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.

Discover how to find the discontinuity of the function f(x) = (x - 3) / (x^2 + 9). Explore why the function is continuous for all real values of x and why the discontinuities occur only at imaginary numbers. Learn more about rational functions and their properties.

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