Problem 19 - 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. $$ f(x)=x^{3}+3 x^{2}-1 \text { on }[-3,2] $$

Step-by-Step Solution

Answer

The absolute maximum value of the function \(f(x) = x^3 + 3x^2 -1\) on the interval \([-3, 2]\) is 19, which occurs at \(x = 2\), and the absolute minimum value is -1, which occurs at \(x = -3\) and \(x = 0\).

Step 1: Find the critical points

First, we need to find the first derivative of the function and set it equal to zero or undefined to find the critical points. The function is \(f(x) = x^3 + 3x^2 -1\). Differentiate the function with respect to x: \(f'(x) = 3x^2 + 6x\). Now we need to find the critical points by setting the first derivative equal to zero: \[3x^2 + 6x = 0\]

Step 2: Solve for the critical points

To find the critical points, we need to solve the equation: \[3x^2 + 6x = 0\] Factor out the common term 3x: \(3x(x + 2) = 0\) Now, set each factor equal to zero and solve for x: \(3x = 0 \Rightarrow x = 0\) \(x + 2 = 0 \Rightarrow x = -2\) So, the critical points are \(x = -2\) and \(x = 0\).

Step 3: Evaluate the function at critical points and endpoints

Now, we need to evaluate the function at the critical points and the endpoints of the interval to find the absolute maximum and minimum values. Evaluate the function at the critical points and endpoints: \(f(-3) = (-3)^3 + 3(-3)^2 -1 = -27 + 27 -1 = -1\) \(f(-2) = (-2)^3 + 3(-2)^2 -1 = -8 +12 -1 = 3\) \(f(0) = 0^3 + 3(0)^2 -1 = -1\) \(f(2) = 2^3 + 3(2)^2 -1 = 8 + 12 - 1 = 19\)

Step 4: Compare the values to determine the absolute maximum and minimum values

Now, we compare the values of the function to determine the absolute maximum and minimum values. \(f(-3) = -1\) \(f(-2) = 3\) \(f(0) = -1\) \(f(2) = 19\) The absolute maximum value is 19, which occurs at \(x = 2\), and the absolute minimum value is -1, which occurs at \(x = -3\) and \(x = 0\).

Key Concepts

Critical Points

Critical points are essential for finding where a function may reach its highest or lowest values. These occur where the derivative of the function is zero or undefined. Consider a function as a path taken by a hiker. The critical points represent the tops of hills or the bottoms of valleys along the path.

To find critical points, you first need to compute the first derivative of the function. Then, solve for when this derivative equals zero. This is because a zero derivative implies a flat slope - like reaching the peak of a hill or the bottom of a valley. For example, in our function, the derivative is given by:

• \[ f'(x) = 3x^2 + 6x \]

Set it equal to zero to find critical points:

• \[ 3x(x + 2) = 0 \]

Here, the critical points are at \( x = -2 \) and \( x = 0 \).

Don't confuse critical points with absolute max/min values yet. They simply indicate where these could potentially be. Further analysis is required.

Absolute Maximum and Minimum

Once critical points are found, you need to determine which ones actually represent the highest or lowest values of the function within a specific interval. This is the concept of absolute maximum and minimum.

To find an absolute maximum or minimum, you need to evaluate the original function at these critical points and at the endpoints of the interval. In our example, the interval is \([-3, 2]\). Hence, you assess the function at:

• Endpoints: \( x = -3 \) and \( x = 2 \)
• Critical Points: \( x = -2 \) and \( x = 0 \)

Calculating these, we find:

• \( f(-3) = -1 \)
• \( f(-2) = 3 \)
• \( f(0) = -1 \)
• \( f(2) = 19 \)

Therefore, the absolute maximum value of the function is \(19\), happening at \(x = 2\), and the absolute minimum value is \(-1\), which occurs at both \(x = -3\) and \(x = 0\).

Reviewing these values helps to pinpoint the function's behavior over the specified interval.

Derivative

The derivative is a powerful tool for understanding how a function behaves. Conceptually, it gives you the rate of change of the function's value as the input changes. Think about a car's speedometer that shows speed at any moment - that's what a derivative is for functions: it shows the function's rate of change at any point.

Mathematically, finding the derivative of a function involves differentiating it with respect to its variable. For instance, our function is \( f(x) = x^3 + 3x^2 - 1 \). Its derivative is computed as:

• \[ f'(x) = 3x^2 + 6x \]

This derivative can then be used to locate critical points by solving for zero. It also helps determine the concavity and points of inflection. Derivatives are at the heart of calculus and have broad applications in science, engineering, economics, and much more.

Polynomial Functions

Polynomial functions are among the simplest functions to understand and work with. They are composed of variables raised to whole number powers, with each term having a constant multiplier. A typical polynomial function looks something like \( a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0 \).

Our specific example is \( f(x) = x^3 + 3x^2 -1 \), making it a cubic function because the highest power of \( x \) is 3.

Characteristics of polynomial functions:

• They are continuous everywhere. No breaks, jumps, or asymptotes.
• The degree of the polynomial (highest exponent on \( x \)) indicates the potential number of roots and the function’s end behavior.
• They can be easily differentiated to find rates of change, as seen in our exercise.

Cubic functions, like in our example, often have points of inflection and can exhibit varying behavior such as hills and valleys within their graph, which are precisely the focus points when exploring critical points or absolute extremum.