The function given in the image is:

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

Part (a) - Where is $f(x)$ discontinuous?

To find where $f(x)$ is discontinuous, we need to examine the denominator $x^2 + 3x - 4$. Discontinuities occur when the denominator equals zero because division by zero is undefined.

So, let's solve:

$x^2 + 3x - 4 = 0$

Factoring the quadratic:

$(x + 4)(x - 1) = 0$

This gives us the critical points $x = -4$ and $x = 1$, where $f(x)$ is potentially discontinuous.

Part (b) - Classify the type of discontinuity

To classify the discontinuity, we must check the behavior of the function around these points.

1. At $x = 1$:

The numerator is $x^2 - 1 = (x - 1)(x + 1)$, so at $x = 1$, the numerator also becomes zero. This creates a removable discontinuity at $x = 1$ because the common factor $(x - 1)$ cancels out.

After canceling the common factor, the function simplifies to:

$f(x) = \frac{x + 1}{x + 4} \quad \text{for} \, x \neq 1$

Thus, the discontinuity at $x = 1$ is removable.

2. At $x = -4$:

At $x = -4$, the denominator becomes zero, but the numerator does not. This causes a non-removable (infinite) discontinuity, also known as a vertical asymptote, at $x = -4$.

Conclusion:

• $f(x)$ is discontinuous at $x = 1$ (removable discontinuity).
• $f(x)$ has an infinite discontinuity (vertical asymptote) at $x = -4$.

Would you like more details on any of the steps? Here are some related questions:

1. How do we determine the limit of a function around a removable discontinuity?
2. What is the difference between removable and non-removable discontinuities?
3. How can you factor more complex polynomials to find discontinuities?
4. How does the behavior near a vertical asymptote affect the function graph?
5. What happens to the simplified function $\frac{x+1}{x+4}$ as $x \to -4$?

Tip: When checking for discontinuities, always verify whether the numerator and denominator share a common factor to identify removable discontinuities.