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To find the domain of \( f(x) = \sqrt{\frac{x^2 - 5x - 36}{x - 2}} \), we need to ensure two conditions are met:

1. **The radicand (inside the square root) must be non-negative**: \( \frac{x^2 - 5x - 36}{x - 2} \geq 0 \).
2. **The denominator cannot be zero**: \( x - 2 \neq 0 \), which implies \( x \neq 2 \).

### Step 1: Factor the Numerator
The expression in the numerator, \( x^2 - 5x - 36 \), factors as:
\[
x^2 - 5x - 36 = (x - 9)(x + 4)
\]
So, the function becomes:
\[
f(x) = \sqrt{\frac{(x - 9)(x + 4)}{x - 2}}
\]

### Step 2: Determine When \( \frac{(x - 9)(x + 4)}{x - 2} \geq 0 \)
To find the intervals where this expression is non-negative, identify the critical points where the expression is zero or undefined:
- **Zeros** of the numerator: \( x = 9 \) and \( x = -4 \).
- **Zero of the denominator** (undefined point): \( x = 2 \).

### Step 3: Analyze Intervals
Using the critical points \( x = -4 \), \( x = 2 \), and \( x = 9 \), we test intervals:
1. **Interval \( (-\infty, -4) \)**: Pick a point like \( x = -5 \). Substitute into \( \frac{(x - 9)(x + 4)}{x - 2} \); the result is positive.
2. **Interval \( (-4, 2) \)**: Pick a point like \( x = 0 \); the result is negative.
3. **Interval \( (2, 9) \)**: Pick a point like \( x = 3 \); the result is positive.
4. **Interval \( (9, \infty) \)**: Pick a point like \( x = 10 \); the result is positive.

### Step 4: Combine Intervals
The intervals where \( \frac{(x - 9)(x + 4)}{x - 2} \geq 0 \) and \( x \neq 2 \) are:
\[
(-\infty, -4] \cup (2, 9] \cup [9, \infty)
\]

### Final Answer
The domain of \( f(x) \) is:
\[
(-\infty, -4] \cup (2, 9]
\]

Would you like further explanation on any step?

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#### Related Questions:
1. How would the domain change if the square root were removed?
2. What happens to the domain if the denominator is a square rather than linear?
3. How do critical points affect the sign of a rational expression?
4. How can interval notation help in visualizing domains of functions?
5. How do you find the range once the domain is established?

#### Tip:
When working with rational functions inside square roots, always analyze sign changes across critical points to ensure valid domains.

Let f(x) = √((x² - 5x - 36) / (x - 2)). Write the domain of f as an interval.

Learn how to determine the domain of the function f(x) = √((x² - 5x - 36) / (x - 2)). This solution involves analyzing the behavior of square roots and rational expressions, using factoring and interval testing. Suitable for students in Grades 10-12.

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