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?

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.