## Precalculus (6th Edition) Blitzer

The domain of the function $f\left( x \right)=\frac{x}{{{x}^{2}}+4x-21}$ is $\left( -\infty,-7 \right)\cup \left( -7,3 \right)\cup \left( 3,\infty \right)$.
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 $\left( -\infty,-7 \right)\cup \left( -7,3 \right)\cup \left( 3,\infty \right)$.