Problem 24 - Essential Calculus

Question

Locate the absolute extrema of the function on the closed interval. $$ h(t)=\frac{t}{t-2},[3,5] $$

Step-by-Step Solution

Answer

The absolute maximum of \( h(t) = \frac{t}{t-2} \) on the interval [3,5] is 3, which occurs at \( t = 3 \), and the absolute minimum is \( \frac{5}{3} \), which occurs at \( t = 5 \).

Step 1: Determine the derivative

Before identifying critical points, the derivative of the function \( h(t)=\frac{t}{t-2} \) needs to be found. We will use the quotient rule. The derivative function \( h'(t) \) is given as \( h'(t) = \frac{ (t-2) - t}{(t-2)^2} = \frac{-2}{(t-2)^2} \).

Step 2: Identify critical points

To find the critical points, we need to set the derivative equal to zero and solve for \( t \). However, since the numerator of \( h'(t) \) is a constant (-2), there are no critical points (since -2 never equals 0). Therefore, the only possible points where the maximum or minimum values can occur are at the endpoints of the interval, which are \( t = 3 \) and \( t=5 \).

Step 3: Calculate function values at endpoints

Substitute the endpoint values \( t = 3 \) and \( t = 5 \) into the original function \( h(t) \) to get \( h(3) = \frac{3}{3-2} = 3 \) and \( h(5) = \frac{5}{5-2} = \frac{5}{3} \) respectively.

Step 4: Identify the extrema

Compare the function values at the endpoints \( t = 3 \) and \( t = 5 \) to determine the maximum and minimum values. Here, \( h(3) = 3 \) is greater than \( h(5) = \frac{5}{3} \), thus the maximum value is 3, which occurs at \( t = 3 \) and the minimum value is \( \frac{5}{3} \), occurring at \( t = 5 \)

Key Concepts

Quotient Rule

The quotient rule in calculus is a technique used to differentiate functions that are the ratio of two other functions. If you have a function represented as a division, like the one in our exercise, it becomes essential to apply the quotient rule to find its derivative. The rule states that if you have a function in the form of \( \frac{u(t)}{v(t)} \), where \( u(t) \) and \( v(t) \) are differentiable functions, the derivative \( h'(t) \) is:

• \( h'(t) = \frac{u'(t)v(t) - u(t)v'(t)}{[v(t)]^2} \)

To apply the quotient rule correctly:

• First, determine \( u(t) \) and \( v(t) \). In our function \( h(t)=\frac{t}{t-2} \), \( u(t) = t \) and \( v(t) = t-2 \).
• Next, find the derivatives \( u'(t) \) and \( v'(t) \). For this example, \( u'(t) = 1 \) and \( v'(t) = 1 \).
• Substitute these into the quotient rule formula to get the derivative of the function.

This method helps us find how the function changes at any given point, leading to critical insights into its behavior.

Critical Points

Critical points are key in understanding the maximum and minimum values that a function can attain. These points occur where the derivative of the function is zero or undefined. In finding critical points, the first crucial step is to derive the function. In our problem, after applying the quotient rule, we determined the derivative was \( h'(t) = \frac{-2}{(t-2)^2} \).
The next step is to set this derivative equal to zero to find possible points of extremum. However, when the derivative has a constant numerator like -2, it implies there are no values of \( t \) that satisfy this equation because -2 will never equal zero.
Thus, in the absence of critical points in our interval \([3,5]\), the endpoints become our only focus. It's crucial to not overlook endpoints in closed intervals, as they could contain our extrema.

Closed Interval

A closed interval is a set of numbers that includes all numbers between two endpoints, and it also contains the endpoints themselves. Mathematically, this is expressed using square brackets, like \([3, 5]\), which includes 3 and 5.
When analyzing absolute extrema, especially for functions on closed intervals, always check the values of the function at these endpoints. This is important because, even if no critical points exist within the interval, the highest or lowest function values may still occur at its ends.

• In this exercise, since no critical points existed within the interval, the extrema had to be at the endpoints \( t=3 \) and \( t=5 \).
• Evaluating \( h(t) \) at these endpoints, we found \( h(3) = 3 \) and \( h(5) = \frac{5}{3} \).

Understanding closed intervals ensures that every possible option for extrema is considered, maximizing accuracy in identifying the absolute minimum and maximum of a function.