Answer
Over the specified interval, the function has an absolute maximum equal to $-\dfrac{1}{2}$ and an absolute minimum equal to $-1.$
Work Step by Step
$h'(s)=-\dfrac{1}{(s-2)^2}$
$h'(s)$ 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 $-1.$
