n\pi}$, where $n$ is a non-zero integer. 
In this example, $f$ is differentiable at $0$ but the derivative is not continuous at $0$. We can fix that if we want by using $x^3\sin(1/x)$. 
Added: The question has been clarified. Perhaps it asks now whether we can have a function which is everywhere differentiable, is not identically $0$, but is identically $0$ in some interval. The answer is yes, and such functions are even useful.
Let $g(x)=e^{-1/x^2}$ when $x\gt 0$, and let $g(x)=0$ for $x\lt 0$. Then $g(x)$ is everywhere differentiable, infinitely often.
Using $g(x)$, we can construct a function $f(x)$ which is is everywhere differentiable infinitely often, is $0$ in the interval $[-1,1]$, and non-zero when $|x|\gt 1$. 
