Answer
The values of the functions are as follows:
$f+g=6,f-g=6-\frac{2}{x},f\cdot g=\frac{6x-1}{{{x}^{2}}}$ and $\frac{f}{g}=6x-1$.
The domain of the functions $\frac{f}{g},f-g,fg$ , and $f+g$ 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,
$\begin{align}
& f+g=f\left( x \right)+g\left( x \right) \\
& =6-\frac{1}{x}+\frac{1}{x} \\
& =6\ \ \ \
\end{align}$
Calculate the value of $f-g$ as shown below to get,
$\begin{align}
& f-g=f\left( x \right)-g\left( x \right) \\
& =6-\frac{1}{x}-\frac{1}{x} \\
& =6-\frac{2}{x}
\end{align}$
Calculate the value of $fg$ as shown below to get,
$\begin{align}
& fg=f\left( x \right)g\left( x \right) \\
& =\left( 6-\frac{1}{x} \right)\left( \frac{1}{x} \right) \\
& =\frac{6}{x}-\frac{1}{{{x}^{2}}} \\
& =\frac{6x-1}{{{x}^{2}}}
\end{align}$
Calculate the value of $f+g$ as shown below to get,
$\begin{align}
& \frac{f}{g}=\frac{f\left( x \right)}{g\left( x \right)} \\
& =\frac{6-\frac{1}{x}}{\frac{1}{x}} \\
& =\frac{\left( \frac{6x-1}{x} \right)}{\left( \frac{1}{x} \right)} \\
& =6x-1
\end{align}$
If the functions $f-g$ and $fg$ contain division by zero, they will be undefined. So, put the denominator $x$ equal to zero.
$\begin{align}
& {{x}^{2}}=0 \\
& x=0
\end{align}$
Now, the domain of functions $f-g$ and $fg$ is all the real numbers except 0.
Therefore, the domain of functions $f-g$ and $fg$ is $\left( -\infty ,0 \right)\cup \left( 0,\infty \right)$.
The functions $f-g$ and $\frac{f}{g}$ do not involve any division or do not contain even roots, so the domain of the functions $f+g$ and $\frac{f}{g}$ is the set of real numbers that is $\left( -\infty ,\infty \right)$.
