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Magazine Chapter 5: Problem 25
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) \((f / g)(x)\) $$ f(x)=\frac{x^{2}-3 x+2}{x+1}, \quad g(x)=\frac{x^{2}-1}{x-2} $$
Short Answer
Expert verified
(f+g)(x): Domain is all x except -1, 2. (f-g)(x): Domain is all x except -1, 2. (f路g)(x): Domain is all x except -1, 2. (f/g)(x): Domain is all x except -1, 1, 2.
Step by step solution
01
Identify the expressions
First, understand what each given expression represents. The expressions are: 1. \((f+g)(x)\), which represents the sum of the two functions.2. \((f-g)(x)\), which represents the difference of the two functions.3. \((f\cdot g)(x)\), which represents the product of the two functions.4. \((f/g)(x)\), which represents the division of the two functions.
02
Add the functions f(x) and g(x)
The sum \((f+g)(x)\) is given by:\[(f+g)(x) = f(x) + g(x) = \frac{x^{2}-3x+2}{x+1} + \frac{x^{2}-1}{x-2}\]To add these rational functions, find a common denominator \((x+1)(x-2)\) and combine the numerators:\[(f+g)(x) = \frac{(x^{2}-3x+2)(x-2) + (x^{2}-1)(x+1)}{(x+1)(x-2)}\]
03
Subtract the functions f(x) and g(x)
The difference \((f-g)(x)\) is given by:\[(f-g)(x) = f(x) - g(x) = \frac{x^{2}-3x+2}{x+1} - \frac{x^{2}-1}{x-2}\]Find a common denominator \((x+1)(x-2)\) and subtract the numerators:\[(f-g)(x) = \frac{(x^{2}-3x+2)(x-2) - (x^{2}-1)(x+1)}{(x+1)(x-2)}\]
04
Multiply the functions f(x) and g(x)
The product \((f\cdot g)(x)\) is given by:\[(f \cdot g)(x) = f(x) \cdot g(x) = \frac{x^{2}-3x+2}{x+1} \cdot \frac{x^{2}-1}{x-2}\]Multiply the numerators together and the denominators together:\[(f \cdot g)(x) = \frac{(x^{2}-3x+2)(x^{2}-1)}{(x+1)(x-2)}\]
05
Divide the functions f(x) by g(x)
The division \((f/g)(x)\) is given by:\[(f/g)(x) = \frac{f(x)}{g(x)} = \frac{\frac{x^{2}-3x+2}{x+1}}{\frac{x^{2}-1}{x-2}}\]This is simplified to:\[(f/g)(x) = \frac{x^{2}-3x+2}{x^{2}-1} \cdot \frac{x-2}{x+1}\]Resulting in:\[(f/g)(x) = \frac{(x^{2}-3x+2)(x-2)}{(x^{2}-1)(x+1)}\]
06
Finding the domain for (f+g)(x) and (f-g)(x)
The domain is where both functions are defined without making any denominators zero:- For \((f+g)(x)\) and \((f-g)(x)\): Denominators are \((x+1)(x-2)\), so the domain is all real numbers except where \(x+1 = 0\) and \(x-2 = 0\), i.e., \(x eq -1\) and \(x eq 2\).
07
Finding the domain for (f路g)(x)
The product \((f \cdot g)(x)\) requires both denominators to not be zero: - The domain is all real numbers except where \(x+1 = 0\) and \(x-2 = 0\), i.e., \(x eq -1\) and \(x eq 2\).
08
Finding the domain for (f/g)(x)
The division \((f/g)(x)\) involves additional exclusions:- Besides \(x eq -1\) and \(x eq 2\), \(x^2 - 1 eq 0\) must be considered, which implies \(x eq \pm 1\).- The domain is all real numbers except \(x = -1, 1, 2\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Rational Expressions
Rational expressions are essentially fractions where both the numerator and the denominator are polynomials. They are a critical part of algebraic functions due to their unique characteristics and behavior.
- Just like fractions, rational expressions can be added, subtracted, multiplied, and divided, but additional care must be taken due to the presence of variables.
- Operations involving these expressions require a common denominator similar to how fractions are managed. This often involves finding the least common multiple (LCM) of the polynomial denominators before performing any arithmetic operation on them.
- One of the main considerations when working with rational expressions is ensuring that the denominator never equals zero, as division by zero is undefined.
Function Domain
The domain of a function is the set of all possible input values (usually represented as \( x \)) that will produce a valid output from a given function. In rational expressions, domains are affected primarily by their denominators.
- To find the domain of a rational function, consider the values of \( x \) that would make any denominator zero, as these would cause the function to be undefined.
- For functions such as \( (f+g)(x) \), \( (f-g)(x) \), and \( (f \cdot g)(x) \), the domain excludes any \( x \) that makes the polynomial of either denominator zero. Thus, in this context, exclude \( x = -1 \) and \( x = 2 \).
- In division, such as \( (f/g)(x) \), you must also consider zeros in the denominator of \( g(x) \) as divided terms introduce further restrictions. This means excluding \( x = 1 \) along with other exclusions.
Arithmetic Operations on Functions
Arithmetic operations on functions include addition, subtraction, multiplication, and division, and they allow the creation of new functions from existing ones.
- Addition \((f+g)(x)\) and subtraction \((f-g)(x)\): These operations involve combining or separating the outputs of two functions. To properly execute these, a common denominator is required for rational expressions, where numerators are also combined or subtracted accordingly.
- Multiplication \((f \cdot g)(x)\): When multiplying, each numerator is multiplied together and each denominator is multiplied together. Ensure each polynomial expression in the denominator is not zero.
- Division \((f/g)(x)\): This is managed by multiplying with the reciprocal of the dividing function \( g(x) \). The rule for rational functions dictates that the recursive denominator's zeros further constrict the domain to a more complex scenario.
