/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 You are given a pair of function... [FREE SOLUTION] | 91Ó°ÊÓ

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You are given a pair of functions, \(f\) and \(g .\) In each case, find \((f+g)(x),(f-g)(x),(f \cdot g)(x),\) and \((f / g)(x)\) and the domains of each. $$ f(x)=x^{3}, g(x)=x-1 $$

Short Answer

Expert verified
(f+g)(x)=x^3+x-1; (f-g)(x)=x^3-x+1; (f·g)(x)=x^4-x^3; (f/g)(x)=x^3/(x-1); domains are all real numbers except for (f/g)(x) at x=1.

Step by step solution

01

Find \((f + g)(x)\)

To find \((f + g)(x)\), simply add the expressions for \(f(x)\) and \(g(x)\). Given:\[f(x) = x^3\] \[g(x) = x - 1\] \((f + g)(x) = f(x) + g(x) = x^3 + (x - 1) = x^3 + x - 1\). The domain of \((f + g)(x)\) is all real numbers since both \(f(x)\) and \(g(x)\) are polynomials, which are defined everywhere.
02

Find \((f - g)(x)\)

To find \((f - g)(x)\), subtract the expression for \(g(x)\) from \(f(x)\). \((f - g)(x) = f(x) - g(x) = x^3 - (x - 1) = x^3 - x + 1\). The domain of \((f - g)(x)\) is all real numbers since both functions are defined everywhere.
03

Find \((f \cdot g)(x)\)

To find \((f \cdot g)(x)\), multiply the expressions for \(f(x)\) and \(g(x)\).\((f \cdot g)(x) = f(x) \cdot g(x) = x^3 \cdot (x - 1) = x^4 - x^3\). The domain of \((f \cdot g)(x)\) is all real numbers, similar to the previous steps, as polynomial multiplication is defined everywhere.
04

Find \((f / g)(x)\) and its Domain

To find \((f / g)(x)\), divide the expression for \(f(x)\) by \(g(x)\). \((f / g)(x) = \frac{f(x)}{g(x)} = \frac{x^3}{x - 1}\). The domain of \((f / g)(x)\) is all real numbers except where \(g(x) = 0\). Since \(g(x) = x - 1\), we exclude \(x = 1\) from the domain. Thus, the domain is all real numbers except \(x = 1\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Polynomial Operations
Polynomial operations involve fundamental arithmetic actions such as addition, subtraction, and multiplication on polynomial functions. These operations are quite straightforward due to the rules of algebra, and they can be performed similarly to operations with numbers.

- **Addition**: When you add two polynomials, like \(f(x) = x^3\) and \(g(x) = x-1\), you simply combine their terms. This is done by adding the coefficients of like terms, which results in \((f + g)(x) = x^3 + x - 1\).

- **Subtraction**: Subtracting \(g(x)\) from \(f(x)\) means you subtract the coefficients of like terms. The subtraction \((f - g)(x) = x^3 - x + 1\) is straightforward as long as you carefully distribute the negative sign across all terms in \(g(x)\).

- **Multiplication**: To multiply two polynomials \(f(x)\) and \(g(x)\), you distribute each term in the first polynomial across every term in the second polynomial. For example, \((f \cdot g)(x) = x^3 \cdot (x - 1) = x^4 - x^3\). This requires careful application of the distributive property.
Function Domains
The domain of a function is all about finding the set of input values (usually referred to as \(x\)) for which the function is defined. For polynomial functions like \(f(x) = x^3\) and \(g(x) = x - 1\), the domain typically includes all real numbers because polynomials are defined for every real number.

- In arithmetic operations like addition and subtraction, the domain of the resultant function will also be all real numbers since both functions involved are polynomials. Hence, \((f + g)(x)\) and \((f - g)(x)\) are defined everywhere.

- Similar reasoning applies to multiplication. Multiplying two polynomials gives another polynomial, so the domain of \((f \cdot g)(x)\) remains all real numbers.

- **Division**: Here, things get tricky. Division of polynomials can lead to restrictions. Specifically, the divisibility function \((f / g)(x)\) is undefined wherever the denominator, \(g(x)\), is zero. For \(g(x) = x-1\), \(g(x)\) equals zero when \(x = 1\). Thus, the domain of \((f / g)(x)\) excludes \(x = 1\). This is crucial in determining where the function can or cannot be evaluated.
Division of Functions
When we talk about dividing functions, we are really talking about the division of the output values of one function by another, where possible. For our functions \(f(x) = x^3\) and \(g(x) = x - 1\), division is expressed as \((f / g)(x) = \frac{x^3}{x - 1}\). Division of functions involves understanding both algebraic expressions and constraints set by division itself.

- **Expressing the Division**: Division is simply putting one function as the numerator and the other as the denominator, given by \(\frac{f(x)}{g(x)}\).

- **Restricting the Domain**: The most important part is understanding when this expression is undefined. This occurs when the denominator equals zero. Hence, for \(g(x) = x - 1\), the function \(f / g\) doesn't exist at \(x = 1\).

- **Behavior of the Function**: It's also helpful to consider how the function behaves near undefined points. For instance, as \(x\) approaches 1 from the left and right, the value of \((f/g)(x)\) changes dramatically, which may lead to limits or discontinuities in behavior.

Understanding these nuances is crucial for comprehending not just the mechanics, but also the real-world implications of dividing functions.

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