Problem 1 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Let \(f(x)=x+2 \sin x\). (a) Find all of the critical points. (b) Where is \(f(x)\) increasing? Decreasing? (c) Where does \(f(x)\) have local maxima? Local minima? (d) Does \(f(x)\) have global maxima? Global minima? If so, what are the absolute maximum and minimum values? (e) Where is \(f(x)\) concave up? Concave down? (f) Sketch a graph of \(f(x)\).

Step-by-Step Solution

Answer

The critical points are \(\pm \frac{2\pi}{3} + 2n \pi\). The function is increasing on \(\left(-\frac{2\pi}{3},\frac{2\pi}{3}\right)\) and decreasing on \(\left(-\infty,-\frac{2\pi}{3}\right) \cup \left(\frac{2\pi}{3},\infty\right)\). Local maximum at \(x = -\frac{2\pi}{3}\) and local minimum at \(x = \frac{2\pi}{3}\). There are no global maxima or minima. Concave up on \((- \infty, -\pi) \cup (0, \pi) \cup (2\pi, 3\pi) \cup ...\); concave down on \((- \pi, 0) \cup (\pi, 2\pi) \cup (3\pi, 4\pi) \cup ...\)

Step 1: Find the critical points

The critical points are found by setting the derivative of the function equal to zero. First, find the derivative of the function \(f(x) = x + 2 \sin x\) which is \(f'(x) = 1 + 2 \cos x\). Setting this equal to zero gives \(1 + 2 \cos x = 0\) which solves to \(x = \pm \frac{2\pi}{3} + 2n \pi\) where \(n\) is an integer.

Step 2: Determine increasing and decreasing intervals

To find the intervals where \(f(x)\) is increasing and decreasing, plot the critical points on a number line and test points in each interval. If \(f'(x) > 0\), then \(f(x)\) is increasing and if \(f'(x) < 0\), then \(f(x)\) is decreasing. Therefore, the function \(f(x) = x + 2 \sin x\) is decreasing on \(\left(- \infty, - \frac{2\pi}{3}\right) \cup \left(\frac{2\pi}{3}, \infty\right)\) and is increasing on \(\left(- \frac{2\pi}{3}, \frac{2\pi}{3}\right)\).

Step 3: Find local maxima and minima

The function \(f(x)\) will have local maxima and minima where \(f'(x)\) changes sign. From these critical points and the intervals of increasing and decreasing, conclude that the function has a local maximum at \(x = -\frac{2\pi}{3}\) and a local minimum at \(x = \frac{2\pi}{3}\).

Step 4: Find the absolute maximum and minimum values

The function \(f(x) = x + 2 \sin x\) does not have global maxima or minima because the function is not bounded as \(x \to \pm \infty\). Therefore, there are no absolute maximum and minimum values.

Step 5: Determine concavity

The second derivative of the function determines the concave up and down intervals. Take the derivative of \(f'(x)\) to find the second derivative: \(f''(x) = -2 \sin x\). When \(f''(x) > 0\), the function is concave up and when \(f''(x) < 0\), it's concave down. Therefore, the function is concave up on intervals \((- \infty, -\pi) \cup (0, \pi) \cup (2\pi, 3\pi) \cup ...\), and concave down on \((- \pi, 0) \cup (\pi, 2\pi) \cup (3\pi, 4\pi) \cup ...\)

Step 6: Sketching a graph of the function

Based on the analysis of the function \(f(x)\), you can sketch a graph which shows the critical points, increasing and decreasing intervals, local maxima and minima, concavity changes and direction of the function both towards \(\pm \infty\). As the functions involves periodic function (\(sin x\)), the graph will also exhibit periodic pattern. Also, the function has an increasing trend generally due to the \(x\) term in the equation of function

Key Concepts

Critical Points

In calculus, finding critical points is essential for understanding a function's behavior. Critical points occur where the first derivative of the function is either zero or undefined. For the function \(f(x) = x + 2 \sin x\), the first derivative is \(f'(x) = 1 + 2 \cos x\).

By setting \(f'(x) = 0\), we get:

• \(1 + 2 \cos x = 0\)
• \(\cos x = -\frac{1}{2}\)

Solving for \(x\) gives us \(x = \pm \frac{2\pi}{3} + 2n\pi\), where \(n\) is an integer. These points are crucial as they are the places where the function changes its direction or rate.

Increasing and Decreasing Intervals

Determining where a function is increasing or decreasing involves analyzing the sign of its first derivative. When \(f'(x) > 0\), the function is increasing, and when \(f'(x) < 0\), it's decreasing.

For \(f(x) = x + 2 \sin x\), the intervals are:

• Decreasing on \((-\infty, -\frac{2\pi}{3}) \cup (\frac{2\pi}{3}, \infty)\)
• Increasing on \((-\frac{2\pi}{3}, \frac{2\pi}{3})\)

This analysis divides the graph into sections where the function moves upwards or downwards, giving you insights into its behavior.

Local and Global Extrema

Local extrema refer to the peaks (maxima) and valleys (minima) within an interval of a function. To find local extrema, we look at where the derivative changes its sign. For the function \(f(x) = x + 2 \sin x\), the derivative changes sign at:

• Local maximum at \(x = -\frac{2\pi}{3}\)
• Local minimum at \(x = \frac{2\pi}{3}\)

However, the function doesn't have a global maximum or minimum because as \(x\) approaches \(\pm \infty\), the function is unbounded. Thus, there are no absolute extreme values.

Concavity

Concavity tells us how a function curves. A function is concave up when its second derivative is positive, and concave down when it's negative. For \(f(x) = x + 2 \sin x\), the second derivative is \(f''(x) = -2 \sin x\).

From here, we find:

• Concave up on \((-\infty, -\pi) \cup (0, \pi) \cup (2\pi, 3\pi)\)
• Concave down on \((-\pi, 0) \cup (\pi, 2\pi) \cup (3\pi, 4\pi)\)

This information about concavity provides insights into the shape and direction of the curve, making it easier to visualize the graph.