Problem 20 - Intermediate Algebra

Question

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

Step-by-Step Solution

Answer

\( g \cdot f(x) = 2x^2 - 5x - 3 \), with domain \( \mathbb{R} \).

Step 1: Understanding the Functions

We begin by understanding the given functions. We have two functions:1. \( f(x) = 2x + 1 \)2. \( g(x) = x - 3 \). Each of these are linear functions and they are defined for all real numbers.

Step 2: Define the Domain

Since both \( f(x) \) and \( g(x) \) are linear functions, their domain is all real numbers, i.e., \( \text{Domain of } f(x) = \text{Domain of } g(x) = \mathbb{R} \). This means they can take any real number as input.

Step 3: Determine \( g \cdot f \)

To find \( g \cdot f \), we need to multiply the two functions:\[ (g \cdot f)(x) = g(x) \cdot f(x) = (x - 3) \cdot (2x + 1) \].

Step 4: Expand \( g \cdot f \)

Expand the expression \((x - 3)(2x + 1)\) using the distributive property:\[ (x - 3)(2x + 1) = x \cdot 2x + x \cdot 1 - 3 \cdot 2x - 3 \cdot 1 \].

Step 5: Simplify the Expression

Perform the multiplications:\[ 2x^2 + x - 6x - 3 \].Combine like terms:\[ 2x^2 - 5x - 3 \].Thus, the function \( g \cdot f \) is \( 2x^2 - 5x - 3 \).

Step 6: Establish the Domain of \( g \cdot f \)

Since \( g \cdot f \) is a quadratic function, its domain is also all real numbers, \( \mathbb{R} \), because a quadratic function is defined everywhere on the real number line.

Key Concepts

Domain of Linear Functions

When dealing with linear functions, it's important to understand what the domain means. The domain of a function describes the set of input values that the function can accept. For linear functions like \( f(x) = 2x + 1 \) and \( g(x) = x - 3 \), the domain is all real numbers, denoted by \( \mathbb{R} \). The reason lies in the nature of linear functions:

• There are no restrictions such as divisions by zero or square roots of negative numbers, which might otherwise limit the domain.
• Linear functions are continuous, meaning they form a straight line across the entire number line when graphed.

Thus, both \( f(x) \) and \( g(x) \) can take any real number as input without any restrictions.

Quadratic Functions

Quadratic functions are polynomials of degree 2, commonly expressed in the standard form \( ax^2 + bx + c \). In the context of this exercise, after multiplying the two linear functions \( (x - 3) \) and \( (2x + 1) \), we obtain a quadratic function: \[ g \cdot f(x) = 2x^2 - 5x - 3 \] Here are some important aspects of quadratic functions:

• The graph of a quadratic function is a parabola. It can open either upwards or downwards depending on the sign of \( a \).
• Since quadratic functions are polynomials, their domain is also all real numbers, \( \mathbb{R} \), similar to linear functions.
• This means you can input any real number into the function, and it will yield a corresponding output.

Distributive Property in Algebra

The distributive property is key to understanding how to multiply expressions like the ones found in our function multiplication problem. The distributive property states that for any numbers \( a \), \( b \), and \( c \), the equation \( a(b + c) = ab + ac \) holds. It allows you to distribute the multiplication of the outer term to each of the terms within the parentheses. This property simplifies expressions and is especially handy in polynomial multiplication.In our exercise, when we found \( (g \cdot f)(x) = (x - 3)(2x + 1) \), we used the distributive property as follows:

• First, distribute \( x \) across \( (2x + 1) \), resulting in \( 2x^2 + x \).
• Next, distribute \( -3 \) across \( (2x + 1) \), yielding \( -6x - 3 \).
• Finally, combine the terms \( 2x^2 + x - 6x - 3 \) by adding like terms to simplify to \( 2x^2 - 5x - 3 \).

For this reason, the distributive property is a fundamental tool in algebra, crucial for expanding and simplifying expressions during multiplication.