find the absolute extrema of the function on the closed interval.\n$h(t)=\\frac{t}{t - 1}$, $2,6$\nminimum…

Explanation:

Step1: Find the derivative of (h(t))

Use the quotient rule ((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}). Here (u = t), (u^\prime=1), (v=t - 1), (v^\prime = 1). [ \begin{align*} h^\prime(t)&=\frac{1\times(t - 1)-t\times1}{(t - 1)^{2}}\ &=\frac{t-1 - t}{(t - 1)^{2}}\ &=\frac{-1}{(t - 1)^{2}} \end{align*} ] Since (h^\prime(t)=\frac{-1}{(t - 1)^{2}}\lt0) for all (t\neq1) in the domain. The function (h(t)) is decreasing on the interval ([2,6]).

Step2: Evaluate the function at the endpoints

Evaluate (h(t)) at (t = 2) and (t=6). When (t = 2), (h(2)=\frac{2}{2 - 1}=2). When (t = 6), (h(6)=\frac{6}{6 - 1}=\frac{6}{5}=1.2).

Answer:

minimum ((t,h)=(6,\frac{6}{5})) maximum ((t,h)=(2,2))