Problem 23 - Contemporary Precalculus

Question

Find the rule of the function \(f \circ g,\) the domain of \(f \circ g,\) the rule of \(g \cdot f,\) and the domain of \(g \circ f\) $$f(x)=\frac{1}{2 x+1}, \quad g(x)=x^{2}-1$$

Step-by-Step Solution

Answer

In conclusion, for the given functions \(f(x) = \frac{1}{2 x + 1}\) and \(g(x) = x^2 - 1\), we found the following: 1. The rule of \(f \circ g\) is \(f \circ g(x) = \frac{1}{2(x^2 - 1) + 1}\), and its domain is all real numbers. 2. The rule of \(g \circ f\) is \(g \circ f(x) = \left(\frac{1}{2x+1}\right)^2 -1\), and its domain is \(x \in \mathbb{R} \setminus \left\{-\frac{1}{2}\right\}\), meaning all real numbers except \(x = -\frac{1}{2}\).

Step 1: Find the rule of \(f \circ g\)

First, we need to find \(f(g(x))\). Since \(f(x)=\frac{1}{2 x+1}\) and \(g(x) = x^2 - 1\), we can substitute \(g(x)\) into \(f(x)\): $$f(g(x)) = f(x^2 - 1) = \frac{1}{2(x^2 - 1) + 1}$$ So the rule of \(f \circ g\) is: $$f \circ g(x) = \frac{1}{2(x^2 - 1) + 1}$$

Step 2: Find the domain of \(f \circ g\)

The domain of \(f \circ g\) is the set of all \(x\) values that make \(f(g(x))\) defined. We notice that the function \(f(g(x))\) is undefined when the denominator is zero, or when: $$2(x^2 - 1) + 1 = 0$$ Solve the equation to find the \(x\) values when the function is undefined, $$2x^2 - 2 + 1 = 0$$ $$2x^2 -1 = 0$$ However, this equation does not have any real solutions for \(x\). Therefore, the domain of \(f \circ g\) is all real numbers.

Step 3: Find the rule of \(g \circ f\)

Now, we need to find \(g(f(x))\). Since \(f(x)=\frac{1}{2 x+1}\) and \(g(x) = x^2 - 1\), we can substitute \(f(x)\) into \(g(x)\): $$g(f(x)) = g\left(\frac{1}{2x+1}\right) = \left(\frac{1}{2x+1}\right)^2 -1$$ So the rule of \(g \circ f\) is: $$g \circ f(x) = \left(\frac{1}{2x+1}\right)^2 -1$$

Step 4: Find the domain of \(g \circ f\)

The domain of \(g \circ f\) is the set of all \(x\) values that make \(g(f(x))\) defined. We notice that the function \(g(f(x))\) is undefined when the denominator of \(f(x)\) is zero, or when: $$2x + 1 = 0$$ Solve the equation to find the \(x\) value when the function is undefined, $$2x = -1$$ $$x = -\frac{1}{2}$$ So the domain of \(g \circ f\) is all real numbers except \(x = -\frac{1}{2}\), which can be expressed as: $$x \in \mathbb{R} \setminus \left\{-\frac{1}{2}\right\}$$

Key Concepts

Function Composition

Function composition is a fundamental concept in mathematics where you combine two functions to form a new function. When you see notation like \(f \circ g\), it means you're creating a new function by applying \(g\) first and then applying \(f\) to the result. This is written as \(f(g(x))\).
Let's break this down with an example. If you have \(f(x) = \frac{1}{2x+1}\) and \(g(x) = x^2 - 1\), the composite function \(f \circ g(x)\) would mean plugging \(g(x)\) into \(f(x)\). Specifically, use the result of \(g(x)\) as the input for \(f(x)\).

• First, calculate \(g(x): x^2 - 1\).
• Next, use this result in \(f(x): f(g(x)) = \frac{1}{2(x^2-1)+1}\).

The beauty of function composition is that it allows for more complex functions to be formed from simpler ones. It’s widely used across different fields of mathematics and science.

Domain of a Function

The domain of a function refers to the set of possible input values (\(x\)) that the function can accept without resulting in undefined or non-real outputs. Understanding the domain is essential when working with composite functions. For composite functions like \(f \circ g(x)\) or \(g \circ f(x)\), you need to ensure both the input function and the output function are defined.
For example, for \(f \circ g(x)\), you found that \(f(g(x))\) is defined for all real numbers because the equation \(2(x^2-1)+1 = 0\) has no real solutions. Hence, the domain is all real numbers, \(x \in \mathbb{R}\).
In contrast, for \(g \circ f(x)\), since \(f(x)\) involves a division, the denominator \(2x+1\) must not be zero to avoid undefined expressions. Solving \(2x + 1 = 0\) yields \(x = -\frac{1}{2}\), so the domain is all real numbers except \(-\frac{1}{2}\). This is notated as \(x \in \mathbb{R} \setminus \{-\frac{1}{2}\}\).

• Always check for values that make denominators zero, cause negative arguments under even roots, or lead to undefined expressions.
• Consider each function's individual domain and how they interact in composition.

Algebraic Functions

Algebraic functions are types of functions constructed using polynomials, rational expressions, or roots, and they are the building blocks in many areas of science and engineering. In the context of the exercise, both \(f(x) = \frac{1}{2x+1}\) and \(g(x) = x^2 - 1\) are algebraic functions. These functions are essential because they allow computations of various values through algebraic manipulations.
When dealing with algebraic functions, it’s important to remember the characteristics that define them:

• Polynomials: Functions like \(g(x) = x^2 - 1\) are simple polynomial functions, constructed by adding, subtracting, or multiplying constants and variables raised to whole number powers.
• Rational Expressions: The function \(f(x) = \frac{1}{2x+1}\) is a rational expression, representing a ratio of polynomial functions.

These types of functions form the foundation for more advanced mathematical concepts, such as limits, derivatives, and integrals in calculus. Understanding how to combine them through function composition expands their application across different mathematical topics.