Answer
The function $ f\left( x \right)=\frac{\left( x+1 \right)}{\left( x+1 \right)\left( x-4 \right)}$ is discontinuous for the points $-1\text{ and }4$.
Work Step by Step
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 $-1\text{ and }4$.
