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 $\forall x \in [a,b] x\sin(\frac1x) = 0$
 A: 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 $|x|\gt 1$. 
A: 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 $[0,1]$.  
A: If I have understood your question correctly, then this is wrong. Counter-example: $\sin(x)$. It is non-constant, differentiable everywhere, and is $0$ at $x_0=0$. Your question apparently asks us to prove that this is the only place where the function is zero, which in fact is false, because $f(x)=0 \quad\forall x=n\pi ,n\in \mathbb N$.
EDIT: Following the edit, the question now seems to be that is it possible to find $f(x)$ such that $f(x)$ is zero for $x_0$ and also zero for some $x\in[x_0 - \epsilon,x_0+\epsilon]$ for arbitrarily small $\epsilon$. Clearly yes, since 
$$f(x)=\begin{cases}
x^2\cdot\sin(\frac{1}{x})&x\ne0\\
0&x=0\\
\end{cases}$$
fits the condition, as many people have observed.
EDIT 2:
It seems the OP's question is "Is it possible that a non-constant function be zero on an entire interval?"
Indeed it is. Consider the function 
$$f(x)=\begin{cases}
(x+1)^2&x\lt-1\\
0&x \in [-1,1]\\
(x-1)^2&x\gt1\\
\end{cases}$$
This function is non constant, differentiable everywhere, and is zero $\forall x\in [-1,1]$ 
A: No, because of bolzano, it only tells that exist one solution but maybe it can have more for example sin(x) is a diferentiable function and you ha 0 is a root but 180 are root too.
