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

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

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

$\"\\frac{d\\$

Quotient rule of derivatives : $\"\"$ .

$\"\\\\h\'(t)=\\frac{(t-2)(1)-t(1)}{(t-2)^2}\\\\$

To find the critical numbers of $\"h(t)\"$ , equate $\"h\'(t)\"$ to zero.

$\"\\\\h\'(t)=0\\\\$

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

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

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

$\"\"$

Left end point $\"(3,\\$ is maximum point over the interval $\"[$ .

Right end point $\"\\left$ is minimum point over the interval $\"[$ .

$\"\"$

Minimum point is $\"\\left$ .

Maximum point is $\"(3,\\$ .