Problem 28 - Calculus of a Single Variable

Question

Finding Extrema on a Closed Interval In Exercises \(17-36\) , find the absolute extrema of the function on the closed interval. $$ h(t)=\frac{t}{t+3},[-1,6] $$

Step-by-Step Solution

Answer

The maximum value of the function \(h(t)\) over the interval \([-1,6]\) is \(\frac{2}{3}\) at point \(6\) and the minimum value is \(-\frac{1}{2}\) at point \(-1\).

Step 1: Find the derivative of the function

The derivative of the function \(h(t)\) is obtained to determine critical points where the function may achieve a maximum or minimum value. Using the quotient rule for differentiation, we find that the derivative is \(h'(t)=\frac{3}{(t+3)^2}\)

Step 2: Find the critical points

Critical points are values at which the derivative does not exist or equals zero. However, for this function, \(h'(t)\) is always positive for \(t != -3\) and does not exist at \(t = -3\). Therefore, there are no critical points within the given closed interval \([-1,6]\).

Step 3: Evaluate the function at the endpoints

By substituting the end values of the interval, \(-1\) and \(6\), into the original function, the values at these points can be determined. Thus, \(h(-1)=-\frac{1}{2}\) and \(h(6)=\frac{6}{9} = \frac{2}{3}\).

Step 4: Compare the function values

By comparing the function values calculated, \(h(-1)\) is the lowest and \(h(6)\) is the greatest. Therefore, \(-\frac{1}{2}\) and \(\frac{2}{3}\) are the absolute minimum and maximum values respectively for the function \(h(t)\) over the interval \([-1,6]\).

Key Concepts

Closed Interval

When finding extrema of a function, we often consider a closed interval. A closed interval, like [-1, 6], includes its endpoints. In mathematical notation, this is shown with square brackets.
This means:

• Every point from -1 to 6 is included.
• We pay special attention to the endpoints -1 and 6.

Evaluating the behavior of a function on this interval is crucial since extrema can occur at endpoints or critical points within.

Critical Points

Critical points help us find where extrema might occur. They are points where the function's derivative is zero or undefined. Checking these points tells us where the function's slope is flat or changing direction.
For this function:

• The derivative is always positive and not zero, so no critical points where the derivative equals zero.
• There’s a point, -3, where the derivative is undefined, but it doesn't affect our interval [-1,6].

In this specific exercise, endpoints hold more significance as there are no interior critical points in the interval.

Derivative

Understanding derivatives is key to analyzing a function's rate of change. The derivative represents the slope at any given point. For the function \(h(t) = \frac{t}{t+3}\), we use derivatives to locate potential extrema.To find the derivative:

• Apply the quotient rule.
• Determine that \(h'(t) = \frac{3}{(t+3)^2}\).

This derivative helps identify the function's behavior. Here, since \(h'(t)\) is always positive, the function is always increasing on the given interval.

Quotient Rule

The quotient rule is a technique for finding the derivative of a quotient of two functions. Given two functions, \(u(t)\) and \(v(t)\), where \(h(t) = \frac{u(t)}{v(t)}\), the quotient rule is \[ h'(t) = \frac{u'(t)v(t) - u(t)v'(t)}{(v(t))^2} \]For \(h(t) = \frac{t}{t+3}\),

• \(u(t) = t\) with derivative \(u'(t) = 1\)
• \(v(t) = t+3\) with derivative \(v'(t) = 1\)

Using the formula, we get \(h'(t) = \frac{3}{(t+3)^2}\). This simple yet powerful rule aids in solving problems of finding extrema by deriving a function's slope effectively.