Rolle's Theorem Explained: Analyzing The Function's Behavior

Hey guys! Let's dive into the fascinating world of calculus and explore Rolle's Theorem. This theorem is a cornerstone concept, helping us understand the behavior of functions within specific intervals. We'll break down the theorem, analyze its application to a particular function, and clear up any confusion. So, buckle up, and let's get started!

Understanding Rolle's Theorem: The Basics

Rolle's Theorem is a fundamental result in calculus that provides insights into the relationship between a function's values at the endpoints of an interval and the existence of a point where the function's derivative is zero. Basically, it helps us find critical points where the function might have a maximum or minimum value. It's like a detective tool for analyzing function behavior! To get a grip on Rolle's Theorem, you need to know its essential requirements. First, the function, let's call it f(x), must be continuous on a closed interval, say [a, b]. This means that the function has no jumps, breaks, or holes within that interval. It's a smooth, unbroken curve. Second, the function must be differentiable on the open interval (a, b). Differentiability means that the function has a well-defined derivative at every point within the interval. In simpler terms, the function has no sharp corners or cusps. It has a tangent line at every point. Finally, and this is the crucial part, the function values at the endpoints of the interval must be equal. That is, f(a) = f(b). If all these conditions are met, then Rolle's Theorem guarantees that there exists at least one point, let's call it c, within the open interval (a, b) such that the derivative of the function at that point is zero, or f'(c) = 0. This point c is where the tangent line to the function is horizontal. Think of it like this: if a continuous and differentiable function starts and ends at the same height, there must be at least one point in between where it momentarily stops changing (has a zero derivative) before changing direction. Pretty neat, right? This theorem gives us a powerful tool to find critical points, which are essential in optimization problems and curve sketching. Remember, it's all about continuity, differentiability, and equal function values at the endpoints. If these conditions are not met, then Rolle's Theorem cannot be applied, and we can't guarantee the existence of a point with a zero derivative.

Applying Rolle's Theorem to the Function: $f(x) = x^3 - x + 3$ over $[-1, 0]$

Now, let's get our hands dirty and apply Rolle's Theorem to the function f(x) = x^3 - x + 3 over the interval [-1, 0]. The question here is whether Rolle's Theorem can be used with this specific function and, if so, what we can deduce from it. First, we need to check the conditions of the theorem. Is f(x) continuous on the closed interval [-1, 0]? Absolutely! Polynomial functions, like the one we have here (x^3 - x + 3), are continuous everywhere. There are no breaks or jumps. It's a smooth curve. Second, is f(x) differentiable on the open interval (-1, 0)? Yes, again! The derivative of f(x), which is f'(x) = 3x^2 - 1, exists for all values of x within the interval. There are no sharp corners or cusps. Lastly, do we have f(-1) = f(0)? Let's find out! When x = -1, f(-1) = (-1)^3 - (-1) + 3 = -1 + 1 + 3 = 3. When x = 0, f(0) = (0)^3 - (0) + 3 = 3. Bingo! f(-1) = f(0) = 3. All three conditions of Rolle's Theorem are satisfied. Since all the conditions are met, we can now apply Rolle's Theorem. This means there must be at least one point c in the open interval (-1, 0) where f'(c) = 0. To find this point c, we need to solve the equation f'(x) = 0. We already know that f'(x) = 3x^2 - 1. Setting this equal to zero, we have 3x^2 - 1 = 0. Solving for x, we get x^2 = 1/3, and therefore, x = ±√(1/3). Now, we must make sure that the value of c that we get is between -1 and 0, if it is not, then we cannot apply Rolle's Theorem. Given that √(1/3) is approximately 0.577 and -√(1/3) is approximately -0.577. The point c = -√(1/3) falls within the open interval (-1, 0). This tells us that at x = -√(1/3), the tangent line to the function f(x) is horizontal. This is a critical point where the function could have a local maximum or minimum. So, in the context of the provided function and interval, Rolle's Theorem does indeed apply, and we have found a critical point within the specified interval.

Analyzing the Statements and Determining Their Truth

Okay, let's break down the statements regarding Rolle's Theorem and determine which ones are correct for our function f(x) = x^3 - x + 3 over the interval [-1, 0]. Based on our analysis in the previous sections, we know that Rolle's Theorem does apply because the function is continuous, differentiable, and the function values at the interval's endpoints are equal. Now let's evaluate some potential claims.

• Statement A: