Problem 529 - Precalculus

Question

For the following exercises, identify the removable discontinuity. $$f(x)=\frac{x^{2}-4}{x-2}$$

Step-by-Step Solution

Answer

The removable discontinuity is at \(x = 2\).

Step 1: Identify Points of Discontinuity

The function \(f(x) = \frac{x^2 - 4}{x - 2}\) is undefined where the denominator is 0. Set \(x - 2 = 0\) to find this point. Solving gives \(x = 2\). Therefore, there is a discontinuity at \(x = 2\).

Step 2: Factor the Numerator

The numerator \(x^2 - 4\) is a difference of squares and can be factored as \((x - 2)(x + 2)\). This will help us simplify the function.

Step 3: Simplify the Function

After factoring, the function becomes \(f(x) = \frac{(x-2)(x+2)}{x-2}\). We can cancel out the common factor \(x-2\) from the numerator and the denominator, giving us \(f(x) = x + 2\), except when \(x = 2\).

Step 4: Confirm Removable Discontinuity

Since the discontinuity at \(x = 2\) is removed after cancellation, it is called a removable discontinuity. The simplified function \(f(x) = x + 2\) is continuous everywhere except at \(x = 2\), where the removable discontinuity is located.

Key Concepts

Difference of Squares

The term "difference of squares" refers to a specific algebraic expression: when you have two perfect squares separated by a subtraction sign, like \(x^2 - 4\). This expression is special because it can be easily factored using the formula \(a^2 - b^2 = (a - b)(a + b)\). Here, \(x^2\) is the square of \(x\), and \(4\) is the square of \(2\). Hence, \(x^2 - 4\) can be factored into \((x - 2)(x + 2)\).

Recognizing a difference of squares helps simplify polynomial expressions, making it easier to solve equations or analyze functions, especially when addressing issues like discontinuities in rational functions.

Factoring

Factoring is a crucial algebraic technique used to simplify expressions and solve equations. When we talk about factoring, we're looking for numbers or expressions that can be multiplied to produce the original expression. In our exercise, we factored \(x^2 - 4\) into \((x-2)(x+2)\) by recognizing it as a difference of squares.

This method isn't limited to simple squares—almost any polynomial can be factored if you spot the right pattern or use techniques like grouping. Factoring is especially useful in simplifying rational functions, like in our exercise, where removing common factors can eliminate discontinuities and help in finding simpler equivalent expressions.

Rational Functions

Rational functions are expressions written as the quotient of two polynomials, such as \(f(x) = \frac{x^2 - 4}{x-2}\). Understanding rational functions is essential because they can have unique features such as vertical asymptotes and discontinuities.

In rational functions, discontinuities often occur where the denominator equals zero. Our task in simplifying these functions is typically finding and canceling out similar factors between the numerator and the denominator to remove what are known as removable discontinuities, which can make the function easier to analyze and graph.

Limits and Continuity

Limits and continuity are critical concepts in calculus that help us understand the behavior of functions at specific points, especially around discontinuities. A limit asks what value a function approaches as the input nears a certain point. For our simplified function \(f(x) = x + 2\), we find that as \(x\) approaches 2, \(f(x)\) approaches 4.

A point of discontinuity can be removable if the limit of the function as it approaches a certain point exists and is finite, but the function itself is not defined at that point. By simplifying \(f(x) = \frac{(x-2)(x+2)}{x-2}\) to \(f(x) = x + 2\) (while noting \(x eq 2\)), we remove the discontinuity, showing the function is continuous everywhere else. This technique is crucial for accurately describing and understanding the behavior of complex functions across their domain.