Problem 28 - College Algebra with Modeling and Visualization

Question

Exercises \(11-30:\) Use \(f(x)\) and \(g(x)\) to find a formula for each expression. Identify its domain. (a) \((f+g)(x)\) (b) \((f-g)(x)\) (c) \((f g)(x)\) (d) \((f / g)(x)\) $$ f(x)=\frac{1}{x+2}, \quad g(x)=x^{2}+x-2 $$

Step-by-Step Solution

Answer

Expressions: \((f+g)(x)\), \((f-g)(x)\), \((fg)(x)\), \((f/g)(x)\). Domain: \(x \neq -2\), and for \(f/g\), \(x \neq 1\).

Step 1: Understand the Functions

First, understand each of the given functions. Function \( f(x) = \frac{1}{x+2} \) and function \( g(x) = x^2 + x - 2 \). These definitions help to find each required operation such as sum, difference, product, and quotient.

Step 2: Compute \((f+g)(x)\)

To find \((f+g)(x)\), add the functions: \((f+g)(x) = \frac{1}{x+2} + (x^2+x-2)\). The domain is all \(x\) such that each function's domain is considered, thus \(x eq -2\).

Step 3: Compute \((f-g)(x)\)

To find \((f-g)(x)\), subtract the functions: \((f-g)(x) = \frac{1}{x+2} - (x^2+x-2)\). The domain remains \(x eq -2\).

Step 4: Compute \((fg)(x)\)

For the product, multiply the functions: \((fg)(x) = \frac{1}{x+2} \times (x^2+x-2) = \frac{x^2+x-2}{x+2}\). The domain is \(x eq -2\).

Step 5: Compute \((f/g)(x)\)

For the quotient, divide the functions: \((f/g)(x) = \frac{\frac{1}{x+2}}{x^2+x-2}\). Simplified: \(\frac{1}{(x+2)(x^2+x-2)} \). The domain is where the denominator is not zero, so \(x eq -2, x eq 1, x eq -2\).

Step 6: Evaluate the Domain

Organize each domain: For all expressions, the domain excludes points where any function is undefined. Hence: - \((f+g)(x)\), \((f-g)(x)\), and \((fg)(x)\) exclude \(-2\).- \((f/g)(x)\) excludes \(-2, 1, -2\).

Key Concepts

Domain Identification

When working with functions, it's important to determine the domain. The domain of a function includes all the possible values of \( x \) for which the function is defined. In other words, these are the values that you can safely substitute into the function without making it undefined.

For function \( f(x) = \frac{1}{x+2} \), the denominator \( x+2 \) cannot be zero, because division by zero is undefined. Therefore, the domain of \( f(x) \) includes all real numbers except \( x = -2 \).

The function \( g(x) = x^2 + x - 2 \) is a quadratic function, which is defined for all real numbers. However, when combining \( f(x) \) and \( g(x) \) through operations such as addition, subtraction, multiplication, or division, the resultant domain needs to consider the restrictions of each operation. When multiplying or dividing, additional restrictions might emerge if the result simplifies further. Always pay attention to these nuances.

Sum and Difference of Functions

When you add or subtract two functions, you're essentially creating a new function. The sum \( (f+g)(x) = f(x) + g(x) \) and the difference \( (f-g)(x) = f(x) - g(x) \) are defined at every point within the intersection of the domains of \( f(x) \) and \( g(x) \).

For our specific functions, let's add and subtract them:

• Sum: \( (f+g)(x) = \frac{1}{x+2} + (x^2 + x - 2) \).
• Difference: \( (f-g)(x) = \frac{1}{x+2} - (x^2 + x - 2) \).

The domain for both operations excludes \( x = -2 \) because, for \( f(x) = \frac{1}{x+2} \), it implies division by zero. So, the domain is all real numbers except \( x = -2 \), making sure each function component has a valid outcome.

Product and Quotient of Functions

The product and quotient of functions involve multiplying and dividing functions, which can introduce additional domain restrictions due to the nature of multiplication and division.

For the product, you multiply the expressions for \( f(x) \) and \( g(x) \):

• Product: \( (fg)(x) = \frac{1}{x+2} \times (x^2 + x - 2) = \frac{x^2 + x - 2}{x+2} \).

The domain here is the same as that of \( f(x) \), excluding \( x = -2 \). As the operation involves division by \( x+2 \), the term in the denominator cannot equal zero.

The quotient operation involves dividing one function by another:

• Quotient: \( (f/g)(x) = \frac{\frac{1}{x+2}}{x^2 + x - 2} = \frac{1}{(x+2)(x^2 + x - 2)} \).

For the quotient \( (f/g)(x) \), the domain is further restricted by ensuring the denominator is not zero. This means not only can \( x = -2 \) make the function undefined, but also the roots of \( x^2 + x - 2 \), calculated as solving for zero in the quadratic equation, need to be considered. Upon solving, you find additional restrictions such as \( x = 1 \). Thus, the domain is all real numbers except \( x = -2 \) and \( x = 1 \), where the function does not exist.