Question

the function f is continuous on the interval -1 < x < 3 and is not continuous on the interval -1 ≤ x ≤ 3. which of the following could not be an expression for f(x)? a $\frac{x + 1}{x - 3}$ b $\frac{x - 3}{x + 1}$ c (x + 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 $x = 3$. The open - interval $-1\lt x\lt3$ does not include $x = 3$, and the closed - interval $-1\leq x\leq3$ includes $x = 3$. This function can satisfy the given continuity conditions.

Step3: Analyze option B

For $y=\frac{x - 3}{x + 1}$, the function is discontinuous at $x=-1$. The open - interval $-1\lt x\lt3$ does not include $x=-1$, but the closed - interval $-1\leq x\leq3$ includes $x=-1$. This function can satisfy the given continuity conditions.

Step4: Analyze option C

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

Answer:

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