Find the domain of the following ` y = tan^(-1) ( sqrt (x^(2) - 1)) `

Options

A. `(-infty,-2]cup[1, infty)`
B. `(-infty,-1]
C. `(-infty,-1]cup[2, infty)`
D. `(-infty,-1]cup[1, infty)`

Answer

To find the domain of the function \( y = \tan^{-1}(\sqrt{x^2 - 1}) \), we need to determine the values of \( x \) for which the expression inside the inverse tangent function is defined.

### Step 1: Set up the inequality for the square root
The expression inside the square root, \( x^2 - 1 \), must be non-negative for the square root to be defined. Therefore, we set up the inequality:
\[ x^2 - 1 \geq 0 \]

### Step 2: Solve the inequality
Rearranging the inequality gives:
\[ x^2 \geq 1 \]
This can be factored as:
\[ (x - 1)(x + 1) \geq 0 \]

### Step 3: Determine the critical points
The critical points from the factors \( (x - 1) \) and \( (x + 1) \) are:
\[ x = -1 \quad \text{and} \quad x = 1 \]

### Step 4: Test intervals around the critical points
We will test the intervals defined by the critical points to determine where the product \( (x - 1)(x + 1) \) is non-negative.

1. **Interval \( (-\infty, -1) \)**: Choose \( x = -2 \)
\[ (-2 - 1)(-2 + 1) = (-3)(-1) = 3 \quad (\text{positive}) \]

2. **Interval \( (-1, 1) \)**: Choose \( x = 0 \)
\[ (0 - 1)(0 + 1) = (-1)(1) = -1 \quad (\text{negative}) \]

3. **Interval \( (1, \infty) \)**: Choose \( x = 2 \)
\[ (2 - 1)(2 + 1) = (1)(3) = 3 \quad (\text{positive}) \]

### Step 5: Combine the results
From the testing, we find that the expression \( (x - 1)(x + 1) \) is non-negative in the intervals:
- \( (-\infty, -1] \)
- \( [1, \infty) \)

Thus, the domain of the function \( y = \tan^{-1}(\sqrt{x^2 - 1}) \) is:
\[ \text{Domain} = (-\infty, -1] \cup [1, \infty) \]

### Summary of Steps:
1. Set up the inequality for the square root.
2. Solve the inequality.
3. Identify critical points.
4. Test intervals around the critical points.
5. Combine results to find the domain.