Problem 16 - 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. $$ g(x)=x^{2}-2 x-3 \text { on }[0,4] $$

Step-by-Step Solution

Answer

The absolute maximum value of the function \(g(x) = x^2 - 2x - 3\) on the interval [0, 4] is \(g(4) = 5\), and the absolute minimum value is \(g(1) = -4\).

Step 1: Compute the first derivative of the function.

First, let's find the first derivative of the given function: \(g(x) = x^2 - 2x - 3\) Differentiate both sides of the equation with respect to x: \(g'(x) = \frac{d}{dx}(x^2) - \frac{d}{dx}(2x) - \frac{d}{dx}(3) = 2x - 2\)

Step 2: Identify critical points of the function.

A critical point of the function is a point where the first derivative of the function is either 0 or undefined. In this case, the first derivative is a linear function and is defined everywhere, so we only need to find the points where the first derivative is equal to 0: \(2x - 2 = 0\) Solve for x to find the critical point(s): \(x = 1\)

Step 3: Compute the function values at the critical points and the endpoints of the interval.

Next, we need to compare the values of the function at the critical point(s) and the endpoints of the given interval, in order to determine the absolute maximum and minimum values. Calculate the function values at these points: - \(g(0) = (0)^2 - 2(0) - 3 = -3\) - \(g(1) = (1)^2 - 2(1) - 3 = -4\) - \(g(4) = (4)^2 - 2(4) - 3 = 5\)

Step 4: Determine the absolute maximum and minimum values on the interval [0, 4].

Based on our calculations in the previous step, we can observe the following: - The absolute maximum value of the function on the interval [0, 4] is 5, which occurs at \(x = 4\). - The absolute minimum value of the function on the interval [0, 4] is -4, which occurs at \(x = 1\). Therefore, the absolute maximum value of the function is \(g(4)= 5\), and the absolute minimum value of the function is \(g(1) = -4\).

Key Concepts

Critical Points

When working with a function, critical points are essential to understand because they offer key insights into the behavior of the function. Let me break it down for you. A critical point occurs where the derivative of a function is zero or undefined. These points are vital because they can help identify where the function reaches a local maximum or minimum. Once you find the derivative, simply set it to zero to find these critical points.

For instance, with our function \(g(x) = x^2 - 2x - 3\), we first found the derivative to be \(g'(x) = 2x - 2\). Setting this derivative equal to zero gives us \(2x - 2 = 0\). Solve for \(x\), and we find a critical point at \(x=1\).

This critical point is one potential location for an absolute extremum (a max or min), but remember, in calculus optimization, checking these points along with the interval boundaries is essential.

Derivative

The derivative is a crucial tool in calculus. It represents the rate of change of a function. When dealing with optimization problems, derivatives help find both critical points and, subsequently, maximum and minimum values.
In basic terms, the derivative tells us how the function behaves:

• When the derivative is positive, the function is increasing.
• When the derivative is negative, the function is decreasing.
• When the derivative equals zero, it may indicate a maximum, minimum, or even a saddle point.

To calculate the derivative of a polynomial like \(g(x) = x^2 - 2x - 3\), apply standard differentiation rules. By differentiating each term, we find \(g'(x) = 2x - 2\). This derivative clearly outlines the slope behavior of the original function \(g(x)\).
Understanding derivatives enables you to dive deeper into function analysis, making them invaluable for solving extremum problems. Once you have the derivative, you can explore its zero points, like in our example, where solving \(2x - 2 = 0\) gave us \(x = 1\).

Absolute Maximum and Minimum Values

Finding absolute maximum and minimum values is a central task in optimization. When dealing with continuous functions over a closed interval, these values give us the highest and lowest points the function can reach within the interval.
For example, given an interval like \([0, 4]\) and a function such as \(g(x) = x^2 - 2x - 3\), the absolute maximum or minimum can occur at critical points or at the endpoints of the interval. Here’s what you do:

• Calculate the function values at all critical points from the derivative step.
• Compute the values of the function at the interval's endpoints.
• Compare all these values to determine the absolute extremes.

In our exercise, we computed \(g(0) = -3\), \(g(1) = -4\), and \(g(4) = 5\), leading us to conclude:

• The absolute maximum value is \(5\), located at \(x = 4\).
• The absolute minimum value is \(-4\), located at \(x = 1\).

This comparison helps ensure that you don't miss any possible extremum points, as the highest and lowest points carry significant insight into the function’s behavior over specific intervals.