The function is $\"h(s)=\\frac{1}{s-2}\"$ and the interval is $\"[$ .

$\"h(s)=\\frac{1}{s-2}\"$

Apply derivative on each side with respect to $\"s\"$ .

$\"\\frac{d\\$

$\"h\'(s)=\\frac{-1}{(s-2)^2}\"$

To find the critical numbers of $\"h(s)\"$ , equate $\"h\'(s)\"$ to zero.

$\"\\\\h\'(s)=0\\\\$

The function $\"h(s)\"$ has critical value at $\"s=2\"$ .

$\"s=2\"$ is not in the region $\"[$ .

To find the absolute extreme of function in given region $\"[$ , sketch the function .

$\"\"$

Left end point $\"(0,\\$ is maximum point over the interval $\"[$ .

Right end point $\"(1,\\$ is minimum point over the interval $\"[$ .

$\"\"$

Minimum point is $\"(1,\\$ .

Maximum point is $\"(0,\\$ .