Answer
The values of the functions are:
$f+g=\frac{9{{x}^{2}}+79x-28}{\left( x-4 \right)\left( x+8 \right)}$ , $f-g=\frac{9{{x}^{2}}+65x+28}{\left( x-4 \right)\left( x+8 \right)}$ , $fg=\frac{63x}{\left( x-4 \right)\left( x+8 \right)},$ and $\frac{f}{g}=\frac{9{{x}^{2}}+72x}{7x-28}$. The domain of $\frac{f}{g}$ is $\left( -\infty ,-8 \right)\cup \left( -8,4 \right)\cup \left( 4,\infty \right)$ and that of $f-g,f+g$, and $fg$ is $\left( -\infty ,-8 \right)\cup \left( -8,4 \right)\cup \left( 4,\infty \right)$.
Work Step by Step
Calculate the value of $f+g$ as shown below to get,
$\begin{align}
& f+g=f\left( x \right)+g\left( x \right) \\
& =\frac{9x}{x-4}+\frac{7}{x+8} \\
& =\frac{9x\left( x+8 \right)+7\left( x-4 \right)}{\left( x-4 \right)\left( x+8 \right)} \\
& =\frac{9{{x}^{2}}+72x+7x-28}{\left( x-4 \right)\left( x+8 \right)}
\end{align}$
$=\frac{9{{x}^{2}}+79x-28}{\left( x-4 \right)\left( x+8 \right)}$
Calculate the value of $f-g$ as shown below to get,
$\begin{align}
& f-g=f\left( x \right)-g\left( x \right) \\
& =\frac{9x}{x-4}-\frac{7}{x+8} \\
& =\frac{9x\left( x+8 \right)-7\left( x-4 \right)}{\left( x-4 \right)\left( x+8 \right)} \\
& =\frac{9{{x}^{2}}+72x-7x+28}{\left( x-4 \right)\left( x+8 \right)}
\end{align}$
$=\frac{9{{x}^{2}}+65x+28}{\left( x-4 \right)\left( x+8 \right)}$
Calculate the value of $fg$ as shown below to get,
$\begin{align}
& fg=f\left( x \right)g\left( x \right) \\
& =\left( \frac{9x}{x-4} \right)\left( \frac{7}{x+8} \right) \\
& =\frac{63x}{\left( x-4 \right)\left( x+8 \right)}
\end{align}$
Calculate the value of $\frac{f}{g}$ as shown below to get,
$\begin{align}
& \frac{f}{g}=\frac{f\left( x \right)}{g\left( x \right)} \\
& =\frac{\left( \frac{9x}{x-4} \right)}{\left( \frac{7}{x+8} \right)} \\
& =\frac{9x}{x-4}\times \frac{x+8}{7} \\
& =\frac{9x\left( x+8 \right)}{7\left( x-4 \right)} \\
& =\frac{9{{x}^{2}}+72x}{7x-28}
\end{align}$
If the functions $f+g$ , $f-g$ and $fg$ contain division by zero, they will be undefined. So, put the denominator $\left( x-4 \right)\left( x+8 \right)$ equal to zero.
$\left( x-4 \right)\left( x+8 \right)=0$
$\begin{align}
& \left( x-4 \right)\left( x+8 \right)=0 \\
& x-4=0,x+8=0 \\
\end{align}$
This implies that $x=4$ and $x=-8$.
Now, the domain of the functions $f+g$ $f-g$ and $fg$ is all the real numbers except $-8$ and 4.
Therefore, the domain of functions $f-g$ and $fg$ is $\left( -\infty ,-8 \right)\cup \left( -8,4 \right)\cup \left( 4,\infty \right)$.
In the function $\frac{f}{g}$, we check for division by zero. So, put the denominator $7x-28$ equal to zero.
$\begin{align}
& 7x-28=0 \\
& 7x=28 \\
& x=4
\end{align}$
Now, the domain of the function $\frac{f}{g}$ is all the real numbers except $4$.
Therefore, the domain of the functions $\frac{f}{g}$ is $\left( -\infty ,-8 \right)\cup \left( -8,4 \right)\cup \left( 4,\infty \right)$.
