Problem 86 - Precalculus : Building Concepts and Connections

Question

In Exercises \(67-86,\) find expressions for \((f \circ g)(x)\) and \((g \circ f)(x)\) Give the domains of \(f \circ g\) and \(g \circ f\). $$f(x)=\frac{-x+1}{2 x+3} ; g(x)=\frac{1}{x^{2}+1}$$

Step-by-Step Solution

Answer

The composition \(f \circ g(x)\) simplifies to \(-\frac{1}{2x^2+3x+4}\), and its domain is all real numbers. The composition \(g \circ f(x)\) simplifies to \( \frac{4(2x+3)^2}{4x^2-4x+5} \), and its domain is all real numbers except \(x = -\frac{3}{2}\).

Step 1: Find \(f \circ g(x)\)

Start by substituting \(g(x)\) into the equation for \(f(x)\). You will get: \(f(g(x))=\frac{-g(x)+1}{2g(x)+3} = \frac{-\frac{1}{x^{2}+1}+1}{2*\frac{1}{x^{2}+1}+3}\). Now simplify this expression to obtain the resultant of \(f \circ g(x)\)

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

Substitute \(f(x)\) into the equation for \(g(x)\) to get \(g(f(x))=\frac{1}{f(x)^{2}+1}=\frac{1}{(\frac{-x+1}{2x+3})^{2}+1}\). Simplify this expression to get \(g \circ f(x)\).

Step 3: Find the domains of \(f \circ g(x)\) and \(g \circ f(x)\)

The domains are the set of all real numbers \(x\) for which the functions are defined. For \(f \circ g(x)\), \(x\) can be anything, but for \(g \circ f(x)\), the denominator \(2x+3\) cannot be 0, which means \(x\) cannot be -3/2. Thus, the domain of \(f \circ g(x)\) can be said to be all reals and the domain of \(g \circ f(x)\) is all reals except \(x = -3/2\).

Key Concepts

Domain of a Function

When dealing with functions, the domain is a critical aspect. The domain of a function is essentially the set of all possible input values (usually noted as "x") for which a function is defined. For example, if a function has a fraction, like \(f(x) = \frac{1}{x}\), the domain includes all numbers except those that make the denominator zero.
This is important because dividing by zero is undefined in the realm of mathematics. To find the domain, begin by identifying values that make the expression invalid, such as:

• Fractions where the denominator can become zero
• Square roots involving negative numbers

In the context of composite functions, you must consider the domains of both the individual functions \(f\) and \(g\), as well as their composition. Hence, for functions like \(f(x)=\frac{-x+1}{2x+3}\) and \(g(x)=\frac{1}{x^2+1}\), we check each part. The domain of \(f(x)\) excludes \(x=\frac{-3}{2}\) due to the zero denominator, while \(g(x)\) is valid for all real numbers since the denominator \(x^2+1\) is never zero.

Composite Functions

Composite functions are an intriguing concept, involving the combination of two functions in a way that the output of one function becomes the input for another. In mathematical terms, if you have two functions \(f\) and \(g\), their composite is denoted as \((f \circ g)(x) = f(g(x))\). Similarly, \((g \circ f)(x) = g(f(x))\).
This means taking \(g(x)\), calculating it, and then using that result as \(x\) in \(f(x)\) when forming \(f \circ g(x)\). Consider the following bullet points for clarity:

• Determine \(g(x)\) first and use this result as the input for \(f(x)\).
• When constructing \(g \circ f(x)\), switch the order: compute \(f(x)\) first and then use its result as input for \(g(x)\).
• Always simplify the resulting expression for clarity.

For example, given \(f(x)=\frac{-x+1}{2x+3}\) and \(g(x)=\frac{1}{x^2+1}\), for \(f \circ g(x)\) you substitute \(g(x)\) into \(f\), and for \(g \circ f(x)\), you substitute \(f(x)\) into \(g\). Recognizing these patterns helps solve composite tasks efficiently by simplifying and ensuring the functions remain within their domains.

Rational Functions

Rational functions are a staple concept in algebra, defined as functions represented by the ratio of two polynomials. They typically appear in the form \(\frac{P(x)}{Q(x)}\) where both \(P(x)\) and \(Q(x)\) are polynomials.
An integral aspect of rational functions is to determine where they are valid, focusing on ensuring the denominator \(Q(x)\) is never zero. This aligns with the need for finding function domains. Key points about rational functions:

• The zeros of \(Q(x)\) (where the function is undefined) are critical when determining the domain.
• Rational functions can exhibit horizontal or slant asymptotes, impacting their graphs.
• Simplifying expressions is essential to properly understand their behavior.

Using \(f(x)=\frac{-x+1}{2x+3}\) as an example, the function is rational as it consists of polynomials, and the domain excludes values where \(2x+3=0\). This example reflects typical traits of rational functions, such as carefully handling exclusions for valid inputs, while seeking understanding of their overall behavior through simplification and graph analysis.