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Let's analyze each function to determine its domain and range.

### a) \( f(x) = \frac{x + 2}{x^2 - 1} \)

- **Domain**: We need to find where the denominator is not zero. Set \( x^2 - 1 = 0 \), which gives \( x = \pm 1 \). The function is undefined at these points.
  - Domain: \( x \in (-\infty, -1) \cup (-1, 1) \cup (1, \infty) \).

- **Range**: Since this is a rational function (a ratio of polynomials), determining the range analytically is complex, but the function can approach any real value due to the form of the numerator and denominator. The vertical asymptotes at \( x = \pm 1 \) suggest the range is all real numbers:
  - Range: \( y \in (-\infty, \infty) \).

### b) \( g(x) = \sqrt{x^2 - 2x - 8} \)

- **Domain**: The square root function is only defined when the expression inside is non-negative. Solve the inequality \( x^2 - 2x - 8 \geq 0 \).
  - Factor: \( (x - 4)(x + 2) \geq 0 \).
  - Solve: \( x \in (-\infty, -2] \cup [4, \infty) \).

- **Range**: The square root function only gives non-negative values, so the range starts at 0 and goes to infinity.
  - Range: \( y \in [0, \infty) \).

### c) \( g(x) = \frac{\sqrt{2 + x^2}}{3 - x} \)

- **Domain**: The square root function is always defined because \( 2 + x^2 \geq 2 \) for all real \( x \). However, the denominator cannot be zero, so solve \( 3 - x \neq 0 \), giving \( x \neq 3 \).
  - Domain: \( x \in (-\infty, 3) \cup (3, \infty) \).

- **Range**: The numerator \( \sqrt{2 + x^2} \) is always greater than or equal to \( \sqrt{2} \), and the function becomes undefined as \( x \) approaches 3. The overall range is hard to find without more advanced techniques, but generally:
  - Range: \( (-\infty, 0) \cup (0, \infty) \) (due to vertical asymptote at \( x = 3 \)).

### d) \( f(x) = \frac{x}{\sqrt[4]{9 - x^2}} \)

- **Domain**: The expression inside the fourth root, \( 9 - x^2 \), must be non-negative. Solve \( 9 - x^2 > 0 \), which gives \( -3 < x < 3 \).
  - Domain: \( x \in (-3, 3) \).

- **Range**: The fourth root function is always non-negative, but \( \frac{x}{\text{something positive}} \) means the function can take on any real value, except at the boundaries \( x = -3 \) and \( x = 3 \), where it is undefined.
  - Range: \( (-\infty, \infty) \).

Let me know if you'd like further details on any of these solutions or have any questions! 

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Here are 5 related questions you might find helpful:
1. How do you find the vertical asymptotes of rational functions?
2. What is the difference between square root and fourth root functions in terms of domain and range?
3. How can we determine the range of a function graphically?
4. How do you solve quadratic inequalities like \( (x - 4)(x + 2) \geq 0 \)?
5. Why does a rational function have undefined points at the zeros of the denominator?

**Tip**: When solving for the domain of a function with square roots, always ensure the expression inside the root is non-negative to avoid undefined regions!

Find the domain and range of the functions: a) f(x) = (x + 2) / (x^2 - 1), b) g(x) = sqrt(x^2 - 2x - 8), c) g(x) = sqrt(2 + x^2) / (3 - x), d) f(x) = x / sqrt(9 - x^2).

Find the domain and range of functions including rational functions like f(x) = (x + 2) / (x^2 - 1) and square root functions like g(x) = sqrt(x^2 - 2x - 8). This problem explores key concepts in algebra, including inequalities and the behavior of square root functions. Suitable for high school students in Grades 10-12.

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