Problem 37 - Calculus A New Horizon

Question

Find the relative extrema in the interval \(0

Step-by-Step Solution

Answer

Critical points are \(x = \pi\) (local max), \(x = \frac{\pi}{2}\) and \(x = \frac{3\pi}{2}\) (local mins), consistent with the graph.

Step 1: Find the derivative

To find the relative extrema, first, differentiate the function. The given function is \(f(x) = \cos^2 x\). Using the chain rule, the derivative is: \(f'(x) = 2\cos x (-\sin x) = -2\cos x \sin x\). Simplifying further, we have \(f'(x) = -\sin(2x)\) using the double angle identity.

Step 2: Solve for critical points

To find the critical points, set the derivative equal to zero: \(-\sin(2x) = 0\). This implies \(\sin(2x) = 0\). The solutions for \(2x = n\pi\) where \(n\) is an integer are \(x = \frac{n\pi}{2}\). Since we need \(0 < x < 2\pi\), the valid points are \(x = \frac{\pi}{2}, \pi, \frac{3\pi}{2}\).

Step 3: Determine nature of critical points

Use the second derivative test to classify these critical points. First, compute the second derivative: \(f''(x) = -2\cos(2x)\). Evaluate the second derivative at each critical point:- \(x = \frac{\pi}{2}\): \(f''\left(\frac{\pi}{2}\right) = 0\), so the test is inconclusive.- \(x = \pi\): \(f''(\pi) = -2\), which indicates a local maximum.- \(x = \frac{3\pi}{2}\): \(f''\left(\frac{3\pi}{2}\right) = 0\), so the test is inconclusive.For the inconclusive cases, observe the sign changes in \(f'(x)\) around the points or use a graphing utility.

Step 4: Verify results with a graph

Graph \(f(x) = \cos^2 x\) using a graphing utility to visually confirm the extrema:- Around \(x = \frac{\pi}{2}\), the graph indicates a local minimum.- Around \(x = \pi\), the graph confirms a local maximum.- Around \(x = \frac{3\pi}{2}\), the graph indicates a local minimum.

Key Concepts

Derivative

In calculus, finding the derivative of a function is a fundamental process. The derivative provides information about the rate at which a function is changing at any given point. For the function \(f(x) = \cos^2 x\), we begin by differentiating using the chain rule. The chain rule is helpful when dealing with composite functions, allowing us to take the derivative of the outer function and multiply it by the derivative of the inner function. In this case, the outer function is \(x^2\) and the inner function is \(\cos x\), giving us the derivative:

• \(f'(x) = 2\cos x(-\sin x) = -2\cos x\sin x\).

Utilizing the trigonometric identity for the double angle, \(\sin(2x) = 2\sin(x)\cos(x)\), simplifies this to:

• \(f'(x) = -\sin(2x)\).

This derivative will help us find the critical points where the slope of the tangent is zero or undefined.

Critical Points

Critical points of a function occur where the derivative equals zero or is undefined. These points can indicate local maxima, minima, or points of inflection. For \(f(x) = \cos^2 x\), the critical points are found by solving the equation \(-\sin(2x) = 0\). This is

• equivalent to \(\sin(2x) = 0\),

which has solutions

• \(2x = n\pi\)

for any integer \(n\). Solving for \(x\), we have:

• \(x = \frac{n\pi}{2}\).

Given the interval \(0 < x < 2\pi\), the relevant critical points are \(x = \frac{\pi}{2}, \pi, \frac{3\pi}{2}\). These points will be further analyzed to determine whether they correspond to local maxima, minima, or saddle points.

Second Derivative Test

The second derivative test is a method used to classify critical points determined by the first derivative. It involves taking the second derivative of the function and evaluating it at the critical points. For our function, the second derivative is:

• \(f''(x) = -2\cos(2x)\).

Using these calculations at each critical point, we test:

• At \(x = \frac{\pi}{2}\): \(f''\left(\frac{\pi}{2}\right) = 0\), the result is inconclusive.
• At \(x = \pi\): \(f''(\pi) = -2\), indicating a local maximum since the second derivative is negative.
• At \(x = \frac{3\pi}{2}\): \(f''\left(\frac{3\pi}{2}\right) = 0\), also inconclusive.

When the second derivative is zero, the test doesn't provide clear results, so looking at the first derivative or using a graph can help further interpret these points.

Graphing Utility

A graphing utility, such as a graphing calculator or software, is a valuable tool for visualizing functions. It helps corroborate analytic findings from derivatives. By graphing \(f(x) = \cos^2 x\), we can observe the behavior of the function around the identified critical points:

• Near \(x = \frac{\pi}{2}\), the graph shows a local minimum.
• At \(x = \pi\), it confirms a local maximum.
• Close to \(x = \frac{3\pi}{2}\), another local minimum is evident.

Using a graphing utility provides a visual check for our calculations, allowing us to see the symmetry and periodic nature of trigonometric functions like this one. This tool offers an intuitive understanding of the mathematical problem and confirms the results obtained from calculus techniques.