Problem 3 - College Algebra

Question

In Exercises \(1-8,\) find the domain of each rational function. $$g(x)=\frac{3 x^{2}}{(x-5)(x+4)}$$

Step-by-Step Solution

Answer

The domain of the function \(g(x)=\frac{3 x^{2}}{(x-5)(x+4)}\) is all real numbers except \(5\) and \(-4\).

Step 1: Identify the denominator

Start by identifying the denominator of the rational function. In this case, the denominator is \((x-5)(x+4)\)

Step 2: Set denominator equal to zero

Set the denominator equal to zero and solve for \(x\) to determine which values to exclude from the domain. This gives us two equations \(x - 5 = 0\) and \(x + 4 = 0\). Solving these gives \(x = 5\) and \(x = -4\) respectively.

Step 3: Define the Domain

The domain of a function is all the possible values that \(x\) can take. Since the only values \(x\) can't take in this function are \(5\) and \(-4\), then everything else is the domain. The domain is therefore defined as \(x \in R\), \(x≠5\), and \(x≠-4\)

Key Concepts

Rational Functions

Rational functions are expressions that involve the division of two polynomials. The form of a rational function is typically written as \( R(x) = \frac{P(x)}{Q(x)} \), where \( P(x) \) and \( Q(x) \) are polynomials, and \( Q(x) \) is not equal to zero. This type of function is called "rational" because it resembles a fraction, which is also known as a "ratio" of two numbers. Like other functions, rational functions can be graphed to show the relationship between the dependent and independent variables. The graph of a rational function can have various features, such as vertical asymptotes, horizontal asymptotes, or holes, which are influenced by the polynomials in the numerator and denominator. Understanding the behavior of rational functions involves analyzing their algebraic structure and observing how the function values change as \( x \) varies. This is crucial for accurately determining the range and restrictions of rational functions.

Denominator

The denominator is the part of the rational function that appears below the fraction line. In simple terms, it's the divisor of the rational function. For the function \( g(x) = \frac{3x^2}{(x-5)(x+4)} \), the denominator is \((x-5)(x+4)\). The denominator is key in identifying restrictions for the rational function, as it determines which values of \( x \) make the entire function undefined. It is crucial to remember that division by zero is undefined. Hence, any value of \( x \) that results in a denominator of zero must be excluded from the domain. This leads us to our next topic, which is why understanding the denominator is so essential in finding the function's domain.

Domain Exclusion

Domain exclusion refers to the process of eliminating certain values from the set of possible values \( x \) can take in a rational function. These exclusions are necessary because certain values can make the function undefined. To determine which values to exclude, you need to focus on the denominator of the rational function. Specifically, find the values of \( x \) for which the denominator equals zero. These are the points where the function cannot be evaluated.For the function \( g(x) = \frac{3x^2}{(x-5)(x+4)} \), we identify that the denominator is \((x-5)(x+4)\). To find the exclusions:

• Set \( x-5 = 0 \) and solve for \( x \), giving \( x = 5 \).
• Set \( x+4 = 0 \) and solve for \( x \), giving \( x = -4 \).

Both \( x = 5 \) and \( x = -4 \) are values that must be excluded from the domain of the function.

Solve for x

"Solve for \( x \)" involves finding the values of \( x \) that satisfy a given equation. In the context of rational functions, this often involves setting the denominator equal to zero and finding which values of \( x \) should be excluded from the domain because they lead to division by zero.Consider the function \( g(x) = \frac{3x^2}{(x-5)(x+4)} \). To solve for \( x \) in this scenario, identify where the denominator becomes zero:

• Set \( (x-5) = 0 \) and solve, thus yielding \( x = 5 \).
• Set \( (x+4) = 0 \) and solve, thus yielding \( x = -4 \).

These solutions demonstrate the points where the function is undefined and thus should not be included in the domain.Once we solve for these particular values of \( x \), we understand which parts of the real number line are accessible to the function. The domain is then all real numbers except \( x=5 \) and \( x=-4 \).