Answer
The values of
$f+g,f-g,fg$ $\text{ and }\frac{f}{g}$
are $f+g=3x+2\ ,f-g=x+4,fg=\ 2{{x}^{2}}+x-3\ ,\frac{f}{g}=\frac{2x+3}{x-1}$
and domain of the functions $f+g,f-g,fg$ is:
$\left( -\infty ,\infty \right)$
and that of $\frac{f}{g}$ is
$\left( -\infty ,1 \right)\cup \left( 1,\infty \right)$.
Work Step by Step
Calculate the value of $f+g$ as shown below to get:
$\begin{align}
& f+g=f\left( x \right)+g\left( x \right) \\
& =\ 2x+3+x-1 \\
& =3x+2
\end{align}$
Calculate the value of $f-g$ as shown below to get:
$\begin{align}
& f-g=f\left( x \right)-g\left( x \right) \\
& =\left( 2x+3 \right)-\left( x-1 \right) \\
& =2x+3-x+1 \\
& =x+4
\end{align}$
Calculate the value of $fg$ as shown below to get:
$\begin{align}
& fg=f\left( x \right)g\left( x \right) \\
& =\left( 2x+3 \right)\left( x-1 \right) \\
& =2x\left( x-1 \right)+3\left( x-1 \right) \\
& =2{{x}^{2}}-2x+3x-3
\end{align}$
$=2{{x}^{2}}+x-3$
Calculate the value of $\frac{f}{g}$ as shown below to get:
$\begin{align}
& \frac{f}{g}=\frac{f\left( x \right)}{g\left( x \right)} \\
& =\frac{2x+3}{x-1}
\end{align}$
If the function $\frac{f}{g}$ is divided by zero, it will be undefined. So, put the denominator $x-1$ equal to zero.
Therefore,
$\begin{align}
& x-1=0 \\
& x=1
\end{align}$
Now, the domain of the function $\frac{f}{g}$ is all real numbers except 1.
Therefore, the domain of $\frac{f}{g}$ is $\left( -\infty ,1 \right)\cup \left( 1,\infty \right)$
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, $\left( -\infty ,\infty \right)$
