## Calculus 10th Edition

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