After reviewing the problem again, the domain of the rational function $f(x) = \frac{x + 1}{2x + 9}$ is indeed all real numbers except where the denominator equals zero.

Following the solution steps:

1. Identify the restriction in the domain:
   Set the denominator $2x + 9$ equal to zero: $2x + 9 = 0$
2. Solve for $x$: $2x = -9$ $x = -\frac{9}{2}$

This means that $x$ cannot be $-\frac{9}{2}$.

Therefore, the correct domain is: $\{ x \mid x \neq -\frac{9}{2} \}$

Answer

The correct option is (C): $\{ x \mid x \neq -\frac{9}{2} \}$.

Would you like further explanation on this topic?

Here are some related questions to deepen understanding:

1. How would you determine the domain if the function had multiple fractions?
2. What if the function was $f(x) = \frac{x + 1}{2x^2 + 9x}$; how would that change the solution?
3. How does the domain of a function affect its graph?
4. What is the process for finding the domain of a function with square roots in the denominator?
5. How would you define the domain of a polynomial function?

Tip: For rational functions, always check for any values that make the denominator zero. These are the values excluded from the domain.