The domain of the function of defined by `f(x)=sqrt(4-x) +(1)/(x^(2)-1)` is equal to:

Options

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

Answer

To find the domain of the function \( f(x) = \sqrt{4 - x} + \frac{1}{x^2 - 1} \), we need to determine the restrictions on \( x \) that will keep the function defined.

### Step 1: Analyze the square root part
The expression under the square root, \( 4 - x \), must be greater than or equal to zero for the square root to be defined. Therefore, we set up the inequality:

\[ 4 - x \geq 0 \]

### Step 2: Solve the inequality
Rearranging the inequality gives:

\[ -x \geq -4 \]

Multiplying both sides by -1 (which reverses the inequality):

\[ x \leq 4 \]

This means \( x \) can take any value less than or equal to 4.

### Step 3: Analyze the rational part
The denominator of the rational part, \( x^2 - 1 \), cannot be zero because division by zero is undefined. We set up the equation:

\[ x^2 - 1 \neq 0 \]

### Step 4: Solve the equation
Factoring gives:

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

This implies:

\[ x \neq 1 \quad \text{and} \quad x \neq -1 \]

### Step 5: Combine the restrictions
From the square root, we found \( x \leq 4 \). From the rational part, we found that \( x \) cannot be 1 or -1. Therefore, we can express the domain as:

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

### Conclusion
The domain of the function \( f(x) = \sqrt{4 - x} + \frac{1}{x^2 - 1} \) is:

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

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