Answer
The domain of the function $f\left( x \right)=\frac{2x+7}{{{x}^{3}}-5{{x}^{2}}-4x+20}$ is $\left( -\infty ,-2 \right)\cup \left( -2,2 \right)\cup \left( 2,5 \right)\cup \left( 5,\infty \right)$.
Work Step by Step
In the given function
$f\left( x \right)=\frac{2x+7}{{{x}^{3}}-5{{x}^{2}}-4x+20}$
if the denominator is zero then it will make the function undefined. So, put the denominator ${{x}^{3}}-5{{x}^{2}}-4x+20$ equal to zero.
Therefore, ${{x}^{3}}-5{{x}^{2}}-4x+20=0$
Factorize the equation ${{x}^{3}}-5{{x}^{2}}-4x+20=0$
$\begin{align}
{{x}^{2}}\left( x-5 \right)-4\left( x-5 \right)=0 & \\
\left( {{x}^{2}}-4 \right)\left( x-5 \right)=0 & \\
\end{align}$
Case 1:
$\begin{align}
& {{x}^{2}}-4=0\ \ \ \ \ \ \ \ \\
& \left( x-2 \right)\left( x+\text{2} \right)\text{=}\ \text{0}
\end{align}$
This implies $x=2$ and $x=-2$.
Case 2:
$\begin{align}
x-5=0 & \\
x=5 & \\
\end{align}$
The factors of the equation are $2\ ,-2$ and $5$. Exclude the factors from the domain of the function $f\left( x \right)=\frac{2x+7}{{{x}^{3}}-5{{x}^{2}}-4x+20}$.
Hence, the domain of the function $f\left( x \right)=\frac{2x+7}{{{x}^{3}}-5{{x}^{2}}-4x+20}$ is $\left( -\infty ,-2 \right)\cup \left( -2,2 \right)\cup \left( 2,5 \right)\cup \left( 5,\infty \right)$.
