Understanding the Maximum and Minimum Values of a Cubic Function

In calculus, determining the maximum and minimum values of a function is essential for optimizing problems and understanding the behavior of functions. This article will guide you through a detailed analysis of the function f(x) x^3 - x^2 - x, demonstrating how to identify its critical points and evaluate the maximum and minimum values.

Introduction to the Function

Let's start by considering the function f(x) x^3 - x^2 - x. Understanding its behavior will involve identifying where it reaches its maximum and minimum values. This is achieved through a series of steps involving differentiation and solving for critical points.

Step 1: Find the Derivative

The first step is to find the derivative of the given function f(x). The derivative is given by:

f'(x) 3x^2 - 2x - 1

Step 2: Set the Derivative to Zero

To find the critical points, we set the derivative equal to zero:

3x^2 - 2x - 1 0

Step 3: Solve the Quadratic Equation

We can solve the quadratic equation using the quadratic formula x -b ± sqrt(b^2 - 4ac) / 2a. Here, a 3, b -2, and c -1. Plugging these values into the formula gives:

x (2 ± sqrt((-2)^2 - 4 * 3 * -1)) / (2 * 3)

x (2 ± sqrt(4 12)) / 6

x (2 ± sqrt(16)) / 6

x (2 ± 4) / 6

This simplifies to two solutions:

x (2 4) / 6 6 / 6 1 x (2 - 4) / 6 -2 / 6 -1/3

Step 4: Evaluate the Function at the Critical Points

Next, we evaluate the function f(x) at the critical points x 1 and x -1/3 to find the corresponding function values:

f(1) 1^3 - 1^2 - 1 1 - 1 - 1 -1 f(-1/3) (-1/3)^3 - (-1/3)^2 - (-1/3) -1/27 - 1/9 1/3

To simplify f(-1/3):

f(-1/3) -1/27 - 3/27 9/27 (9 - 3 - 1) / 27 5 / 27

Step 5: Determine the Maximum and Minimum Values

By comparing the function values at the critical points, we find:

Function value at x 1: -1 (minimum) Function value at x -1/3: 5/27 (maximum)

Thus, the minimum value of f(x) is -1 at x 1, and the maximum value is 5/27 at x -1/3.

Conclusion

In summary, the maximum value of the function f(x) x^3 - x^2 - x is 5/27 at x -1/3, and the minimum value is -1 at x 1.

It's worth noting that while the function has a local maximum and minimum, it does not have an absolute maximum or minimum because as x → ±∞, f(x) → ±∞. This indicates that the function extends to infinity in both positive and negative directions.