Find the domain of `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 determine the values of \( x \) for which the function is defined. This involves ensuring that the expressions under the square roots are non-negative and that the denominator is not zero.

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

The expression \( 4 - x \) must be non-negative:
\[ 4 - x \geq 0 \]
Solving this inequality:
\[ x \leq 4 \]

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

The expression \( x^2 - 1 \) must be positive (since it is in the denominator):
\[ x^2 - 1 > 0 \]
Factoring gives:
\[ (x - 1)(x + 1) > 0 \]
To solve this inequality, we find the critical points where \( x^2 - 1 = 0 \):
\[ x = -1 \quad \text{and} \quad x = 1 \]

### Step 3: Determine the intervals for \( (x - 1)(x + 1) > 0 \)

We analyze the sign of \( (x - 1)(x + 1) \) in the intervals defined by the critical points:
- For \( x < -1 \): both factors are negative, so the product is positive.
- For \( -1 < x < 1 \): one factor is negative and the other is positive, so the product is negative.
- For \( x > 1 \): both factors are positive, so the product is positive.

Thus, the solution to \( (x - 1)(x + 1) > 0 \) is:
\[ x < -1 \quad \text{or} \quad x > 1 \]

### Step 4: Combine the results

Now we combine the results from Step 1 and Step 3:
1. From \( \sqrt{4 - x} \): \( x \leq 4 \)
2. From \( \frac{1}{\sqrt{x^2 - 1}} \): \( x < -1 \) or \( x > 1 \)

### Step 5: Identify the valid intervals

We need to find the intersection of these conditions:
- For \( x < -1 \): This interval is valid since it is also less than or equal to 4.
- For \( x > 1 \): The upper limit is 4, so we have \( 1 < x \leq 4 \).

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

### Final Answer:
The domain of the function \( f(x) = \sqrt{4 - x} + \frac{1}{\sqrt{x^2 - 1}} \) is:
\[ (-\infty, -1) \cup (1, 4] \]