Problem 13 - University Calculus: Early Transcendentals

Question

Answer the following questions about the functions whose derivatives are given. 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: \(\frac{\pi}{2}, \frac{2\pi}{3}, \frac{4\pi}{3}\). Increase: \((\frac{\pi}{2}, \frac{2\pi}{3})\) and \((\frac{4\pi}{3}, 2\pi)\). Decrease: \((0, \frac{\pi}{2})\) and \((\frac{2\pi}{3}, \frac{4\pi}{3})\). Max: \(\frac{2\pi}{3}\). Min: \(\frac{\pi}{2}, \frac{4\pi}{3}\).

Step 1: Identify Critical Points

To find critical points of a function, set its derivative equal to zero. Given that \(f'(x) = (\sin x - 1)(2\cos x + 1)\), we solve two equations: \[ \sin x - 1 = 0 \] and \[ 2\cos x + 1 = 0 \] For \(\sin x - 1 = 0\), we have \(\sin x = 1\) which gives \(x = \frac{\pi}{2}\). For \(2\cos x + 1 = 0\), \(\cos x = -\frac{1}{2}\) yielding solutions \(x = \frac{2\pi}{3}\) and \(x = \frac{4\pi}{3}\) within the interval \([0, 2\pi]\). Thus, the critical points are \(x = \frac{\pi}{2}, \frac{2\pi}{3}, \text{and } \frac{4\pi}{3}\).

Step 2: Determine Intervals of Increase or Decrease

To determine where \(f\) is increasing or decreasing, analyze the sign of \(f'(x)\) on intervals between the critical points found in Step 1. - For \(0 < x < \frac{\pi}{2}\), choose \(x = \frac{\pi}{4}\): \(f'(\frac{\pi}{4}) < 0\), so \(f\) is decreasing.- For \(\frac{\pi}{2} < x < \frac{2\pi}{3}\), choose \(x = \frac{3\pi}{4}\): \(f'(\frac{3\pi}{4}) > 0\), so \(f\) is increasing.- For \(\frac{2\pi}{3} < x < \frac{4\pi}{3}\), choose \(x = \pi\): \(f'(\pi) < 0\), so \(f\) is decreasing.- For \(\frac{4\pi}{3} < x < 2\pi\), choose \(x = \frac{3\pi}{2}\): \(f'(\frac{3\pi}{2}) > 0\), so \(f\) is increasing.

Step 3: Identify Local Maxima and Minima

A local maximum occurs when \(f'(x)\) changes from positive to negative, and a local minimum occurs when \(f'(x)\) changes from negative to positive.- At \(x = \frac{\pi}{2}\), \(f'(x)\) changes from negative to positive, indicating a local minimum.- At \(x = \frac{2\pi}{3}\), \(f'(x)\) changes from positive to negative, indicating a local maximum.- At \(x = \frac{4\pi}{3}\), \(f'(x)\) changes from negative to positive, indicating a local minimum.

Key Concepts

Derivative

In calculus, the derivative of a function is a measure of how a function's output changes with respect to its input. Think of it as the function's rate of change or its "slope" at any point along the curve. The derivative provides valuable insights into the behavior of the function, allowing us to gauge how the function acts within a particular domain.
To find critical points in a function, which are points where the derivative is either zero or undefined, we focus on solving the derivative equation set to zero. In the exercise provided, we analyze the derivative function: \[ f'(x) = (\sin x - 1)(2\cos x + 1) \] To locate critical points, each expression within the product is set to zero:

• \( \sin x - 1 = 0 \Rightarrow \sin x = 1 \)
• \( 2\cos x + 1 = 0 \Rightarrow \cos x = -\frac{1}{2} \)

Solving these equations gives us the critical points \( x = \frac{\pi}{2}, \frac{2\pi}{3}, \text{and } \frac{4\pi}{3} \). These points are essential to understanding how the function behaves and finding where it might be at a peak or trough.

Increasing and Decreasing Intervals

Once critical points are identified, determining whether a function is increasing or decreasing on certain intervals requires analyzing the sign of its derivative. This tells us if the function's output is rising or falling as we move along the x-axis.
In our exercise, the derivative helps us break the interval \([0, 2\pi]\) into smaller intervals based on the critical points found:

• From \(0 < x < \frac{\pi}{2}\), test point \(x = \frac{\pi}{4}\): The derivative is negative, so the function is decreasing.
• From \(\frac{\pi}{2} < x < \frac{2\pi}{3}\), test point \(x = \frac{3\pi}{4}\): The derivative is positive, meaning the function is increasing.
• From \(\frac{2\pi}{3} < x < \frac{4\pi}{3}\), test point \(x = \pi\): The derivative turns negative again, indicating a decreasing interval.
• From \(\frac{4\pi}{3} < x < 2\pi\), test point \(x = \frac{3\pi}{2}\): Here, the derivative is positive, showing the function is increasing.

This analysis provides a map of where the function rises or falls, which is crucial for identifying peaks and valleys (local maximum and minimum points).

Local Maxima and Minima

Local maxima and minima are the "hills" and "valleys" in the graph of a function, where, locally, the function's output is respectively higher or lower than at nearby points. To find these points, we observe how the derivative's sign changes across critical points. A shift from positive to negative at a critical points indicates a local max, and a change from negative to positive indicates a local min.
In the exercise example:

• At \(x = \frac{\pi}{2}\), the derivative shifts from negative to positive, thus indicating a local minimum since the function bottoms out before rising again.
• At \(x = \frac{2\pi}{3}\), the derivative changes from positive to negative. Here, the function reaches a local maximum as it peaks before declining.
• At \(x = \frac{4\pi}{3}\), we observe a change from negative to positive in the derivative, suggesting a local minimum where the function dips before climbing again.

Recognizing these local extrema helps in understanding a function's behavior, especially for optimizing situations where maxima or minima are sought within a set range.