The values of x for which the function $${\text{f}}\left( {\text{x}} \right) = \frac{{{{\text{x}}^2} – 3{\text{x}} – 4}}{{{{\text{x}}^2} + 3{\text{x}} – 4}}$$ is NOT continuous are A. 4 and -1 B. 4 and 1 C. -4 and 1 D. -4 and -1

The function $f(x) = \frac{x^2 – 3x – 4}{x^2 + 3x – 4}$ is not continuous at $x = 1$ and $x = -1$. This is because the two-sided limit of $f(x)$ as $x$ approaches $1$ from the left and from the right are not equal.

To see this, let’s consider the limit of $f(x)$ as $x$ approaches $1$ from the left. We can write this as:

$$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} \frac{x^2 – 3x – 4}{x^2 + 3x – 4} = \lim_{x \to 1^-} \frac{(x – 1)(x + 4)}{(x – 1)(x + 4)} = 1$$

On the other hand, the limit of $f(x)$ as $x$ approaches $1$ from the right is:

$$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} \frac{x^2 – 3x – 4}{x^2 + 3x – 4} = \lim_{x \to 1^+} \frac{(x – 1)(x – 4)}{(x – 1)(x + 4)} = -1$$

Since the two-sided limit of $f(x)$ as $x$ approaches $1$ does not exist, $f(x)$ is not continuous at $x = 1$.

Similarly, we can show that $f(x)$ is not continuous at $x = -1$. Therefore, the values of $x$ for which the function $f(x)$ is not continuous are $x = 1$ and $x = -1$.