Problem 15 - Calculus of a Single Variable

Question

Intervals on Which \(f\) Is Increasing or Decreasing In Exercises \(9-16\) , identify the open intervals on which the function is increasing or decreasing. $$ y=x-2 \cos x, \quad 0 < x < 2 \pi $$

Step-by-Step Solution

Answer

The function \(y=x-2 \cos x\) is increasing on the intervals \(0 < x < 7/6 \pi\) and \(11/6 \pi < x < 2\pi\), and decreasing on the interval \(7/6 \pi< x < 11/6 \pi\).

Step 1: Find the derivative of the function

The first step is to find the derivative of \(y=x-2 \cos x\). The derivative of 'x' is 1 and the derivative of \(-2\cos x\) is \(2\sin x\). Hence, the derivative of the given function, denoted as \(y'\), is \(1+2 \sin x\).

Step 2: Solve for critical points

Next, solve for 'x' when the derivative equals zero, these points are called critical points. Equate \(1+2 \sin x\) to zero, then solve for x. \(\sin x = -1/2\). The solutions to this equation in the interval \(0 < x < 2 \pi\) are \(x = 7/6 \pi, 11/6 \pi\).

Step 3: Apply the first derivative test

The interval \(0 < x < 2 \pi\) is split into three intervals by the critical points: \(0 < x < 7/6 \pi\), \(7/6 \pi < x < 11/6 \pi\), and \(11/6 \pi < x < 2\pi\). Check the sign of the derivative at a test point chosen from each of the three intervals: if the derivative is positive, the function is increasing, if it is negative, the function is decreasing. Upon testing, we find that \(y'=1+2\sin x>0\) when \(0 < x < 7/6 \pi\) and \(11/6 \pi < x < 2\pi\), \(y'=1+2\sin x<0\) when \(7/6 \pi< x < 11/6 \pi\).

Key Concepts

Increasing and Decreasing Functions

In calculus, we often want to find out where a function increases or decreases along its graph. This means looking for intervals where the slope of the tangent to the curve is positive (increasing) or negative (decreasing). This is hugely important as it helps in sketching the curve and understanding the behavior of the function.

To find these intervals, we first need the original function, in this case, it’s given as \(y = x - 2\cos x\), and then find its derivative. The derivative helps us know the slope of the tangent.

• If the derivative \(y' > 0\), the function is said to be increasing on that interval.
• If the derivative \(y' < 0\), the function is decreasing.

With this particular exercise, you're looking at the interval \(0 < x < 2\pi\), and you use the critical points to determine these intervals more precisely.

Critical Points

Critical points are very interesting parts of a function's graph. A critical point occurs where the derivative of the function is zero or does not exist. These points are significant because they often indicate a change from increasing to decreasing or vice versa.

In this problem, after finding the derivative \(y' = 1 + 2\sin x\), we set it to zero to find the critical points. Solving \(1 + 2\sin x = 0\), leads us to the solutions \(x = \frac{7\pi}{6}\) and \(x = \frac{11\pi}{6}\) within the interval \(0 < x < 2\pi\).

These critical points split the interval into smaller segments, helping to analyze each segment to see where the function’s behavior changes.

First Derivative Test

The first derivative test is an effective tool in determining whether a function is increasing or decreasing on certain intervals between critical points. This test involves simply checking the sign of the derivative within these intervals.

For this exercise, once you've identified the critical points \(x = \frac{7\pi}{6}\) and \(x = \frac{11\pi}{6}\), you can divide the domain \(0 < x < 2\pi\) into intervals: \(0 < x < \frac{7\pi}{6}\), \(\frac{7\pi}{6} < x < \frac{11\pi}{6}\), and \(\frac{11\pi}{6} < x < 2\pi\).

Select a test point in each of these intervals:

• In \(0 < x < \frac{7\pi}{6}\), you might pick \(x = \frac{\pi}{2}\). If \(y' > 0\), the function is increasing.
• In \(\frac{7\pi}{6} < x < \frac{11\pi}{6}\), try \(x = \pi\); if \(y' < 0\), the function is decreasing.
• In \(\frac{11\pi}{6} < x < 2\pi\), you might choose \(x = \frac{3\pi}{2}\), and if \(y' > 0\), the function is increasing.

These assessments using the first derivative test give you the behavior of the function over the specified intervals.