Answer
The values of the functions are as follows:
$f+g=\frac{9x-1}{{{x}^{2}}-9}$ , $f-g=\frac{1}{x-3}$ , $fg=\frac{20{{x}^{2}}-6x-2}{\left( {{x}^{2}}-9 \right)\left( {{x}^{2}}-9 \right)}$ , and $\frac{f}{g}=\frac{5x+1}{4x-2}$.
The domain of $f-g$ , $f+g$, and $fg$ is $\left( -\infty ,-3 \right)\cup \left( -3,3 \right)\cup \left( 3,\infty \right)$ and that of $\frac{f}{g}$ is $\left( -\infty ,-3 \right)\cup \left( -3,\frac{1}{2} \right)\cup \left( \frac{1}{2},3 \right)\cup \left( 3,\infty \right)$.
Work Step by Step
Calculate the value of $f+g$ as shown below to get,
$\begin{align}
& \left( f+g \right)\left( x \right)=f\left( x \right)+g\left( x \right) \\
& =\frac{5x+1}{{{x}^{2}}-9}+\frac{4x-2}{{{x}^{2}}-9} \\
& =\frac{5x+1+4x-2}{{{x}^{2}}-9} \\
& =\frac{9x-1}{{{x}^{2}}-9}
\end{align}$
Calculate the value of $f-g$ as shown below to get,
$\begin{align}
& \left( f-g \right)\left( x \right)=f\left( x \right)-g\left( x \right) \\
& =\frac{5x+1}{{{x}^{2}}-9}-\frac{4x-2}{{{x}^{2}}-9} \\
& =\frac{5x+1-4x+2}{{{x}^{2}}-9} \\
& =\frac{x+3}{{{x}^{2}}-9} \\
& =\frac{x+3}{\left( x+3 \right)\left( x-3 \right)} \\
& =\frac{1}{x-3}
\end{align}$
Calculate the value of $fg$ as shown below to get,
$\begin{align}
& \left( fg \right)\left( x \right)=f\left( x \right)g\left( x \right) \\
& =\frac{5x+1}{{{x}^{2}}-9}\left( \frac{4x-2}{{{x}^{2}}-9} \right) \\
& =\frac{\left( 5x+1 \right)\left( 4x-2 \right)}{\left( {{x}^{2}}-9 \right)\left( {{x}^{2}}-9 \right)} \\
& =\frac{20{{x}^{2}}-10x+4x-2}{\left( {{x}^{2}}-9 \right)\left( {{x}^{2}}-9 \right)} \\
& =\frac{20{{x}^{2}}-6x-2}{\left( {{x}^{2}}-9 \right)\left( {{x}^{2}}-9 \right)}
\end{align}$
Calculate the value of $\frac{f}{g}$ as shown below to get,
$\begin{align}
& \left( \frac{f}{g} \right)\left( x \right)=\frac{f\left( x \right)}{g\left( x \right)} \\
& =\frac{\left( \frac{5x+1}{{{x}^{2}}-9} \right)}{\left( \frac{4x-2}{{{x}^{2}}-9} \right)} \\
& =\frac{5x+1}{4x-2}
\end{align}$
If the functions $f+g$ , $f-g$ and $fg$ contain division by zero, they will be undefined. So, put the denominator ${{x}^{2}}-9$ equal to zero.
$\begin{align}
& {{x}^{2}}-9=0 \\
& x=\pm \sqrt{9} \\
& x=\pm 3
\end{align}$
Now, the domain of the function $f+g$ , $f-g$ and $fg$ is all the real numbers except 3 and -3.
Therefore, the domain of functions $f-g$ and $fg$ is $\left( -\infty ,-3 \right)\cup \left( -3,3 \right)\cup \left( 3,\infty \right)$.
In the function $\frac{f}{g}$, division by zero is undefined. So, put the denominator $4x-2$ equal to zero.
$\begin{align}
& 4x-2=0 \\
& 4x=2 \\
& x=\frac{2}{4} \\
& x=\frac{1}{2}
\end{align}$
Now, the domain of the function $\frac{f}{g}$ is all the real numbers except $\frac{1}{2}$.
Therefore, the domain of the functions $\frac{f}{g}$ is $\left( -\infty ,-3 \right)\cup \left( -3,\frac{1}{2} \right)\cup \left( \frac{1}{2},3 \right)\cup \left( 3,\infty \right)$.
