Question

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

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 \(u = t\), \(u^\prime=1\), \(v=t - 1\), \(v^\prime = 1\).

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Since \(h^\prime(t)=\frac{-1}{(t - 1)^{2}}\lt0\) for all \(t
eq1\) 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 \((t,h)=(2,2)\)