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Let's break down the domain finding process for each function.

1. **Function \( f(x) = \frac{x+6}{x^2 + 2x - 24} \)**:
   - The domain of a rational function excludes values of \(x\) that make the denominator equal to zero.
   - Set the denominator equal to zero and solve for \(x\):
     \[
     x^2 + 2x - 24 = 0
     \]
   - Factor the quadratic equation:
     \[
     (x + 6)(x - 4) = 0
     \]
   - So, \(x = -6\) and \(x = 4\) are the values that make the denominator zero.
   - Therefore, the domain of \( f(x) \) excludes \(x = -6\) and \(x = 4\).
   - The domain is: 
     \[
     (-\infty, -6) \cup (-6, 4) \cup (4, \infty)
     \]

2. **Function \( g(x) = \frac{x}{x^2 + 9} \)**:
   - Again, the domain excludes values of \(x\) that make the denominator zero.
   - Set the denominator equal to zero:
     \[
     x^2 + 9 = 0
     \]
   - Solving for \(x\), we get:
     \[
     x^2 = -9
     \]
   - This has no real solution because a square cannot be negative.
   - Therefore, the denominator never equals zero for real numbers, meaning the function is defined for all real numbers.
   - The domain is: 
     \[
     (-\infty, \infty)
     \]

Would you like more details or further clarification?

Here are 5 related questions:
1. How do we handle undefined points in rational functions?
2. Can you explain how to factor quadratic expressions like \(x^2 + 2x - 24\)?
3. What happens

The functions f(x) = (x + 6) / (x^2 + 2x - 24) and g(x) = x / (x^2 + 9) are defined. Find the domain of each function and write each answer as an interval or union of intervals.

Learn how to find the domain of rational functions f(x) = (x + 6) / (x^2 + 2x - 24) and g(x) = x / (x^2 + 9). Detailed explanation of factoring quadratics and determining where the function is defined. Suitable for high school students in Grades 9-12.

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