- The difference is \((f - g)(x) = 2x^2 - 5x + 2\).
- The sum is \((f + g)(x) = -2x^2 + 9x - 4\).
- The quotient is \(\left(\frac{g}{f}\right)(x) = 3 - x\) with domain \(\{x \in \mathbb{R} \mid x
eq \frac{1}{2}\}\).
- The product is \((f \cdot g)(x) = -4x^3 + 16x^2 - 13x + 3\).
Final answer: \(\boxed{(f-g)(x)=2x^2-5x+2, (f+g)(x)=-2x^2+9x-4, (\frac{g}{f})(x)=3-x \text{ (for } x
eq \frac{1}{2}\text{)}, (f \cdot g)(x)=-4x^3+16x^2-13x+3}\)
Explanation
- Problem Analysis
We are given two functions, \(f(x) = 2x - 1\) and \(g(x) = -2x^2 + 7x - 3\). Our objective is to perform standard operations on these functions, namely finding the sum, difference, product, and quotient, and to determine the domain of the resulting rational function. These operations are essential in algebra for understanding how functions interact and combine.
- Calculating Sum and Difference of Functions
To find \((f - g)(x)\) and \((f + g)(x)\), we perform the corresponding arithmetic operations on the expressions for \(f(x)\) and \(g(x)\).
For the difference \((f - g)(x)\):
\[(f - g)(x) = f(x) - g(x) = (2x - 1) - (-2x^2 + 7x - 3)\]
Distributing the negative sign gives:
\[2x - 1 + 2x^2 - 7x + 3 = 2x^2 - 5x + 2\]
For the sum \((f + g)(x)\):
\[(f + g)(x) = f(x) + g(x) = (2x - 1) + (-2x^2 + 7x - 3)\]
Combining like terms results in:
\[-2x^2 + 9x - 4\]
- Calculating Quotient and Domain
To find the quotient \(\left(\frac{g}{f}\right)(x)\), we divide \(g(x)\) by \(f(x)\):
\[\left(\frac{g}{f}\right)(x) = \frac{-2x^2 + 7x - 3}{2x - 1}\]
To simplify, we factor the numerator \(g(x) = -2x^2 + 7x - 3\). Setting \(g(x) = 0\), we find roots using the quadratic formula:
\[x = \frac{-7 \pm \sqrt{7^2 - 4(-2)(-3)}}{2(-2)} = \frac{-7 \pm \sqrt{49 - 24}}{-4} = \frac{-7 \pm \sqrt{25}}{-4} = \frac{-7 \pm 5}{-4}\]
The roots are \(x_1 = 0.5 = \frac{1}{2}\) and \(x_2 = 3\). Thus, \(g(x) = -2(x - 0.5)(x - 3) = -(2x - 1)(x - 3)\).
Substituting this back into the quotient:
\[\left(\frac{g}{f}\right)(x) = \frac{-(2x - 1)(x - 3)}{2x - 1} = -(x - 3) = 3 - x \text{ for } x
eq \frac{1}{2}\]
The domain of \(\left(\frac{g}{f}\right)(x)\) is the set of all real numbers \(x\) such that the denominator is not zero. Setting \(2x - 1
eq 0\) gives \(x
eq \frac{1}{2}\). Thus, the domain is \(\{x \in \mathbb{R} \mid x
eq \frac{1}{2}\}\).
- Calculating Product of Functions
To find the product \((f \cdot g)(x)\), we multiply the polynomials:
\[(f \cdot g)(x) = (2x - 1)(-2x^2 + 7x - 3)\]
Expanding this expression:
\[= 2x(-2x^2 + 7x - 3) - 1(-2x^2 + 7x - 3)\]
\[= -4x^3 + 14x^2 - 6x + 2x^2 - 7x + 3\]
Combining like terms yields:
\[= -4x^3 + 16x^2 - 13x + 3\]
Examples
In economics, revenue and cost functions are often combined to calculate profit. If you have a revenue function \(R(x)\) and a cost function \(C(x)\), the profit function is defined as \(P(x) = (R - C)(x)\). Similarly, understanding how to divide functions is helpful when determining average costs, where the average cost is \(\bar{C}(x) = \frac{C(x)}{x}\). These algebraic operations allow you to derive new models from existing ones to analyze various scenarios in finance and business.