Problem 21 - Thomas Calculus in SI Units

Question

Find the absolute maximum and minimum values of each function on the given interval. Then graph the function. Identify the points on the graph where the absolute extrema occur, and include their coordinates. $$f(x)=\frac{2}{3} x-5, \quad-2 \leq x \leq 3$$

Step-by-Step Solution

Answer

Absolute maximum: -3 at \(x=3\), Absolute minimum: \(-\frac{19}{3}\) at \(x=-2\).

Step 1: Determine the Critical Points

Since the function \(f(x) = \frac{2}{3}x - 5\) is a linear function, it has a constant slope and therefore no critical points where the derivative equals zero. We proceed by analyzing the endpoints of the interval.

Step 2: Evaluate the Function at the Endpoints

To find the absolute maximum and minimum, evaluate the function at the endpoints of the interval. Calculating at \(x = -2\) and \(x = 3\) gives:\[f(-2) = \frac{2}{3}(-2) - 5 = -\frac{4}{3} - 5 = -\frac{19}{3}\]\[f(3) = \frac{2}{3}(3) - 5 = 2 - 5 = -3\]Thus, \(f(-2) = -\frac{19}{3}\) and \(f(3) = -3\).

Step 3: Compare and Identify Extrema

Compare the values of \(f(x)\) at the endpoints: \(-\frac{19}{3}\) and \(-3\).The absolute maximum value of the function in this interval is \(-3\) at \(x=3\), and the absolute minimum value is \(-\frac{19}{3}\) at \(x=-2\).

Step 4: Graph the Function and Identify Points

Graph \(f(x) = \frac{2}{3}x - 5\) on the interval \([-2, 3]\). Mark the points where the extrema occur:- Point \((-2, -\frac{19}{3})\) for the minimum value.- Point \((3, -3)\) for the maximum value.By plotting these on the graph, you can clearly see the linear nature of the function and the extrema at the endpoints provided.

Key Concepts

Linear Functions

Linear functions are fundamental in understanding various mathematical concepts. A linear function can be generally expressed as \(f(x) = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept. These functions are characterized by a constant rate of change, which means the slope \(m\) is the same at every point on the graph.
Linear functions graph as straight lines, making them relatively easy to analyze. They do not have turns or bends, and as a result, do not possess critical points where the derivative is zero. This consistency simplifies the process of finding absolute extrema, as we'll see by focusing on endpoints of a given interval.

Interval Evaluation

Interval evaluation is crucial for determining absolute extrema of functions, especially linear ones. In this context, it's about assessing the function's value strictly within a finite interval rather than the entire set of real numbers.
For linear functions, since there are no critical points to consider, we evaluate the function at the given interval's endpoints. For example, in the function \(f(x) = \frac{2}{3}x - 5\) and interval \([-2, 3]\), the values of the function are calculated at \(x = -2\) and \(x = 3\).
This endpoint evaluation gives us critical insight into identifying where absolute maximum and minimum values occur by comparing the function's outputs at these specific points.

Graphing Functions

Graphing functions is a visual means to understand the behavior of expressions over intervals. For linear functions like \(f(x) = \frac{2}{3}x - 5\), graphing can be straightforward, requiring just two points to determine the entire line.
To graph on the interval \([-2,3]\), start by plotting the calculated endpoints:

• \((-2, -\frac{19}{3})\)
• \((3, -3)\)

Connect these points with a straight line. The line's gradient indicates how the function increases or decreases, emphasizing its linearity.
Through the graph, you can visually confirm the positions of the absolute extrema at the specified points - validating calculations made about endpoint evaluations.

Endpoint Analysis

Endpoint analysis involves the investigation of a function's behavior at the extremes of its interval. This method is particularly important for linear functions, where extrema can occur only at these boundaries.
For \(f(x) = \frac{2}{3}x - 5\) within \([-2, 3]\), endpoint analysis shows:

• At \(-2\), the function reaches its absolute minimum value of \(-\frac{19}{3}\).
• At \(3\), it attains its absolute maximum value of \(-3\).

The absence of internal critical points shifts the focus solely to these boundaries. With endpoint analysis, you confirm the function's limits and overall comprehension of absolute extrema in the given interval.