Problem 13 - Thomas Calculus in SI Units

Question

Answer the following questions about the functions whose derivatives are given in Exercises \(1-14:\) a. What are the critical points of \(f ?\) b. On what open intervals is \(f\) increasing or decreasing? c. At what points, if any, does \(f\) assume local maximum and minimum values? $$f^{\prime}(x)=(\sin x-1)(2 \cos x+1), 0 \leq x \leq 2 \pi$$

Step-by-Step Solution

Answer

Critical points: \(x=\frac{\pi}{2}, \frac{2\pi}{3}, \frac{4\pi}{3}\). Increasing on \((\frac{\pi}{2}, \frac{4\pi}{3})\), decreasing on \((0, \frac{\pi}{2})\) and \((\frac{4\pi}{3}, 2\pi)\). Local minimum at \(x=\frac{\pi}{2}\), local maximum at \(x=\frac{4\pi}{3}\).

Step 1: Identify Critical Points

Critical points of a function occur where the derivative is zero or undefined. Set \( f'(x) = 0 \) to find critical points:\[(\sin x - 1)(2 \cos x + 1) = 0\]This gives two equations:1. \( \sin x - 1 = 0 \), so \( \sin x = 1 \).2. \( 2 \cos x + 1 = 0 \), so \( \cos x = -\frac{1}{2} \).For \( \sin x = 1 \), \( x = \frac{\pi}{2} \).For \( \cos x = -\frac{1}{2} \), \( x = \frac{2\pi}{3} \) and \( x = \frac{4\pi}{3} \).So, the critical points are \( x = \frac{\pi}{2}, \frac{2\pi}{3}, \text{and} \frac{4\pi}{3} \).

Step 2: Determine Increasing or Decreasing Intervals

Use the critical points to test intervals and determine whether \( f(x) \) is increasing or decreasing.Test intervals:1. \( (0, \frac{\pi}{2}) \): Choose \( x = \frac{\pi}{4} \). - \( f'(\frac{\pi}{4}) = (\sin \frac{\pi}{4} - 1)(2 \cos \frac{\pi}{4} + 1) < 0 \), decreasing.2. \( (\frac{\pi}{2}, \frac{2\pi}{3}) \): Choose \( x = \pi \). - \( f'(\pi) = (\sin \pi - 1)(2 \cos \pi + 1) > 0 \), increasing.3. \( (\frac{2\pi}{3}, \frac{4\pi}{3}) \): Choose \( x = \pi \). - \( f'(\pi) = (\sin \pi - 1)(2 \cos \pi + 1) > 0 \), increasing.4. \( (\frac{4\pi}{3}, 2\pi) \): Choose \( x = \frac{3\pi}{2} \). - \( f'(\frac{3\pi}{2}) = (\sin \frac{3\pi}{2} - 1)(2 \cos \frac{3\pi}{2} + 1) < 0 \), decreasing.Thus, increasing on \((\frac{\pi}{2}, \frac{2\pi}{3})\) and \((\frac{2\pi}{3}, \frac{4\pi}{3})\), decreasing on \((0, \frac{\pi}{2})\) and \((\frac{4\pi}{3}, 2\pi)\).

Step 3: Identify Local Maximum and Minimum Points

Evaluate slope changes at critical points for maxima or minima:1. At \( x = \frac{\pi}{2} \), \( f'(x) \) changes from negative to positive: local minimum.2. At \( x = \frac{2\pi}{3} \), \( f'(x) \) continues positive: neither maximum nor minimum.3. At \( x = \frac{4\pi}{3} \), \( f'(x) \) changes from positive to negative: local maximum.

Key Concepts

Derivative Test

The Derivative Test is a handy tool to determine the behavior of functions around critical points. Critical points occur where the derivative of a function equals zero or is undefined. To apply the derivative test to the function from the original exercise, start by taking the derivative provided, \[ f'(x) = (\sin x - 1)(2 \cos x + 1), \]which needs to be set to zero to find the critical points. Solve the equation by separating into two:

• \( \sin x - 1 = 0 \) gives you \sin x = 1, leading to \ x = \frac{\pi}{2} \.
• \(2 \cos x + 1 = 0 \) provides \ cos x = -\frac{1}{2}\, which results in x = \frac{2\pi}{3} \ and \ x = \frac{4\pi}{3} \.

These values are identified as critical points. Understanding and finding these critical points is the first step in analyzing the function's behavior.

Increasing and Decreasing Intervals

Once critical points are identified, determining the intervals where the function is increasing or decreasing follows. We use these critical points to test different intervals:

• For \( (0, \frac{\pi}{2})\), testing shows the derivative is negative, indicating that the function is decreasing.
• In the interval \((\frac{\pi}{2}, \frac{2\pi}{3})\), the derivative is positive, so the function is increasing.
• Between \( \frac{2\pi}{3} \) and \( \frac{4\pi}{3} \) it stays positive, remaining increasing.
• Finally, from \( \frac{4\pi}{3}, 2\pi\) \, the derivative turns negative again, signaling a decrease.

Visualizing these intervals can clarify how the function's slope changes across the graph, where it climbs or falls based on the sign of the derivative. Identifying these intervals helps pinpoint where the function changes its increasing to decreasing trends and vice versa.

Local Maximum and Minimum Values

At this stage, we use the information from the derivative's sign changes to determine local maxima and minima. At each critical point, observe how the derivative transitions:

• At \( x = \frac{\pi}{2} \), the derivative transitions from negative to positive, implying a local minimum.
• At \ x = \frac{2\pi}{3} \, there is no sign change in the derivative, suggesting no local extremum here.
• For \ x = \frac{4\pi}{3} \, the derivative shifts from positive to negative, indicating a local maximum.

Understanding these points allows us to map the peaks and valleys of a function, giving insight into the function's overall shape. Knowing where a function achieves its highest or lowest values can be crucial in various practical applications, reflecting the function's efficiency or output within specific boundaries.