Problem 77 - College Algebra with Corequisite Support

Question

For the following exercises, identify the removable discontinuity. $$ f(x)=\frac{x^{2}+x-6}{x-2} $$

Step-by-Step Solution

Answer

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

Step 1: Identify the Factorable Parts

First, observe the numerator in the function \( f(x) = \frac{x^2 + x - 6}{x - 2} \). We can factor \( x^2 + x - 6 \). Look for two numbers that multiply to \(-6\) and add to \(1\). These numbers are \(3\) and \(-2\).

Step 2: Rewrite the Function

Rewrite the numerator using its factors: \( f(x) = \frac{(x + 3)(x - 2)}{x - 2} \).

Step 3: Cancel Common Factors

Notice that \((x - 2)\) is present in both the numerator and the denominator. Cancel these common factors to simplify the function: \( f(x) = x + 3 \), excluding the point where the cancellation was done.

Step 4: Identify the Removable Discontinuity

The factor \((x - 2)\) that was canceled represents a removable discontinuity at \( x = 2 \), since this was originally in the denominator and made the original function undefined at \( x = 2 \).

Key Concepts

Factoring Polynomials

Factoring polynomials is an essential skill in algebra that helps simplify expressions and solve equations. When working with polynomials, the goal is to express them as a product of their simplest polynomial factors. To factor a polynomial, such as the numerator in the given function \(x^2 + x - 6\), look for two numbers that simultaneously multiply to the constant term, and add to the linear coefficient.

This concept is rooted in the reverse application of the distributive property, commonly known as the FOIL method (First, Outside, Inside, Last) used in multiplying binomials. In this context:

• The constant term is \(-6\).
• The linear coefficient in \( x \) is \(1\).

To factor \(x^2 + x - 6\), the numbers \(3\) and \(-2\) work because they multiply to \(-6\) and add to \(1\). Thus, the polynomial factors into \((x + 3)(x - 2)\). Factoring is often the first step in simplifying expressions that might look complex at first glance but have a simpler structure beneath.

Rational Functions

Rational functions are expressions that can be expressed as the quotient of two polynomials \(\frac{P(x)}{Q(x)}\), where \(Q(x)eq0\). These functions are significant in understanding calculus concepts like limits and continuity. In the problem, the function \(f(x) = \frac{x^2 + x - 6}{x - 2}\) is a rational function because it has a polynomial in the numerator and a polynomial in the denominator.

The capability to simplify rational functions by canceling shared factors between the numerator and the denominator is a critical aspect of these functions. However, it is crucial to remember that this cancellation only holds when the function is not undefined.
In the given problem, the cancellation of \(x-2\) from both the numerator and the denominator transforms our function into \(f(x) = x + 3\), except at the point where \(x = 2\). This point is where the function becomes undefined and highlights the concept of removable discontinuity.

Simplifying Expressions

Simplifying expressions is one of the fundamental tasks in algebra and calculus, helping to make complex expressions easier to work with. The process involves reducing expressions to their simplest form. In rational functions, like those discussed here, simplifying often involves factoring polynomials and canceling out common factors.

For example, the given rational function \(f(x) = \frac{(x + 3)(x - 2)}{x - 2}\) can be simplified by canceling the \(x-2\) terms from both the numerator and the denominator, resulting in a simpler linear expression \(f(x) = x + 3\).
This step might seem straightforward; however, it's important to understand that although the expression is simplified, the "removed" factor still plays a crucial role as it signifies where the function was originally undefined. At \(x = 2\), \(f(x)\) has a removable discontinuity — a point that was undefined in the original function but doesn't affect the rest of the function's continuity. Understanding this principle allows for better handling of expressions and addressing potential discontinuities effectively.