The problem involves two functions, $f(x)$ and $g(x)$, defined as follows: $f(x) = \sqrt{x - 12}$ $g(x) = \sqrt{x + 12}$

Part a: Find the domain of $f$, $g$, $f + g$, $f - g$, $fg$, $\frac{f}{g}$, and $\frac{g}{f}$.

1. Domain of $f(x)$: $f(x) = \sqrt{x - 12}$ implies $x - 12 \geq 0$. Hence, the domain is $x \geq 12$, or in interval notation, $[12, \infty)$.

2. Domain of $g(x)$: $g(x) = \sqrt{x + 12}$ implies $x + 12 \geq 0$. Hence, the domain is $x \geq -12$, or in interval notation, $[-12, \infty)$.

3. Domain of $f + g$, $f - g$, and $fg$: These operations require both $f(x)$ and $g(x)$ to be defined at a particular $x$. The intersection of their domains gives $x \geq 12$. Thus, the domain for $f + g$, $f - g$, and $fg$ is $[12, \infty)$.

4. Domain of $\frac{f}{g}$ and $\frac{g}{f}$: Each ratio requires the denominator to be non-zero in addition to the individual functions being defined.

   • For $\frac{f}{g}$, $g(x) \neq 0$ which is true for $x > -12$ but we also need $x \geq 12$, so the domain is $[12, \infty)$.
   • For $\frac{g}{f}$, $f(x) \neq 0$ which is true for $x > 12$, and $x \geq -12$ also holds, so the domain is $(12, \infty)$ (notice it's exclusive of 12 because $f(12) = 0$).

Part b: Find $f(g(x))$, $g(f(x))$, $f(f(x))$, $g(g(x))$.

1. $f(g(x))$: $f(g(x)) = f(\sqrt{x + 12}) = \sqrt{\sqrt{x + 12} - 12}$ For $f(g(x))$ to exist, $\sqrt{x + 12} - 12 \geq 0$ which leads to $\sqrt{x + 12} \geq 12$ or $x + 12 \geq 144$, thus $x \geq 132$. Therefore, $f(g(x))$ is defined for $x \geq 132$.

2. $g(f(x))$: $g(f(x)) = g(\sqrt{x - 12}) = \sqrt{\sqrt{x - 12} + 12}$ This expression is valid for all $x \geq 12$, the domain of $f(x)$.

3. $f(f(x))$: $f(f(x)) = f(\sqrt{x - 12}) = \sqrt{\sqrt{x - 12} - 12}$ This requires $\sqrt{x - 12} - 12 \geq 0$, leading to $\sqrt{x - 12} \geq 12$, or $x - 12 \geq 144$, so $x \geq 156$.

4. $g(g(x))$: $g(g(x)) = g(\sqrt{x + 12}) = \sqrt{\sqrt{x + 12} + 12}$ This is valid for all $x \geq -12$, the domain of $g(x)$.

Would you like more details or help with any other questions? Here are some further exploration ideas:

1. How does changing the constants in $f(x)$ and $g(x)$ affect their domains?
2. What happens to the domain of $\frac{f}{g}$ and $\frac{g}{f}$ if the operations inside the square roots are changed?
3. How would the solutions to these functions graphically look like?
4. Can you solve for $x$ where $f(x) = g(x)$?
5. What are the practical applications of finding the domain in real-world problems?

Tip: Remember, when solving for compositions of functions involving square roots, always ensure the expression under each square root is non-negative.