the function f is continuous on the interval -1 < x < 3 and is not continuous on the interval -1 ≤ x ≤ 3…

Explanation:

Step1: Recall continuity condition

A rational - function $\frac{g(x)}{h(x)}$ is discontinuous where $h(x)=0$. A polynomial function is continuous everywhere.

Step2: Analyze option A

For $y = \frac{x + 1}{x - 3}$, the function is discontinuous at $x = 3$. It is continuous on $-1<x<3$ and discontinuous on $-1\leq x\leq3$ since $x = 3$ is in the closed - interval.

Step3: Analyze option B

For $y=\frac{x - 3}{x + 1}$, the function is discontinuous at $x=-1$. It is continuous on $-1<x<3$ and discontinuous on $-1\leq x\leq3$ since $x=-1$ is in the closed - interval.

Step4: Analyze option C

The function $y=(x + 1)(x - 3)=x^{2}-2x - 3$ is a polynomial function. Polynomial functions are continuous everywhere, including on the interval $-1\leq x\leq3$. So it cannot be the function $f(x)$.

Step5: Analyze option D

For $y=\frac{1}{(x + 1)(x - 3)}$, the function is discontinuous at $x=-1$ and $x = 3$. It is continuous on $-1<x<3$ and discontinuous on $-1\leq x\leq3$ since $x=-1$ and $x = 3$ are in the closed - interval.

Answer:

C. $(x + 1)(x - 3)$