Problem 75 - Elementary and Intermediate Algebra

Question

Let \(f(x)=x^{2}-1\) and \(g(x)=x^{2}-4 .\) Find each function and give its domain. $$ g / f $$

Step-by-Step Solution

Answer

The function \( \frac{g}{f} \) is \( \frac{(x-2)(x+2)}{(x-1)(x+1)} \), and its domain is \((-\infty, -1) \cup (-1, 1) \cup (1, \infty)\).

Step 1: Define Functions

The functions are defined as follows: \( f(x) = x^2 - 1 \) and \( g(x) = x^2 - 4 \).

Step 2: Determine the Expression for g/f

The expression for \( \frac{g}{f} \) is given by \( \frac{g(x)}{f(x)} = \frac{x^2 - 4}{x^2 - 1} \).

Step 3: Factor the Numerator and Denominator

Factor the quadratic expressions: \( x^2 - 4 = (x - 2)(x + 2) \) and \( x^2 - 1 = (x - 1)(x + 1) \).

Step 4: Simplify the Expression

The simplified form of \( \frac{g(x)}{f(x)} \) remains \( \frac{(x - 2)(x + 2)}{(x - 1)(x + 1)} \) since there are no common factors to cancel between numerator and denominator.

Step 5: Determine the Domain

For the function \( \frac{g(x)}{f(x)} \), the domain excludes the values that make the denominator zero. Set \(f(x) = 0\) to find restricted values: \((x - 1)(x + 1) = 0\) giving \(x = 1\) and \(x = -1\) as restrictions. Therefore, the domain of \( \frac{g}{f} \) is \( (-\infty, -1) \cup (-1, 1) \cup (1, \infty) \).

Key Concepts

Domain of a Function

The domain of a function refers to the set of all possible input values (usually denoted as \( x \)) for which the function is defined. When considering rational functions, such as \( \frac{g(x)}{f(x)} \), it's essential to identify values that would make the denominator zero, as division by zero is undefined.
To find the domain of \( \frac{g}{f} = \frac{x^2 - 4}{x^2 - 1} \), we first need to determine when the denominator \( f(x) = x^2 - 1 \) equals zero.

• Set the denominator \( x^2 - 1 \to (x - 1)(x + 1) = 0 \).
• Solve for \( x \) to get \( x = 1 \) and \( x = -1 \).

These are the restricted values, so the function is not defined at these points. Hence, the domain of \( \frac{g}{f} \) is all real numbers except \( x = 1 \) and \( x = -1 \). This is expressed in interval notation as \( (-\infty, -1) \cup (-1, 1) \cup (1, \infty) \).

Quadratic Equations

Quadratic equations are polynomial equations of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. Solving these equations often involves finding the values of \( x \) that make the equation true.
In the context of rational functions like \( \frac{g}{f} = \frac{x^2 - 4}{x^2 - 1} \), you are dealing with quadratic expressions both in the numerator and the denominator.

• The expression \( x^2 - 4 \) is a quadratic which can be factored as \( (x - 2)(x + 2) \).
• The expression \( x^2 - 1 \) is another quadratic that factors to \( (x - 1)(x + 1) \).

Factoring is crucial because it simplifies the expression and helps identify restrictions in the domain by showing which values would make the denominator zero.

Function Simplification

Function simplification involves reducing a function to its simplest form, making it easier to analyze and interpret. In rational functions, simplification can often involve factoring and canceling out common factors.
In the exercise, we started with \( \frac{g(x)}{f(x)} = \frac{x^2 - 4}{x^2 - 1} \). After factoring, we found

• Numerator: \((x - 2)(x + 2)\)
• Denominator: \((x - 1)(x + 1)\)

Here, no common factors exist between the numerator and the denominator that can be canceled out. Thus, the expression is already in its simplest form. Simplification helps us to better understand the behavior of the function, especially its domain and potential asymptotes, and ensures we are aware of any undefined points.