Answer
The value of $f+g=3{{x}^{2}}+x-5,f-g=-3{{x}^{2}}+x-5,fg=\ 3{{x}^{3}}-15{{x}^{2}}$ and $\frac{f}{g}=\frac{x-5}{3{{x}^{2}}}$. The domain of the functions $f+g,f-g$ , and $fg$ is $\left( -\infty ,\infty \right)$ and $\frac{f}{g}$ 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,
$\begin{align}
& f+g=f\left( x \right)+g\left( x \right) \\
& =\ x-5+3{{x}^{2}} \\
& =3{{x}^{2}}+x-5
\end{align}$
Calculate the value of $f-g$ as shown below to get,
$\begin{align}
& f-g=f\left( x \right)-g\left( x \right) \\
& =\ x-5-3{{x}^{2}} \\
& =-3{{x}^{2}}+x-5
\end{align}$
Calculate the value of $fg$ as shown below to get,
$\begin{align}
& fg=f\left( x \right)g\left( x \right) \\
& =\ \left( x-5 \right)\left( 3{{x}^{2}} \right) \\
& =3{{x}^{3}}-15{{x}^{2}}
\end{align}$
Calculate the value of $\frac{f}{g}$ as shown below to get,
$\begin{align}
& \ \frac{f}{g}=\frac{f\left( x \right)}{g\left( x \right)} \\
& \ \ \ \ \ =\frac{x-5}{3{{x}^{2}}} \\
\end{align}$
If the function $\frac{f}{g}$ is divided by zero, it will be undefined. So, put the denominator $3{{x}^{2}}$ equal to zero.
$\begin{align}
& 3{{x}^{2}}=0 \\
& x=0
\end{align}$
Now, the domain of the function $\frac{f}{g}$ is all the real numbers except 0.
Therefore, the domain of $\frac{f}{g}$ is $\left( -\infty ,0 \right)\cup \left( 0,\infty \right)$.
In the functions $f+g,f-g,fg$, there are no divisions or even roots, so the domain of the functions $f+g,f-g,fg$ is the set of real numbers -- that is, $\left( -\infty ,\infty \right)$.
