Problem 19 - Calculus of a Single Variable

Question

Determine whether Rolle's Theorem can be applied to \(f\) on the closed interval \([a, b] .\) If Rolle's Theorem can be applied, find all values of \(c\) in the open interval \((a, b)\) such that \(f^{\prime}(c)=0 .\) If Rolle's Theorem cannot be applied, explain why not. \(f(x)=\sin 3 x, \quad\left[0, \frac{\pi}{3}\right]\)

Step-by-Step Solution

Answer

Rolle's Theorem can be applied to the function. The only solution in the given open interval (0, \(\frac{\pi}{3}\)) is \(c = \frac{\pi}{6}\).

Step 1: Check \(f(a) = f(b)\) condition

First, substitute \(a = 0\) and \(b = \frac{\pi}{3}\) into \(f(x) = \sin 3x\). This gives \(f(a) = \sin(3 * 0) = 0\) and \(f(b) = \sin(3 * \frac{\pi}{3}) = \sin(\pi) = 0\). So, \(f(a) = f(b)\), which satisfies this condition of Rolle's Theorem.

Step 2: Calculate the derivative of \(f(x)\)

The derivative of \(f(x)\) is calculated as the derivative of \(sin 3x\), which is \(f'(x) = 3\cos 3x\).

Step 3: Find the value(s) of \(c\)

To find the \(c\) values, set \(f'(c) = 0\) which gives \(3\cos 3c = 0\). Solving, we obtain \(cos 3c = 0\). From the cosine function, we know that this condition is met at \(c = \frac{\pi}{6} + n\pi\). Since we are dealing with the open interval \((0, \frac{\pi}{3})\), the only solution in this range when \(n = 0\) is \(c = \frac{\pi}{6}\).

Key Concepts

Calculus

Calculus is a branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series. This discipline constitutes a major part of modern mathematics education. It has two main branches: Differential Calculus and Integral Calculus. Differential Calculus studies the rate at which quantities change, while Integral Calculus focuses on the accumulation of quantities.

Understanding calculus is vital to grasp the changes occurring within a system and it extensively applies to various fields such as physics, engineering, economics, statistics, and medicine. The core concept exercises like Rolle's Theorem meld both branches, using derivatives to ultimately discuss the existence of specific points related to the function's behavior on a certain interval.

Derivative of Trigonometric Functions

The derivative of a function at a point is the rate at which the function value changes with respect to a change in the input value. In trigonometric functions, derivatives follow specific rules that differ from polynomials.

The sine, cosine, tangent, and other trig functions each have their own derivatives which must be memorized for practical use. For example, the derivative of \( \sin(x) \) is \( \cos(x) \) and that of \( \cos(x) \) is \( -\sin(x) \). Therefore, the derivative of \( \sin(3x) \), as seen in the exercise, is \( 3\cos(3x) \) due to the chain rule, which involves differentiating the composite of the functions.

Critical Points

Critical points are key in the study of the behavior of functions. In the context of calculus, a critical point occurs where the derivative of a function is either zero or undefined. It's where the function's graph changes direction, creating a peak, trough or an inflection point.

Finding the critical points of a function is an essential step in understanding the function’s overall behavior, its maxima and minima, and in optimizing real-world scenarios. This is demonstrated in our exercise where finding the value of \(c\) that makes \(f'(c) = 0\) essentially locates a critical point in the interval.

Closed Interval Method

The closed interval method is a technique used to find the absolute maximum and minimum values of a continuous function on a closed interval \([a, b]\). This method involves evaluating the function at its critical points within the interval and at the endpoints \(a\) and \(b\).

The main steps in this process are to identify the critical points, evaluate the function at those points and the endpoints, and then compare those values to determine the absolute extremum on the interval. In the context of Rolle's Theorem, the closed interval method helps ensure that the function satisfies the conditions needed for the theorem to apply - that the function is differentiable in \((a, b)\) and continuous in \([a, b]\).