Answer
The value of $\left( f+g \right)=4x-6$, $\left( f-g \right)=2\left( x+2 \right)$, $\left( fg \right)=3{{x}^{2}}-16x+5$, and $\left( \frac{f}{g} \right)=\frac{3x-1}{x-5}$. The domain of $f+g,f-g,fg,\text{ and }\frac{f}{g}$ are $\left( -\infty,\infty \right)$, $\left( -\infty,\infty \right)$, $\left( -\infty,\infty \right)$, and $\left( -\infty,5 \right)\cup \left( 5,\infty \right)$ respectively.
Work Step by Step
Consider the function,
$\begin{align}
& f+g=3x-1+x-5 \\
& =4x-6
\end{align}$
Since, the function contains neither division nor the square root, the domain of this function is the set of all real numbers.
So, $\text{domain of }\left( f+g \right)=\left( -\infty,\infty \right)$.
Now, consider the function,
$\begin{align}
& f-g=3x-1-x+5 \\
& =2x+4 \\
& =2\left( x+2 \right)
\end{align}$
Since, the function contains neither division nor the square root, the domain of this function is the set of all real numbers.
So, $\text{domain of }\left( f-g \right)=\left( -\infty,\infty \right)$.
Consider the function,
$\begin{align}
& fg=\left( 3x-1 \right)\left( x-5 \right) \\
& =3{{x}^{2}}-15x-x+5 \\
& =3{{x}^{2}}-16x+5
\end{align}$
Since the function contains neither division nor the square root, the domain of this function is the set of all real numbers.
So, $\text{domain of }\left( fg \right)=\left( -\infty,\infty \right)$.
Consider the function,
$\frac{f}{g}=\frac{3x-1}{x-5}$
As this function contains division and division by 0 is not defined, exclude those values of $x$ from the domain that cause the denominator to be zero.
For this, set the denominator equal to 0, that is, $x-5=0$.
This implies that $x=5$.
Therefore, the domain of the function $\frac{f}{g}$ is the set of all real numbers excluding the number 5. So, $\text{domain of }\left( \frac{f}{g} \right)=\left( -\infty,5 \right)\cup \left( 5,\infty \right)$.
Hence, the value of $\left( f+g \right)=4x-6$, $\left( f-g \right)=2\left( x+2 \right)$, $\left( fg \right)=3{{x}^{2}}-16x+5$, and $\left( \frac{f}{g} \right)=\frac{3x-1}{x-5}$. The domain of $f+g,f-g,fg,\text{ and }\frac{f}{g}$ are $\left( -\infty,\infty \right)$, $\left( -\infty,\infty \right)$, $\left( -\infty,\infty \right)$, and $\left( -\infty,5 \right)\cup \left( 5,\infty \right)$ respectively.
