The function you've provided is:

$f(x) = \frac{x - 3}{x^2 + 9}$

To find the discontinuities of the function, we look for points where the function is undefined. For rational functions like this one, the function is undefined where the denominator equals zero.

Step 1: Set the denominator equal to zero

$x^2 + 9 = 0$

Step 2: Solve for $x$

$x^2 = -9$

$x = \pm \sqrt{-9}$

$x = \pm 3i$

Conclusion

The solutions $x = \pm 3i$ are imaginary numbers, meaning there are no real values of $x$ that make the denominator zero. Therefore, the function $f(x)$ has no real discontinuities.

This function is continuous for all real values of $x$ because the denominator never equals zero for any real number.

Would you like more details or have any questions?

Here are 5 related questions:

1. What are the types of discontinuities in a function?
2. How do you determine if a function is continuous at a point?
3. Can a function be continuous but not differentiable? Explain.
4. How do you find the limits of a function as it approaches a point of discontinuity?
5. What is the difference between a removable discontinuity and an infinite discontinuity?

Tip: When solving for discontinuities, always check if the points where the denominator is zero correspond to real or imaginary numbers, as only real numbers lead to discontinuities in real-valued functions.