Problem 19 - Intermediate Algebra

Question

Let \(f(x)=2 x+1\) and \(g(x)=x-3 .\) Find each function and give its domain. See Example 1. $$ g / f $$

Step-by-Step Solution

Answer

The function \( h(x) = \frac{x-3}{2x+1} \) with domain \( x \in \mathbb{R} \setminus \{-\frac{1}{2}\} \).

Step 1: Understanding the Composition

The expression \( g/f \) represents the function \( h(x) = \frac{g(x)}{f(x)} \). The goal is to find this function and its domain.

Step 2: Calculate the Function h(x)

Substitute the given functions into the expression for \( h(x) \). We have \( f(x) = 2x + 1 \) and \( g(x) = x - 3 \). Thus, the function is \( h(x) = \frac{x - 3}{2x + 1} \).

Step 3: Determine the Domain of h(x)

The domain of \( h(x) = \frac{x - 3}{2x + 1} \) is all real numbers except where the denominator is zero. Solve \( 2x + 1 = 0 \) to find the values that are not part of the domain: \( x = -\frac{1}{2} \).

Step 4: Conclude the Domain

The domain of \( h(x) \) is all real numbers except \( x = -\frac{1}{2} \). Thus, the domain is \( x \in \mathbb{R} \setminus \left\{ -\frac{1}{2} \right\} \).

Key Concepts

Domain of a Function

The domain of a function is the set of all possible input values (usually represented as \(x\)) that will produce a valid output in a function. To determine the domain, consider any restrictions that might occur due to mathematical operations you can't perform.

• Infinite Possibilities: For many functions, the domain includes all real numbers, which means there are no restrictions on the inputs.
• Specific Operations: With operations like division, the denominator cannot be zero, as division by zero is undefined in mathematics. To find where the division breaks down, set the denominator equal to zero and solve for \(x\).
• In our Exercise: For the function \(h(x) = \frac{x-3}{2x+1}\), the domain excludes the value that makes \(2x+1=0\). Solve this: \(2x+1=0\) yields \(x=-\frac{1}{2}\).

Therefore, the domain of \(h(x)\) in this exercise is simply all real numbers except \(-\frac{1}{2}\).

Rational Functions

Rational functions are fractions made up of polynomials. Understanding them involves recognizing both the numerator and the denominator of these functions.

• Form: A rational function looks like \(\frac{P(x)}{Q(x)}\), where both \(P(x)\) and \(Q(x)\) are polynomials.
• Key Concept: The denominator \(Q(x)\) cannot be zero, as this would make the function undefined. Therefore, finding where \(Q(x) = 0\) tells us where the rational function is undefined.
• Example in Exercise: In the function \(h(x)=\frac{x-3}{2x+1}\), the numerator is \(x-3\) and the denominator is \(2x+1\). Resolving \(2x+1=0\) helped us determine that \(x=-\frac{1}{2}\) is not a valid input for this function.

Rational functions are crucial in analyzing many natural phenomena modeled mathematically by such expressions.

Algebraic Expressions

Algebraic expressions are combinations of numbers, variables, and arithmetic operations like addition, subtraction, multiplication, and division. Understanding them is key to mastering more complex mathematics.

• Simple to Complex: An algebraic expression can be as simple as \(x + 2\) or more complex, like \(2x^2 + 3x + 4\).
• Manipulating Expressions: You can perform various operations like addition or substitution. For example, substituting \(x = 1\) into \(2x + 1\) gives \(3\).
• Applying Expressions: In a function like \(f(x)=2x+1\), this is considered an algebraic expression. It's useful in understanding how inputs turn into outputs.

In our exercise, substituting the given expressions into a new form, \(h(x) = \frac{x-3}{2x+1}\), allowed us to both determine \(h(x)\) and its domain.