Problem 531 - Precalculus

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: Factor the Numerator

First, let's factor the numerator of the function \( f(x) = \frac{x^2 + x - 6}{x - 2} \). We need to express \( x^2 + x - 6 \) as a product of two binomials. It factors into \( (x + 3)(x - 2) \).

Step 2: Rewrite the Function

Substitute the factored form of the numerator into the function:\[ f(x) = \frac{(x + 3)(x - 2)}{x - 2} \]

Step 3: Identify the Removable Discontinuity

Notice that the factor \( (x - 2) \) is present in both the numerator and the denominator. This indicates that there is a removable discontinuity at \( x = 2 \).

Step 4: Simplify the Function

By canceling the common factor \( (x - 2) \) from both the numerator and the denominator, the function simplifies to:\[ f(x) = x + 3 \] for \( x eq 2 \).

Step 5: State the Function and Discontinuity

The simplified function is \( f(x) = x + 3 \) for \( x eq 2 \), and there's a removable discontinuity at \( x = 2 \).

Key Concepts

Factoring Polynomials

Factoring polynomials is a fundamental skill in algebra that allows us to break down complex polynomial expressions into simpler parts. This is like splitting a river into smaller streams, making it easier to manage. In an equation, such as the one in the original exercise, the expression \( x^2 + x - 6 \) can be a bit daunting at first.
But with practice, you can learn to factor it into the product of simpler binomials \( (x + 3)(x - 2) \). To factor this polynomial, you should first find numbers that multiply to \(-6\) (the constant term) and add up to \(1\) (the coefficient of the linear term \(x\)). In this case, the numbers \(3\) and \(-2\) fit the bill, thus allowing us to write the polynomial as \( (x + 3)(x - 2) \).
Factoring allows us to identify common factors in expressions, which is crucial for simplifying complex rational functions.

Rational Functions

Rational functions are expressions that involve fractions of polynomials. These functions, like \( f(x) = \frac{(x + 3)(x - 2)}{x - 2} \), often have interesting properties. They may have discontinuities or undefined points in their domain, due to the nature of division.In this specific exercise, the denominator \(x - 2\) can potentially make the function undefined when \(x = 2\). Such points are usually where the complexity of rational functions surfaces.
It's essential to determine whether these discontinuities are removable or not. Removable discontinuities occur when factors in the numerator and the denominator can cancel each other out, making the function simpler and helping to "remove" potential undefined points.

Simplification of Functions

The simplification of functions involves the reduction of expressions to their simplest forms by eliminating any removable discontinuities. When a factor appears both in the numerator and the denominator, as \((x-2)\) does in our example, it can be canceled, leading to a much simpler expression.By canceling \((x-2)\), the function \( f(x) \) simplifies to \( x+3 \), albeit with the condition it does not hold at \( x = 2 \), due to its original form. Simplifying allows for an easier evaluation of the function at most points, though it's vital to remember that the cancelling process does not exist at the discontinuity itself.
The process of simplification highlights the power of algebraic manipulation in making complex expressions more manageable and less prone to errors when further processed or analyzed.