#### Answer

Over the specified interval, the function has an absolute maximum equal to $5$ and an absolute minimum equal to $0.$

#### Work Step by Step

Using the rule $\dfrac{d}{dx}|z(x)|=z'(x)\times\dfrac{|z(x)|}{z(x)}\to$
$g'(x)=\dfrac{|x+4|}{x+4}$ which is undefined for $x=-4.$
The interval's boundaries and $x=-4$ are possible candidates for absolute extrema.
$g(-7)=3.$
$g(-4)=0.$
$g(1)=5.$
Over the specified interval, the function has an absolute maximum equal to $5$ and an absolute minimum equal to $0.$