Find the absolute extrema of the function on the closed interval.

minimum
maximum

Ask by Ingram Gray. in the United States

Mar 14,2025

Question

Find the absolute extrema of the function on the closed interval. $$ h(t)=\frac{t}{t-4},[5,9] $$ minimum $$ \quad(t, h)=(9,1.8 $$ maximum $$ \quad(t, h)=(\square) $$

UpStudy AI · Accepted Answer

**Step 1. Evaluate the Function at the Endpoints**

The given function is 
$$
h(t)=\frac{t}{t-4}
$$
and the interval is $$[5,9]$$.

Evaluate at $$ t=5 $$:
$$
h(5)=\frac{5}{5-4}=\frac{5}{1}=5.
$$

Evaluate at $$ t=9 $$:
$$
h(9)=\frac{9}{9-4}=\frac{9}{5}=1.8.
$$

**Step 2. Find Critical Points**

To find critical points, we first calculate the derivative using the quotient rule. The derivative of $$ h(t) $$ is:
$$
h'(t)=\frac{(t-4)(1)-t(1)}{(t-4)^2}=\frac{t-4-t}{(t-4)^2}=\frac{-4}{(t-4)^2}.
$$

Since $$(t-4)^2$$ is always positive (except when $$ t=4 $$, but $$ t=4 $$ is not in our interval), the derivative does not equal zero:
$$
\frac{-4}{(t-4)^2}=0 \quad \text{has no solution.}
$$

Thus, there are no critical points in $$[5,9]$$.

**Step 3. Determine the Absolute Extrema**

Since there are no critical points in the interior of the interval, the absolute extrema must occur at the endpoints.

- At $$ t=5 $$, $$ h(5)=5 $$.
- At $$ t=9 $$, $$ h(9)=1.8 $$.

The maximum value is $$ 5 $$ at $$ t=5 $$ and the minimum value is $$ 1.8 $$ at $$ t=9 $$.

**Answer**

The absolute maximum is 
$$
(t, h)=(5, 5)
$$
and the absolute minimum is 
$$
(t, h)=(9, 1.8).
$$
The absolute maximum is at $$ (5, 5) $$ and the absolute minimum is at $$ (9, 1.8) $$.

Find the absolute extrema of the function on the closed interval.

minimum
maximum

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Find the absolute extrema of the function on the closed interval. minimum maximum