Answer
The values of the functions are as follows:
$f+g=\sqrt{x}+x-4,f-g=\sqrt{x}-x+4,fg=\sqrt{x}\left( x-4 \right)$ and $\frac{f}{g}=\frac{\sqrt{x}}{x-4}$.
The domain of $f+g,f-g$ , and $fg$ is $\left[ 0,\infty \right)$ and that of $\frac{f}{g}$ is $\left[ 0,4 \right)\cup \left( 4,\infty \right)$.
Work Step by Step
Calculate the value of $f+g$ as shown below to get:
$\begin{align}
& f+g=f\left( x \right)+g\left( x \right) \\
& =\sqrt{x}+x-4\
\end{align}$
Calculate the value of $f-g$ as shown below to get:
$\begin{align}
& f-g=f\left( x \right)-g\left( x \right) \\
& =\sqrt{x}-x+4
\end{align}$
Calculate the value of $fg$ as shown below to get:
$\begin{align}
& fg=f\left( x \right)g\left( x \right) \\
& =\sqrt{x}\left( x-4 \right)\
\end{align}$
Calculate the value of $\frac{f}{g}$ as shown below to get:
$\begin{align}
& \frac{f}{g}=\frac{f\left( x \right)}{g\left( x \right)} \\
& =\frac{\sqrt{x}}{x-4}
\end{align}$
In the function $f\left( x \right)=\sqrt{x}$ only positive numbers have square roots that are real number, so the expression under the square root sign, $x$, must be positive. Therefore, $\begin{align}
& \sqrt{x}\ge 0 \\
& \ \ x\ge 0
\end{align}$
If the function $\frac{f}{g}$ is divided by zero, it will be undefined. So, put the denominator $x-4$ equal to zero.
$\begin{align}
& x-4=0 \\
& x=4
\end{align}$
This implies that $x=0$ and $x=4$.
Now, the domain of the function $\frac{f}{g}$ is all the real numbers except 0 and 4.
Therefore, the domain of $\frac{f}{g}$ is $\left[ 0,4 \right)\cup \left( 4,\infty \right)$.
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, $\left[ 0,\infty \right)$.