To find the domain of the function $f(x) = \frac{(x + 9)(x + 1)}{(x + 3)(x + 8)}$, we need to determine when the denominator is equal to zero, as division by zero is undefined.

The denominator is $(x + 3)(x + 8)$. Set this equal to zero and solve for $x$:

$(x + 3)(x + 8) = 0$

This gives two solutions:

$x + 3 = 0 \quad \Rightarrow \quad x = -3$ $x + 8 = 0 \quad \Rightarrow \quad x = -8$

Thus, the function is undefined at $x = -3$ and $x = -8$.

The domain of $f(x)$ is all real numbers except $x = -3$ and $x = -8$. In interval notation, this is written as:

$(-\infty, -8) \cup (-8, -3) \cup (-3, \infty)$

Would you like more details or have any questions?

Here are 5 related questions for further exploration:

1. How do you find vertical asymptotes for a rational function?
2. What is the range of the function $f(x) = \frac{(x + 9)(x + 1)}{(x + 3)(x + 8)}$?
3. How do you determine the horizontal asymptote of a rational function?
4. What is the behavior of $f(x)$ near the excluded values $x = -3$ and $x = -8$?
5. How would you find the zeros of $f(x)$?

Tip: When finding the domain of a rational function, always check where the denominator equals zero.