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To determine the domains of the functions \( f(x) = \frac{x}{x^2 + 64} \) and \( g(x) = \frac{x - 3}{x^2 - 9} \), we need to identify where these functions are undefined (i.e., where their denominators equal zero).

### 1. Domain of \( f(x) \):
For the function \( f(x) = \frac{x}{x^2 + 64} \), the denominator is \( x^2 + 64 \).

- \( x^2 + 64 = 0 \) has no real solutions because \( x^2 \geq 0 \) for all real \( x \), and adding 64 means this expression is always positive.
- Therefore, the function \( f(x) \) is defined for all real numbers.

**Domain of \( f(x) \):** \( (-\infty, \infty) \)

### 2. Domain of \( g(x) \):
For the function \( g(x) = \frac{x - 3}{x^2 - 9} \), the denominator is \( x^2 - 9 \), which can be factored as \( (x - 3)(x + 3) \).

- The function is undefined when \( x^2 - 9 = 0 \), or equivalently when \( x = 3 \) or \( x = -3 \).
- Therefore, the function \( g(x) \) is undefined at \( x = 3 \) and \( x = -3 \).

**Domain of \( g(x) \):** \( (-\infty, -3) \cup (-3, 3) \cup (3, \infty) \)

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### Summary:
- **Domain of \( f(x) \):** \( (-\infty, \infty) \)
- **Domain of \( g(x) \):** \( (-\infty, -3) \cup (-3, 3) \cup (3, \infty) \)

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### Related questions:
1. How do you find the domain of a rational function?
2. Why does the function \( f(x) \) have a domain of all real numbers?
3. How do you factor the denominator of \( g(x) \)?
4. What happens to a function when its denominator equals zero?
5. Can the domain of a rational function ever include the points where the denominator is zero?

**Tip:** Always check for values that make the denominator zero, as these values are excluded from the domain of the function.

For each function, find the domain. f(x) = \frac{x}{x^2 + 64} and g(x) = \frac{x - 3}{x^2 - 9}. Write each answer as an interval or union of intervals.

Learn how to find the domain of rational functions f(x) = x/(x^2 + 64) and g(x) = (x - 3)/(x^2 - 9). The solution includes factoring quadratics and checking where denominators equal zero. Suitable for students in Grades 9-12.

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