To find the domain of the function \( f(x) = \log_3{\log_{\frac{1}{2}}(x^2 + 4x + 4)} \), we need to ensure that the arguments of both logarithmic functions are valid.
### Step 1: Analyze the inner logarithm
The inner function is \( \log_{\frac{1}{2}}(x^2 + 4x + 4) \). For this logarithm to be defined, its argument must be positive:
\[
x^2 + 4x + 4 > 0
\]
### Step 2: Factor the quadratic expression
The quadratic expression can be factored:
\[
x^2 + 4x + 4 = (x + 2)^2
\]
Thus, we need to solve:
\[
(x + 2)^2 > 0
\]
### Step 3: Determine when the quadratic is positive
The expression \( (x + 2)^2 \) is equal to zero when \( x + 2 = 0 \), which gives:
\[
x = -2
\]
Since \( (x + 2)^2 \) is a square term, it is always non-negative and is greater than zero for all \( x \) except \( x = -2 \). Therefore, the solution to this inequality is:
\[
x \in (-\infty, -2) \cup (-2, \infty)
\]
### Step 4: Analyze the outer logarithm
Next, we need to ensure that the result of the inner logarithm is positive:
\[
\log_{\frac{1}{2}}(x^2 + 4x + 4) > 0
\]
Since the base \( \frac{1}{2} < 1 \), the logarithm will be positive when its argument is less than 1:
\[
x^2 + 4x + 4 < 1
\]
### Step 5: Solve the inequality
Rearranging gives:
\[
x^2 + 4x + 3 < 0
\]
Factoring this expression:
\[
(x + 1)(x + 3) < 0
\]
### Step 6: Determine the intervals for the product
To find the intervals where this product is negative, we can test the intervals defined by the roots \( x = -1 \) and \( x = -3 \):
- For \( x < -3 \), both factors are negative, so the product is positive.
- For \( -3 < x < -1 \), one factor is negative and the other is positive, so the product is negative.
- For \( x > -1 \), both factors are positive, so the product is positive.
Thus, the solution to the inequality is:
\[
x \in (-3, -1)
\]
### Step 7: Combine the intervals
Now, we combine the intervals from the two conditions:
1. From \( (x + 2)^2 > 0 \): \( x \in (-\infty, -2) \cup (-2, \infty) \)
2. From \( (x + 1)(x + 3) < 0 \): \( x \in (-3, -1) \)
The intersection of these two sets gives us the final domain:
\[
x \in (-3, -2) \cup (-2, -1)
\]
### Final Domain
Thus, the domain of the function \( f(x) = \log_3{\log_{\frac{1}{2}}(x^2 + 4x + 4)} \) is:
\[
\boxed{(-3, -2) \cup (-2, -1)}
\]
