Answer
The value of $\left( \frac{f}{g} \right)\left( x \right)$ and its domain with the functions $f\left( x \right)={{x}^{2}}-x-4$ and $g\left( x \right)=2x-6$ is $\left( -\infty ,3 \right)\cup \left( 3,\infty \right)$.
Work Step by Step
We know that the function $\left( \frac{f}{g} \right)\left( x \right)$ can also be defined as $\frac{f\left( x \right)}{g\left( x \right)}$:
$\begin{align}
& \left( \frac{f}{g} \right)\left( x \right)=\frac{{{x}^{2}}-x-4}{2x-6} \\
& =\frac{{{x}^{2}}-x-4}{2x-6}
\end{align}$
Hence,
$\left( \frac{f}{g} \right)\left( x \right)=\frac{{{x}^{2}}-x-4}{2x-6}$.
We can see that $\left( \frac{f}{g} \right)\left( x \right)$ is not defined for $x=3$.
Therefore, the domain of $\left( \frac{f}{g} \right)\left( x \right)$ is $\left( -\infty ,3 \right)\cup \left( 3,\infty \right)$.
Hence, the value of $\left( \frac{f}{g} \right)\left( x \right)$ is $\frac{{{x}^{2}}-x-4}{2x-6}$ and its domain is $\left( -\infty ,3 \right)\cup \left( 3,\infty \right)$.
