Problem 5 - College Algebra and Calculus: An Applied Approach

Question

Find (a) \((f+g)(x)\), (b) \((f-g)(x)\), (c) \((f g)(x)\), and (d) \((f / g)(x)\). What is the domain of \(f / g\) ? \(f(x)=x+1, \quad g(x)=x-1\)

Step-by-Step Solution

Answer

The results are (a) \((f+g)(x) = 2x\), (b) \((f-g)(x) = 2\), (c) \((fg)(x) = x^2 -1\), and (d) \((f/g)(x) = (x+1) / (x-1)\), domain: \(x \neq 1\).

Step 1: Find (f+g)(x)

Recognize that \((f+g)(x) = f(x) + g(x)\). Substitute \(f(x)=x+1\) and \(g(x)=x-1\) into the equation, and get \((f+g)(x) = (x+1) + (x-1)\). Simplify to get \(2x\). So, \((f+g)(x) = 2x\)

Step 2: Find (f-g)(x)

Likewise, \((f-g)(x) = f(x) - g(x)\). Substitute \(f(x)=x+1\) and \(g(x)=x-1\) into the equation, and get \((f-g)(x) = (x+1) - (x-1)\). Simplify to get \(2\). So, \((f-g)(x) = 2\)

Step 3: Find (fg)(x)

Similarly, \((fg)(x) = f(x) * g(x)\). Substitute \(f(x)=x+1\) and \(g(x)=x-1\) into the equation, and get \((fg)(x) = (x+1) * (x-1)\). Simplify to get \(x^2 -1\). So, \((fg)(x) = x^2 -1\)

Step 4: Find (f/g)(x) and the domain

\((f/g)(x) = f(x) / g(x)\). Substitute \(f(x)=x+1\) and \(g(x)=x-1\) into the equation, and get \((f/g)(x) = (x+1) / (x-1)\). Recognize that the function has a undefined point at \(x=1\) as division by zero is undefined. So, the domain of \(f/g\) is all real numbers except \(x=1\). This gives \((f/g)(x) = (x+1) / (x-1)\), domain: \(x \neq 1\)

Key Concepts

Function Operations

When studying functions in algebra, we often encounter various operations that can be performed on them. Function operations include standard arithmetic operations applied to functions: addition, subtraction, multiplication, and division. Understanding these concepts is essential in exploring how complex functions can be constructed from simpler ones.

Let’s consider the functions given in the exercise, where \(f(x) = x + 1\) and \(g(x) = x - 1\). When we combine these functions through operation, we get different results. For addition \((f+g)(x)\), both functions’ values at \(x\) are added together, yielding \(2x\). Subtraction \((f-g)(x)\) involves taking the value of \(g\) from \(f\) at any given point, which results in a constant \(2\). Multiplication \((fg)(x)\) leads to the product of values, and division \((f/g)(x)\) gives us a ratio of one function's value to another's. However, with division, one must be careful to avoid dividing by zero, as it leads to undefined results.

Function Domain

The domain of a function is the set of all possible input values (commonly represented as \(x\)) for which the function is defined. In other words, the domain includes all the real numbers that you can put into the function without causing any mathematical inconsistencies, such as division by zero.

In the context of the given exercise, the function \(f/g\) corresponds to \((f/g)(x) = (x+1)/(x-1)\). Here, the function is defined as long as the denominator \(g(x) = x - 1\) is not zero. Hence, the domain of \(f/g\) would be all real numbers except where \(x = 1\), since at this value of \(x\), the denominator becomes zero, and the expression becomes undefined. Identifying the domain is crucial because it informs us about the limitations and boundaries within which we can work with a given function.

Algebraic Expressions

Algebraic expressions are combinations of variables, numbers, and operations that represent a particular value. Expressions can consist of terms, which are the separate entities joined by addition or subtraction signs. They can also include coefficients, which are the numerical factors that multiply the variables.

For example, in the operation \((fg)(x)\) from the exercise, we multiply the functions \(f(x)\) and \(g(x)\), which are both algebraic expressions, and we get a new algebraic expression \(x^2 - 1\). This product is a polynomial, an algebraic expression with several terms, in this case, \(x^2\) and \(-1\). Understanding how to manipulate algebraic expressions is a fundamental stepping stone in algebra that helps students solve equations and simplify complex mathematical scenarios.