Problem 16 - College Algebra

Question

Use each pair of functions to find \(f(g(x))\) and \(g(f(x))\). Simplify your answers. $$ f(x)=\frac{1}{x-6}, g(x)=\frac{7}{x}+6 $$

Step-by-Step Solution

Answer

\( f(g(x)) = \frac{x}{7} \) and \( g(f(x)) = 7x - 36 \).

Step 1: Identify the Functions

First, let's clearly identify the given functions. We have two functions: \[ f(x) = \frac{1}{x-6} \] and \[ g(x) = \frac{7}{x} + 6 \]. Our goal is to find and simplify the compositions: \( f(g(x)) \) and \( g(f(x)) \).

Step 2: Find f(g(x))

To find \( f(g(x)) \), we will substitute \( g(x) \) into \( f(x) \). This means: \[ f(g(x)) = f\left(\frac{7}{x} + 6\right) = \frac{1}{\left(\frac{7}{x} + 6\right) - 6} \]Simplify the expression inside the denominator: \[ \left(\frac{7}{x} + 6\right) - 6 = \frac{7}{x} \]Thus, \[ f(g(x)) = \frac{1}{\frac{7}{x}} = \frac{x}{7} \].

Step 3: Find g(f(x))

To find \( g(f(x)) \), substitute \( f(x) \) into \( g(x) \). This means:\[ g(f(x)) = g\left(\frac{1}{x-6}\right) = \frac{7}{\frac{1}{x-6}} + 6 \]Simplify \( \frac{7}{\frac{1}{x-6}} \):\[ \frac{7}{\frac{1}{x-6}} = 7(x-6) = 7x - 42 \]Thus, \[ g(f(x)) = 7x - 42 + 6 = 7x - 36 \].

Step 4: Simplify and Conclude

We have found:\[ f(g(x)) = \frac{x}{7} \] and \[ g(f(x)) = 7x - 36 \]. These are the simplified forms of the compositions.

Key Concepts

Algebraic Functions

Algebraic functions are mathematical expressions built using algebraic operations such as addition, subtraction, multiplication, division, and taking roots. Understanding these functions is fundamental as they form the basis for creating more complex mathematical expressions.
Let's break it down:

• Basic Operations: These include the traditional operations like addition (+), subtraction (-), multiplication (×), and division (÷). For example, in the function \( f(x) = x^2 + 3x - 5 \), each term results from algebraic operations on \( x \).
• Combining Functions: You can create new functions by combining existing ones using algebraic operations. For example, combining \( f(x) = x + 1 \) and \( g(x) = 2x \) gives us a new function \( h(x) = f(x) + g(x) = (x + 1) + 2x = 3x + 1 \).
• Polynomials: A special class of algebraic functions, formed by the sum of terms each consisting of a variable raised to an integer power multiplied by a coefficient. For instance, \( x^2 + 3x - 5 \) is a polynomial.

These building blocks form the foundation of more complex functions, setting the stage for understanding composite and rational functions in the given exercise.

Composite Functions

Composite functions involve creating new functions by applying one function to the results of another. This concept is crucial when analyzing combinations of two or more functions, such as in the task of finding \( f(g(x)) \) and \( g(f(x)) \).
Here's how to handle composite functions:

• Substitution: To compute \( f(g(x)) \), substitute \( g(x) \) into \( f(x) \). This means wherever there is an \( x \) in \( f(x) \), replace it with the expression for \( g(x) \). For example, in the exercise, \( f(g(x)) = f\left(\frac{7}{x} + 6\right) \). Use careful arithmetic and simplification.
• Order of Operations: The order matters; \( f(g(x)) \) differs from \( g(f(x)) \). First, resolve the innermost function before applying the outer function.
• Domain Considerations: Determine the domain for which both \( f \) and \( g \) are applicable. This ensures the composition is meaningful and defined.

By applying these steps, you can effectively solve problems involving composite functions, understanding how one function transforms under the action of another.

Rational Functions

Rational functions are essentially the ratio of two polynomials. They are expressed in the form \( \frac{P(x)}{Q(x)} \), where \( P(x) \) and \( Q(x) \) are polynomials. The given exercise involves rational functions and their compositions.
Let's consider key points:

• Structure: In \( f(x) = \frac{1}{x-6} \), \( f(x) \) is a rational function because it can be seen as dividing the constant \( 1 \) by the linear polynomial \( x - 6 \).
• Domain: The domain excludes values that make the denominator zero. For \( f(x) \), \( x = 6 \) is excluded because it makes the denominator zero.
• Simplification: Rational expressions often require simplification. In the exercise, when simplifying \( f(g(x)) \), you're essentially simplifying a rational function: \( \frac{1}{\frac{7}{x}} = \frac{x}{7} \).

Understanding these properties helps when dealing with rational functions, especially when forming compositions like \( f(g(x)) \) or \( g(f(x)) \). Confidents in these basics allows for tackling more complex algebraic tasks.