The domain of the function `f(x)=(1)/(sqrt(x^(2)-3x+2))` is

Options

A. `(-oo, 1)`
B. `(-oo, 1) cup (2, oo)`
C. `(-oo, 1] cup [ 2, oo)`
D. `(2, oo)`

Answer

To find the domain of the function \( f(x) = \frac{1}{\sqrt{x^2 - 3x + 2}} \), we need to ensure that the expression under the square root is positive, since the square root of a negative number is not defined in the real number system, and the denominator cannot be zero.

### Step-by-step solution:

1. **Identify the expression under the square root:**
\[ x^2 - 3x + 2 \]

2. **Set the expression greater than zero:**
\[ x^2 - 3x + 2 > 0 \]

3. **Factor the quadratic expression:**
\[ x^2 - 3x + 2 = (x - 1)(x - 2) \]

4. **Determine where the product is greater than zero:**
We need to find the intervals where \( (x - 1)(x - 2) > 0 \). The critical points are \( x = 1 \) and \( x = 2 \).

5. **Test intervals around the critical points:**
- For \( x < 1 \): Choose \( x = 0 \):
\[ (0 - 1)(0 - 2) = 1 \cdot 2 = 2 > 0 \quad \text{(True)} \]
- For \( 1 < x < 2 \): Choose \( x = 1.5 \):
\[ (1.5 - 1)(1.5 - 2) = 0.5 \cdot (-0.5) = -0.25 < 0 \quad \text{(False)} \]
- For \( x > 2 \): Choose \( x = 3 \):
\[ (3 - 1)(3 - 2) = 2 \cdot 1 = 2 > 0 \quad \text{(True)} \]

6. **Combine the intervals where the product is positive:**
The expression \( (x - 1)(x - 2) > 0 \) holds true for:
\[ (-\infty, 1) \cup (2, \infty) \]

7. **Conclusion:**
Therefore, the domain of the function \( f(x) \) is:
\[ (-\infty, 1) \cup (2, \infty) \]