Calculus 10th Edition

Over the specified interval, the function has an absolute maximum equal to $3$ and an absolute minimum equal to $-1.$
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 $-1.$