Points of Discontinuity - Algebra-2

1. Fundamental Concepts

Definition: Points of discontinuity in rational functions are values of \(x\) where the function is undefined, typically due to division by zero.
Types: There are two main types of discontinuities: removable and non-removable (infinite or jump).
Removable Discontinuity: Occurs when a factor in the numerator cancels out with a factor in the denominator, leaving a hole in the graph.
Non-Removable Discontinuity: Occurs when there is no cancellation, resulting in an asymptote.

2. Key Concepts

Identifying Discontinuities: To find points of discontinuity, set the denominator equal to zero and solve for \(x\).

Evaluating Limits: Evaluate the limit as \(x\) approaches the point of discontinuity to determine if it is removable or non-removable.

Graphical Interpretation: Graphically, removable discontinuities appear as holes, while non-removable discontinuities appear as vertical asymptotes.

3. Examples

Example 1 (Basic)

Problem: Identify the points of discontinuity for the function \(f(x) = \frac{x^2 - 4}{x - 2}\).

Step-by-Step Solution:

Set the denominator equal to zero: \(x - 2 = 0\), so \(x = 2\).
Factor the numerator: \(x^2 - 4 = (x + 2)(x - 2)\).
The function simplifies to \(f(x) = x + 2\) for \(x \neq 2\).
At \(x = 2\), the function has a removable discontinuity because the factor \((x - 2)\) cancels out.

Validation: Substitute \(x = 2\) into the simplified form \(f(x) = x + 2\) → Simplified: \(2 + 2 = 4\). The original expression is undefined at \(x = 2\), confirming a removable discontinuity.

Example 2 (Intermediate)

Problem: Determine the points of discontinuity for the function \(g(x) = \frac{x^2 - 5x + 6}{x - 3}\).

Step-by-Step Solution:

Set the denominator equal to zero: \(x - 3 = 0\), so \(x = 3\).
Factor the numerator: \(x^2 - 5x + 6 = (x - 2)(x - 3)\).
The function simplifies to \(g(x) = x - 2\) for \(x \neq 3\).
At \(x = 3\), the function has a removable discontinuity because the factor \((x - 3)\) cancels out.

Validation: Substitute \(x = 3\) into the simplified form \(g(x) = x - 2\) → Simplified: \(3 - 2 = 1\). The original expression is undefined at \(x = 3\), confirming a removable discontinuity.

4. Problem-Solving Techniques

Factorization: Always start by factoring both the numerator and the denominator to identify common factors.
Limit Evaluation: Evaluate the limit of the function as \(x\) approaches the point of discontinuity to determine its nature.
Graphical Analysis: Use graphs to visualize the behavior of the function around points of discontinuity.