Product of functions is $0$ but none of the functions is identically $0$? 
Given that $f(x) \cdot g(x) = 0$ for all $x$ is it true that at least one of the functions is $0$ for all $x$?

The correct answer is this doesn't necessarily hold true. Can you give such example? I have a feeling this had something to do with piecewise functions.
 A: As a further fun example, the functions
$$
f(x) = \begin{cases}\exp\left[\frac{1}{x(x-2)}\right] & 0 < x < 2 \\ 0 & \mathrm{else}\end{cases}\;\;\;\; ;\;\;\;\; g(x) = \begin{cases}\exp\left[\frac{1}{x(x+2)}\right] & -2 < x < 0 \\ 0 & \mathrm{else}\end{cases}
$$
satisfy $f(x)g(x) = 0$ for all real $x$ and are infinitely differentiable everywhere.
A: You can go even more basic, not worrying about continuity, holomorphicity or using modulus signs. Let $f,g : \{0,1\} \to \{0,1\}$ with
$$ f(0) = 0, f(1) = 1, g(0) = 1, g(1) = 0.$$
We then have $f(0)g(0) = 0*1 = 0 = 1*0 = f(1)g(1)$. (This is in essence the same as quasi's accepted answer, but giving an even more 'basic' function.)
A: $$f(x)\cdot g(x)=0\iff f(x)=0\lor g(x)=0.$$
Then take two distinct values $x_1,x_2$ and define the functions in such a way that $f(x_1)\ne0$ and $g(x_2)\ne0$ and zeroes everywhere else.
A: Inspired by Arthur's example, here's a differentiable example:

Let $f(x) = \text{max}(x^3,0)$ and let $g(x) = f(-x)$.
A: Let $f(x)$ be any function whose range is $\{0,1\}$, and let $g(x) = 1 - f(x)$.

For example, let $f$ be such that $f(0) = 1$ and $f(x) = 0$ for all $x \ne 0$.
A: One continuous example: $f(x)=|x|+x, g(x)=|x|-x$.
For any $x$, at least one of the functions must be $0$, but there is nothing stopping them from "sharing" the $x$-axis between them.
A: Rather than “piecewise” functions, this has to do with functions which are not analytic.
If $f$ and $g$ are analytic in a neighborhood $U$ of $x_0$, which means that their Taylor series at $x_0$ exist and converge to the functions in a neighborhood of $x_0$, then from $f(x)g(x)=0$ for $x\in U$, then either $f(x)=0$ for every $x\in U$ or $g(x)=0$ for every $x\in U$.
This is easy to prove in the context of holomorphic complex functions: if the set of zeros of a function which is holomorphic in an open connected set has an accumulation point, then the function is identically zero.
Thus all counterexamples to the assertion must not be analytic and the easiest way to produce nonanalytic functions is by using “piecewise definitions”.
A $C^{\infty}$ example:
$$
f(x)=\begin{cases}
0 & x\le 0 \\
e^{-1/x^2} & x>0
\end{cases}
\qquad
g(x)=\begin{cases}
e^{-1/x^2} & x<0 \\
0 & x\ge 0
\end{cases}
$$
The function
$$
F(x)=\begin{cases}
e^{-1/x^2} & x\ne0 \\
0 & x=0
\end{cases}
$$
is the prime example of a $C^{\infty}$ function which is not the sum of its Taylor series at $0$ (all derivatives at $0$ are $0$).
Of course, if you don't care about differentiability, simpler examples can be found:
$$
f(x)=x+|x|\qquad g(x)=x-|x|
$$
needs no "piecewise definition”.
