Answer
The domain of the function $f\left( x \right)=\frac{x}{{{x}^{2}}+4x-21}$ is $\left( -\infty,-7 \right)\cup \left( -7,3 \right)\cup \left( 3,\infty \right)$.
Work Step by Step
Consider the function given $f\left( x \right)=\frac{x}{{{x}^{2}}+4x-21}$.
We can see that this function contains division and division by 0 is not defined, excluding those values of $x$ from the domain that cause the denominator to be zero.
For this, set the denominator equal to 0, that is, ${{x}^{2}}+4x-21=0$.
Now, if we factorize it as follows:
$\begin{align}
& {{x}^{2}}+4x-21=0 \\
& {{x}^{2}}+7x-3x-21=0 \\
& x\left( x+7 \right)-3\left( x+7 \right)=0 \\
& \left( x-3 \right)\left( x+7 \right)=0
\end{align}$
Now, $\left( x-3 \right)\left( x+7 \right)=0$ implies that either $\left( x-3 \right)=0$ or $\left( x+7 \right)=0$.
That is, either $x=3$ or $x=-7$.
Therefore, the domain of the given function is the set of all real numbers excluding the numbers $3$ and $-7$.
Hence, the domain of the function $f\left( x \right)=\frac{x}{{{x}^{2}}+4x-21}$ is $\left( -\infty,-7 \right)\cup \left( -7,3 \right)\cup \left( 3,\infty \right)$.
