Problem 18 - Precalculus

Question

In Exercises \(11-20\), use the pair of functions \(f\) and \(g\) to find the domain of the indicated function then find and simplify an expression for it. $$ f(x)=x-1 \text { and } g(x)=\frac{1}{x-1} $$

Step-by-Step Solution

Answer

Domain: \((-\infty, 1) \cup (1, \infty)\). Simplified expression: \(\frac{x(x-2)}{x-1}\), \(x \neq 1\).

Step 1: Determine the Domains of f and g

First, examine the domains of both \( f(x) \) and \( g(x) \). The function \( f(x) = x - 1 \) is a linear function that is defined for all real numbers, so its domain is \( (-\infty, \infty) \).

Step 2: Determine the Domain of g(x)

For \( g(x) = \frac{1}{x-1} \), we need to ensure that the denominator does not equal zero. Therefore, the function is undefined when \( x-1 = 0 \), or when \( x = 1 \). Thus, the domain of \( g(x) \) is all real numbers except 1: \( (-\infty, 1) \cup (1, \infty) \).

Step 3: Combine Domains for Operations

We now need the domain for any compositions or combinations of \( f(x) \) and \( g(x) \). The operations of interest, such as \( f + g \), \( f - g \), \( f \cdot g \), or \( \frac{f}{g} \), require where both functions are defined at the same time. Use the intersection of \( (-\infty, \infty) \) and \( (-\infty, 1) \cup (1, \infty) \) to get the combined domain: \((-\infty, 1) \cup (1, \infty)\).

Step 4: Simplify the Expression for the Indicated Operation

Assuming the indicated function is a basic operation or a specific expression, let's choose subtraction: \( (f-g)(x) = (x-1) - \frac{1}{x-1} \). Simplify this expression: find a common denominator which is \(x-1\). The simplified form is \(\frac{(x-1)(x-1)-1}{x-1} = \frac{x^2-2x}{x-1}\).

Step 5: Further Simplify and State Restrictions

Further simplification gives \( \frac{x(x-2)}{x-1} \), maintaining the domain where \( x eq 1 \).

Key Concepts

Linear Function

Linear functions are among the simplest forms of functions we explore in mathematics. They follow the standard form of a straight line and are defined by the general equation \( f(x) = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept. In this form, the variable \( x \) is raised to the first power, making the graph a straight line.

The domain of a linear function is all real numbers, denoted as \( (-\infty, \infty) \), because there are no restrictions on the values that \( x \) can take. For example, in our exercise, the function \( f(x) = x - 1 \) has a linearly defined equation. It extends infinitely in both the positive and negative directions along the x-axis, thus accepting any real number as a valid input for \( x \).

Understanding linear functions is crucial as they are foundational in algebra and calculus, setting the stage for more complex types of functions, like quadratic and polynomial functions.

Rational Function

Rational functions take the form \( g(x) = \frac{P(x)}{Q(x)} \), where both \( P(x) \) and \( Q(x) \) are polynomials. The domain of a rational function excludes values of \( x \) that make the denominator zero, as division by zero is undefined.

In our exercise, the function \( g(x) = \frac{1}{x-1} \) is a simple rational function. Here, \( Q(x) \) is \( x-1 \), implying that the function is undefined when \( x = 1 \). Thus, the domain of \( g(x) \) is all real numbers except \( x = 1 \), which can be expressed as \((-\infty, 1) \cup (1, \infty)\).

Working with rational functions often involves identifying these restrictions at the outset. It’s vital to recognize these interruptions when calculating domains and preparing functions for composition or any operations like adding, subtracting, or multiplying.

Function Composition

Function composition involves creating a new function by combining two existing functions. In this process, the output of one function becomes the input to another. Mathematically, this is denoted as \( (f \circ g)(x) \), meaning you apply \( g \) first, then \( f \).

For function composition to be valid, the domain of the inner function \( g(x) \) must be a subset of the domain of the outer function \( f(x) \). This ensures that every output from \( g \) is a valid input for \( f \).

In our example, combining functions \( f(x) = x - 1 \) and \( g(x) = \frac{1}{x-1} \) requires determining where both functions are defined together. We take the intersection of their domains. Here, \( (-\infty, \infty) \cap ((-\infty, 1) \cup (1, \infty)) = (-\infty, 1) \cup (1, \infty) \) provides the domain for the composite function.

Simplifying Expressions

Simplifying expressions allows us to rewrite them in a form that is easier to understand or work with, often preparing them for further operations or evaluations.

Consider the operation \( (f-g)(x) = (x-1) - \frac{1}{x-1} \). To simplify, a common denominator, \( x-1 \), is necessary to combine the terms:

• Write \( (x-1)(x-1) \) as \( \frac{(x-1)(x-1)}{x-1} \),
• Subtract \( \frac{1}{x-1} \) from it,
• This results in \( \frac{x^2 - 2x}{x-1} \).

Further simplification shows how the expression can be broken down into \( \frac{x(x-2)}{x-1} \). Even in this simplified form, it's important to restate any restrictions, such as \( x eq 1 \), to ensure that the domain remains consistent with the original functions.