The function $f(x) = \frac{7}{x^2 - 3}$ is a rational function, which is continuous wherever the denominator is not equal to zero. To identify any discontinuities, we need to solve for $x$ where the denominator equals zero:

$x^2 - 3 = 0$

Solving for $x$:

$x^2 = 3$ $x = \pm \sqrt{3}$

Thus, the function has discontinuities at $x = \sqrt{3}$ and $x = -\sqrt{3}$.

Interval of Continuity:

The function is continuous on the intervals where the denominator is non-zero. The function is continuous everywhere except at $x = \pm \sqrt{3}$. So, the intervals of continuity are:

$(-\infty, -\sqrt{3}) \cup (-\sqrt{3}, \sqrt{3}) \cup (\sqrt{3}, \infty)$

Discontinuities:

The function has discontinuities at $x = -\sqrt{3}$ and $x = \sqrt{3}$.

Final Answer:

• Interval(s) of continuity: $(-\infty, -\sqrt{3}) \cup (-\sqrt{3}, \sqrt{3}) \cup (\sqrt{3}, \infty)$
• Discontinuities: $x = -\sqrt{3}, \sqrt{3}$

Would you like further clarification on this process, or do you have any other questions? Here are some related questions to explore:

1. How does the presence of a discontinuity affect the graph of a rational function?
2. What are the different types of discontinuities that can occur in rational functions?
3. How can you find the domain of a rational function?
4. What is the difference between removable and non-removable discontinuities?
5. How can you check the continuity of a function using limits?

Tip: To check the continuity of a function at a specific point, you can evaluate the left-hand and right-hand limits at that point and ensure they match the function's value there.