Mean Value Theorem Application For F(x) = 8 - X² On [-1, 2]

Jul 8, 2025 by Admin 60 views

Introduction

The Mean Value Theorem (MVT) is a cornerstone of calculus, providing a vital link between the average rate of change of a function over an interval and the instantaneous rate of change at a specific point within that interval. This theorem has profound implications in various fields, including physics, engineering, and economics, making it essential for students and professionals alike to grasp its intricacies. In this article, we delve into the application of the Mean Value Theorem to the function f(x) = 8 - x² on the interval [-1, 2]. Our primary goal is to rigorously determine whether the conditions of the MVT are met for this particular function and interval. If these conditions are satisfied, we will then proceed to find the point(s) guaranteed to exist by the theorem. This process involves a careful examination of the function's properties, such as continuity and differentiability, and a systematic approach to solving the resulting equations. By working through this example, we aim to provide a comprehensive understanding of how the Mean Value Theorem works in practice and to highlight its significance in mathematical analysis. This exploration will not only enhance your problem-solving skills but also deepen your appreciation for the elegance and power of calculus theorems. The Mean Value Theorem is more than just a theoretical concept; it's a practical tool that helps us understand the behavior of functions and their rates of change. So, let's embark on this journey to unravel the mysteries of the MVT and its application to the function at hand.

Understanding the Mean Value Theorem

Before we dive into the specifics of our function, it's crucial to have a solid grasp of what the Mean Value Theorem (MVT) actually states. In essence, the MVT provides a bridge between the average rate of change of a function over an interval and its instantaneous rate of change at a specific point within that interval. More formally, the theorem states that if a function f(x) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one point c in the interval (a, b) such that the derivative of the function at c, denoted as f'(c), is equal to the average rate of change of the function over the interval [a, b]. Mathematically, this is expressed as:

f'(c) = [f(b) - f(a)] / [b - a]

This equation is the heart of the MVT, and it tells us that at some point c, the slope of the tangent line to the function's graph is equal to the slope of the secant line connecting the points (a, f(a)) and (b, f(b)). In simpler terms, there's a point where the instantaneous rate of change matches the average rate of change. The MVT is not just a standalone result; it's deeply connected to other fundamental theorems in calculus, such as Rolle's Theorem. Rolle's Theorem is actually a special case of the MVT, where f(a) = f(b). In this case, the average rate of change is zero, and the MVT guarantees the existence of a point c where the derivative is also zero. Understanding the MVT is crucial for solving a wide range of problems in calculus and related fields. It allows us to make predictions about the behavior of functions, estimate values, and prove other important results. The MVT's conditions of continuity and differentiability are not arbitrary; they are essential for the theorem to hold true. If a function fails to meet these conditions, the MVT may not apply, and the conclusion may not be valid. Therefore, it's important to carefully check these conditions before attempting to apply the MVT.

Part a: Verifying the Conditions of the Mean Value Theorem for f(x) = 8 - x² on [-1, 2]

In this section, we turn our attention to the specific function f(x) = 8 - x² and the interval [-1, 2]. Our first task is to meticulously verify whether the conditions of the Mean Value Theorem (MVT) are satisfied for this function on the given interval. As we've discussed, the MVT requires two key conditions: continuity on the closed interval and differentiability on the open interval. Let's begin by addressing continuity. The function f(x) = 8 - x² is a polynomial function, specifically a quadratic function. Polynomial functions are known for their well-behaved properties; they are continuous everywhere. This means that there are no breaks, jumps, or holes in the graph of f(x). Therefore, f(x) is indeed continuous on the closed interval [-1, 2]. This satisfies the first condition of the MVT. Next, we need to check for differentiability. Differentiability implies that the function has a derivative at every point in the interval. To determine this, we find the derivative of f(x). The derivative of f(x) = 8 - x² is f'(x) = -2x. The derivative f'(x) = -2x is also a polynomial function, a linear function in this case. Linear functions are differentiable everywhere, meaning that f'(x) exists for all x. Thus, f(x) is differentiable on the open interval (-1, 2). We have now confirmed that both conditions of the MVT are met for the function f(x) = 8 - x² on the interval [-1, 2]. The function is continuous on the closed interval and differentiable on the open interval. This is a crucial step because it allows us to confidently apply the MVT and proceed to find the point(s) guaranteed to exist by the theorem. If either of these conditions were not satisfied, we would not be able to use the MVT, and our analysis would have to take a different direction. However, since the conditions are met, we can move forward with the next part of our exploration.

