Answer
The domain of the function
$f\left( x \right)=\frac{7x+2}{{{x}^{3}}-2{{x}^{2}}-9x+18}$
is:
$\left( -\infty ,-3 \right)\cup \left( -3,2 \right)\cup \left( 2,3 \right)\cup \left( 3,\infty \right)$.
Work Step by Step
In the given function
$f\left( x \right)=\frac{7x+2}{{{x}^{3}}-2{{x}^{2}}-9x+18}$
if the denominator is zero, then it will make the function undefined.
So, put the denominator ${{x}^{3}}-2{{x}^{2}}-9x+18$ equal to zero.
Therefore, ${{x}^{3}}-2{{x}^{2}}-9x+18=0$
Factorize the equation ${{x}^{3}}-2{{x}^{2}}-9x+18=0$
${{x}^{2}}\left( x-2 \right)\ -9\left( x-2 \right)=0$
$\left( {{x}^{2}}-9 \right)\left( x-2 \right)=0$
Case 1:
$\begin{align}
& {{x}^{2}}-9=\text{ }0\ \ \ \ \ \ \ \\
& \left( x-3 \right)\left( x+3 \right)=\text{ }0 \\
\end{align}$
This implies that $x=3$ and $x=-3$
Case 2:
$\begin{align}
& x-2=0 \\
& x=2 \\
\end{align}$
The factors of the equation are $-3,2$ , and $3$. Exclude the factors $-3,2$ , and $3$ from the domain of the function $f\left( x \right)=\frac{7x+2}{{{x}^{3}}-2{{x}^{2}}-9x+18}$.
Thus, the domain is $\left( -\infty ,-3 \right)\cup \left( -3,2 \right)\cup \left( 2,3 \right)\cup \left( 3,\infty \right)$.
Hence, the domain of the function $f\left( x \right)=\frac{7x+2}{{{x}^{3}}-2{{x}^{2}}-9x+18}$ is $\left( -\infty ,-3 \right)\cup \left( -3,2 \right)\cup \left( 2,3 \right)\cup \left( 3,\infty \right)$.
