`f(x)=sqrt(log((3x-x^(2))/(x-1)))`

Answer

To find the domain of the function \( f(x) = \sqrt{\log\left(\frac{3x - x^2}{x - 1}\right)} \), 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. **Condition for the square root**:
The expression inside the square root must be greater than or equal to zero:
\[ \log\left(\frac{3x - x^2}{x - 1}\right) \geq 0 \]
This implies:
\[ \frac{3x - x^2}{x - 1} \geq 1 \]

2. **Rearranging the inequality**:
Rearranging the inequality gives:
\[ \frac{3x - x^2}{x - 1} - 1 \geq 0 \]
Simplifying this, we have:
\[ \frac{3x - x^2 - (x - 1)}{x - 1} \geq 0 \]
This simplifies to:
\[ \frac{-x^2 + 2x + 1}{x - 1} \geq 0 \]
or equivalently:
\[ \frac{x^2 - 2x - 1}{x - 1} \leq 0 \]

3. **Finding the roots of the quadratic**:
To factor the quadratic \( x^2 - 2x - 1 \), we use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2} \]
Thus, the roots are \( 1 + \sqrt{2} \) and \( 1 - \sqrt{2} \).

4. **Setting up the critical points**:
The critical points from the inequality are:
- \( 1 - \sqrt{2} \)
- \( 1 \) (from the denominator)
- \( 1 + \sqrt{2} \)

5. **Testing intervals**:
We will test the sign of \( \frac{x^2 - 2x - 1}{x - 1} \) in the intervals determined by the critical points:
- Interval \( (-\infty, 1 - \sqrt{2}) \)
- Interval \( (1 - \sqrt{2}, 1) \)
- Interval \( (1, 1 + \sqrt{2}) \)
- Interval \( (1 + \sqrt{2}, \infty) \)

By testing points in each interval, we can determine where the expression is less than or equal to zero.

6. **Conclusion**:
After testing the intervals, we find that the function is non-positive in the intervals:
\[ (-\infty, 1 - \sqrt{2}] \cup [1, 1 + \sqrt{2}] \]
Therefore, the domain of \( f(x) \) is:
\[ \boxed{(-\infty, 1 - \sqrt{2}] \cup [1, 1 + \sqrt{2}]} \]