College Algebra, Chapter 3, 3.6, Section 3.6, Problem 6

Evaluate $f + g$, $f - g$, $fg$ and $\displaystyle \frac{f}{g}$ of the function $f(x) = x^2 + 2x$ and $g(x) = 3x^2 - 1$ and find their domain

For $f+g$,

$\begin{equation} \begin{aligned} f+g &= f(x) + g(x)\\ \\ f+g &= x^2 + 2x + 3x^2 - 1 && \text{Substitute } f(x) = x^2 + 2x \text{ and } g(x) = 3x^2 - 1\\ \\ f+g &= 4x^2 + 2x - 1 \end{aligned} \end{equation}$

The domain of $f(x) + g(x)$ is $(-\infty,\infty)$

For $f-g$

$\begin{equation} \begin{aligned} f-g &= f(x) - g(x) \\ \\ f-g &= x^2 + 2x - \left( 3x^2 - 1 \right) && \text{Apply Distributive rule}\\ \\ f-g &= x^2 + 2x - 3x^2 + 1 && \text{Simplify}\\ \\ f-g &= -2x^2 + 2x + 1 \end{aligned} \end{equation}$

The domain of $f(x) - g(x)$ is $(-\infty, \infty)$

For $fg$

$\begin{equation} \begin{aligned} fg &= f(x) \cdot g(x) \\ \\ fg &= \left( x^2 + 2x \right) \left( 3x^2 + 1 \right) && \text{Apply Distributive property}\\ \\ fg &= 3x^4 + x^2 + 6x^3 + 2x \text{ or } fg = 3x^4 + 6x^3 + x^2 + 2x \end{aligned} \end{equation}$

The domain of $f(x) \cdot g(x)$ is $(-\infty, \infty)$

For $\displaystyle \frac{f}{g}$

$\begin{equation} \begin{aligned} \frac{f}{g} &= \frac{f(x)}{g(x)}\\ \\ \frac{f}{g} &= \frac{x^2 + 2x}{3x^2 - 1} \end{aligned} \end{equation}$

The domain of $\displaystyle \frac{f(x)}{g(x)} \text{ is } \left( -\infty, -\frac{1}{\sqrt{3}} \right) \bigcup \left( -\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right)\bigcup \left( \frac{1}{\sqrt{3}}, \infty \right)$