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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.

The functions f and g are defined as f(x) = √(x - 12) and g(x) = √(x + 12). Find the domain of f, g, f + g, f - g, fg, f/g, and g/f. Also, find the domain of the composite functions f(g(x)), g(f(x)), f(f(x)), and g(g(x)).

Learn how to determine the domains of functions involving square roots, including individual functions, combined operations, and composite functions. This problem covers essential algebraic concepts, suitable for college algebra students, with step-by-step explanations.

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