Problem 78 - College Algebra

Question

For the following exercises, identify the removable discontinuity. $$ f(x)=\frac{2 x^{2}+5 x-3}{x+3} $$

Step-by-Step Solution

Answer

The removable discontinuity is at \( x = -3 \).

Step 1: Factor the Numerator

The first step in identifying a removable discontinuity is to factor the numerator of the function if possible. In this case, the numerator is \( 2x^2 + 5x - 3 \). We should try to express it as a product of binomials. After factoring, we have:\[ 2x^2 + 5x - 3 = (2x - 1)(x + 3) \].

Step 2: Simplify the Function

Once factored, the function \( f(x) = \frac{2x^2 + 5x - 3}{x + 3} \) can be rewritten using the new factors:\[ f(x) = \frac{(2x - 1)(x + 3)}{x + 3} \].We can then simplify the function by canceling out the common factor \( x + 3 \) in the numerator and the denominator, giving:\[ f(x) = 2x - 1 \], for \(x eq -3 \).

Step 3: Identify the Removable Discontinuity

The removable discontinuity occurs where a factor is canceled from the numerator and denominator. Here, the factor \( x + 3 \) was canceled, indicating a discontinuity at \( x = -3 \). However, because the discontinuity can be removed by simplifying the expression, it is classified as removable.

Key Concepts

Factoring Polynomials

Factoring polynomials is like breaking down a complex problem into simpler parts. When you have a polynomial expression such as \(2x^2 + 5x - 3\), the goal is to express it as a product of simpler polynomials, usually binomials. This method requires recognizing patterns and using methods like grouping or trial and error.
To factor \(2x^2 + 5x - 3\), we look for two numbers that multiply to give the constant term (-3) and add to give the middle coefficient (5). This can sometimes be an intuitive process or done by systematically trying possible factor pairs.
After finding the factors, we end up with \((2x - 1)(x + 3)\), which means the polynomial can be expressed as the product of these two binomials. Successful factoring sets the stage for simplification and further analysis.

Simplifying Rational Expressions

Simplifying rational expressions involves reducing fractions to their simplest form. In the case of a rational expression like \(f(x) = \frac{2x^2 + 5x - 3}{x + 3}\), simplification requires canceling out common factors in the numerator and the denominator.
Once the polynomial in the numerator is factored, you might find common terms in both the numerator and denominator. These common terms can be canceled out. In this example, \(x + 3\) is a common factor, so it is removed from both parts of the fraction.
The simplified expression becomes \(f(x) = 2x - 1\). It's important to note that this simplification only applies for values of \(x\) that do not make the original denominator zero, in this case, \(x eq -3\). This helps to accurately define the domain of the function after simplification.

Identifying Discontinuities

Discontinuities occur in rational functions where the denominator equals zero, causing the function to be undefined. A removable discontinuity happens if the discontinuity can be "removed" by simplifying the function.
In \(f(x) = \frac{2x^2 + 5x - 3}{x + 3}\), the factor \(x + 3\) in the denominator suggests a possible discontinuity at \(x = -3\). Upon simplification, \(x + 3\) is canceled, indicating that this discontinuity is removable.
Even though the expression simplifies to \(2x - 1\), the original function is still undefined at \(x = -3\), marking the location of the removable discontinuity. Understanding these points ensures that the scope and behavior of the function are accurately comprehended in its specific context.