The domain of the function `sqrt(log_(1/3) log_4 ([x]^2 - 5 ))` is (where [x] denotes greatest integer function)

Explanation

To find the domain of the function \( f(x) = \sqrt{\log_{1/3} \log_4 ([x]^2 - 5)} \), we need to ensure that the expression inside the square root is non-negative, and that the logarithmic functions are 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:
\[ \log_{1/3} \log_4 ([x]^2 - 5) \geq 0 \]

2. **Convert the Logarithm**:
Since the base of the logarithm is \( \frac{1}{3} \), which is less than 1, the inequality reverses when we take the antilogarithm:
\[ \log_4 ([x]^2 - 5) \leq 1 \]

3. **Solve the Logarithmic Inequality**:
Now, we convert the logarithmic inequality to its exponential form:
\[ [x]^2 - 5 \leq 4^1 \]
This simplifies to:
\[ [x]^2 - 5 \leq 4 \implies [x]^2 \leq 9 \]

4. **Find the Range for \([x]\)**:
Taking the square root gives us:
\[ -3 \leq [x] \leq 3 \]
This means that \([x]\) can take integer values from -3 to 3.

5. **Determine the Corresponding Values of \(x\)**:
Since \([x]\) denotes the greatest integer function, we can find the intervals for \(x\):
- If \([x] = -3\), then \( -3 \leq x < -2 \)
- If \([x] = -2\), then \( -2 \leq x < -1 \)
- If \([x] = -1\), then \( -1 \leq x < 0 \)
- If \([x] = 0\), then \( 0 \leq x < 1 \)
- If \([x] = 1\), then \( 1 \leq x < 2 \)
- If \([x] = 2\), then \( 2 \leq x < 3 \)
- If \([x] = 3\), then \( 3 \leq x < 4 \)

6. **Combine the Intervals**:
The values of \(x\) corresponding to \([x]\) from -3 to 3 can be combined:
\[ (-3, -2) \cup [-2, -1) \cup [-1, 0) \cup [0, 1) \cup [1, 2) \cup [2, 3) \cup [3, 4) \]

7. **Final Domain**:
Therefore, the domain of the function \( f(x) \) is:
\[ (-3, -2) \cup [-2, 4) \]