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