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

Options

A. `(-oo,-1)uu(1,4]`
B. `(-oo,-1]uu(1,4]`
C. `(-oo,-1)uu[1,4]`
D. `(-oo,-1)uu[1,4)`

Explanation

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 analyzing both terms in the function.

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

For the square root to be defined, the expression inside the square root must be non-negative:

\[ 4 - x \geq 0 \]

This simplifies to:

\[ x \leq 4 \]

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

For this term to be defined, the expression inside the square root must be positive (since we cannot divide by zero):

\[ x^2 - 1 > 0 \]

This can be factored as:

\[ (x - 1)(x + 1) > 0 \]

### Step 3: Solve the inequality \( (x - 1)(x + 1) > 0 \)

To solve this inequality, we find the critical points by setting the factors equal to zero:

\[ x - 1 = 0 \quad \Rightarrow \quad x = 1 \]
\[ x + 1 = 0 \quad \Rightarrow \quad x = -1 \]

Now, we test the intervals determined by these critical points:

1. \( (-\infty, -1) \)
2. \( (-1, 1) \)
3. \( (1, \infty) \)

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

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

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

Thus, the solution to the inequality \( (x - 1)(x + 1) > 0 \) is:

\[ x \in (-\infty, -1) \cup (1, \infty) \]

### Step 4: Combine the conditions

Now we have two conditions:

1. \( x \leq 4 \)
2. \( x \in (-\infty, -1) \cup (1, \infty) \)

We need to find the intersection of these two conditions.

- From \( x \leq 4 \), we can see that it does not restrict the intervals \( (-\infty, -1) \) and \( (1, \infty) \) since both intervals extend to \( -\infty \) and \( 4 \) is greater than \( 1 \).

Thus, the valid intervals that satisfy both conditions are:

\[ (-\infty, -1) \cup (1, 4] \]

### Final Domain

The domain of the function \( f(x) \) is:

\[ \boxed{(-\infty, -1) \cup (1, 4]} \]