Find the domain of `f(x)=1/sqrt(x^2 -x -6)`

Answer

To find the domain of the function \( f(x) = \frac{1}{\sqrt{x^2 - x - 6}} \), we need to ensure that the expression under the square root is positive and that the denominator is not zero. Let's go through the steps to find the domain.

### Step 1: Identify the expression under the square root
The expression under the square root is \( x^2 - x - 6 \).

### Step 2: Set the expression greater than zero
For the function to be defined, we need:
\[ x^2 - x - 6 > 0 \]

### Step 3: Factor the quadratic expression
We can factor \( x^2 - x - 6 \):
\[ x^2 - x - 6 = (x - 3)(x + 2) \]
Thus, we need to solve:
\[ (x - 3)(x + 2) > 0 \]

### Step 4: Find the critical points
The critical points occur when the expression equals zero:
\[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \]
\[ x + 2 = 0 \quad \Rightarrow \quad x = -2 \]
So, the critical points are \( x = -2 \) and \( x = 3 \).

### Step 5: Test intervals around the critical points
We will test the sign of the expression \( (x - 3)(x + 2) \) in the intervals defined by the critical points:
1. **Interval 1:** \( (-\infty, -2) \)
- Choose \( x = -3 \):
\[ (-3 - 3)(-3 + 2) = (-6)(-1) = 6 > 0 \]
2. **Interval 2:** \( (-2, 3) \)
- Choose \( x = 0 \):
\[ (0 - 3)(0 + 2) = (-3)(2) = -6 < 0 \]
3. **Interval 3:** \( (3, \infty) \)
- Choose \( x = 4 \):
\[ (4 - 3)(4 + 2) = (1)(6) = 6 > 0 \]

### Step 6: Determine the intervals where the expression is positive
From our tests, we find:
- The expression is positive in the intervals \( (-\infty, -2) \) and \( (3, \infty) \).

### Step 7: Exclude the critical points
Since the expression cannot equal zero (as it would make the denominator undefined), we exclude \( x = -2 \) and \( x = 3 \).

### Final Domain
Thus, the domain of the function \( f(x) = \frac{1}{\sqrt{x^2 - x - 6}} \) is:
\[ \text{Domain: } (-\infty, -2) \cup (3, \infty) \]