Problem 15 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find the absolute maximum value and the absolute minimum value, if any, of each function. $$ f(x)=x^{2}-2 x-3 \text { on }[-2,3] $$

Step-by-Step Solution

Answer

The absolute maximum value of the function is 5 at \(x = -2\) and the absolute minimum value of the function is -4 at \(x = 1\).

Step 1: Identify the function and interval

First, we are given the function: $$ f(x) = x^2 - 2x - 3 $$ and the closed interval: $$ [-2, 3] $$

Step 2: Find the first derivative of the function

We need to find the critical points of this function. Critical points occur where the first derivative is equal to 0 or is undefined. The derivative of \(f(x) = x^2 - 2x - 3\) using the power rule is: $$ f'(x) = 2x - 2 $$

Step 3: Find the critical points

To find the critical points of the function, set the first derivative equal to 0 and solve for x: $$ 2x - 2 = 0 $$ Adding 2 to both sides and then dividing by 2, we get the critical point: $$ x = 1 $$

Step 4: Evaluate the function at the endpoints and critical points

Now, we need to find the value of the function at the critical point \(x = 1\), as well as at the endpoints of the interval, \(x = -2\) and \(x = 3\). For \(x = -2\): $$ f(-2) = (-2)^2 - 2(-2) - 3 = 4 + 4 - 3 = 5 $$ For \(x = 1\): $$ f(1) = (1)^2 - 2(1) - 3 = 1 - 2 - 3 = -4 $$ For \(x = 3\): $$ f(3) = (3)^2 - 2(3) - 3 = 9 - 6 - 3 = 0 $$

Step 5: Identify the absolute minimum and maximum values

By comparing the values of the function at the critical point and the endpoints, we can determine the absolute minimum and maximum values. The highest value is \(f(-2) = 5\), so the absolute maximum value is 5 and occurs at \(x = -2\). The lowest value is \(f(1) = -4\), so the absolute minimum value is -4 and occurs at \(x = 1\). In conclusion, the absolute maximum value of the function is 5 at \(x = -2\), and the absolute minimum value of the function is -4 at \(x = 1\).

Key Concepts

Understanding Critical Points

Critical points are specific points on the graph of a function where the function's derivative is either zero or undefined. These points are significant because they indicate where a function might have a local maximum, minimum, or a point of inflection.
This means that the function changes its direction of slope at critical points.

• If the derivative is zero at a point, the graph of the function has a horizontal tangent line at that point.
• If the derivative is undefined, then the tangent line at that point could be vertical, or the nature of the derivative changes abruptly.

For our problem, the derivative of the function, \(f'(x) = 2x - 2\), is zero at \(x = 1\), which makes \(x = 1\) a critical point. This calculation is fundamental to evaluating potential points of interest for maximum and minimum values within a given interval.

Finding the Absolute Maximum

The absolute maximum of a function on a closed interval is the highest value that the function attains at any point within that interval. This includes endpoints as well as any critical points within the interval. To identify the absolute maximum:

• Evaluate the function at each critical point.
• Evaluate the function at the endpoints of the interval.
• Compare these values to see which is the largest.

For the function \(f(x) = x^2 - 2x - 3\) evaluated on the interval \([-2, 3]\), the calculations show that the maximum value of the function is 5 at \(x = -2\). Comparing \(f(-2) = 5\), \(f(1) = -4\), and \(f(3) = 0\), 5 is the largest.

Locating the Absolute Minimum

Similarly, the absolute minimum of a function within a closed interval is the smallest value that the function reaches on that interval, again considering both endpoints and critical points. To find this:

• Evaluate the function at all critical points and the interval endpoints.
• Identify the smallest value from these evaluations.

In our problem, we calculated \(f(x)\) at \(x = 1\), \(x = -2\), and \(x = 3\) and found the smallest value to be \(-4\) at \(x = 1\). This means that the absolute minimum value of the function on the interval \([-2, 3]\) is \(-4\).

Understanding Derivatives

A derivative represents the rate of change or slope of a function at any given point. In calculus, finding the derivative of a function is a crucial step in identifying critical points where maximum and minimum values may occur.
To find the derivative of \(f(x) = x^2 - 2x - 3\), we apply the power rule within differentiation.

• The derivative of \(x^2\) is \(2x\).
• The derivative of \(-2x\) is \(-2\).
• The derivative of a constant, like \(-3\), is \(0\).

By summing these, the derivative is \(f'(x) = 2x - 2\). This function helps us in determining critical points by setting it to zero and solving for \(x\). Understanding how to take derivatives is essential for solving problems involving slopes and optimization.

Evaluating the Function on Interval

Interval evaluation is a process used in calculus to determine the values at which a function achieves its extreme points, such as absolute maximum and minimum values. It involves:

• Assessing the function at all endpoints of the given interval.
• Evaluating the function at all critical points within the interval.
• Comparing these values to find the least and greatest among them.

In our example for \([-2, 3]\), we evaluated \(f(x)\) at both the endpoints and the critical point. These checks guarantee that all possible maximum and minimum values within that interval have been considered.