Finding The Minimum Value Of A Function F(x)=x^3-3x^2+2

Jul 14, 2025 by ADMIN 56 views

Finding the Minimum Value of a Function An In-Depth Guide

Introduction

In the realm of calculus and mathematical analysis, identifying the minimum value of a function is a fundamental task with far-reaching applications. From optimization problems in engineering and economics to curve sketching and data analysis, the ability to pinpoint the minimum of a function is an indispensable skill. In this comprehensive guide, we will delve into the techniques and concepts involved in finding the minimum value of a function, using the specific example of the function f(x) = x³ - 3x² + 2 as a case study. This exploration will not only provide a step-by-step solution to the given problem but also offer a broader understanding of the underlying principles and their relevance in various contexts.

The function we will be focusing on is a cubic polynomial, a type of function that exhibits interesting behaviors, including local minima and maxima. To find the minimum value, we will employ the tools of differential calculus, which provide a systematic way to analyze the rate of change of a function. By understanding how the derivative of a function relates to its increasing and decreasing intervals, we can accurately locate points where the function attains its minimum value. This guide will walk you through the process, explaining each step in detail and providing insights into the mathematical reasoning behind it. Whether you are a student learning calculus or a professional applying these techniques in your field, this guide will serve as a valuable resource for mastering the art of finding minimum values of functions.

Understanding the Problem

Before diving into the solution, let's clearly define the problem at hand. We are given the function f(x) = x³ - 3x² + 2 and tasked with finding the x-value at which this function attains its minimum value. It's crucial to understand that we are looking for the x-coordinate of the minimum point, not the minimum value of the function itself. This distinction is important because the x-coordinate tells us where the minimum occurs, while the function value at that point tells us what the minimum value is.

To approach this problem effectively, we need to utilize the concepts of differential calculus. The key idea is that the derivative of a function, denoted as f'(x), gives us the slope of the tangent line to the function's graph at any given point. At a minimum point, the tangent line is horizontal, meaning the slope is zero. Therefore, we need to find the points where f'(x) = 0. These points are called critical points and are potential locations of minima or maxima. However, not all critical points are minima; they could also be maxima or saddle points. To determine whether a critical point is a minimum, we can use the second derivative test or analyze the sign of the first derivative around the critical point.

In the context of our function, f(x) = x³ - 3x² + 2, we will first find the derivative f'(x). Then, we will set f'(x) equal to zero and solve for x to find the critical points. Once we have the critical points, we will use the second derivative test or the first derivative test to determine which of these points correspond to a minimum value. This systematic approach will ensure that we accurately identify the x-value at which the function f(x) has a minimum. Understanding this process is crucial not only for solving this specific problem but also for tackling a wide range of optimization problems in calculus and beyond.

Finding the First Derivative

The cornerstone of finding the minimum value of a function using calculus lies in the concept of derivatives. The derivative of a function, often denoted as f'(x), provides a powerful tool for understanding the function's behavior, particularly its rate of change. In the context of finding minima and maxima, the first derivative plays a crucial role in identifying critical points, which are potential locations of these extreme values. For our function, f(x) = x³ - 3x² + 2, the first step is to determine its derivative.

To find the derivative, we apply the power rule, a fundamental rule in calculus that states that the derivative of x^n is nx^(n-1). This rule allows us to differentiate each term of the polynomial function separately. Applying the power rule to f(x) = x³ - 3x² + 2, we get:

The derivative of x³ is 3x².
The derivative of -3x² is -6x.
The derivative of the constant term 2 is 0, as the derivative of any constant is always zero.

Combining these results, we find that the first derivative of f(x) is f'(x) = 3x² - 6x. This new function, f'(x), represents the slope of the tangent line to the original function f(x) at any given point x. Understanding this relationship is key to finding the minimum value. At the minimum point, the tangent line is horizontal, and thus the slope is zero. Therefore, we need to find the x-values where f'(x) = 0. These x-values are the critical points of the function and are potential locations for minima or maxima. The next step is to solve the equation 3x² - 6x = 0 to find these critical points, which will lead us closer to identifying the x-value where the minimum of the function occurs.

Identifying Critical Points

With the first derivative, f'(x) = 3x² - 6x, in hand, the next crucial step in our quest to find the minimum value of the function f(x) = x³ - 3x² + 2 is to identify the critical points. Critical points are the x-values where the derivative is either zero or undefined. These points are of significant importance because they represent locations where the function's slope is either horizontal (zero) or has a discontinuity (undefined). In the context of finding minima and maxima, critical points are the prime suspects, as they are the potential locations where the function's extreme values occur.

To find the critical points, we need to solve the equation f'(x) = 0. In our case, this means solving 3x² - 6x = 0. This is a quadratic equation, and there are several ways to solve it. One common method is to factor the equation. We can factor out a 3x from both terms, giving us 3x(x - 2) = 0. Now, we apply the zero-product property, which states that if the product of two factors is zero, then at least one of the factors must be zero. Thus, either 3x = 0 or x - 2 = 0.

