Answer
See the explanation
Work Step by Step
a.
$h(-2)=1$,
$h(0)=-1$,
$h(3)=4$,
b.
$D=\{x|-3\leq x\leq 4\}=[-3, 4]$
$R=[-1, 4]$
c.
$h(x)=3$ at $x=2$ and $x=-3$
d.
$h\leq3$ at $x\in [-3, 2]$
e.
$h(-3)=3$ and $h(3)=4$, thus
$\Delta h=h(3)-h(-3)=4-3=1$
