Answer
The values of the functions are:
$f+g=\frac{8{{x}^{2}}+30x-12}{\left( x-2 \right)\left( x+3 \right)}$ , $f-g=\frac{8{{x}^{2}}+18x+12}{\left( x-2 \right)\left( x+3 \right)}$ , $fg=\frac{48x}{\left( x-2 \right)\left( x+3 \right)},$ and $\frac{f}{g}=\frac{8{{x}^{2}}+24x}{6x-12}$.
The domain of $\frac{f}{g}$ is $\left( -\infty ,-3 \right)\cup \left( -3,2 \right)\cup \left( 2,\infty \right)$ and that of $f-g,f+g$ and $fg$ is $\left( -\infty ,-3 \right)\cup \left( -3,2 \right)\cup \left( 2,\infty \right)$.
Work Step by Step
Calculate the value of $f+g$ as shown below to get,
$\begin{align}
& f+g=f\left( x \right)+g\left( x \right) \\
& =\frac{8x}{x-2}+\frac{6}{x+3} \\
& =\frac{8x\left( x+3 \right)+6\left( x-2 \right)}{\left( x-2 \right)\left( x+3 \right)} \\
& =\frac{8{{x}^{2}}+24x+6x-12}{\left( x-2 \right)\left( x+3 \right)}
\end{align}$
$=\frac{8{{x}^{2}}+30x-12}{\left( x-2 \right)\left( x+3 \right)}$
Calculate the value of $f-g$ as shown below to get,
$\begin{align}
& f-g=f\left( x \right)-g\left( x \right) \\
& =\frac{8x}{x-2}-\frac{6}{x+3} \\
& =\frac{8x\left( x+3 \right)-6\left( x-2 \right)}{\left( x-2 \right)\left( x+3 \right)} \\
& =\frac{8{{x}^{2}}+24x-6x+12}{\left( x-2 \right)\left( x+3 \right)}
\end{align}$
$=\frac{8{{x}^{2}}+18x+12}{\left( x-2 \right)\left( x+3 \right)}$
Calculate the value of $fg$ as shown below to get,
$\begin{align}
& fg=f\left( x \right)g\left( x \right) \\
& =\left( \frac{8x}{x-2} \right)\left( \frac{6}{x+3} \right) \\
& =\frac{48x}{\left( x-2 \right)\left( x+3 \right)}
\end{align}$
Calculate the value of $\frac{f}{g}$ as shown below to get,
$\begin{align}
& \frac{f}{g}=\frac{f\left( x \right)}{g\left( x \right)} \\
& =\frac{\left( \frac{8x}{x-2} \right)}{\left( \frac{6}{x+3} \right)} \\
& =\frac{8x}{x-2}\times \frac{x+3}{6} \\
& =\frac{8x\left( x+3 \right)}{6\left( x-2 \right)} \\
& =\frac{8{{x}^{2}}+24x}{6x-12}
\end{align}$
If the functions $f+g$ , $f-g$ and $fg$ contain division by zero, they will be undefined. So, put the denominator $\left( x-2 \right)\left( x+3 \right)$ equal to zero.
$\left( x-2 \right)\left( x+3 \right)=0$
This implies that $x=2$ and $x=-3$.
Now, the domain of the functions $f+g$ $f-g$ and $fg$ is all the real numbers except $-3$ and $2$.
Therefore, the domain of the functions $f+g$ $f-g$ and $fg$ is $\left( -\infty ,-3 \right)\cup \left( -3,2 \right)\cup \left( 2,\infty \right)$.
If the function $\frac{f}{g}$ is divided by zero, it will be undefined. So, put the denominator $6x-12$ equal to zero.
$\begin{align}
& 6x-12=0 \\
& 6x=12 \\
& x=2
\end{align}$
Now, the domain of the function $\frac{f}{g}$ is all the real numbers except $2$.
Therefore, the domain of the function $\frac{f}{g}$ is $\left( -\infty ,-3 \right)\cup \left( -3,2 \right)\cup \left( 2,\infty \right)$.
