Answer
The domain of the given function is $\left( -\infty ,-6 \right)\bigcup \left( -6,6 \right)\bigcup \left( 6,\infty \right)$
Work Step by Step
The domain in which the given function
$f\left( x \right)=\frac{\left( x-6 \right)}{\left( {{x}^{2}}-36 \right)}$
is defined is given as:
${{x}^{2}}-36\ne 0$ because $\frac{a}{b}$ , where $b\ne 0$
Using property the ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$, we simplify ${{x}^{2}}-36\ne 0$:
${{x}^{2}}-36=\left( x+6 \right)\left( x-6 \right)\ne 0$
And from the zero-factor property:
$\left( x+6 \right)\ne 0,\left( x-6 \right)\ne 0$
Therefore, $\left( x+6 \right)\ne 0$.
Add $-6$ to both sides:
$\begin{align}
& \left( x+6-6 \right)\ne 0-6 \\
& x=-6
\end{align}$
And $\left( x-6 \right)\ne 0$.
Add $+6$ to both sides:
$\begin{align}
& \left( x-6+6 \right)\ne 0+6 \\
& x=+6
\end{align}$
So, the domain of the function $f\left( x \right)=\frac{\left( x-6 \right)}{\left( {{x}^{2}}-36 \right)}$ is $x\in \text{R}-\left( 6,-6 \right)$.
This means that it includes all the real numbers excluding (6,-6).
This can represented as:
Therefore, $\left( -\infty ,-6 \right)\bigcup \left( -6,6 \right)\bigcup \left( 6,\infty \right)$.
Hence, the domain of the function $f\left( x \right)=\frac{\left( x-6 \right)}{\left( {{x}^{2}}-36 \right)}$ is $\left( -\infty ,-6 \right)\bigcup \left( -6,6 \right)\bigcup \left( 6,\infty \right)$.
