Answer
Over the specified interval, the function has an absolute maximum equal to $3$ and an absolute minimum equal to $-1.$
Work Step by Step
Using the rule ($\dfrac{d}{dx}|z(x)|=z'(x)\times\dfrac{|z(x)|}{z(x)}$) $\to$
$y'=-\dfrac{|t-3|}{t-3}.$
$y'$ is undefined for $t=3$ which is in the specified interval; hence, $t=3$, along with the interval's endpoints, are possible candidates for an absolute extremum.
$y(-1)=-1.$
$y(3)=3.$
$y(5)=1.$
Over the specified interval, the function has an absolute maximum equal to $3$ and an absolute minimum equal to $-1.$
