What is the domain of `f(x) = sqrt((4-x)^(2)-5)` ?

Options

A. `x le -3` or `x ge 7`
B. `2-sqrt(5)le x le 2+sqrt(5)`
C. `x le 4 - sqrt(5)` or `x ge 4 + sqrt(5)`
D. `4 - sqrt(5)le x le 4 + sqrt(5)`

Explanation

To find the domain of the function \( f(x) = \sqrt{(4 - x)^2 - 5} \), we need to ensure that the expression inside the square root is non-negative, as the square root is only defined for non-negative values.

### Step-by-Step Solution:

1. **Set the expression inside the square root greater than or equal to zero:**
\[ (4 - x)^2 - 5 \geq 0 \]

2. **Rearrange the inequality:**
\[ (4 - x)^2 \geq 5 \]

3. **Take the square root of both sides:**
\[ |4 - x| \geq \sqrt{5} \]

4. **Split the absolute value into two cases:**
- Case 1:
\[ 4 - x \geq \sqrt{5} \implies x \leq 4 - \sqrt{5} \]
- Case 2:
\[ 4 - x \leq -\sqrt{5} \implies x \geq 4 + \sqrt{5} \]

5. **Combine the results from both cases:**
The domain of \( f(x) \) is:
\[ x \leq 4 - \sqrt{5} \quad \text{or} \quad x \geq 4 + \sqrt{5} \]

### Conclusion:
The domain of the function \( f(x) = \sqrt{(4 - x)^2 - 5} \) is:
\[ (-\infty, 4 - \sqrt{5}] \cup [4 + \sqrt{5}, \infty) \]