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Magazine Chapter 2: Problem 12
Find \(f+g, f-g, f g,\) and \(f / g\) and their domains. $$f(x)=x^{2}+2 x, \quad g(x)=3 x^{2}-1$$
Short Answer
Expert verified
For each operation with domains:
1. \(f+g = 4x^2 + 2x - 1\), Domain: \( \mathbb{R} \).
2. \(f-g = -2x^2 + 2x + 1\), Domain: \( \mathbb{R} \).
3. \(fg = 3x^4 + 6x^3 - x^2 - 2x\), Domain: \( \mathbb{R} \).
4. \(\frac{f}{g}\), Domain: \( \mathbb{R} \setminus \{ \pm \frac{1}{\sqrt{3}} \} \).
Step by step solution
01
Find \( f+g \)
To find \( f+g \), we need to add the functions \( f(x) \) and \( g(x) \):\[ f(x) + g(x) = (x^2 + 2x) + (3x^2 - 1) \]Combine like terms:\[ (x^2 + 3x^2) + 2x - 1 = 4x^2 + 2x - 1 \]So, \( f+g = 4x^2 + 2x - 1 \).
02
Find Domain of \( f+g \)
Since both \( f(x) = x^2 + 2x \) and \( g(x) = 3x^2 - 1 \) are polynomials, they are defined for all real numbers. Thus, the domain of \( f+g \) is all real numbers, \( \mathbb{R} \).
03
Find \( f-g \)
To find \( f-g \), we subtract \( g(x) \) from \( f(x) \):\[ f(x) - g(x) = (x^2 + 2x) - (3x^2 - 1) \]Simplify by distributing and combining like terms:\[ x^2 + 2x - 3x^2 + 1 = -2x^2 + 2x + 1 \]So, \( f-g = -2x^2 + 2x + 1 \).
04
Find Domain of \( f-g \)
Like \( f+g \), the expression \( f-g \) is also a polynomial and therefore defined for all real numbers. The domain is \( \mathbb{R} \).
05
Find \( fg \)
To find \( fg \), multiply the functions \( f(x) \) and \( g(x) \):\[ fg(x) = (x^2 + 2x)(3x^2 - 1) \]Use the distributive property:\[ x^2(3x^2 - 1) + 2x(3x^2 - 1) = 3x^4 - x^2 + 6x^3 - 2x \]Combine the terms to get:\[ fg = 3x^4 + 6x^3 - x^2 - 2x \].
06
Find Domain of \( fg \)
Since \( fg \) is a polynomial, it is defined for all real numbers, so the domain of \( fg \) is \( \mathbb{R} \).
07
Find \( \frac{f}{g} \)
To find \( \frac{f}{g} \), divide \( f(x) \) by \( g(x) \):\[ \frac{f(x)}{g(x)} = \frac{x^2 + 2x}{3x^2 - 1} \]This quotient is defined wherever the denominator is not zero.
08
Find Domain of \( \frac{f}{g} \)
The function \( \frac{f}{g} \) is undefined where \( g(x) = 3x^2 - 1 = 0 \).Solving for \( x \):\[ 3x^2 = 1 \]\[ x^2 = \frac{1}{3} \]\[ x = \pm \frac{1}{\sqrt{3}} \]Therefore, the domain of \( \frac{f}{g} \) is all real numbers except \( x = \pm \frac{1}{\sqrt{3}} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Function Operations
When dealing with functions like polynomials, you can perform several operations, such as addition, subtraction, multiplication, and division. Each operation combines or alters the functions' algebraic expressions via their formulas. Let's explore these operations with an example using.\[f(x) = x^2 + 2x\]and \[g(x) = 3x^2 - 1.\]
- **Addition (\( f+g \)):** Add the corresponding terms of both functions. Here, \((x^2 + 2x) + (3x^2 - 1)\) results in \(4x^2 + 2x - 1\).
- **Subtraction (\( f-g \)):** Subtract the terms of the second function from the first. So, \((x^2 + 2x) - (3x^2 - 1)\) simplifies to \(-2x^2 + 2x + 1\).
- **Multiplication (\( fg \)):** Multiply each term from one function by each term of the other. This can be expressed as \((x^2 + 2x)\cdot(3x^2 - 1)\), which simplifies to \(3x^4 + 6x^3 - x^2 - 2x\).
- **Division (\( \frac{f}{g} \)):** Divide the terms of \( f(x) \) by those of \( g(x) \). It results in a fraction \( \frac{x^2 + 2x}{3x^2 - 1} \), where careful attention to the denominator is necessary.
Domains of Functions
Understanding the domain of a function entails determining all the input values (\(x\) values) for which the function is defined. For polynomial functions, like those in our example, the domain is typically all real numbers because their expressions are smooth and continuous without breaks or undefined values.
- **Sum and Difference:** Since \(f(x)\) and \(g(x)\) are both polynomials, \(f+g\) and \(f-g\) are defined for all real values of \(x\).
- **Product:** Similarly, the product \(fg\) remains a polynomial and hence has the same domain: all real numbers.
- **Quotient:** For the quotient \(\frac{f}{g}\), caution is required because \(g(x)\) should not equal zero, as division by zero creates undefined outcomes. Accordingly, solving \(3x^2 - 1 = 0\) helps exclude values \(x = \pm \frac{1}{\sqrt{3}}\) from the domain.
Rational Functions
Rational functions are quotients where both the numerator and denominator are polynomials. They can seem a bit tricky due to their potential for undefined values. In general, the rational function \(\frac{f}{g} = \frac{x^2 + 2x}{3x^2 - 1}\) demands scrutiny of the denominator. To avoid division by zero, identify points where \(g(x) = 0\).
- **Finding zero points:** Solve \(3x^2 - 1 = 0\) to determine \(x = \pm \frac{1}{\sqrt{3}}\). These values are excluded from the domain of the rational function.
- **Graph behavior:** As \(x\) approaches these excluded points, the function's graph exhibits vertical asymptotes, representing sharp, undefined spikes. Studying these helps predict function behavior and limits.
