Answer
The value of $f+g=4x-2,f-g=2x-6,fg=\ 3{{x}^{2}}+2x-8\ $ and $\frac{f}{g}=\frac{3x-4}{x+2}$. The domain of the functions $f+g,f-g$ , and $fg$ is $\left( -\infty ,\infty \right)$ and that of $\frac{f}{g}$ is $\left( -\infty ,-2 \right)\cup \left( -2,\infty \right)$.
Work Step by Step
Calculate the value of $f+g$ as shown below to get:
$\begin{align}
& f+g=f\left( x \right)+g\left( x \right) \\
& =\ 3x-4+x+2 \\
& =4x-2
\end{align}$
Calculate the value of $f-g$ as shown below to get:
$\begin{align}
& f-g=f\left( x \right)-g\left( x \right) \\
& =\ \left( 3x-4 \right)-\left( x+2 \right) \\
& =3x-4-x-2 \\
& =2x-6
\end{align}$
Calculate the value of $fg$ as shown below to get:
$\begin{align}
& fg=f\left( x \right)g\left( x \right) \\
& =\ \left( 3x-4 \right)\left( x+2 \right) \\
& =3x\left( x+2 \right)-4\left( x+2 \right) \\
& =3{{x}^{2}}+6x-4x-8
\end{align}$
$=3{{x}^{2}}+2x-8$
Calculate the value of $\frac{f}{g}$ as shown below to get:
$\begin{align}
& \frac{f}{g}=\frac{f\left( x \right)}{g\left( x \right)} \\
& =\frac{3x-4}{x+2}
\end{align}$
If the function $\frac{f}{g}$ is divided by zero it will be undefined. So, put the denominator $x+2$ equal to zero.
$\begin{align}
& x+2=0 \\
& x\ =-2
\end{align}$
Now, we see that the domain of function $\frac{f}{g}$ is all real numbers except -2.
Therefore, the domain of $\frac{f}{g}$ is $\left( -\infty ,-2 \right)\cup \left( -2,\infty \right)$.
The functions $f+g,f-g,fg$ do not involve any division and do not contain even roots, so the domain of the functions $f+g,f-g,fg$ is the set of real numbers -- that is, $\left( -\infty ,\infty \right)$.