Solving these two equations, we find two critical points: x = 0 and x = 2. These are the x-values where the slope of the tangent line to the function f(x) is zero, indicating potential minima or maxima. However, we cannot definitively say which of these points (if any) corresponds to a minimum without further analysis. The critical points are merely candidates. To determine whether these points are minima, maxima, or neither, we need to employ additional techniques, such as the second derivative test or analyzing the sign of the first derivative around these points. The identification of these critical points is a pivotal step, narrowing down the possibilities and guiding us towards the final solution.

Determining the Nature of Critical Points

Having identified the critical points x = 0 and x = 2 for the function f(x) = x³ - 3x² + 2, the next crucial step is to determine the nature of these points. In other words, we need to ascertain whether each critical point corresponds to a local minimum, a local maximum, or neither. This determination is vital because it allows us to pinpoint the x-value at which the function attains its minimum value, which is the ultimate goal of our problem.

There are two primary methods for determining the nature of critical points: the second derivative test and the first derivative test. Let's explore both methods and apply them to our critical points.

Second Derivative Test

The second derivative test involves finding the second derivative of the function, denoted as f''(x), and evaluating it at each critical point. The second derivative provides information about the concavity of the function. If f''(x) > 0 at a critical point, the function is concave up, indicating a local minimum. If f''(x) < 0, the function is concave down, indicating a local maximum. If f''(x) = 0, the test is inconclusive, and we need to use another method.

For our function, f'(x) = 3x² - 6x, so the second derivative is f''(x) = 6x - 6. Now, we evaluate f''(x) at each critical point:

At x = 0, f''(0) = 6(0) - 6 = -6. Since f''(0) < 0, the function is concave down at x = 0, indicating a local maximum.
At x = 2, f''(2) = 6(2) - 6 = 6. Since f''(2) > 0, the function is concave up at x = 2, indicating a local minimum.

First Derivative Test

The first derivative test involves analyzing the sign of the first derivative, f'(x), in the intervals around each critical point. If f'(x) changes from negative to positive as we move across a critical point, the function has a local minimum at that point. If f'(x) changes from positive to negative, the function has a local maximum.

To apply this test, we consider the intervals determined by the critical points: (-∞, 0), (0, 2), and (2, ∞). We choose a test value within each interval and evaluate f'(x):

In (-∞, 0), let's choose x = -1. f'(-1) = 3(-1)² - 6(-1) = 9, which is positive.
In (0, 2), let's choose x = 1. f'(1) = 3(1)² - 6(1) = -3, which is negative.
In (2, ∞), let's choose x = 3. f'(3) = 3(3)² - 6(3) = 9, which is positive.

We observe that f'(x) changes from positive to negative at x = 0, indicating a local maximum, and from negative to positive at x = 2, indicating a local minimum. Both the second derivative test and the first derivative test confirm that the function f(x) has a local minimum at x = 2. This consistent result reinforces the accuracy of our analysis and leads us to the final answer.

Final Answer

After a thorough analysis using both the first and second derivative tests, we have confidently determined the nature of the critical points for the function f(x) = x³ - 3x² + 2. Our investigations revealed that:

At x = 0, the function has a local maximum.
At x = 2, the function has a local minimum.

The question specifically asked for the x-value at which f(x) has a minimum. Based on our findings, the function f(x) attains a local minimum at x = 2. Therefore, the final answer to the problem is x = 2.

It is important to note that this minimum is a local minimum, meaning it is the smallest value of the function in a particular neighborhood. To determine if this is also the global minimum (the absolute smallest value of the function over its entire domain), we would need to consider the function's behavior as x approaches positive and negative infinity. However, for the purpose of this problem, we have successfully identified the x-value at which a local minimum occurs.

In summary, by applying the principles of differential calculus, including finding the first and second derivatives, identifying critical points, and using the second derivative test or the first derivative test, we have methodically solved the problem. This process not only provides the answer but also enhances our understanding of the function's behavior and the powerful tools of calculus used to analyze it.

Conclusion

In this comprehensive guide, we embarked on a journey to find the minimum value of the function f(x) = x³ - 3x² + 2. Through a step-by-step approach, we explored the fundamental concepts of calculus, including derivatives, critical points, and the tests used to determine their nature. We successfully identified the x-value at which the function attains its minimum, demonstrating the power and elegance of calculus in solving optimization problems.

The process began with understanding the problem and recognizing the importance of finding the x-coordinate of the minimum point. We then delved into finding the first derivative of the function, f'(x) = 3x² - 6x, which represents the slope of the tangent line at any point on the function's graph. By setting the first derivative equal to zero, we identified the critical points, x = 0 and x = 2, which are the potential locations of minima or maxima.

To determine the nature of these critical points, we employed both the second derivative test and the first derivative test. The second derivative test involved finding f''(x) = 6x - 6 and evaluating it at the critical points. The results indicated a local maximum at x = 0 and a local minimum at x = 2. The first derivative test, which analyzed the sign of f'(x) in the intervals around the critical points, corroborated these findings.

Finally, we concluded that the function f(x) has a local minimum at x = 2, providing the answer to the problem. This process highlights the systematic approach required to solve optimization problems using calculus. The ability to find the minimum or maximum of a function has wide-ranging applications in various fields, from engineering and physics to economics and computer science. This guide serves as a valuable resource for anyone seeking to master these essential techniques and apply them to real-world problems. The journey through this problem has not only provided a solution but also deepened our understanding of the fundamental principles of calculus and their practical significance.