## Calculus 10th Edition

Over the specified interval, the function has an absolute maximum equal to $-\dfrac{1}{2}$ and an absolute minimum equal to $-1.$
$h'(s)=-\dfrac{1}{(s-2)^2}$ $h'(s)$ is never equal to $0$ but is undefined for $s=2$ but since that is not in the specified interval. The only possible candidates for absolute extremum are the interval's endpoints. $h(0)=-\dfrac{1}{2}.$ $h(1)=-1.$ Over the specified interval, the function has an absolute maximum equal to $-\dfrac{1}{2}$ and an absolute minimum equal to $-1.$