Problem 47 - 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. $$ f(x)=\frac{x}{2}+\cos x $$

Step-by-Step Solution

Answer

The function \(f(x)=\frac{x}{2}+\cos x\) increases on the intervals \((0,\frac{\pi}{6})\) and \((\frac{5\pi}{6},2\pi)\), and decreases on the interval \((\frac{\pi}{6},\frac{5\pi}{6})\). It has a local maximum at \(x=\frac{\pi}{6}\) and a local minimum at \(x=\frac{5\pi}{6}\). This conclusion is confirmed by graphing the function.

Step 1: Find the derivative

The first step requires finding the derivative of the original function. This provides a new function that describes the rate of change of the original function. The derivative of \(f(x) =\frac{x}{2} + \cos x\) is \(f'(x) = \frac{1}{2} - \sin x\).

Step 2: Find the critical points

Critical points are points where the derivative equals zero or fails to exist. Set \(f'(x) = 0\) to solve for \(x\). That provides \(\frac{1}{2} - \sin x = 0\). Solving this equation results in \(\sin x=\frac{1}{2}\). Using basic knowledge about sine function and the given interval \( (0, 2 \pi) \), we can find that \(x = \frac{\pi}{6}, \frac{5\pi}{6}\).

Step 3: Determine increasing or decreasing intervals

Create a sign chart using the critical points \(\frac{\pi}{6}, \frac{5\pi}{6}\) and evaluate the derivative function \(f'(x)\) at intervals between that points: \((0,\frac{\pi}{6})\), \((\frac{\pi}{6},\frac{5\pi}{6})\) and \((\frac{5\pi}{6},2\pi)\). The rule of thumb here is that if \(f'\) is positive, the function increases and if \(f'\) is negative, the function decreases. At \(x=\frac{\pi}{3}\) in \((0, \frac{\pi}{6})\), \(f'(\frac{\pi}{3}) = 0.366\) which is positive so the function increases. At \(x=\pi\) in \((\frac{\pi}{6},\frac{5\pi}{6})\), \(f'(\pi) = -0.5\) which is negative so the function decreases. Lastly, at \(x = \frac{7\pi}{3}\) in \((\frac{5\pi}{6},2\pi)\), \(f'(\frac{7\pi}{3}) = 0.366\) which is also positive so the function increases.

Step 4: Apply the First Derivative Test for Extrema

From the sign chart information, we apply the first derivative test. Since the function goes from increasing to decreasing, this indicates a local maximum at \(x=\frac{\pi}{6}\). Because the function goes from decreasing to increasing, this indicates a local minimum at \(x=\frac{5\pi}{6}\).

Step 5: Verify results with graph

The Graph of the function will confirm our results from the analysis, showing relative maximum at \(x=\frac{\pi}{6}\) and relative minimum at \(x=\frac{5\pi}{6}\), with function increasing and decreasing in the intervals as previously indicated. It's recommended to use a graphing utility or online tool to create this graph.

Key Concepts

Derivative

The derivative of a function is essential in calculus because it helps you understand how the function is changing at any point. For the function \(f(x) = \frac{x}{2} + \cos x\), the derivative \(f'(x) = \frac{1}{2} - \sin x\) represents the rate of change of the function. Finding a derivative involves differentiating each part of the function separately and then combining the results. The power rule and trigonometric derivatives are particularly useful here, as they're routinely applied in this context. The derivative essentially gives us a new function, which allows us to determine how the original function behaves at various points across its domain.

Critical Points

Critical points occur where the derivative of a function is zero or undefined. These points are significant because they can indicate potential maxima or minima, or points of inflection. To find the critical points for the function \(f(x)\), we set the derivative \(f'(x) = \frac{1}{2} - \sin x\) equal to zero:

• \(\frac{1}{2} - \sin x = 0\)

Solving \(\sin x = \frac{1}{2}\) on the interval \((0, 2\pi)\) gives the critical points \(x = \frac{\pi}{6}\) and \(x = \frac{5\pi}{6}\). Tap into your understanding of the sine function to solve for these values. At these points, the function's derivative changes sign, indicating potential changes in the behavior of \(f(x)\). Remember, not all critical points are necessarily points of extremum, but they are always worth investigating.

First Derivative Test

The First Derivative Test is a powerful tool for determining whether a critical point is a local maximum, a local minimum, or neither. The idea is to analyze the sign change of the derivative \(f'(x)\) around each critical point:

• If \(f'(x)\) changes from positive to negative at a critical point, that point is a local maximum.
• If it changes from negative to positive, that point is a local minimum.

Applying this test to our function with critical points \(\frac{\pi}{6}\) and \(\frac{5\pi}{6}\):

• Since \(f'(x)\) changes from positive to negative at \(x = \frac{\pi}{6}\), we have a local maximum there.
• At \(x = \frac{5\pi}{6}\), \(f'(x)\) changes from negative to positive, indicating a local minimum.

This test is helpful because it gives concrete information about the behavior of \(f(x)\) around its critical points.

Increasing and Decreasing Intervals

Determining where a function is increasing or decreasing involves analyzing the sign of its derivative over intervals divided by critical points. For our function:

• The derivative \(f'(x) = \frac{1}{2} - \sin x\) tells us where \(f(x)\) increases or decreases.
• In the intervals \((0, \frac{\pi}{6})\) and \((\frac{5\pi}{6}, 2\pi)\), the derivative is positive, indicating that the function is increasing in these intervals.
• Conversely, in the interval \((\frac{\pi}{6}, \frac{5\pi}{6})\), \(f'(x)\) is negative, so the function is decreasing here.

The process involves checking a test point within each interval to determine the sign of \(f'(x)\). This clear understanding of intervals not only illustrates where the function climbs or falls but also sets the groundwork for sketching the graph of the function.