Problem 37 - Algebra and Trigonometry

Question

Find the functions \(f \circ g, g \circ f, f \circ f,\) and \(g \circ g\) and their domains. $$ f(x)=\frac{1}{x}, g(x)=2 x+4 $$

Step-by-Step Solution

Answer

f ∘ g: \(\frac{1}{2x+4}\), domain \(\mathbb{R}\setminus\{-2\}\); g ∘ f: \(\frac{2}{x}+4\), domain \(\mathbb{R}\setminus\{0\}\); f ∘ f: \(x\), domain \(\mathbb{R}\setminus\{0\}\); g ∘ g: \(4x+12\), domain \(\mathbb{R}\).

Step 1: Define the Composite Function f ∘ g

The composite function \(f \circ g\) is defined as \(f(g(x))\). Substitute \(g(x)\) into \(f(x)\): \[f(g(x)) = f(2x + 4) = \frac{1}{2x + 4}\].

Step 2: Determine the Domain of f ∘ g

The domain of \(f \circ g\) is determined by the values of \(x\) for which both \(g(x)\) is defined and \(f(g(x))\) does not lead to division by zero. Since \(f(g(x)) = \frac{1}{2x + 4}\), the expression is undefined for \(2x + 4 = 0\). Solving \(2x + 4 = 0\) gives \(x = -2\). Therefore, the domain is all real numbers except \(x = -2\).

Step 3: Define the Composite Function g ∘ f

The composite function \(g \circ f\) is defined as \(g(f(x))\). Substitute \(f(x)\) into \(g(x)\): \[g(f(x)) = g\left(\frac{1}{x}\right) = 2\left(\frac{1}{x}\right) + 4 = \frac{2}{x} + 4\].

Step 4: Determine the Domain of g ∘ f

The domain of \(g \circ f\) is determined by the values of \(x\) for which both \(f(x)\) is defined and \(g(f(x))\) does not lead to division by zero. Since \(f(x) = \frac{1}{x}\), \(x eq 0\). Therefore, the domain is all real numbers except \(x = 0\).

Step 5: Define the Composite Function f ∘ f

The composite function \(f \circ f\) is defined as \(f(f(x))\). Substitute \(f(x)\) into itself: \[f(f(x)) = f\left(\frac{1}{x}\right) = \frac{1}{\left(\frac{1}{x}\right)} = x\].

Step 6: Determine the Domain of f ∘ f

The domain of \(f \circ f\) is determined by the values of \(x\) for which \(f(x)\) is defined. Since \(f(x) = \frac{1}{x}\), \(x eq 0\). Therefore, the domain is all real numbers except \(x = 0\).

Step 7: Define the Composite Function g ∘ g

The composite function \(g \circ g\) is defined as \(g(g(x))\). Substitute \(g(x)\) into itself: \[g(g(x)) = g(2x + 4) = 2(2x + 4) + 4 = 4x + 8 + 4 = 4x + 12\].

Step 8: Determine the Domain of g ∘ g

Since \(g(x)\) is a polynomial, it is defined for all real numbers. Therefore, \(g \circ g\) is also defined for all real numbers. The domain is all real numbers.

Key Concepts

Domain of a Function

Understanding the domain of a function is essential because it tells us the set of all possible input values, often represented as "x" values, for which the function is defined. For the functions given in our exercise:

• **f(x) = \(\frac{1}{x}\)**: Here, the function is undefined if the denominator is zero. Therefore, the domain is all real numbers except where \(x = 0\), since division by zero is not possible.
• **g(x) = 2x + 4**: This is a linear function, meaning it can take any real number as domain because there are no restrictions like division by zero or square root of a negative number.

When we form composite functions like \(f \circ g\) or \(g \circ f\), we need to find the restrictions separately for both parts of the composite function. This means the domain of a composite function is limited to where all parts of the function are defined.

Function Composition

Function composition means combining two functions to create a third function. Given two functions, \(f\) and \(g\), the composition \(f \circ g\), pronounced as "f composed with g", is defined by applying \(g\) first and then \(f\). In other words, you substitute the entire function \(g(x)\) into \(f\).Let's break down the compositions in our problem:

• **\(f \circ g = f(g(x)) = \frac{1}{2x + 4}\)**: Here, substituting \(g(x)\) into \(f\) results in a rational function, \(\frac{1}{2x + 4}\), highlighting the need for finding a domain where the denominator isn't zero.
• **\(g \circ f = g(f(x)) = \frac{2}{x} + 4\)**: Similarly, inserting \(f(x)\) into \(g\) also results in a rational expression. This helps us see how the composition impacts both the expression of the function and the domain.
• **\(f \circ f\) and \(g \circ g\)**: These compositions demonstrate self-substitution. For \(f \circ f\), since \(f(f(x)) = x\), it simplifies remarkably to just \(x\), and its domain excludes \(x = 0\). For \(g \circ g\), it's a polynomial \(4x + 12\) with no domain restrictions.

Rational Functions

Rational functions are ratios of two polynomials. They take the form \(\frac{P(x)}{Q(x)}\), where \(P(x)\) and \(Q(x)\) are polynomials, and \(Q(x) eq 0\).In our exercise, we dealt with:

• **\(f(x) = \frac{1}{x}\)**: This is the simplest type of rational function where \(P(x) = 1\) and \(Q(x) = x\). The condition \(Q(x) eq 0\) leads to the domain restriction: \(x eq 0\).
• **\(f \circ g = \frac{1}{2x + 4}\)** and **\(g \circ f = \frac{2}{x} + 4\)**: Both of these are rational functions obtained through function composition. They showcase how the function composition can transform linear or other types of functions into rational ones, requiring careful consideration of the new domain.

Rational functions are peculiar because, unlike polynomial functions that are generally defined for all x-values, these functions have defined limits. Paying attention to what makes the denominator zero is the key to finding the domain.