Problem 17 - 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\()\) \(f(x)=\frac{3}{2 x}\) and \(g(x)=\frac{1}{x+1}\)

Step-by-Step Solution

Answer

(f ∘ g)(x) = \(\frac{3x + 3}{2}\), domain: \(x ≠ -1\). (g ∘ f)(x) = \(\frac{2x}{3 + 2x}\), domain: \(x ≠ 0, x ≠ -\frac{3}{2}\).

Step 1: Determine (f ∘ g)(x)

The composition \((f \circ g)(x)\) can be found by substituting \(g(x)\) into \(f(x)\). We start by evaluating \(g(x) = \frac{1}{x+1}\). Now, substitute \(g(x)\) into \(f(x)\): \[f(g(x)) = f\left(\frac{1}{x+1}\right) = \frac{3}{2 \cdot \frac{1}{x+1}} = \frac{3(x + 1)}{2} = \frac{3x + 3}{2}\].Thus, \((f \circ g)(x) = \frac{3x + 3}{2}\).

Step 2: Determine the Domain of (f ∘ g)(x)

The domain of \((f \circ g)(x)\) is determined by the domain of \(g(x)\) and where \(f(g(x))\) is defined. For \(g(x) = \frac{1}{x+1}\), \(x+1 eq 0\), so \(x eq -1\). Since \(f(x) = \frac{3}{2x}\), \(g(x) eq 0\). \\(g(x) = \frac{1}{x+1} eq 0\), implying \(x eq -1\). Therefore, the domain is \(x \in \mathbb{R}, x eq -1\).

Step 3: Determine (g ∘ f)(x)

The composition \((g \circ f)(x)\) is found by substituting \(f(x)\) into \(g(x)\). Start by evaluating \(f(x) = \frac{3}{2x}\). Now substitute \(f(x)\) into \(g(x)\): \[g(f(x)) = g\left(\frac{3}{2x}\right) = \frac{1}{\frac{3}{2x} + 1}\]. Simplifying, we get:\[g(f(x)) = \frac{1}{\frac{3}{2x} + 1} = \frac{2x}{3 + 2x}\].Thus, \((g \circ f)(x) = \frac{2x}{3 + 2x}\).

Step 4: Determine the Domain of (g ∘ f)(x)

The domain of \((g \circ f)(x)\) depends on where both \(f(x)\) and \(g(f(x))\) are defined. For \(f(x) = \frac{3}{2x}\), \(x eq 0\). Additionally, for \(g(f(x)) = \frac{2x}{3 + 2x}\), we must have \(3 + 2x eq 0\). Solving \(3 + 2x = 0\) gives \(x = -\frac{3}{2}\). Thus, the domain is \(x \in \mathbb{R}, x eq 0, x eq -\frac{3}{2}\).

Key Concepts

Domain of a Function

The domain of a function is a fundamental concept in mathematics. It tells us all possible input values ("x" values) that a function can accept. When dealing with real-world applications, the domain represents all feasible situations that can be modeled by the function. For instance, if a function is defined as a fraction, the denominator must not be zero because division by zero is undefined in mathematics.

Let's consider the function examples from our exercise:

• For the function \(f(x) = \frac{3}{2x}\), the expression is undefined at \(x = 0\) because it would make the denominator zero.
• For the function \(g(x) = \frac{1}{x + 1}\), it's undefined at \(x = -1\) for the same reason.

Function compositions add another layer to domain determination, where we need to consider the domains of both individual functions. For example, the domain of \((f \circ g)(x)\) relies on the domains of \(f(x)\) and \(g(x)\). Besides ensuring that each is defined independently, we also make sure that the intermediate outcomes, like \(g(x)\), stay within the usable range of \(f(x)\). Otherwise, the composition itself would not be defined for those values.

Understanding the domain helps prevent mathematical errors and ensures the validity of solutions.

Rational Functions

Rational functions are a type of algebraic expression that are particularly interesting because they include both a numerator and a denominator with variables. This feature means they can behave quite differently compared to basic linear or quadratic functions. A rational function can be written in the form \(\frac{P(x)}{Q(x)}\) where both \(P(x)\) and \(Q(x)\) are polynomials, and the function is undefined wherever \(Q(x) = 0\).

In our problem set, the functions \(f(x) = \frac{3}{2x}\) and \(g(x) = \frac{1}{x+1}\) are both rational functions. Both require careful attention to the potential zeroes in their denominators:

• \(f(x)\) becomes undefined when \(x = 0\).
• \(g(x)\) requires that \(x ot= -1\).

When combining them as \((f \circ g)(x)\) or \((g \circ f)(x)\), the interaction between these restrictions becomes even more intricate. For example, for \((f \circ g)(x)\), it's necessary that \(g(x) ot= 0\) to keep \(f(g(x))\) defined. That extra layer of restrictions leads to a combined domain where all individual restrictions are respected.

This complexity makes rational functions a rich topic to study, as they mirror real-world situations involving division and can illustrate limits and asymptotic behavior.

Algebraic Expressions

Algebraic expressions are critical building blocks in any kind of mathematical operations or functions. An algebraic expression consists of numbers, variables, and operators (like +, -, *, /) that together describe a calculation or a function.

In our example, the functions under consideration, \(f(x) = \frac{3}{2x}\) and \(g(x) = \frac{1}{x + 1}\), are represented entirely with such algebraic expressions:

• Variables: Both functions use the variable \(x\).
• Operations: They involve division, a fundamental aspect of these rational expressions.

When evaluating composites like \((f \circ g)(x)\), the transformation involves substituting one algebraic expression into another, creating entirely new but coherent expressions such as \(\frac{3x + 3}{2}\) or \(\frac{2x}{3 + 2x}\).

These transformations require applying basic algebraic rules, such as distributing multiplication over addition, and simplifying fractions. Mastery of these steps ensures capability in tackling more complex functions and equations encountered in higher mathematics.