Problem 50 - Essential Calculus

Question

Consider the function on the interval \((0,2 \pi)\) For each function, (a) find the open interval(s) on which the function is increasing or decreasing, (b) apply the First Derivative Test to identify all relative extrema, and (c) use a graphing utility to confirm your results. $$ f(x)=x+2 \sin x $$

Step-by-Step Solution

Answer

The function \(f(x)=x+2 \sin x\) is increasing on the intervals \((0, \frac{2}{3} \pi)\) and \((\frac{4}{3} \pi , 2 \pi )\) and decreasing on the interval \((\frac{2}{3} \pi , \frac{4}{3} \pi )\). The function has a relative maxima at \(x= \frac{2}{3}\pi\) and a relative minima at \(x= \frac{4}{3}\pi\).

Step 1: Find the Derivative of the Function

The derivative of the function \(f(x)=x+2 \sin x\) is found by applying the standard rules of differentiation. The derivative of the function is \(f'(x)=1+2 \cos x\).

Step 2: Find the Critical Points

Critical points occur where the derivative is equal to zero or undefined. Solve \(1+2 \cos x=0\). That gives us \(\cos x = -0.5\). Thus, the critical points on the interval \((0,2 \pi)\) are \(x= \frac{2}{3} \pi, \frac{4}{3} \pi\).

Step 3: Identify Intervals of Increase or Decrease

Inspect the sign of the derivative on the intervals \( (0, \frac{2}{3} \pi ), ( \frac{2}{3} \pi , \frac{4}{3} \pi ), ( \frac{4}{3} \pi , 2 \pi ) \). If \(f'(x) > 0\), the function is increasing; if \(f'(x) < 0\), the function is decreasing. Thus, the function increases on \((0, \frac{2}{3} \pi)\) and \((\frac{4}{3} \pi , 2 \pi )\) and decreases on \((\frac{2}{3} \pi , \frac{4}{3} \pi )\).

Step 4: Apply the First Derivative Test

By The First Derivative Test, \(f(x)\) has a relative maxima at \(x= \frac{2}{3}\pi\) and a relative minima at \(x= \frac{4}{3}\pi\).

Step 5: Confirm Results with Graph

A graphical representation will confirm the findings. By plotting the function \(f(x)=x+2 \sin x\) on a graphing tool, you can confirm that the function has relative maxima and minima at the calculated points.

Key Concepts

First Derivative Test

The First Derivative Test is an efficient method to determine whether a critical point of a function is a relative maximum, minimum, or neither. This test involves examining the sign of the derivative (denoted as \(f'(x)\) for a function \(f(x)\)) before and after each critical point.
Here's an easy way to remember the steps of the First Derivative Test:

• Find the critical points by setting \(f'(x) = 0\) or where \(f'(x)\) is undefined.
• Determine the sign of \(f'(x)\) on each interval created by these critical points.
• Apply the test by checking the sign change:
• If \(f'(x)\) changes from positive to negative at the critical point, the function has a relative maximum.
• If \(f'(x)\) changes from negative to positive, the function has a relative minimum.
• If the sign does not change, the critical point is not a relative extremum.

By using the First Derivative Test on \(f(x) = x + 2 \sin x\), the critical points are found at \(x = \frac{2}{3}\pi\) and \(x = \frac{4}{3}\pi\). The test reveals a relative maximum at \(x = \frac{2}{3}\pi\) and a relative minimum at \(x = \frac{4}{3}\pi\). This is due to changes in the sign of \(f'(x)\) as described.

Critical Points

Critical points hold valuable information about the behavior of a function. They are essentially points on the graph where the derivative \(f'(x)\) is zero or undefined, indicating potential relative maxima, minima, or saddle points.
To find critical points, follow these steps:

• Compute the derivative of the function.
• Solve for \(x\) where \(f'(x) = 0\) or \(f'(x)\) does not exist.

For the function \(f(x) = x + 2 \sin x\), the derivative is \(f'(x) = 1 + 2 \cos x\). Setting this equal to zero, we solve \(1 + 2 \cos x = 0\) to find \(\cos x = -0.5\). In the interval \((0, 2\pi)\), this occurs at \(x = \frac{2}{3}\pi\) and \(x = \frac{4}{3}\pi\). These are the critical points of the function.
Analyzing these points helps to understand where the function changes direction—either increasing or decreasing its value—marking regions of potential extrema.

Intervals of Increase/Decrease

Understanding where a function increases or decreases is fundamental to grasping its overall behavior. By examining the sign of the derivative, \(f'(x)\), one can identify these intervals.
Here's how to determine the intervals of increase and decrease for a function:

• Calculate the derivative \(f'(x)\).
• Find all critical points—these will divide the domain into distinct intervals.
• Test the sign of \(f'(x)\) within each interval:
• If \(f'(x) > 0\), the function is increasing.
• If \(f'(x) < 0\), the function is decreasing.

For the function \(f(x) = x + 2 \sin x\), the derivative \(f'(x) = 1 + 2 \cos x\) allows us to determine the function's behavior in the following intervals:
- On \((0, \frac{2}{3}\pi)\), \(f'(x) > 0\), so the function is increasing.
- On \((\frac{2}{3}\pi, \frac{4}{3}\pi)\), \(f'(x) < 0\), indicating the function is decreasing.
- On \((\frac{4}{3}\pi, 2\pi)\), \(f'(x) > 0\), hence it is increasing again.
This analysis helps in sketching the graph of the function and understanding its growth and decay trends across the domain.