Problem 44 - Calculus of a Single Variable

Question

Applying the First Derivative Test In Exercises \(41-48\) , 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, 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 \(f(x) = x + 2 \sin x\) on the interval (0,2\(\pi\)) increases on the interval (\(\frac{2\pi}{3}, \frac{4\pi}{3}\)), decreases on the intervals (0,\(\frac{2\pi}{3}\)) and (\(\frac{4\pi}{3}, 2\pi\)). The function has a local minimum at \(x = \frac{2\pi}{3}\) and a local maximum at \(x = \frac{4\pi}{3}\).

Step 1: Compute the derivative

Start by calculating the derivative of the function \(f(x)=x+2 \sin x\). The derivative of \(x\) is \(1\), and that of \(\sin x\) is \(\cos x\). Therefore, \(f'(x) = 1+2 \cos x\).

Step 2: Find critical points

Critical points occur where the derivative either equals zero or is undefined. This derivative is never undefined as it is a polynomial, so set the derivative equal to zero and solve for \(x\). \n 1 + 2 \cos x=0 \n 2 \cos x=-1 \n \cos x=-1/2. \n By solving this, we find that the critical points, \(x\), are \(\frac{2 \pi}{3}\) and \(\frac{4 \pi}{3}\).

Step 3: The First Derivative Test

We will now use the First Derivative Test to determine where the function is increasing and decreasing, and thus find the relative extrema. Determine the sign of the derivative to the left and the right of each critical point within the given interval.\n\n To the left of \(\frac{2\pi}{3}\), we could pick \(\pi\), and \(f'(\pi)=1+2 \cos(\pi)<0\), so \(f(x)\) is decreasing in this interval.\n\n Between \(\frac{2\pi}{3}\) and \(\frac{4\pi}{3}\), we could pick \(\pi\), and \(f'(\pi)=1+2 \cos(\pi)>0\), so \(f(x)\) is increasing in this interval.\n\n To the right of \(\frac{4\pi}{3}\), we could pick \(2\pi\), and \(f'(2\pi)=1+2 \cos(2\pi)<0\), so \(f(x)\) is decreasing in this interval.\n\nTherefore, by the First Derivative Test, \(f(x)\) has a local minimum at \(x = \frac{2\pi}{3}\) and a local maximum at \(x=\frac{4\pi}{3}\).

Step 4: Confirmation via Plotting

Use a graphing utility to plot \(f(x) = x + 2 \sin x\) and confirm these results visually. The intervals of increase and decrease should be apparent along with the local minimum at \(x = \frac{2 \pi}{3}\) and the local maximum at \(x = \frac{4 \pi}{3}\).

Key Concepts

Critical Points

Critical points are fundamental in calculus, especially when analyzing functions' behaviors. They occur where the derivative of a function equals zero or is undefined. These points are essential because they often indicate where the graph of a function changes direction, like at a hilltop or a valley.

To find the critical points for a function, you first compute the derivative. For our function, \(f(x) = x + 2 \sin x\), the derivative is \(f'(x) = 1 + 2 \cos x\). Next, set this derivative equal to zero: \(1 + 2 \cos x = 0\). Solving this equation gives us the values for \(x\) at which the critical points occur.

In this example, solving gives \(x = \frac{2 \pi}{3}\) and \(x = \frac{4 \pi}{3}\) as critical points for the function within the interval \((0, 2 \pi)\). These are the x-values where the function may have relative extrema.

Relative Extrema

Relative extrema refer to local maximums or minimums within a specific interval of a function. These points show the highest or lowest output value of the function compared to its immediate surroundings.

To identify relative extrema, the First Derivative Test can be applied to the critical points found earlier. After determining the sign of the derivative on either side of each critical point:

• At \(x = \frac{2\pi}{3}\), the function transitions from decreasing to increasing, indicating a local minimum.
• At \(x = \frac{4\pi}{3}\), the function changes from increasing to decreasing, marking a local maximum.

These results allow us to declare that the function \(f(x) = x + 2 \sin x\) has a local minimum at \(x = \frac{2\pi}{3}\) and a local maximum at \(x = \frac{4\pi}{3}\). This understanding helps learners visually and conceptually connect calculus with the shape and turning points of graphs.

Increasing and Decreasing Intervals

Being able to determine where a function increases or decreases is essential for sketching its graph accurately. This is done by examining the sign of the derivative across different intervals.

To find increasing and decreasing intervals, you evaluate the derivative \(f'(x) = 1 + 2 \cos x\) at various points within sections of the domain divided by the critical points:\((0, \frac{2\pi}{3})\), \((\frac{2\pi}{3}, \frac{4\pi}{3})\), and \((\frac{4\pi}{3}, 2\pi)\).

• In the interval \((0, \frac{2 \pi}{3})\), choosing a test point like \(x = \pi/3\), we find \(f'(x) > 0\), indicating the function is increasing.
• In \((\frac{2\pi}{3}, \frac{4\pi}{3})\), picking \(x = \pi\) reveals \(f'(x) > 0\), so this part is also increasing.
• For \((\frac{4 \pi}{3}, 2 \pi)\), selecting \(x = 5\pi/3\) yields \(f'(x) < 0\); thus, the function decreases here.

Clearly outlining these intervals enhances understanding, making it easier to visualize how the function behaves throughout its range. It highlights where the function rises and falls, crucial for graph analysis and real-world applications.