Given $f(x) = x + |x|$ for what values of $x$ is $f$ differentiable 
Problem : Given $f(x) = x + |x|$ for what values of $x$ is $f$ differentiable?

For the sake of generality, let's assume that it is unknown to us that $|x|$ is not differentiable at $x = 0$
Attempted Solution :
Using the definition of differentiability, a function is differentiable over an interval $I$ $\text{iff}$
$$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}, \ \ \forall x\in I$$
Now implicitly this definition of differentiability requires $$\lim_{h\to 0^+} \frac{f(x+h)-f(x)}{h} = \lim_{h\to 0^-} \frac{f(x+h)-f(x)}{h}$$
Therefore $f$ will be differentiable only when
$$\lim_{h\to 0^+} \frac{(x + h + |x+h|)-(x + |x|)}{h} = \lim_{h\to 0^-} \frac{(x - h + |x-h|)-(x + |x|)}{h}$$
But it is unclear what to do next as $x$ is just arbitrary
$$
\begin{equation}
\begin{aligned}
|x+h| &= \begin{cases}x+h & \text{if} & x \geq -h\\ -x-h & \text{if} & x < -h \end{cases} \\
&= \begin{cases}x+h & \text{if} & x \geq 0\\ -x-h & \text{if} & x < 0 \end{cases}
\end{aligned}
\end{equation}$$

A Wrong Solution :
We could take the derivative of $f$, (and evaluate the domain of the derivative $f'(x)$ to values for which it is defined), using the rules of differentiation, but we would get a wrong answer.
$$f'(x) = \frac{|x| + x}{|x|}$$
$$\implies f'(x) = \begin{cases}
2 & \text{if} & x \geq 0 \\
0 & \text{if} & x < 0
\end{cases}$$
This implies the derivative $f'(0)$ exists, when it does not as per the definition of differentiability. Why is that so?

Questions:


*

*How can a solution be found using the definition of differentiablity?

*Why does taking the derivative of $f$, $f'(x)$ and evaluating it's domain, not give the correct values of $x$ for which $f$ is differentiable?

 A: 
Given $f(x) = x + |x|$ for what values of $x$ is $f$ differentiable?

One may observe that
$$
 f(x) =
    \begin{cases}
2x  & \text{if $x\geq 0$,} \\[2ex]
0 & \text{if $x<0$.}
\end{cases}
$$ Then one may apply the definition of differentiability, obtaining easily that $f$ is differentiable over $(-\infty,0)\cup (0,\infty)$.
At $x=0$, one has, as $h \to 0$,
$$
\frac{f(h)-f(0)}{h}=
    \begin{cases}
2  & \text{if $h>0$,} \\[2ex]
0 & \text{if $h<0$,}
\end{cases}
$$ and the given function is not differentiable at $0$.
A: The best thing to do is to simplify the function before you use the definition of the derivative. For $x\geq 0$, $f(x)=x+|x|=x+x=2x$, while, for $x <0$, $f(x)=x-x=0$. Hence, we have the following:
$$
\begin{aligned}
f(x)=
\begin{cases}
2x &\text{ if }x\geq 0\\
0 &\text{ if }x < 0 
\end{cases}.
\end{aligned}
$$
From this, we can see right away that $f$ is differentiable for all $x\not=0$. So, we only need to focus on $x=0$.
$$
\lim_{h\to 0^{-}}\frac{f(0+h)-f(0)}{h}=\lim_{h\to 0^{-}}\frac{f(h)}{h}=\lim_{h\to 0^{-}}\frac{0}{h}=0,
$$
while
$$
\lim_{h\to 0^{+}}\frac{f(0+h)-f(0)}{h}=\lim_{h\to 0^{+}}\frac{f(h)}{h}=\lim_{h\to 0^{+}}\frac{2h}{h}=2.
$$
Since these limits aren't equal, $f$ is not differentiable at $0$.
For your second question: you cannot take the derivative using the usual rules since $|x|$ itself is not differentiable at $0$.
A: If $x\lt 0$ then $f(x)=0$ and if $x\ge 0$ then $f(x)=2x$. Thus the function $f$ is differentiable in all points excepting for $x=0$ where, it is clear, there is not a well defined tangent. Another way is take $f$ as being the sum of $x$ plus $|x|$ and it is well known that $|x|$ is not differentiable in $x=0$
A: $f$ is differentiable in $a \in \mathbb{R}$ if and only if $f-\text{id}$ is differentiable in $a$, so $f$ is differentiable in $a$ if and only if the absolute value function is differentiable in $a$. Now the absolute value function is differentiable in $a$ for $a \neq 0$, basically because if $a>0$, to get $$|\frac{|a+h|-|a|}{h}-1|$$
small, one chooses $|h|$ so small that $a+h$ is positive for all $h$ that small (an analogous argument applies to $a<0$). This trick cannot be done for $a=0$ of course, and one easily checks that 
$$\lim_{h \to 0^+} \frac{|a+h|-|a|}{h}=1 \neq -1 = \lim_{h \to 0^-} \frac{|a+h|-|a|}{h}, $$
so that the limit does not exist for $a =0$.
