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 1: Analyze the quadratic expression
The first part of the function is \( x^2 - 8x + 12 \). We need this to be greater than or equal to 0.
1. Factor the quadratic:
\[
x^2 - 8x + 12 = (x - 6)(x - 2)
\]
2. Set the inequality:
\[
(x - 6)(x - 2) \geq 0
\]
### Step 2: Find the critical points
The critical points from the factors are \( x = 6 \) and \( x = 2 \). We will analyze the sign of the product in the intervals determined by these points:
- \( (-\infty, 2) \)
- \( (2, 6) \)
- \( (6, \infty) \)
### Step 3: Test the intervals
1. For \( x < 2 \) (e.g., \( x = 0 \)):
\[
(0 - 6)(0 - 2) = 12 > 0
\]
So, the expression is positive.
2. For \( 2 < x < 6 \) (e.g., \( x = 4 \)):
\[
(4 - 6)(4 - 2) = -2 \cdot 2 = -4 < 0
\]
So, the expression is negative.
3. For \( x > 6 \) (e.g., \( x = 7 \)):
\[
(7 - 6)(7 - 2) = 1 \cdot 5 = 5 > 0
\]
So, the expression is positive.
### Step 4: Determine the intervals
From our tests, we find that \( (x - 6)(x - 2) \geq 0 \) is satisfied for:
- \( x \in (-\infty, 2] \) (where it is positive)
- \( x \in [6, \infty) \) (where it is positive)
### Step 5: Analyze the logarithmic part
Now we need to ensure that \( \ln(x - 3) \) is defined and non-negative:
1. The argument of the logarithm must be greater than 0:
\[
x - 3 > 0 \implies x > 3
\]
### Step 6: Combine the conditions
Now we combine the intervals:
- From the quadratic: \( (-\infty, 2] \cup [6, \infty) \)
- From the logarithm: \( (3, \infty) \)
The intersection of these two conditions gives us:
- The interval \( [6, \infty) \) is valid since it is greater than 3.
### Final Domain
Thus, the domain of the function \( f(x) \) is:
\[
\boxed{[6, \infty)}
\]
