Answer
$(-\infty,-2)\cup(-2,1)\cup(1,\infty)$
See graphs
Work Step by Step
We have:
$$\begin{align*}
|x-1|=\begin{cases}
-x+1,&x\leq 1\\
x-1,&x>1.
\end{cases}
\end{align*}$$
$$\begin{align*}
|x+2|=\begin{cases}
-x-2,&x\leq -2\\
x+2,&x>-2.
\end{cases}
\end{align*}$$
Rewrite the given function:
$$\begin{align*}
h(x)&=\begin{cases}
-x+1-x-2,&&x\leq -2\\
-x+1+x+2,&&-21
\end{cases}\\
&=\begin{cases}
-2x-1,&&x\leq -2\\
3,&&-21
\end{cases}
\end{align*}$$
Determine the first derivative $h'(x)$:
$$\begin{align*}
h'(x)&=\begin{cases}
-2,&&x< -2\\
0,&&-21
\end{cases}
\end{align*}$$
In the points $x=-2$ and $x=1$ the first derivative is not defined, so $f$ is differentiable on $(-\infty,-2)\cup(-2,1)\cup(1,\infty)$.
