Problem 47 - Precalculus with Limits

Question

In Exercises 41-48, find (a) \(f \circ g\), and (b) \(g \circ f\). Find the domain of each function and each composite function. \(f(x) = \frac{1}{x}\), \(g(x) = x+3\)

Step-by-Step Solution

Answer

\(f \circ g = \frac{1}{x+3}\) with domain \(x ≠ -3\), and \(g \circ f = \frac{1}{x}+3\) with domain \(x ≠ 0\). The function \(f(x)\) has domain \(x ≠ 0\) and the function \(g(x)\) has domain all real numbers.

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

In this step, substitute \(g(x)\) into \(f(x)\). So, \(f \circ g\) can be computed as \(f(g(x)) = f(x+3) = \frac{1}{x+3}\)

Step 2: Compute \(g \circ f\)

Substitute \(f(x)\) into \(g(x)\). This gives \(g \circ f = g(f(x)) = g(\frac{1}{x}) = \frac{1}{x}+3\

Step 3: Determine domains of the composite functions

The domain of the composite function \(f \circ g\) is the set of all \(x\) such that \(x+3 ≠ 0\), so \(x ≠ -3\). The domain of the composite function \(g \circ f\) is the set of all \(x\) such that \(1/x ≠ 0\), so \(x ≠ 0\)

Step 4: Recap original functions' domains

For \(f(x) = \frac{1}{x}\), \(x ≠ 0\). In the case of \(g(x) = x+3\), the function is defined for all real numbers, so the domain is all real numbers.

Key Concepts

Function Composition

Function composition is a process where you combine two functions in a specific order to create a new function. It's like putting one function inside another. For example, with two functions, say, f and g, when we perform f after g, it's written as \(f \circ g\). In our exercise, we have \(f\circg\) and \(g\circf\) with \(f(x)=\dfrac{1}{x}\) and \(g(x)=x+3\).

Firstly, to find \(f\circg\), we substitute \(g(x)\) into \(f(x)\), which yields \(\dfrac{1}{x+3}\), showing how g modifies the input for f. Similarly, \(g\circf\) involves putting \(f(x)\) into \(g(x)\), resulting in \(\dfrac{1}{x}+3\), indicating how f influences the input for g. It's essential to recognize that the order of composition matters because \(f\circg\) is generally not the same as \(g\circf\). Understanding how to compose functions properly is a foundational skill for dealing with more complex mathematical concepts.

Domain of a Function

The domain of a function is the set of all possible input values that don't cause any undefined behavior, such as division by zero or taking the square root of a negative number. When it comes to composite functions, like those in our exercise, determining the domain can be a little tricky because we have to consider the restrictions from both the inside and the outside functions.

For instance, the composite function \(f \circ g\) gives us \(\dfrac{1}{x+3}\), and so we must exclude values that make the denominator zero; in this case, \(x \eq -3\). This is a crucial step in ensuring the function only takes valid inputs. Similarly, for \(g \circ f\), the form \(g\left(\dfrac{1}{x}\right)=\dfrac{1}{x}+3\) requires us to exclude \(x=0\) from the domain. Always remember to check for these constraints to avoid errors when working with functions and their compositions.

Rational Functions

A rational function is a ratio of two polynomials, like \(\dfrac{p(x)}{q(x)}\) where both \(p(x)\) and \(q(x)\) are polynomials, and \(q(x)\) is not the zero polynomial. The function \(f(x) = \dfrac{1}{x}\) from our exercise is a simple example of a rational function, where the numerator is 1 (a constant polynomial) and the denominator is a first-degree polynomial.

Rational functions often have restrictions on their domain because we must avoid situations where the denominator is zero. For \(g\left(\dfrac{1}{x}\right)\), we deal with a rational expression in a slightly more complex situation because the function's composition can introduce additional domain restrictions. Understanding the behavior of these functions — especially how they behave near their vertical asymptotes, where the denominator approaches zero — is essential for graphing them and analyzing their properties.