Answer
The values of the functions are as follows:
$f+g=\sqrt{x}+x-5,f-g=\sqrt{x}-x+5,fg=\sqrt{x}\left( x-5 \right)\ $ and $\frac{f}{g}=\frac{\sqrt{x}}{x-5}$.
The domain of the functions $f+g,f-g$ , and $fg$ is $\left[ 0,\infty \right)$ and that of $\frac{f}{g}$ is $\left[ 0,5 \right)\cup \left( 5,\infty \right)$.
Work Step by Step
Calculate the value of $f+g$ as shown below to get:
$\begin{align}
& f+g=f\left( x \right)+g\left( x \right) \\
& =\sqrt{x}+x-5\ \
\end{align}$
Calculate the value of $f-g$ as shown below to get:
$\begin{align}
& f-g=f\left( x \right)-g\left( x \right) \\
& =\sqrt{x}-x+5
\end{align}$
Calculate the value of $fg$ as shown below to get:
$\begin{align}
& fg=f\left( x \right)g\left( x \right) \\
& =\sqrt{x}\left( x-5 \right)
\end{align}$
Calculate the value of $\frac{f}{g}$ as shown below to get:
$\begin{align}
& \frac{f}{g}=\frac{f\left( x \right)}{g\left( x \right)} \\
& =\frac{\sqrt{x}}{x-5}
\end{align}$
In the function $f\left( x \right)=\sqrt{x}$, only positive numbers have square roots that are real numbers, 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 $\frac{f}{g}$ has division by zero, then it will be undefined. So, put the denominator $x-5$ equal to zero.
$\begin{align}
& x-5=0 \\
& x=5
\end{align}$
This implies that $x=0$ and $x=5$.
Now, the domain of the function $\frac{f}{g}$ is all the real numbers except 0 and 5.
Therefore, the domain of $\frac{f}{g}$ is $\left[ 0,5 \right)\cup \left( 5,\infty \right)$.
The functions $f+g,f-g,fg$ do not involve any division and do not contain even roots, so the domain of the functions $f+g,f-g,fg$ is the set of real numbers -- that is, $\left[ 0,\infty \right)$.
