Problem 4 - Advanced Calculus

Question

For each of the following functions, find all the discontinuities and indicate any that are removable. (a) \(f(x)=\frac{x}{\sin (5 \cos x)}\) (b) \(f(x)=x+\frac{\sin x}{2 x-\frac{4}{x-1}}\) (c) \(f(x)=\frac{1}{1+e^{\sec x}}\)

Step-by-Step Solution

Answer

For the function \(f(x)=\frac{x}{\sin (5 \cos x)}\), the discontinuities occur at \(x = \cos^{-1}\left(\frac{n\pi}{5}\right)\), none of which are removable. For the function \(f(x)=x+\frac{\sin x}{2 x-\frac{4}{x-1}}\), the discontinuities occur at \(x = -1\) and \(x = 2\), none of which are removable. For the function \(f(x)=\frac{1}{1+e^{\sec x}}\), discontinuities occur at \(x = \frac{(2n+1)\pi}{2}\), none of which are removable.

Step 1: Identify possible discontinuities for the first function

For the function \(f(x)=\frac{x}{\sin (5 \cos x)}\), the function will be undefined when the denominator is equal to zero. This happens for \(\sin (5 \cos x) = 0\). The equation \(\sin (u) = 0\) has solutions for \(u = n\pi\), where \(n\) is an integer. Therefore, \(5 \cos x = n\pi\), leading to \(x = \cos^{-1}\left(\frac{n\pi}{5}\right)\).

Step 2: Determine the removable discontinuities for the first function

A discontinuity is removable if the function can be manipulated such that it is defined at that point by cancelling out the factor that caused the undefined value. For this function, there is no algebraic manipulation that would allow for this. Meaning, all the discontinuities are not removable.

Step 3: Identify possible discontinuities for the second function

For the function \(f(x)=x+\frac{\sin x}{2 x-\frac{4}{x-1}}\), it will be undefined when \(2x - \frac{4}{x-1} = 0\). Solving this gives \(x = -1\) and \(x = 2\).

Step 4: Determine the removable discontinuities for the second function

As with the first function, there is no algebraic manipulation that would allow for the discontinuities to be removed. Therefore, in this case, all the discontinuities are also not removable.

Step 5: Identify possible discontinuities for the third function

For the function \(f(x)=\frac{1}{1+e^{\sec x}}\), it will be undefined when \(\sec x = \pm \infty\). This occurs when \(x = \frac{(2n+1)\pi}{2}\), where \(n\) is an integer.

Step 6: Determine the removable discontinuities for the third function

Again, there are no ways to algebraically manipulate this function to cancel out the factor causing the undefined value. Therefore, all discontinuities are not removable.

Key Concepts

Removable Discontinuities

Understanding removable discontinuities is essential when dealing with various functions, including rational, trigonometric, and more. A discontinuity occurs at a certain point in a function's domain where it is not defined. However, if this discontinuity is removable, it implies that through algebraic manipulation, we can redefine the function to "fill in" the missing point.
In practical terms, this means the function could be made continuous at that point by cancelling out problematic terms. For example, if you have a function like \( f(x) = \frac{x^2 - 1}{x - 1}\), you can simplify it to \( f(x) = x + 1\) by factoring the numerator, ultimately removing the discontinuity at \( x = 1\).

• Non-removable discontinuities are often more complex and cannot be addressed by simply altering the algebraic structure of the function.

The solutions provided here revealed that certain discontinuities, particularly in complex trigonometric functions, are indeed non-removable, as no algebraic tricks fit neatly in place.

Trigonometric Functions

Trigonometric functions such as sine, cosine, and tangent often introduce a fascinating array of discontinuities in functions. When evaluating discontinuities, especially removable ones, these functions can be challenging. For instance, consider the trigonometric function present in the exercise \( f(x) = \frac{x}{\sin(5 \cos x)} \).
The key is to find points where the denominator equals zero, causing potential discontinuities. In trigonometric functions:

• \( \sin(u) = 0 \) when \( u = n\pi \), \(n\) being an integer.
• Similarly, \( \sec(x) \) tends to become undefined where \( \cos(x) = 0 \), leading to asymptotic behaviours rather than mere point discontinuities.

Identifying these points is the first step in the analysis, often revealing complex patterns based on periodicity and amplitude changes.

Algebraic Manipulation

Algebraic manipulation is a mathematical technique used to rearrange, simplify, and solve equations and functions. It's a key skill in identifying and addressing discontinuities, especially removable ones. In the problem-solving process, you often rearrange terms in a function to find cancellations which can "remove" discontinuities.
In our exercises, no significant algebraic manipulations were able to eliminate discontinuities, indicating their non-removable nature. However, understanding algebraic operations like factorization and simplification can guide you in other contexts to find removable discontinuities.
Remember:

• Always check for common factors in the numerator and denominator.
• Apply identities in trigonometric expressions, such as \( \sin^2(x) + \cos^2(x) = 1 \).
• Look for possibilities to rewrite complex expressions into simpler forms.

These strategies underline the importance of algebraic methods in calculus problem-solving, simplifying the functions if possible.

Calculus Problem-Solving

Dealing with discontinuities is a part of calculus problem-solving, which involves a mix of theoretical understanding and technical application. Throughout this process, identifying different types of discontinuities and applying suitable strategies is crucial.
For each function, problem-solving involves a series of steps:

• Find where the function is undefined by identifying when the denominator is equal to zero.
• Consider any asymptotic behavior, especially in trigonometric functions.
• Attempt algebraic manipulations to turn these into simpler expressions.

Whether you're dealing with trigonometric identities or more standard algebraic expressions, the ultimate aim is to neatly solve and understand the reasons behind discontinuities.
As demonstrated in the provided solutions, calculus problem-solving involves not just applying formulas or memorized techniques but also understanding the underlying mathematical principles that introduce continuity issues.