Problem 27 - Single Variable Calculus

Question

In Exercises \(5-38\), sketch the graph of the function using the curve- sketching guidelines on page \(348 .\) $$ f(x)=\frac{1}{1-\cos x}, \quad-2 \pi

Step-by-Step Solution

Answer

In summary, to sketch the graph of the function \(f(x) = \frac{1}{1 - \cos x}\) on the interval \(-2\pi < x < 2\pi\), we find: 1. The domain of the function: \(x \ne 2\pi k\), where k is an integer. 2. The function is even, thus symmetric about the y-axis. 3. The critical points are at \(x=0\) and \(x=\pi\). 4. The function is concave up for \(0 < x < \pi\) and concave down for \(\pi < x < 2\pi\). 5. Vertical asymptotes are at \(x = 2\pi k\), where k is an integer, and a horizontal asymptote at \(y=0\). Using this information, we can sketch the graph of the function on the given interval, and due to its symmetry, we can easily extend the graph to its entire domain.

Step 1: Domain of the Function

To find the domain of \(f(x) = \frac{1}{1- \cos x}\), we need to find the values of \(x\) for which the denominator is not equal to \(0\). Since the largest value of \(\cos x\) is \(1\), we can see that the only value that would make the denominator equals to \(0\) is when \(\cos x = 1\). The values of \(x\) where \(\cos x = 1\) will be the multiples of \(2\pi\). Hence, the domain of the function is \(x \ne 2\pi k\) where k is an integer.

Step 2: Symmetry of the Function

To identify the symmetry, we need to check if the function is even, odd, or neither. We will test by plugging in \(-x\) for \(x\): \[ f(-x) = \frac{1}{1 - \cos(-x)} \] Since \(\cos(-x) = \cos(x)\), then \[ f(-x) = \frac{1}{1 - \cos(x)} = f(x) \] Thus, the function is even, and it will have symmetry about the y-axis.

Step 3: First and Second Derivatives of the Function

Let's find the first and second derivatives of the function \(f(x)\): \[ f'(x) = \frac{\mathrm{d} }{\mathrm{d} x} \left(\frac{1}{1- \cos x}\right) \] Using the chain rule and the derivative of \(\cos x\): \[ f'(x) = -\frac{\sin x}{(1 - \cos x)^2} \] Now, for the second derivative: \[ f''(x) = \frac{\mathrm{d} }{\mathrm{d} x} \left(-\frac{\sin x}{(1 - \cos x)^2}\right) \] Using the chain rule and the product rule, we get: \[ f''(x) = \frac{\cos x (1 - \cos x) + 2 \sin^2 x}{(1 - \cos x)^3} \]

Step 4: Local Extrema(Critical Points)

First, we find the critical points by finding where \(f'(x)=0\) or undefined. As the function has \(\sin(x)\) in the numerator, \(f'(x)\) will equal zero when \(\sin(x) = 0\). \[ f'(x) = 0 \implies \sin x = 0 \] The solution for this equation on the range \(-2\pi < x < 2\pi\) is \(x = 0,\pi\). These are our critical points.

Step 5: Intervals of Concavity

We now look at the second derivative, \(f''(x)\), to find where the graph is concave up or concave down. \[ f''(x) = \frac{\cos x (1 - \cos x) + 2 \sin^2 x}{(1 - \cos x)^3} \] Since the denominator is always positive, we only need to consider the numerator. We have \(\cos x = 1\) at \(x = 2\pi k\), which we will not consider since those points are excluded from the domain. Now, the second derivative will be positive when \(\cos x (1 - \cos x) + 2 \sin^2 x > 0\), which implies concave up, and negative when it is concave down. For \(0 < x < \pi\), the second derivative will be positive, meaning the graph is concave up. For \(\pi < x < 2\pi\), the second derivative will be negative, which means the graph is concave down.

Step 6: Asymptotes

Since the points where the function is undefined are a multiple of \(2\pi\), we have vertical asymptotes at \(x = 2\pi k\) where k is an integer. Also, notice that since the function is even, we will have a horizontal asymptote at \(y=0\).

Step 7: Sketch the Graph

To sketch the graph, we can use the information about symmetry, critical points, intervals of concavity, local extrema, and asymptotes: 1. The function has symmetry about the y-axis due to its even nature. 2. The local extrema are at \(x=0\) and \(x=\pi\). 3. The function is concave up for \(0 < x < \pi\), and concave down for \(\pi < x < 2\pi\). 4. Vertical asymptotes are at \(x = 2\pi k\), where k is an integer. 5. There is a horizontal asymptote at \(y=0\). Now, plot the graph using the information above. We first sketch the curve for \(-2\pi < x < 2\pi\) and, because of the symmetry, replicate it for the entire range of the function with its asymptotes.

Key Concepts

Function Domain

When exploring a function, it's vital to determine the permissible set of input values or the domain. For the function \( f(x) = \frac{1}{1- \cos x} \), we should avoid points where the denominator equates to zero. Since \(\cos x\) reaches its peak at 1, the denominator becomes zero when \(\cos x = 1\). This scenario occurs at multiples of \(2\pi\), such as \(2\pi k\), where \(k\) is an integer. Hence, the domain of this function excludes these values, meaning \(x eq 2\pi k\).

Understanding the function's domain is essential for identifying where the expression remains valid, guiding us in sketching its curve. Always take into account which values are excluded to avoid misunderstandings when visualizing the function.

Symmetry

Symmetry in a function can reveal significant insights. For our function \( f(x) = \frac{1}{1- \cos x} \), discovering symmetry involves checking if the function behaves the same when input values are negated. If it does, the function holds y-axis symmetry, suggesting it's even.

Replacing \(x\) with \(-x\) yields the expression \( f(-x) = \frac{1}{1 - \cos(-x)} \). The identity \(\cos(-x) = \cos(x)\) confirms that \(f(-x) = f(x)\). This match indicates the function is indeed even.

When sketching graphs, acknowledging symmetry helps simplify the sketching process, allowing us to extend a partial graph across the y-axis to form a complete picture.

Derivatives

Derivatives offer a deeper understanding of a function's behavior by providing information on slopes, curvature, and local extrema. For \( f(x) = \frac{1}{1- \cos x} \), we first compute the derivative:

• First derivative: \( f'(x) = -\frac{\sin x}{(1-\cos x)^2} \)
• Second derivative: \( f''(x) = \frac{\cos x (1 - \cos x) + 2 \sin^2 x}{(1 - \cos x)^3} \)

The first derivative helps identify where the function's slope is zero, leading to critical points. Here, \(\sin x = 0\) pinpoints the zero slopes at \(x = 0\) and \(x = \pi\).

The second derivative clarifies regions of concavity. Positive values indicate concave up areas, while negative values highlight concave down sections. For \(0 < x < \pi\), the graph is concave up; for \(\pi < x < 2\pi\), it's concave down. These insights enrich our understanding of the graph's shape and guide proper sketching.

Asymptotes

Asymptotes describe lines that a curve approaches but never truly reaches. For \( f(x) = \frac{1}{1- \cos x} \), two types of asymptotes exist: vertical and horizontal.

• Vertical asymptotes occur where the function is undefined, specifically at \(x = 2\pi k\), coinciding with \(\cos x = 1\). These represent the function rapidly increasing or decreasing near these points.
• Horizontal asymptotes are determined by the end behavior of the function. Given that this function is even (symmetric about the y-axis), as \(x\) extends towards positive or negative infinity, \(f(x)\) level off towards 0, establishing a horizontal asymptote at \(y=0\).

Understanding asymptotes is crucial in accurately sketching a curve, as they define the behavior at extreme values and discontinuities.