Find the domain of the following functions:
`y=sqrt("in"(x^(2)-5x+7))`

Answer

To find the domain of the function \( y = \sqrt{\ln(x^2 - 5x + 7)} \), we need to ensure that the expression inside the square root is non-negative. This means that:

1. The argument of the logarithm must be positive:
\[ x^2 - 5x + 7 > 0 \]

2. The logarithm itself must be non-negative:
\[ \ln(x^2 - 5x + 7) \geq 0 \]

From the second condition, we can rewrite it as:
\[ x^2 - 5x + 7 \geq 1 \]

### Step 1: Solve the inequality \( x^2 - 5x + 7 \geq 1 \)

Subtract 1 from both sides:
\[ x^2 - 5x + 6 \geq 0 \]

### Step 2: Factor the quadratic expression

The quadratic \( x^2 - 5x + 6 \) can be factored as:
\[ (x - 2)(x - 3) \geq 0 \]

### Step 3: Determine the critical points

The critical points from the factors are \( x = 2 \) and \( x = 3 \).

### Step 4: Test intervals around the critical points

We will test the sign of the expression \( (x - 2)(x - 3) \) in the intervals determined by the critical points:
- Interval 1: \( (-\infty, 2) \)
- Interval 2: \( (2, 3) \)
- Interval 3: \( (3, \infty) \)

1. **For \( x < 2 \)** (e.g., \( x = 0 \)):
\[ (0 - 2)(0 - 3) = 6 \quad (\text{positive}) \]

2. **For \( 2 < x < 3 \)** (e.g., \( x = 2.5 \)):
\[ (2.5 - 2)(2.5 - 3) = (0.5)(-0.5) = -0.25 \quad (\text{negative}) \]

3. **For \( x > 3 \)** (e.g., \( x = 4 \)):
\[ (4 - 2)(4 - 3) = 2 \quad (\text{positive}) \]

### Step 5: Combine the results

The expression \( (x - 2)(x - 3) \geq 0 \) is satisfied in the intervals:
- \( (-\infty, 2] \)
- \( [3, \infty) \)

### Final Domain

Thus, the domain of the function \( y = \sqrt{\ln(x^2 - 5x + 7)} \) is:
\[ (-\infty, 2] \cup [3, \infty) \]