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 $g(x)$: $(-\infty, -3) \cup (-3, 3) \cup (3, \infty)$

Summary:

• Domain of $f(x)$: $(-\infty, \infty)$
• Domain of $g(x)$: $(-\infty, -3) \cup (-3, 3) \cup (3, \infty)$

Would you like more details on any of these steps?

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.