Making an absolute value function differentiable 
If I define a function as follows:
$$ f(x) = |x| , \: \text{if x $\neq$ 0}, \: \\ \text{and} \\ f(x) = 0
 , \: \text{if $x = 0$}. \\ $$
Is $f$ differentiable?

I think it is. The absolute value function is differentiable everywhere except from $x=0$. The way $f$ is defined above, however, its derivative at $x=0$ is $f' = 0$ and thereby $f$ is differentiable. 
Am I mistaken?
 A: What you have defined is the absolute value function, since $\lvert 0 \rvert = 0$. But the absolute value is not differentiable at $x = 0$, see for instance here.
A: You have some answers already, but let me just point out what might be a source of confusion: if a function is defined piecewise, it can be differentiated piecewise, except at the endpoints which must be checked individually.
For instance, your function is
$$
    f(x) = |x| = \begin{cases} x & x > 0 \\ -x & x < 0 \\ 0 & x = 0\end{cases}
$$
It's tempting to say, “Well, $x$, $-x$, and $0$ are all differentiable functions, so
$$
\color{red}{
    f'(x) = \begin{cases} 1 & x > 0 \\ -1 & x < 0 \\ 0 & x = 0\end{cases}
}
$$
and therefore $f$ is differentiable for all $x$.”
But actually, we can only differentiate on the interior of the pieces.  That is,
$$
    f'(x) = \begin{cases} 1 & x > 0 \\ -1 & x < 0 \end{cases}
$$
is true.  But to check whether $f$ is differentiable at $0$, we need to decide if
$$
    \lim_{h\to 0} \frac{f(0+h) - f(0)}{h} = \lim_{h\to 0} \frac{|h|}{h}
$$
exists.  In fact, this limit does not exist.
The moral of the story is that differentiability requires checking not just a point, but a neighborhood around that point.  This is why when it can fail to be differentiable at the transition points of piecewise-defined functions.
A: For $t \ne 0$ we have $\frac{f(t)-f(0)}{t-0}= \frac{|t|}{t}.$
Hence $\lim_{t \to 0+0}\frac{f(t)-f(0)}{t-0}=1$ and $\lim_{t \to 0-0}\frac{f(t)-f(0)}{t-0}=-1$.
Consequence: $f$ is not differentiable at $0.$
