Problem 35 - Calculus Early Transcendentals

Question

Find the relative extrema using both first and second derivative tests. $$ f(x)=\sin 2 x, \quad 0

Step-by-Step Solution

Answer

The relative maximum is at \(x = \frac{\pi}{4}\) and the relative minimum is at \(x = \frac{3\pi}{4}\).

Step 1: Find the First Derivative

To find the relative extrema, we first need the first derivative of the function. Given the function \( f(x) = \sin 2x \), use the chain rule to differentiate: \( f'(x) = 2\cos 2x \). This derivative will help us find the critical points.

Step 2: Set the First Derivative to Zero

To identify potential relative extrema, set \( f'(x) = 2\cos 2x \) equal to zero to find critical points. \( 2\cos 2x = 0 \) implies \( \cos 2x = 0 \). The angles where cosine is zero are \(2x = \frac{\pi}{2} + n\pi\); solving for \(x\) within \((0, \pi)\), we find critical points at \(x = \frac{\pi}{4}\) and \(x = \frac{3\pi}{4}\).

Step 3: Find the Second Derivative

The next step is to find the second derivative for the second derivative test. Differentiate \( f'(x) = 2\cos 2x \) again: \( f''(x) = -4\sin 2x \). This will help us determine whether each critical point is a relative maximum or minimum.

Step 4: Apply the Second Derivative Test

Evaluate the second derivative at each critical point to determine concavity. For \( x = \frac{\pi}{4} \), \( f''\left(\frac{\pi}{4}\right) = -4\sin\left(\frac{\pi}{2}\right) = -4\) which is less than zero, indicating a relative maximum. For \( x = \frac{3\pi}{4} \), \( f''\left(\frac{3\pi}{4}\right) = -4\sin\left(\frac{3\pi}{2}\right) = 4\) which is greater than zero, indicating a relative minimum.

Step 5: Verify with the First Derivative Test

Confirm findings by checking the sign of \( f'(x) \) around each critical point. For \( x = \frac{\pi}{4} \), choose \( x = \frac{\pi}{6} \) and \( x = \frac{\pi}{3} \). For \( f'(x) \) changes from positive to negative, confirming a maximum. For \( x = \frac{3\pi}{4} \), choose \( x = \frac{2\pi}{3} \) and \( x = \frac{5\pi}{6} \). \( f'(x) \) changes from negative to positive, confirming a minimum.

Key Concepts

Relative Extrema

When studying calculus, one important concept is understanding relative extrema. These extremas are the points on a graph where a function changes direction, and they can either be a peak (maximum) or a trough (minimum). In other words, at these points, the function exhibits either the tallest or the lowest value in some neighborhood around them. Finding relative extrema involves calculus techniques like the first and second derivative tests.

For example, if we examine the function \(f(x) = \sin 2x\) within the interval \(0 < x < \pi\), determining where the relative maxima or minima occur helps in understanding the behavior of the function which can be quite useful in many practical applications like signal processing or physics.

To find relative extrema, we must first locate the critical points. These are points where the derivative equals zero or does not exist. Once we identify these points, we can apply derivative tests to confirm if they are relative maxima or minima.

First Derivative Test

The first derivative test is an important step for identifying relative extrema of a function. This test is applied by examining the behavior of the derivative before and after a critical point.

To perform this test:

• Calculate the first derivative of the given function. For example, with \(f(x) = \sin 2x\), the first derivative is \(f'(x) = 2\cos 2x\).
• Set this derivative to zero to find critical points: \(2\cos 2x = 0\), leading to critical values of \(x = \frac{\pi}{4}\) and \(x = \frac{3\pi}{4}\) within the interval \((0, \pi)\).
• Investigate the intervals around these critical points. Evaluate the sign of the derivative just before and after each critical point.

If the function changes from increasing to decreasing at a critical point (derivative changes from positive to negative), it indicates a relative maximum. Conversely, if it changes from decreasing to increasing (derivative changes from negative to positive), it indicates a relative minimum.

Applying this test to our function \(f(x)\) confirms that \(x = \frac{\pi}{4}\) is a relative maximum and \(x = \frac{3\pi}{4}\) is a relative minimum.

Second Derivative Test

The second derivative test is another reliable method used to determine the nature of critical points found through the first derivative test. By studying the second derivative, we can learn about the concavity of the function around critical points, which helps distinguish maxima from minima.

Here’s how to use it:

• Find the second derivative of the function. For \(f(x) = \sin 2x\), the second derivative is \(f''(x) = -4\sin 2x\).
• Evaluate the second derivative at each critical point. This reveals the concavity at these points.

If \(f''(x) < 0\), the function is concave down at the critical point, indicating a relative maximum. If \(f''(x) > 0\), the function is concave up, suggesting a relative minimum.

For our example, at \(x = \frac{\pi}{4}\), \(f''\left(\frac{\pi}{4}\right) = -4\), indicating a relative maximum. At \(x = \frac{3\pi}{4}\), \(f''\left(\frac{3\pi}{4}\right) = 4\), showing a relative minimum. This test provides a solid confirmation of the extrema obtained from the first derivative test.