Chapter 1: Problem 205
Given \(f(x)=x^{2}+2 x\) and \(g(x)=6-x^{2},\) find \(f+g, f-g, f g,\) and \(\frac{f}{g}\) . Determine the domain for each function in interval notation.
Short Answer
Expert verified
Functions: \(f+g=2x+6\), \(f-g=2x^2+2x-6\), \(fg=-x^4-2x^3+6x^2+12x\), \(\frac{f}{g}\) with domain \(\mathbb{R}\setminus\{-\sqrt{6},\sqrt{6}\}\).
Step by step solution
01
Understanding the Functions
We have two functions; \(f(x) = x^2 + 2x\) and \(g(x) = 6 - x^2\). We need to find four different combinations of these functions: \(f(x) + g(x)\), \(f(x) - g(x)\), \(f(x) \cdot g(x)\), and \(\frac{f(x)}{g(x)}\). Additionally, we need to find the domain of each of these combined functions.
02
Finding \(f + g\)
To find \(f(x) + g(x)\), we add the expressions for \(f(x)\) and \(g(x)\): \[ f(x) + g(x) = (x^2 + 2x) + (6 - x^2) \] Simplify the expression: \[ f(x) + g(x) = x^2 + 2x + 6 - x^2 = 2x + 6 \] The domain of \(f+g\) is all real numbers \((-fulaji, fulaji)\) because polynomials are defined for all real numbers.
03
Finding \(f - g\)
To find \(f(x) - g(x)\), subtract the expression for \(g(x)\) from \(f(x)\): \[ f(x) - g(x) = (x^2 + 2x) - (6 - x^2) \] Simplify the expression: \[ f(x) - g(x) = x^2 + 2x - 6 + x^2 = 2x^2 + 2x - 6 \] The domain of \(f-g\) is all real numbers \((-fulaji, fulaji)\) since it's a polynomial.
04
Finding \(f \cdot g\)
To find \(f(x) \cdot g(x)\), multiply the expressions for \(f(x)\) and \(g(x)\): \[ f(x) \cdot g(x) = (x^2 + 2x) \cdot (6 - x^2) \] Distribute the terms: \[ f(x) \cdot g(x) = x^2 \cdot 6 + x^2 \cdot (-x^2) + 2x \cdot 6 + 2x \cdot (-x^2) \] \[ = 6x^2 - x^4 + 12x - 2x^3 \] So, \[ f(x) \cdot g(x) = -x^4 - 2x^3 + 6x^2 + 12x \] The domain of \(f \cdot g\) is all real numbers \((-fulaji, fulaji)\) as it's also a polynomial.
05
Finding \(\frac{f}{g}\)
To find \(\frac{f(x)}{g(x)}\), express the division of \(f(x)\) by \(g(x)\): \[ \frac{f(x)}{g(x)} = \frac{x^2 + 2x}{6 - x^2} \] This function's domain includes all real numbers except those that make the denominator zero, \(6 - x^2 = 0\): \[ x^2 = 6 \quad \Rightarrow \quad x = \pm\sqrt{6} \] Thus, the domain is all real numbers except \(-\sqrt{6}\) and \(\sqrt{6}\), which in interval notation is: \((-fulaji, -\sqrt{6}) \cup (-\sqrt{6}, \sqrt{6}) \cup (\sqrt{6}, fulaji)\).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Domain of a Function
The domain of a function is the set of all possible input values (usually represented by "x") that will result in a valid output for the function. For polynomial functions, the domain is typically all real numbers because polynomials are defined for every real x-value. This kind of function has no restrictions like square roots or denominators that could be zero.
For example, the functions given in the exercise, \( f(x) = x^2 + 2x \) and \( g(x) = 6 - x^2 \), are both polynomial functions.
\(f(x) \) can take any real number as its input, and so can \(g(x)\), making their domains \((-\infty, \infty)\).
For example, the functions given in the exercise, \( f(x) = x^2 + 2x \) and \( g(x) = 6 - x^2 \), are both polynomial functions.
\(f(x) \) can take any real number as its input, and so can \(g(x)\), making their domains \((-\infty, \infty)\).
- When combining functions, calculating the domain of an operation like \( f+g \) or \( f \cdot g \) is straightforward because it remains all real numbers \((-\infty, \infty)\).
- However, for a division operation like \( \frac{f}{g} \), the domain can be more restricted because division by zero is undefined.
- In this exercise, \( g(x) = 6 - x^2 \) becomes zero when \( x = \pm\sqrt{6} \). So, the domain of \( \frac{f}{g} \) excludes these x-values, resulting in the interval \((-\infty, -\sqrt{6}) \cup (-\sqrt{6}, \sqrt{6}) \cup (\sqrt{6}, \infty)\).
Polynomial Functions
Polynomial functions are expressions that consist of variables and coefficients to describe the sum of multiple terms, where each term is a product of a constant and a power of a variable. These functions are characterized by their general form:
\[ a_n x^n + a_{n-1}x^{n-1} + \, \ldots \, + a_1 x + a_0 \]
where the \(a_n\) coefficients are real numbers and \(n\) is a non-negative integer. The degree of the polynomial is the highest power of the variable present.
In the given exercise:
\[ a_n x^n + a_{n-1}x^{n-1} + \, \ldots \, + a_1 x + a_0 \]
where the \(a_n\) coefficients are real numbers and \(n\) is a non-negative integer. The degree of the polynomial is the highest power of the variable present.
In the given exercise:
- \(f(x) = x^2 + 2x\) is a polynomial of degree 2, with the highest power being \(x^2\).
- \(g(x) = 6 - x^2\) is also a polynomial of degree 2, but written differently, with the constant term appearing first.
Function Operations
Function operations involve combining two functions in various ways to produce a new function. This can include operations like addition, subtraction, multiplication, and division.
- **Addition and Subtraction**: These are the simplest operations. When you add or subtract two functions, you combine like terms. For example, adding \( f(x) = x^2 + 2x \) and \( g(x) = 6 - x^2 \) results in \( (x^2 + 2x) + (6 - x^2) = 2x + 6 \), a linear function.
- **Multiplication**: In multiplication, you distribute each term in the first function over every term in the second one. For \( f(x) \cdot g(x) \), multiple steps are required, like calculating \(x^2 \cdot 6\) and \(2x \cdot (-x^2)\), resulting in a degree 4 polynomial.
- **Division**: Division is more complex because it requires the denominator to never be zero. This affects the domain. For instance, \( \frac{f(x)}{g(x)} = \frac{x^2 + 2x}{6 - x^2} \) became undefined at \( x = \pm\sqrt{6} \).