Question 8: Applying Rolle's Theorem To Find Critical Points

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Introduction

Rolle's Theorem, a cornerstone of calculus, provides a powerful tool for understanding the behavior of differentiable functions. It essentially states that if a function is continuous on a closed interval, differentiable on the open interval, and has the same value at the endpoints, then there exists at least one point within the interval where the derivative of the function is zero. This point represents a critical point, where the function's slope is momentarily horizontal. In this exploration, we will delve into the application of Rolle's Theorem to the quadratic function f(x) = 2x^2 - 16x - 6 over the interval [2, 6]. Our objective is to determine the number of values of c within this interval for which the derivative f'(c) equals zero. By meticulously examining the conditions of Rolle's Theorem and employing differentiation techniques, we will unravel the solution to this problem and gain a deeper appreciation for the theorem's practical implications.

Rolle's Theorem is not just an abstract mathematical concept; it has significant applications in various fields, including optimization problems, physics, and engineering. Understanding this theorem and its implications is crucial for anyone seeking a solid foundation in calculus and its applications. Before we dive into the specifics of the problem, let's briefly recap the essential conditions for Rolle's Theorem to hold. These conditions serve as the bedrock for our analysis and ensure the validity of our conclusions. We must verify that the function is continuous on the closed interval, differentiable on the open interval, and that the function values at the endpoints of the interval are equal. Once these conditions are met, we can confidently apply Rolle's Theorem to locate the critical points of the function.

Verifying Rolle's Theorem Conditions for f(x) = 2x^2 - 16x - 6 on [2, 6]

Before we can apply Rolle's Theorem, we must first verify that the function f(x) = 2x^2 - 16x - 6 satisfies all the necessary conditions on the interval [2, 6]. These conditions are crucial for the theorem's validity and ensure that our subsequent analysis leads to accurate conclusions. Let's examine each condition in detail:

  1. Continuity on the closed interval [2, 6]: The function f(x) = 2x^2 - 16x - 6 is a polynomial function, and polynomial functions are continuous everywhere. This means that the function is continuous on the entire real number line, including the closed interval [2, 6]. Therefore, the first condition of Rolle's Theorem is satisfied. Continuity ensures that there are no breaks or jumps in the function's graph within the interval, which is essential for the theorem to hold. Without continuity, the theorem's conclusion about the existence of a critical point cannot be guaranteed.

  2. Differentiability on the open interval (2, 6): Again, since f(x) is a polynomial function, it is differentiable everywhere. Polynomial functions have derivatives at every point, making them smooth and well-behaved. Consequently, f(x) is differentiable on the open interval (2, 6), satisfying the second condition of Rolle's Theorem. Differentiability is a stronger condition than continuity and implies that the function has a well-defined tangent line at every point within the interval. This tangent line represents the instantaneous rate of change of the function, which is crucial for finding critical points.

  3. Equality of function values at the endpoints: f(2) = f(6): To verify this condition, we need to evaluate the function at the endpoints of the interval:

    • f(2) = 2(2)^2 - 16(2) - 6 = 8 - 32 - 6 = -30
    • f(6) = 2(6)^2 - 16(6) - 6 = 72 - 96 - 6 = -30

    Since f(2) = f(6) = -30, the third condition of Rolle's Theorem is also satisfied. This condition ensures that the function starts and ends at the same height, creating a