Let's analyze the given functions to determine their domains.

1. Function $f(x) = \frac{x^2}{x + 1}$:

   • The domain of a rational function is all real numbers except where the denominator equals zero. For $f(x)$, the denominator is $x + 1$. Setting it to zero: $x + 1 = 0 \implies x = -1$
   • So, the domain of $f(x)$ is all real numbers except $x = -1$. $\text{Domain of } f(x): (-\infty, -1) \cup (-1, \infty)$

2. Function $g(x) = \frac{x - 8}{x^2 - 64}$:

   • Again, the domain is all real numbers except where the denominator equals zero. The denominator $x^2 - 64$ is a difference of squares and can be factored: $x^2 - 64 = (x - 8)(x + 8)$
   • Set each factor equal to zero: $x - 8 = 0 \implies x = 8$ $x + 8 = 0 \implies x = -8$
   • So, the domain of $g(x)$ is all real numbers except $x = 8$ and $x = -8$. $\text{Domain of } g(x): (-\infty, -8) \cup (-8, 8) \cup (8, \infty)$

Summary:

• Domain of $f(x)$: $(-\infty, -1) \cup (-1, \infty)$
• Domain of $g(x)$: $(-\infty, -8) \cup (-8, 8) \cup (8, \infty)$

Would you like further details or clarifications? Here are 5 related questions:

1. How do you handle rational functions with more complex denominators?
2. What happens to the domain if a rational function has both a numerator and denominator that can be simplified?
3. Can we use graphical methods to determine domains?
4. What’s the significance of excluded values in a function's domain?
5. How do vertical asymptotes relate to domain restrictions?

Tip: Always factor the denominator of a rational function to find all possible values to exclude from the domain.