Part b: Finding the Point(s) Guaranteed by the Mean Value Theorem

Now that we've established that the Mean Value Theorem (MVT) applies to the function f(x) = 8 - x² on the interval [-1, 2], we can proceed to the most exciting part: finding the point(s) c that the theorem guarantees. Recall that the MVT states there exists at least one c in the interval (-1, 2) such that:

f'(c) = [f(b) - f(a)] / [b - a]

where a = -1 and b = 2 in our case. Let's break down this equation step by step. First, we need to calculate the average rate of change of f(x) over the interval [-1, 2]. This is given by the right-hand side of the equation. We have:

f(2) = 8 - (2)² = 8 - 4 = 4

f(-1) = 8 - (-1)² = 8 - 1 = 7

So, the average rate of change is:

[f(2) - f(-1)] / [2 - (-1)] = [4 - 7] / [2 + 1] = -3 / 3 = -1

Next, we need to find the derivative of f(x), which we already found in the previous section: f'(x) = -2x. Now, we set the derivative equal to the average rate of change and solve for c:

f'(c) = -2c = -1

Dividing both sides by -2, we get:

c = 1/2

We have found a value for c, which is c = 1/2. The final step is to ensure that this value lies within the open interval (-1, 2). Since -1 < 1/2 < 2, the value c = 1/2 is indeed within the interval. This confirms that the Mean Value Theorem holds true for the function f(x) = 8 - x² on the interval [-1, 2], and the point c = 1/2 is the point guaranteed to exist by the theorem. At this point, the slope of the tangent line to the graph of f(x) is equal to the slope of the secant line connecting the endpoints of the interval. This example beautifully illustrates the power and precision of the Mean Value Theorem in calculus.

Conclusion

In conclusion, we have successfully applied the Mean Value Theorem (MVT) to the function f(x) = 8 - x² on the interval [-1, 2]. Our journey began with a thorough understanding of the MVT itself, emphasizing its conditions and significance in calculus. We then meticulously verified that the function f(x) = 8 - x² satisfies the two crucial conditions of the MVT: continuity on the closed interval [-1, 2] and differentiability on the open interval (-1, 2). This verification was a critical step, as it allowed us to confidently proceed with the application of the theorem. With the conditions met, we embarked on the task of finding the point(s) c guaranteed to exist by the MVT. This involved calculating the average rate of change of f(x) over the interval [-1, 2] and setting it equal to the derivative of f(x) at c. Solving the resulting equation, we found that c = 1/2. We then confirmed that this value lies within the open interval (-1, 2), thereby validating the conclusion of the MVT. The point c = 1/2 represents the point on the graph of f(x) where the tangent line has the same slope as the secant line connecting the endpoints of the interval. This example serves as a powerful illustration of the MVT in action, showcasing its ability to connect the average and instantaneous rates of change of a function. The MVT is not just a theoretical concept; it's a practical tool with wide-ranging applications in mathematics, physics, engineering, and other fields. By mastering the MVT and its applications, you equip yourself with a valuable tool for problem-solving and analysis in various contexts. The Mean Value Theorem is a testament to the beauty and elegance of calculus, providing insights into the behavior of functions and their derivatives. As you continue your exploration of calculus, remember the MVT as a fundamental principle that underpins many other important results and techniques.