## Precalculus (6th Edition) Blitzer

The function $f\left( x \right)=\frac{\left( x+1 \right)}{\left( x+1 \right)\left( x-4 \right)}$ is discontinuous for the points $-1\text{ and }4$.
Consider the rational function $f\left( x \right)=\frac{\left( x+1 \right)}{\left( x+1 \right)\left( x-4 \right)}$, Here, $p\left( x \right)=x+1\text{ and }q\left( x \right)=\left( x+1 \right)\left( x-4 \right)$ Find the zeros of the function $q\left( x \right)=\left( x+1 \right)\left( x-4 \right)$ by $q\left( x \right)=0$, $\left( x+1 \right)\left( x-4 \right)=0$ Solve for the x, \begin{align} & \left( x+1 \right)=0 \\ & x=-1 \end{align} or \begin{align} & \left( x-4 \right)=0 \\ & x=4 \end{align} The zeros of the function $q\left( x \right)=\left( x+1 \right)\left( x-4 \right)$ are $-1\text{ and }4$. Thus, the function $f\left( x \right)=\frac{\left( x+1 \right)}{\left( x+1 \right)\left( x-4 \right)}$ is discontinuous for the points $-1\text{ and }4$.