Problem 31 - Single Variable Calculus

Question

(a) find the intervals on which \(f\) is increasing or decreasing, and (b) find the relative maxima and relative minima of \(\vec{f}\). $$ f(x)=x-2 \sin x, \quad 0

Step-by-Step Solution

Answer

The function \(f(x) = x - 2\sin{x}\) is increasing on the intervals \((0, \frac{\pi}{3})\) and \((\frac{5\pi}{3}, 2\pi)\), and decreasing on the interval \((\frac{\pi}{3}, \frac{5\pi}{3})\). There is a relative maximum at \(x = \frac{\pi}{3}\), with a value of \(f\left(\frac{\pi}{3}\right) = \frac{\pi}{3} - 2\sin\left(\frac{\pi}{3}\right)\), and a relative minimum at \(x = \frac{5\pi}{3}\), with a value of \(f\left(\frac{5\pi}{3}\right) = \frac{5\pi}{3} - 2\sin\left(\frac{5\pi}{3}\right)\).

Step 1: Find the derivative of the function

We must first find the derivative of the function, \(f(x) = x - 2sin(x)\), to help us later finding its critical points. Using the chain rule, we obtain: $$ f'(x) = \frac{d}{dx} (x - 2\sin x) = 1 - 2\cos x $$

Step 2: Find the critical points

The critical points occur where the derivative \(f'(x)\) is either equal to zero or is undefined. In this case, the derivative is defined for all \(x\). We need to solve \(f'(x) = 0\) for \(x\): $$ 1 - 2\cos x = 0 \Rightarrow \cos x = \frac{1}{2} $$ Within the given interval \((0, 2\pi)\), the solutions of the equation are \(x_1 = \frac{\pi}{3}\) and \(x_2 = \frac{5\pi}{3}\). These are the critical points.

Step 3: Determine intervals where the function is increasing or decreasing

By using a number line and the critical points found above, we evaluate the derivative to the left and to the right of each critical point to determine the sign of the derivative. If the derivative is positive, the function is increasing, and if the derivative is negative, the function is decreasing. For \(0 < x < \frac{\pi}{3}\), let's pick \(x = \frac{\pi}{6}\) as a test point. \(f'(\frac{\pi}{6}) = 1 - 2 \cos(\frac{\pi}{6}) > 0\), so the function is increasing on this interval. For \(\frac{\pi}{3} < x < \frac{5\pi}{3}\), let's pick \(x = \pi\) as a test point. \(f'(\pi) = 1 - 2 \cos(\pi) < 0\), so the function is decreasing on this interval. For \(\frac{5\pi}{3} < x < 2\pi\), let's pick \(x = \frac{7\pi}{3}\) as a test point. \(f'(\frac{7\pi}{3}) = 1 - 2 \cos(\frac{7\pi}{3}) > 0\), so the function is increasing on this interval. Therefore, \(f(x)\) is increasing on the intervals \((0, \frac{\pi}{3})\) and \((\frac{5\pi}{3}, 2\pi)\), and decreasing on the interval \((\frac{\pi}{3}, \frac{5\pi}{3})\).

Step 4: Find relative maxima and minima

Now, we'll determine if the function has a relative maximum or minimum at the critical points. At \(x_1 = \frac{\pi}{3}\), the function changes from increasing to decreasing, so there is a relative maximum at this point. Evaluating the function at this point, we find that the relative maximum is \(f\left(\frac{\pi}{3}\right) = \frac{\pi}{3} - 2\sin\left(\frac{\pi}{3}\right)\). At \(x_2 = \frac{5\pi}{3}\), the function changes from decreasing to increasing, so there is a relative minimum at this point. Evaluating the function at this point, we find that the relative minimum is \(f\left(\frac{5\pi}{3}\right) = \frac{5\pi}{3} - 2\sin\left(\frac{5\pi}{3}\right)\). Thus, we have found the intervals where the function is increasing or decreasing and the relative maxima and minima.

Key Concepts

Critical Points

Understanding the concept of critical points is essential in calculus because they help identify where a function's graph may have relative maxima, minima, or points of inflection. Critical points occur when the derivative of a function is either equal to zero or is undefined. In the context of the function f(x) = x - 2sin(x), finding its critical points meant calculating its derivative, f'(x) = 1 - 2cos(x), and solving the equation f'(x) = 0. Within the interval (0, 2π), the critical points were found to be at x = π/3 and x = 5π/3. These points can potentially be where the function's behavior changes from increasing to decreasing or vice versa.

Determining the nature of these critical points—whether they correspond to relative maxima or minima or neither—requires further investigation into the function's behavior around these points, which leads us to analyze the intervals of increase and decrease.

Relative Maxima and Minima

When analyzing a function, it's useful to know where its peaks and troughs are, which are called relative maxima and minima, respectively. After locating the critical points, we examine the function's behavior around these points to classify them. If the function switches from increasing to decreasing at a critical point, that point is a relative maximum; conversely, if the function goes from decreasing to increasing, you've found a relative minimum.

In the example exercise focusing on f(x) = x - 2sin(x), we found that at x = π/3, the function has a relative maximum, and at x = 5π/3, there is a relative minimum. These are not just random points but locations where the function's rate of change (derivative) switches sign, indicating a shift in the graph's direction, which is a crucial insight for understanding a function's overall behavior.

Derivative of a Function

The derivative of a function represents the rate of change or the slope of the function at any given point. It's a cornerstone concept in calculus and helps determine many aspects of a function's graph, such as increasing/decreasing intervals and curvature. For the function f(x) = x - 2sin(x), we calculated the derivative to be f'(x) = 1 - 2cos(x). This derivative function tells us how f(x) changes as x varies. Where the derivative is positive, f(x) is increasing; where it's negative, f(x) is decreasing.

Analyzing the derivative further, we divide the domain into intervals separated by the critical points and test the sign of the derivative within each interval. This step is why finding the derivative is vital—it informs us about the behavior of the function and allows us to construct a more complete picture of its behavior, such as identifying increasing (e.g., (0, π/3) and (π/5, 2π)) and decreasing (e.g., (π/3, 5π/3)) intervals for the function f(x).