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The task in the provided images is to find the domain of functions algebraically for exercises 9 to 16 and verify the conclusions graphically.

Let's solve for the domains of each function algebraically:

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### 9. \( f(x) = x^2 + 4 \)
- The function \( x^2 + 4 \) is a polynomial, which is defined for all real values of \( x \).
- **Domain:** \( (-\infty, \infty) \)

---

### 10. \( h(x) = \frac{5}{x - 3} \)
- The function has a rational form. The denominator cannot be zero. Solve:
  \[
  x - 3 = 0 \implies x = 3
  \]
  So, the function is undefined at \( x = 3 \).
- **Domain:** \( (-\infty, 3) \cup (3, \infty) \)

---

### 11. \( f(x) = \frac{3x - 1}{(x + 3)(x - 1)} \)
- This is also a rational function. The denominator cannot be zero. Set the denominator equal to zero and solve:
  \[
  (x + 3)(x - 1) = 0
  \]
  So, \( x = -3 \) or \( x = 1 \).
- **Domain:** \( (-\infty, -3) \cup (-3, 1) \cup (1, \infty) \)

---

### 12. \( f(x) = \frac{1}{x} + \frac{5}{x - 3} \)
- There are two rational terms. Each denominator must not be zero. Solve:
  \[
  x = 0 \quad \text{and} \quad x - 3 = 0 \implies x = 3
  \]
  So, the function is undefined at \( x = 0 \) and \( x = 3 \).
- **Domain:** \( (-\infty, 0) \cup (0, 3) \cup (3, \infty) \)

---

### 13. \( g(x) = \frac{x}{x^2 - 5x} \)
- Factor the denominator:
  \[
  x^2 - 5x = x(x - 5)
  \]
  Set the denominator equal to zero:
  \[
  x(x - 5) = 0 \implies x = 0 \quad \text{or} \quad x = 5
  \]
- **Domain:** \( (-\infty, 0) \cup (0, 5) \cup (5, \infty) \)

---

### 14. \( h(x) = \frac{\sqrt{4 - x^2}}{x - 3} \)
- For the square root in the numerator, \( 4 - x^2 \geq 0 \) (the expression inside the square root must be non-negative):
  \[
  4 - x^2 \geq 0 \implies -2 \leq x \leq 2
  \]
  Also, the denominator cannot be zero:
  \[
  x - 3 = 0 \implies x = 3
  \]
  Since \( x = 3 \) is outside the interval \( [-2, 2] \), this doesn't affect the domain.
- **Domain:** \( [-2, 2] \)

---

### 15. \( h(x) = \frac{\sqrt{4 - x}}{(x + 1)(x^2 + 1)} \)
- The square root requires \( 4 - x \geq 0 \), so:
  \[
  4 - x \geq 0 \implies x \leq 4
  \]
  The denominator \( (x + 1)(x^2 + 1) \) never equals zero because \( x^2 + 1 > 0 \) for all \( x \), and \( x + 1 = 0 \) only at \( x = -1 \).
- **Domain:** \( (-\infty, -1) \cup (-1, 4] \)

---

### 16. \( f(x) = \sqrt{x^4 - 16x^2} \)
- Factor the expression inside the square root:
  \[
  x^4 - 16x^2 = x^2(x^2 - 16) = x^2(x - 4)(x + 4)
  \]
  The expression inside the square root must be non-negative:
  \[
  x^2(x - 4)(x + 4) \geq 0
  \]
  Solving this inequality gives:
  \[
  x \in (-\infty, -4] \cup [0, 4]
  \]
- **Domain:** \( (-\infty, -4] \cup [0, 4] \)

---

Would you like me to elaborate on any particular step or graph these functions to verify their domains?

### Follow-up Questions:
1. How would you graphically verify the domain of a rational function?
2. What methods can be used to solve domain restrictions in square root functions?
3. Can you explain how inequalities are solved for domain in functions involving square roots?
4. How would the domain change if the function involved both square roots and rational expressions?
5. Why is factoring the denominator important when finding the domain of rational functions?

### Tip:
When finding the domain of a function, always check for restrictions such as division by zero and taking the square root of negative numbers.

Find the domain of the functions algebraically and verify the conclusions graphically for exercises 9 to 16.

Learn how to find the domain of functions, including polynomial, rational, and square root functions, from exercises 9 to 16. Includes step-by-step algebraic solutions and verification methods. Suitable for students in Grades 10-12.

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