Problem 63 - Calculus Early Transcendentals

Question

Find the relative extrema in the interval \(0

Step-by-Step Solution

Answer

Relative extrema are at \( x = \frac{\pi}{2} \) (maximum) and \( x = \frac{3\pi}{2} \) (minimum).

Step 1: Understand What Relative Extrema Means

Relative extrema refer to the local maximum or minimum points in a certain interval, meaning in this case the values of \( x \) within \( 0 < x < 2\pi \) where \( f(x) \) changes from increasing to decreasing or vice versa.

Step 2: Express the Function to Find Critical Points

The given function is a piecewise function because it involves an absolute value. We express it as: \[ f(x) = \left\{ \begin{array}{ll} \sin 2x & \text{if } \sin 2x \geq 0 \ -\sin 2x & \text{if } \sin 2x < 0 \end{array} \right. \] We need to find points where \( \sin 2x = 0 \) or the derivative is zero or undefined to identify potential extrema.

Step 3: Identify Points where \( \sin 2x \) Changes Sign

The function \( \sin 2x \) is zero at points of the form \( x = \frac{n\pi}{2} \) with \( n \) being an integer. Within the interval \( 0 < x < 2\pi \), these points are \( \frac{\pi}{2}, \pi, \frac{3\pi}{2} \). Check changes in the sign of \( \sin 2x \) at these points.

Step 4: Evaluate \( f(x) \) at Critical Points

Evaluate \( f(x) \) at these key points: * At \( x = \frac{\pi}{2}, \sin 2x = 0 \), so \( f\left(\frac{\pi}{2}\right) = 0.\) * At \( x = \pi, \sin 2x = 0 \), so \( f(\pi) = 0.\) * At \( x = \frac{3\pi}{2}, \sin 2x = 0 \), so \( f\left(\frac{3\pi}{2}\right) = 0.\)

Step 5: Analyze Intervals for Increasing/Decreasing Behavior

To find relative extrema, check the intervals between the critical points: * For \( 0 < x < \frac{\pi}{2} \), \( \sin 2x > 0 \), so \( f(x) = \sin 2x \) is increasing. * For \( \frac{\pi}{2} < x < \pi \), \( \sin 2x < 0 \), so \( f(x) = -\sin 2x \) is decreasing. * For \( \pi < x < \frac{3\pi}{2} \), \( \sin 2x < 0 \), so \( f(x) = -\sin 2x \) is increasing. * For \( \frac{3\pi}{2} < x < 2\pi \), \( \sin 2x > 0 \), so \( f(x) = \sin 2x \) is decreasing.

Step 6: Determine Relative Extrema from Sign Analysis

Using the increasing/decreasing analysis: * \( x = \frac{\pi}{2} \) is a local maximum because \( f(x) \) changes from increasing to decreasing. * \( x = \pi \) does not lead to an extremum as there is no change in trend. * \( x = \frac{3\pi}{2} \) is a local minimum because \( f(x) \) changes from decreasing to increasing.

Step 7: Validate with Graphing Utility

Using a graphing tool, plot the function \( f(x) = |\sin 2x| \) over \( 0 < x < 2\pi \). Confirm that the graph shows a local maximum at \( x = \frac{\pi}{2} \) and a local minimum at \( x = \frac{3\pi}{2} \).

Key Concepts

Relative Extrema

Understanding relative extrema is essential when dealing with calculus problems. Relative extrema are the points in the domain of a function where the function changes behavior from increasing to decreasing and vice versa. These points are classified into two categories: local maxima and local minima. To find these points, we primarily look for critical points. A critical point is where a function's derivative is zero or undefined.

Here's how you can identify relative extrema:

• First, find the derivative of the function and solve for where the derivative equals zero. These are the critical points.
• Analyze the intervals between these critical points to determine where the function is increasing or decreasing.
• Relative maxima occur when the function changes from increasing to decreasing.
• Relative minima occur when the function changes from decreasing to increasing.

By applying these steps to the piecewise function in the exercise, the relative extrema are identified within the specified interval.

Piecewise Functions

Piecewise functions are a type of function composed of multiple sub-functions, each defined over certain intervals. Understanding piecewise functions is like solving a puzzle; we look at different parts of the function separately. The function given in the exercise, \( f(x) = |\sin 2x| \), is indeed a piecewise function due to the absolute value. We rewrite it as:

• \( f(x) = \sin 2x \) when \( \sin 2x \geq 0 \)
• \( f(x) = -\sin 2x \) when \( \sin 2x < 0 \)

This allows us to treat the function by parts. Such functions often have different behaviors to be analyzed in each segment to find critical points or extrema.

When analyzing a piecewise function, it's crucial to:

• Identify where each sub-function applies by determining the points of change, known here as critical points.
• Test the behavior of the function on each side of these points to check for changes in increasing or decreasing trends.

The segmented approach in piecewise functions is particularly useful when dealing with complex graphs, as it simplifies the problem-solving process.

Trigonometric Functions

Trigonometric functions are a significant part of calculus and appear frequently in various types of problems. In this exercise, the function \( f(x) = |\sin 2x| \) involves the trigonometric sine function. Sine functions have a periodic nature and are typically analyzed through their cycles.

For \( \sin 2x \):

• It has a period of \( \pi \), meaning it completes one full cycle every \( \pi \) units.
• The sine function oscillates between -1 and 1, crossing zero at multiples of \( \frac{\pi}{2} \).

When dealing with absolute values, the sine wave is "reflected" upwards, creating a series of peaks that generate the piecewise function.

To better understand the trigonometric functions:

• Examine key features such as amplitude, period, and phase shifts.
• Consider changes in sign since they determine how the function behaves within given intervals.

With this knowledge, you can predict and analyze the behavior of trigonometric functions, enhancing your ability to solve related problems effectively.