Answer
The value of the functions is as follows:
$f+g=5{{x}^{2}}+x-6,f-g=-5{{x}^{2}}+x-6,fg=\ 5{{x}^{3}}-30{{x}^{2}},\frac{f}{g}=\frac{x-6}{5{{x}^{2}}}$. The domain of the functions $f+g,f-g$ , and $\ fg$ is $\left( -\infty ,\infty \right)$ and that of $\frac{f}{g}$ is $\left( -\infty ,0 \right)\cup \left( 0,\infty \right)$.
Work Step by Step
Calculate the value of $f+g$ as shown below to get:
$\begin{align}
& f+g=f\left( x \right)+g\left( x \right) \\
& =\ x-6+5{{x}^{2}} \\
& =5{{x}^{2}}+x-6
\end{align}$
Calculate the value of $f-g$ as shown below to get:
$\begin{align}
& f-g=f\left( x \right)-g\left( x \right) \\
& =\ x-6-5{{x}^{2}} \\
& =-5{{x}^{2}}+x-6
\end{align}$
Calculate the value of $fg$ as shown below to get:
$\begin{align}
& fg=f\left( x \right)g\left( x \right) \\
& =\ \left( x-6 \right)\left( 5{{x}^{2}} \right) \\
& =5{{x}^{3}}-30{{x}^{2}}
\end{align}$
Calculate the value of $f+g$ as shown below to get:
$\begin{align}
& \ \frac{f}{g}=\frac{f\left( x \right)}{g\left( x \right)} \\
& =\frac{x-6}{5{{x}^{2}}}
\end{align}$
If the function $\frac{f}{g}$ is divided by zero, it will be undefined. So, put the denominator $5x$ equal to zero.
$\begin{align}
& 5{{x}^{2}}=0 \\
& x=0
\end{align}$
Now, the domain of the function $\frac{f}{g}$ is all the real numbers except 0.
Therefore, the domain of $\frac{f}{g}$ is $\left( -\infty ,0 \right)\cup \left( 0,\infty \right)$.
The functions $f+g,f-g,fg$ do not involve any division or do not contain even roots, so the domain of the functions $f+g,f-g,fg$ is the set of real numbers -- that is, $\left( -\infty ,\infty \right)$.