Find the domain of the function `f(x) = sqrt( 4 - x) + (1)/( sqrt( x^(2) - 1))`

Answer

To find the domain of the function \( f(x) = \sqrt{4 - x} + \frac{1}{\sqrt{x^2 - 1}} \), we need to ensure that both components of the function are defined for real numbers.

### Step 1: Analyze the first component \( \sqrt{4 - x} \)

The expression inside the square root must be non-negative:
\[ 4 - x \geq 0 \]
This simplifies to:
\[ x \leq 4 \]

### Step 2: Analyze the second component \( \frac{1}{\sqrt{x^2 - 1}} \)

For this component, the expression inside the square root must be positive (since we cannot take the square root of a negative number and cannot divide by zero):
\[ x^2 - 1 > 0 \]
This can be factored as:
\[ (x - 1)(x + 1) > 0 \]
To solve this inequality, we find the critical points where \( x - 1 = 0 \) and \( x + 1 = 0 \), which gives us \( x = 1 \) and \( x = -1 \).

### Step 3: Test intervals around the critical points

We will test the intervals determined by the critical points \( -1 \) and \( 1 \):
1. For \( x < -1 \) (e.g., \( x = -2 \)):
\[ (-2 - 1)(-2 + 1) = (-3)(-1) = 3 > 0 \quad \text{(valid)} \]

2. For \( -1 < x < 1 \) (e.g., \( x = 0 \)):
\[ (0 - 1)(0 + 1) = (-1)(1) = -1 < 0 \quad \text{(invalid)} \]

3. For \( x > 1 \) (e.g., \( x = 2 \)):
\[ (2 - 1)(2 + 1) = (1)(3) = 3 > 0 \quad \text{(valid)} \]

From this analysis, we find that \( x^2 - 1 > 0 \) for:
\[ x < -1 \quad \text{or} \quad x > 1 \]

### Step 4: Combine the results

Now we combine the results from both components:
1. From \( \sqrt{4 - x} \): \( x \leq 4 \)
2. From \( \frac{1}{\sqrt{x^2 - 1}} \): \( x < -1 \) or \( x > 1 \)

Thus, the valid intervals for \( x \) are:
- From \( -\infty \) to \( -1 \) (from the second component)
- From \( 1 \) to \( 4 \) (from both components)

### Final Domain

The domain of the function \( f(x) \) is:
\[ (-\infty, -1) \cup (1, 4] \]