Problem 64 - College Algebra with Modeling and Visualization

Question

Exercises \(57-72:\) Use the given \(f(x)\) and \(g(x)\) to find each of the following. Identify its domain. $$ \begin{array}{llll} \text { (a) }(f \circ g)(x) & \text { (b) }(g \circ f)(x) & \text { (c) }(f \circ f)(x) \end{array} $$ $$ f(x)=\frac{1}{3 x}, \quad g(x)=\frac{2}{x-1} $$

Step-by-Step Solution

Answer

(a) \((f \circ g)(x) = \frac{x-1}{6}\), domain: \(x \neq 1\); (b) \((g \circ f)(x) = \frac{6x}{1-3x}\), domain: \(x \neq 0, \frac{1}{3}\); (c) \((f \circ f)(x) = x\), domain: \(x \neq 0\).

Step 1: Compute \((f \, \circ \, g)(x)\)

To find \((f \, \circ \, g)(x)\), substitute \(g(x)\) into \(f(x)\). The expression becomes \(f(g(x)) = f\left(\frac{2}{x-1}\right) = \frac{1}{3\left(\frac{2}{x-1}\right)} = \frac{x-1}{6}\).

Step 2: Determine the domain of \((f \, \circ \, g)(x)\)

The function \(g(x)\) introduces the restriction that \(x eq 1\) because \(x-1 eq 0\). The overall function is \((f \, \circ \, g)(x) = \frac{x-1}{6}\), which has no additional restrictions since division by a non-zero constant is defined everywhere. Therefore, the domain is \(x eq 1\).

Step 3: Compute \((g \, \circ \, f)(x)\)

To find \((g \, \circ \, f)(x)\), substitute \(f(x)\) into \(g(x)\). It becomes \(g(f(x)) = g\left(\frac{1}{3x}\right) = \frac{2}{\frac{1}{3x} - 1}\), which simplifies to \(\frac{2}{\frac{1-3x}{3x}} = \frac{6x}{1-3x}\).

Step 4: Determine the domain of \((g \, \circ \, f)(x)\)

The expression \(\frac{1}{3x} - 1 eq 0\), leading to \(3x eq 1\) or \(x eq \frac{1}{3}\). Additionally, for \(f(x)\), \(x eq 0\) to avoid division by zero. Therefore, the domain is \(x eq 0\) and \(x eq \frac{1}{3}\).

Step 5: Compute \((f \, \circ \, f)(x)\)

For \((f \, \circ \, f)(x)\), substitute \(f(x)\) into itself: \(f(f(x)) = f\left(\frac{1}{3x}\right) = \frac{1}{3\left(\frac{1}{3x}\right)} = x\).

Step 6: Determine the domain of \((f \, \circ \, f)(x)\)

Since \(f(x)\) requires \(x eq 0\), the domain of \((f \, \circ \, f)(x)\) is the same: \(x eq 0\).

Key Concepts

Domain of a Function

When we talk about the domain of a function, we mean all possible input values (typically expressed as "x") for which the function is defined. In algebra, certain operations can limit the domain because they create undefined conditions. One such operation is division, where division by zero is undefined.

When you are working with functions, it becomes crucial to determine where the function operates smoothly. For instance, in rational functions like \( f(x) = \frac{1}{3x} \), the domain excludes any value of \( x \) that makes the denominator zero. Here:

• For \( f(x) = \frac{1}{3x} \), the domain is \( x eq 0 \).
• For \( g(x) = \frac{2}{x-1} \), the domain is \( x eq 1 \).

Importantly, when these functions are composed, we need to consider the domain restrictions from each involved function. For example, in the composition \((f \circ g)(x)\), you take the restrictions from both \(f\) and \(g\) into account to determine where \( (f \circ g)(x) \) is defined.

Rational Functions

Rational functions are algebraic expressions represented as the ratio of two polynomials. A typical rational function looks like \( \frac{P(x)}{Q(x)} \), where both \(P(x)\) and \(Q(x)\) are polynomials. Due to their nature, rational functions inherit the traits and restrictions of polynomials and division.

A key property of rational functions is that their domain is constrained by the denominator since it cannot be zero. Consider the example of \( f(x) = \frac{1}{3x} \) and \( g(x) = \frac{2}{x-1} \):

• The function \( f(x) \) becomes undefined for \( x = 0 \).
• The function \( g(x) \) becomes undefined for \( x = 1 \).
• Thus, when composing such functions, the resultant expression must account for all these restrictions.

In operations involving these functions, such as finding compositions \((f \circ g)(x) \) or \((g \circ f)(x) \), it's crucial to handle each rational expression carefully to maintain accuracy in understanding where they are defined.

Algebraic Expressions

Algebraic expressions form the foundation of algebra and mathematics in general. These expressions include constants, variables, and operations like addition, subtraction, multiplication, and division.

In particular, when dealing with the composition of functions, algebraic manipulation becomes pivotal. You often substitute one function into another, transforming the initial expressions. For instance:

• In \((f \circ g)(x)\), substituting \(g(x)\) into \(f(x)\) led to the expression \(\frac{x-1}{6}\).
• Similarly, \((g \circ f)(x)\) after substitution and simplification results in \(\frac{6x}{1-3x}\).
• This illustrates the use of algebraic processes like simplifying complex fractions.

Understanding these algebraic instruments not only aids in rewriting functions but also in understanding their domains and other mathematical properties. It's the knack for this manipulation that turns complex expressions into interpretable results.