Problem 13 - Intermediate Algebra

Question

Determine \((f \circ g)(x)\) and \((g \circ f)(x)\) for each pair of functions. Also specify the domain of \((f \circ g)(x)\) and \((g \circ f)(x)\). (Objective 1\()\) \(f(x)=\frac{1}{x}\) and \(g(x)=\frac{1}{x-4}\)

Step-by-Step Solution

Answer

\((f \circ g)(x) = x - 4\) with domain \((-\infty, 4) \cup (4, \infty)\). \((g \circ f)(x) = \frac{x}{1 - 4x}\) with domain \((-\infty, 0) \cup (0, \frac{1}{4}) \cup (\frac{1}{4}, \infty)\).

Step 1: Understand the Composition of Functions

When we write \((f \circ g)(x)\), this means \(f(g(x))\). Similarly, \((g \circ f)(x)\) means \(g(f(x))\). We are required to find both compositions and their domains.

Step 2: Find \((f \circ g)(x)\)

To find \((f \circ g)(x)\), substitute \(g(x)\) into \(f(x)\). We have:\[(f \circ g)(x) = f\left(g(x)\right) = f\left(\frac{1}{x-4}\right)\]Substitute \(\frac{1}{x-4}\) into \(f(x) = \frac{1}{x}\):\[(f \circ g)(x) = \frac{1}{\left(\frac{1}{x-4}\right)} = x-4\]

Step 3: Determine the Domain of \((f \circ g)(x)\)

For the composition \((f \circ g)(x) = x-4\), we need to consider where \(g(x)\) is defined, since that's inside the function. \(g(x) = \frac{1}{x-4}\) is undefined when \(x = 4\) because it causes division by zero.Thus, the domain of \((f \circ g)(x)\) is all real numbers except \(x = 4\). In interval notation, this is:\((-\infty, 4) \cup (4, \infty)\).

Step 4: Find \((g \circ f)(x)\)

To find \((g \circ f)(x)\), substitute \(f(x)\) into \(g(x)\). We have:\[(g \circ f)(x) = g\left(f(x)\right) = g\left(\frac{1}{x}\right)\]Substitute \(\frac{1}{x}\) into \(g(x) = \frac{1}{x-4}\):\[(g \circ f)(x) = \frac{1}{\left(\frac{1}{x} - 4\right)} = \frac{x}{1 - 4x}\]

Step 5: Determine the Domain of \((g \circ f)(x)\)

For \((g \circ f)(x) = \frac{x}{1 - 4x}\), ensure \(f(x)\) and the final result are defined:1. \(f(x) = \frac{1}{x}\) is undefined at \(x = 0\).2. The expression \(1 - 4x\) cannot be equal to zero (since it would make the denominator zero), so solve \(1 - 4x eq 0\):\[1 eq 4x \, \Rightarrow \, x eq \frac{1}{4}\]Thus, the domain of \((g \circ f)(x)\) is all real numbers except \(x = 0\) and \(x = \frac{1}{4}\). In interval notation, this is:\((-\infty, 0) \cup (0, \frac{1}{4}) \cup (\frac{1}{4}, \infty)\).

Key Concepts

Domains of Functions

The domain of a function defines the set of all possible input values (often x-values) for which the function is defined.
For rational functions, this typically means identifying where the denominator is not zero.
When dealing with composite functions like \((f \circ g)(x)\), determining the domain involves:

• Finding where each individual function, \( f(x) \) and \( g(x) \), is defined.
• Considering the domain of \( g(x) \) because \( g(x) \) is the input to the function \( f \).

For the given exercise, \(f(x) = \frac{1}{x}\) is undefined at \(x = 0\), and \(g(x) = \frac{1}{x-4}\) is undefined at \(x = 4\).
By substituting \(g(x)\) into \(f(x)\), the domain of \( (f \circ g)(x) \) is all real numbers except \( x = 4 \).
Similarly, for \( (g \circ f)(x) \), it's important to ensure that neither \(f(x)\) nor the new resulting expression introduces division by zero.

Rational Functions

Rational functions are quotients involving polynomials. A simple example is \(f(x) = \frac{1}{x}\).
The key characteristic of rational functions is that they become undefined when their denominators are zero.
In context, for \( f(x) = \frac{1}{x} \), the function is undefined at \(x = 0\), since division by zero is not allowed. Additionally, when compositions like \( (f \circ g)(x) \) and \( (g \circ f)(x) \) involve rational functions, it’s crucial to check where these compositions are undefined.
Using the example:

• \( f(g(x)) = \frac{1}{(x-4)} \) simplifies to \( x-4 \), a polynomial with no further restrictions aside from \( x eq 4 \).
• \( g(f(x)) = \frac{1}{(\frac{1}{x} - 4)} \), simplifies to \( \frac{x}{1 - 4x} \), having restrictions at \(x = 0\) and \(x = \frac{1}{4}\).

Understanding these essential concepts are fundamental when dealing with rational functions and their transformations.

Function Operations

Function operations refer to the various ways of combining functions.
This can include addition, subtraction, multiplication, division, and most intriguingly, composition, denoted as \((f \circ g)(x)\). With function composition, the output of one function becomes the input of another.
This requires accurate substitution to determine the correct expression.
In the exercise:

• For \((f \circ g)(x)\), substitute \(g(x)\) into \(f(x)\) to get \( \frac{1}{(x-4)} \).
• For \((g \circ f)(x)\), \(f(x)\) is substituted into \(g(x)\), resulting in \( \frac{x}{1 - 4x} \).

Each of these operations reveals new expressions, demonstrating how functions interact and combine.
It's essential to not only perform these operations correctly but also to understand any limitations in their resultant domains.