The domain of the function `f(x)=sqrt((x^(2)-8x+12).ln^(2)(x-3))` is

Options

A. `[3, oo)`
B. `[4, oo)`
C. `[6, oo)uu{2, 4}`
D. `[6, oo)uu{4}`

Answer

To find the domain of the function \( f(x) = \sqrt{(x^2 - 8x + 12) \ln^2(x - 3)} \), we need to ensure that the expression inside the square root is non-negative and that the logarithm is defined.

### Step-by-Step Solution:

1. **Identify the conditions for the square root:**
The expression inside the square root must be greater than or equal to zero:
\[ (x^2 - 8x + 12) \ln^2(x - 3) \geq 0 \]

2. **Factor the quadratic expression:**
We can factor \( x^2 - 8x + 12 \):
\[ x^2 - 8x + 12 = (x - 2)(x - 6) \]

3. **Determine when \( (x - 2)(x - 6) \) is non-negative:**
The product \( (x - 2)(x - 6) \) is non-negative when:
- \( x < 2 \) (both factors are negative)
- \( x > 6 \) (both factors are positive)
- \( x = 2 \) or \( x = 6 \) (the product is zero)

Thus, the intervals where this expression is non-negative are:
\[ (-\infty, 2] \cup [6, \infty) \]

4. **Determine the conditions for the logarithm:**
The logarithm \( \ln(x - 3) \) is defined when:
\[ x - 3 > 0 \implies x > 3 \]

5. **Combine the conditions:**
We need to find the intersection of the intervals from the quadratic and the logarithm:
- From the quadratic: \( (-\infty, 2] \cup [6, \infty) \)
- From the logarithm: \( (3, \infty) \)

The intersection gives us:
- From \( (-\infty, 2] \) and \( (3, \infty) \): No overlap.
- From \( [6, \infty) \) and \( (3, \infty) \): The overlap is \( [6, \infty) \).

6. **Check the boundary points:**
- At \( x = 6 \): \( \ln(6 - 3) = \ln(3) \) is defined and positive, and \( (6 - 2)(6 - 6) = 0 \), so \( f(6) = 0 \).
- At \( x = 4 \): \( \ln(4 - 3) = \ln(1) = 0 \), so \( f(4) = 0 \).

Both \( x = 6 \) and \( x = 4 \) are included in the domain since they yield valid outputs.

### Conclusion:
The domain of the function \( f(x) \) is:
\[ [6, \infty) \cup \{4\} \]

### Final Answer:
The correct option is \( [6, \infty) \cup \{4\} \).

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