#### Answer

h'(x) = $\frac{|x+2|}{x+2}$ + $\frac{|x-1|}{x-1}$
On the graph below:
The black graph is h(x)
The blue graph is h'(x)

#### Work Step by Step

h(x) = |x-1| + |x+2| = $\Bigg\{
\begin{array}{l}
-(x-1) - (x+2) \quad for\thinspace x \leq -2\\
-(x-1)+(x+2) \quad for\thinspace x\thinspace-2\lt x\leq 1\\
+(x-1)+(x+2) \quad for\thinspace x\gt 1
\end{array}
$
h(x) = $\Bigg\{
\begin{array}{l}
-2x-1 \quad for\thinspace x \leq -2\\
\qquad 3 \qquad for\thinspace x\thinspace-2\lt x\leq 1\\
\thinspace2x+1 \qquad for\thinspace x\gt 1
\end{array}
$
Differentiate h(x) to get
h'(x) = $\Bigg\{
\begin{array}{l}
-2 \quad for\thinspace x \leq -2\\
\space 0 \quad for\thinspace x\thinspace-2\lt x\leq 1 \rightarrow \alpha\\
2 \quad for\thinspace x\gt 1
\end{array}
$
Note that h(x) has an edge at x = -2 and x = 1
Therefore, the derivate does not exist at x = -2 and x = 1
We are asked to give a formula for h'(x)
We already know h'(x) for diffrent values of x, now we need to find a formula which satisfies $\alpha$
You can come up with
h'(x) = $\frac{|x+2|}{x+2}$ + $\frac{|x-1|}{x-1}$