Problem 29 - Calculus of a Single Variable

Question

In Exercises 25–34, use a computer algebra system to analyze and graph the function. Identify any relative extrema, points of inflection, and asymptotes. $$ f(x)=2 x-4 \sin x, \quad 0 \leq x \leq 2 \pi $$

Step-by-Step Solution

Answer

The relative minimum is at \(x = \frac{\pi}{3}\) and the relative maximum is at \(x = \frac{5\pi}{3}\). The function does not have any asymptotes. Graphing the function within the range from \(0\) to \(2\pi\) will visually confirm these findings.

Step 1: Find the First Derivative

The first derivative of the function represents the slope at any point of the function, and thus indicates where the function reaches local minimums and maximums. Calculate the first derivative of \(f(x)=2x-4\sin x\). Using properties of derivatives, the derivative of \(2x\) is \(2\) and the derivative of \(-4\sin x\) is \(-4\cos x\). So the first derivative \(f'(x)=2-4\cos x\).

Step 2: Find the Critical Points

Find the critical points by setting the derivative equal to zero and solving for \(x\). Solving \(2 - 4\cos x = 0\) leads to \(\cos x = 0.5\). The solutions between \(0 \leq x \leq 2 \pi\) are \(x = \frac{\pi}{3}\) and \(x = \frac{5\pi}{3}\). Therefore, \(x = \frac{\pi}{3}\) and \(x = \frac{5\pi}{3}\) are the critical points.

Step 3: Find the Second Derivative

The second derivative provides information about the concavity of the function and thus points of inflection. Find the second derivative by differentiating \(f'(x)\) = \(2-4\cos x\). This gives \(f''(x) = 4\sin x\).

Step 4: Evaluate the Second Derivative at Critical Points

Evaluate \(f''(x) = 4\sin x\) at the critical points \(x = \frac{\pi}{3}\) and \(x = \frac{5\pi}{3}\). We find \(f''(\frac{\pi}{3}) = 4\sin(\frac{\pi}{3}) = 2\sqrt{3}\) and \(f''(\frac{5\pi}{3}) = 4\sin(\frac{5\pi}{3}) = -2\sqrt{3}\). Since \(f''(\frac{\pi}{3})\) is positive, \(\frac{\pi}{3}\) is a point of local minimum and since \(f''(\frac{5\pi}{3})\) is negative, \(\frac{5\pi}{3}\) is a point of local maximum.

Step 5: Find Asymptotes

Since the given function \(f(x) = 2x - 4\sin(x)\) is defined for all real numbers and its domain is all real numbers, the function does not have any vertical asymptotes. Also, because the function is not a rational function, it has no horizontal or oblique asymptotes.

Step 6: Graph the Function

Graph the function considering the range from \(0\) to \(2\pi\), including the relative extrema and points of inflection found. The relative minimum occurs at \(x = \frac{\pi}{3}\) and the relative maximum occurs at \(x = \frac{5\pi}{3}\). The function does not have any asymptotes.

Key Concepts

First Derivative

Understanding the first derivative is essential when studying calculus. It helps us determine where a function increases or decreases. In simple terms, for a function like \( f(x) = 2x - 4\sin x \), the first derivative, denoted as \( f'(x) \), tells us the slope at any point on the graph of the function. In this case, the derivative is found by differentiating each term separately. The derivative of \( 2x \) is \( 2 \), and the derivative of \( -4\sin x \) is \(-4\cos x\). Hence, \( f'(x) = 2 - 4\cos x \). This smoothening of the original function gives us the rate at which the function value changes concerning \( x \). By understanding how the derivative operates, you can predict how the function behaves over an interval.

Critical Points

Critical points are points on the function's graph where the first derivative is zero or undefined, indicating potential local minimums or maximums. To find these points for \( f(x) = 2x - 4\sin x \), you set the derived function \( f'(x) = 2 - 4\cos x \) equal to zero and solve for \( x \). This results in \( 2 - 4\cos x = 0 \), or \( \cos x = 0.5 \). Within the interval from \(0\) to \(2\pi\), the solutions are \( x = \frac{\pi}{3} \) and \( x = \frac{5\pi}{3} \). These values are where the function changes direction, marking potential peaks and troughs in the curve.

Second Derivative

The second derivative of a function provides insights into the curvature of the graph, known as concavity. For any function, the second derivative \(f''(x)\) is the derivative of the first derivative. In our example, \( f''(x) = 4\sin x \). This tells us how quickly the slope of the function's curve is changing. Evaluating the second derivative at the critical points helps determine whether these points are maximums, minimums, or points of inflection. A positive \( f''(x) \) at a point suggests the function is concave up, indicating a local minimum. Conversely, a negative result points to concave down, suggesting a local maximum.

Relative Extrema

Relative extrema refer to the local maximum or minimum values of a function. To identify these, we analyze the critical points using the second derivative. For \( f(x) = 2x - 4\sin x \), the critical points \( x = \frac{\pi}{3} \) and \( x = \frac{5\pi}{3} \) were evaluated using the second derivative \( f''(x) = 4\sin x \). At \( x = \frac{\pi}{3} \), \( f''(x) \) is positive, denoting a relative minimum, meaning the function dips at this point as the surrounding values are higher. Conversely, \( x = \frac{5\pi}{3} \) yields a negative second derivative, highlighting a relative maximum where the function peaks. Recognizing and understanding these points is crucial for interpreting the behavior of a function along its curve.

Asymptotes

Asymptotes are lines that a graph approaches but never touches. They typically indicate trends in rational functions as \( x \to \infty \) or near points of discontinuity. However, in the function \( f(x) = 2x - 4\sin x \), there are no asymptotes because the function is continuous for all real numbers over the given interval from \(0\) to \(2\pi\). This implies that the graph doesn't have any vertical, horizontal, or oblique asymptotes. Asymptotes are more common in functions like fractions or those with logarithms, where undefined values exist. In this case, the smooth flow of the sine and linear terms results in a graph without these boundaries.