Problem 5 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Each of the functions in Problems 5 through 10 is either continuous on \((-\infty, \infty)\) or has a point of discontinuity at some point \((s) x=a .\) Determine any point \((s)\) of discontinuity. Is the point of discontinuity removable? In other words, can the function be made continuous by defining or redefining the function at the point of discontinuity? $$ f(x)=\frac{x^{2}-4}{x+2} $$

Step-by-Step Solution

Answer

The function \(f(x)=\frac{x^{2}-4}{x+2}\) has a point of discontinuity at \(x=-2\). However, this point of discontinuity is removable by defining the function value at this point as -4.

Step 1: Factorization

Factor the numerator \(x^2 - 4\) to get \((x - 2)(x + 2)\). Thus, the function simplifies to \(f(x)=\frac{(x - 2)(x + 2)}{x + 2}\).

Step 2: Simplification

Simplify the function by cancelling common terms, we get \(f(x) = x - 2\), this is valid for all \(x\) values except \(x = -2\).

Step 3: Analysis of Discontinuity

As per our simplification, it is clear that the function is undefined at \(x = -2\) as it would create a denominator of zero. Hence, the point of discontinuity is \(x = -2\).

Step 4: Determination of Removability

The discontinuity at \(x = -2\) is removable as we can define \(f(-2) = -2 - 2 = -4\), thus making the function continuous at \(x = -2\)

Key Concepts

Removable Discontinuity

A removable discontinuity occurs in a function when there is a gap or hole in the graph that can be "filled in" by properly redefining the function at that specific point. In the original problem, we analyzed the function:

• Given function: \( f(x) = \frac{x^2 - 4}{x + 2} \)
• Discontinuity occurs at the point \( x = -2 \)

By factorizing the numerator, a common factor appeared, indicating a potential removable discontinuity. This discontinuity is termed "removable" because we can redefine the function at this point to make it continuous. By plugging in and setting \( f(-2) = -4 \), we effectively eliminate the gap from the graph of the function, ensuring continuity at \( x = -2 \). This is the essence of a removable discontinuity: it can be "removed" by appropriately defining the function value at the problematic point.

Factorization

Factorization is a crucial step in analyzing functions, especially when searching for discontinuities. In the function given, the numerator \( x^2 - 4 \) is a quadratic expression that can be rewritten using the difference of squares:

• Expression: \( x^2 - 4 = (x - 2)(x + 2) \)

By doing this, the function \( f(x) = \frac{x^2 - 4}{x + 2} \) transforms into \( f(x) = \frac{(x - 2)(x + 2)}{x + 2} \). Factorization revealed a common factor in the numerator and denominator, informing us about the removable discontinuity at \( x = -2 \). This step lets us see more clearly how the expression can simplify, laying the groundwork for subsequent simplification and analysis.

Function Simplification

Simplification is the process of reducing a mathematical expression to its simplest form. In the context of our function analysis, once the expression \( f(x) = \frac{(x - 2)(x + 2)}{x + 2} \) has been factorized, it can be simplified by canceling out the common term \((x + 2)\) present in both the numerator and the denominator.
After canceling, we are left with the simplified expression:

• \( f(x) = x - 2 \text{ for all } x eq -2 \)

This simplification clarifies that the function behaves like a simple linear function with a gap at \( x = -2 \). Despite initially appearing complex, its behavior simplifies noticeably except at its discontinuity, helping us determine how to make \( f(x) \) continuous by redefining it appropriately.

Limits and Continuity

Understanding limits is fundamental to analyzing continuity and discontinuity within functions. A function is continuous at a point if the limit as \( x \) approaches that point is equal to the function's value at that point. In our function, to determine if the discontinuity is removable, we examine:

• The limit of \( f(x) = x - 2 \) as \( x \) approaches \( -2 \) is \(-4\)
• This limit suggests the value that would "fill" the gap in continuity

If the limit exists and matches a defined value of the function at the discontinuity, the function can be described as continuous at that point. By recognizing this, we inserted a value at \( x = -2 \) that ensures \( f(x) \) is smooth without disruptions, illustrating the power of limits in addressing gaps in continuity.