Find the domain of definitions of the functions
(Read the symbols [*] and {*} as greatestintegers and fractional part functions respectively.)
`f(x) = sqrt(log_((1)/(2)) (x)/(x^(2)- 1))`

Answer

To find the domain of the function \( f(x) = \sqrt{\frac{\log_{(1/2)}(x)}{x^2 - 1}} \), we need to ensure that the expression inside the square root is non-negative. This requires two conditions to be satisfied:

1. The argument of the logarithm must be positive.
2. The entire fraction must be non-negative.

Let's break this down step by step.

### Step 1: Determine when the logarithm is defined and positive

The logarithm \( \log_{(1/2)}(x) \) is defined for \( x > 0 \). Since the base \( \frac{1}{2} \) is less than 1, the logarithm is positive when \( x < 1 \).

**Condition 1:**
\[ x > 0 \quad \text{and} \quad x < 1 \]

This gives us the interval:
\[ 0 < x < 1 \]

### Step 2: Analyze the denominator \( x^2 - 1 \)

The expression \( x^2 - 1 \) must not be zero, and it must also ensure that the fraction \( \frac{\log_{(1/2)}(x)}{x^2 - 1} \) is non-negative.

The expression \( x^2 - 1 = 0 \) when \( x = 1 \) or \( x = -1 \). Thus, we need to consider the sign of \( x^2 - 1 \):

- For \( x < -1 \), \( x^2 - 1 > 0 \)
- For \( -1 < x < 1 \), \( x^2 - 1 < 0 \)
- For \( x > 1 \), \( x^2 - 1 > 0 \)

Since we are interested in the interval \( 0 < x < 1 \), we find that \( x^2 - 1 < 0 \) in this range.

### Step 3: Determine the overall sign of the fraction

Now we analyze the fraction \( \frac{\log_{(1/2)}(x)}{x^2 - 1} \):

- In the interval \( 0 < x < 1 \):
- \( \log_{(1/2)}(x) > 0 \) (since \( x < 1 \))
- \( x^2 - 1 < 0 \)

Thus, the fraction \( \frac{\log_{(1/2)}(x)}{x^2 - 1} < 0 \).

### Step 4: Consider the case when both numerator and denominator are negative

Next, we consider the case where both the numerator and denominator are negative:

- The logarithm \( \log_{(1/2)}(x) < 0 \) when \( x > 1 \).
- The denominator \( x^2 - 1 < 0 \) when \( -1 < x < 1 \).

However, there is no overlap in the intervals where both conditions can be satisfied simultaneously. Therefore, we cannot find valid \( x \) values in this case.

### Conclusion

The only valid interval for \( x \) is:
\[ x \in (0, 1) \]

Thus, the domain of the function \( f(x) \) is:
\[ \text{Domain} = (0, 1) \]