Problem 12 - Algebra 2 and Trigonometry

Question

In \(8-13\) , the domain of \(f(x)=4-2 x\) and of \(g(x)=x^{2}\) is the set of real numbers and the domain of \(h(x)=\frac{1}{x}\) is the set of non-zero real numbers. a. Write each function value in terms of \(x\) . b. Find the domain of each function. \((g+3 f)(x)\)

Step-by-Step Solution

Answer

The function \((g+3f)(x) = x^2 - 6x + 12\) with domain all real numbers.

Step 1: Identify the Functions

We're given three different functions: \(f(x)=4-2x\), \(g(x)=x^2\), and \(h(x)=\frac{1}{x}\). We aim to find \((g+3f)(x)\).

Step 2: Express the Function

We need to express \((g+3f)(x)\). Start by calculating \(3f(x)\):\[3f(x) = 3(4 - 2x) = 12 - 6x.\]Now, we can express \((g+3f)(x)\) as:\[(g+3f)(x) = g(x) + 3f(x) = x^2 + (12 - 6x).\]

Step 3: Simplify the Expression

Let's simplify \((g+3f)(x)\):\[(g+3f)(x) = x^2 + 12 - 6x.\]Combining like terms, the function becomes:\[x^2 - 6x + 12.\]

Step 4: Determine the Domain of the Function

The domain of \(g(x) = x^2\) is all real numbers, and the domain of \(f(x) = 4 - 2x\) is also all real numbers. Therefore, the domain of \((g+3f)(x) = x^2 - 6x + 12\) is all real numbers because there are no restrictions (like division by zero) in this expression.

Key Concepts

Understanding Function Domains

The concept of a function domain is fundamental in algebraic functions. Simply put, the domain of a function is the set of all possible input values (usually represented by \( x \)) for which the function is defined. A function might not be defined for every possible real number, due to restrictions like division by zero or taking the square root of a negative number.

For instance, the domain of a polynomial function like \( g(x) = x^2 \) is the set of all real numbers, since you can substitute any real number into \( x^2 \) and get a valid result. Similarly, \( f(x) = 4 - 2x \) is also defined for all real numbers because subtraction and multiplication with real numbers pose no restrictions.

However, when dealing with a function like \( h(x) = \frac{1}{x} \), we identify a restriction. This function is not defined when \( x = 0 \) as division by zero is undefined. Therefore, its domain is all real numbers except zero. Understanding these rules allows you to identify function domains with confidence.

Combining Functions

Combining functions is a technique that involves creating a new function by adding, subtracting, multiplying, or dividing two or more functions. This exercise involves combining \( g(x) \) and \( f(x) \) to find \((g+3f)(x)\).

Essentially, combining functions means performing algebraic operations on the functions’ expressions. In this case, \( 3f(x) \) is first calculated separately as \( 3(4 - 2x) = 12 - 6x \). The new function \( (g+3f)(x) \) is then found by adding \( g(x) = x^2 \) and \( 3f(x) \). Therefore, the operation can be expressed as:\[ (g+3f)(x) = x^2 + (12 - 6x). \]

Being adept at combining functions is crucial when manipulating and solving algebraic expressions, especially in problems that require more than just simple function evaluation.

Simplifying Expressions

Simplifying expressions is the process of making an algebraic expression easier to work with by combining like terms and reducing it to a more manageable form. It is a vital skill in algebra that helps in solving equations or evaluating functions.

Consider the function \( (g+3f)(x) = x^2 + 12 - 6x \). This expression involves terms that can be rearranged and combined. To simplify, align terms by their degree: square terms, linear terms, and constant terms. In this instance, you rearrange the expression as follows:\[ x^2 - 6x + 12. \]

Here, two modifications were made: like terms were combined (i.e., the constant terms and linear terms), producing a neat, quadratic expression. Simplification like this enhances clarity and ease of computation in further operations or evaluations. Additionally, it assists in better understanding the function's behavior, such as its growth, shrinking, or turning points.