The domain of `y=(x)/(sqrt(x^(2)-3x+2))`

Options

A. `(-oo,oo)`
B. `[0,oo)`
C. `(-oo, 1]uu[2,oo)`
D. none of these

Answer

To find the domain of the function \( y = \frac{x}{\sqrt{x^2 - 3x + 2}} \), we need to ensure that the denominator is defined and not equal to zero. Here are the steps to determine the domain:

### Step 1: Identify the conditions for the denominator
The denominator is \( \sqrt{x^2 - 3x + 2} \). For this expression to be defined, the expression inside the square root must be greater than zero:
\[ x^2 - 3x + 2 > 0 \]

### Step 2: Factor the quadratic expression
We can factor the quadratic expression \( x^2 - 3x + 2 \):
\[ x^2 - 3x + 2 = (x - 1)(x - 2) \]
Thus, we need to solve the inequality:
\[ (x - 1)(x - 2) > 0 \]

### Step 3: Determine the critical points
The critical points from the factors are \( x = 1 \) and \( x = 2 \). These points will divide the number line into intervals that we can test.

### Step 4: Test the intervals
We will test the intervals determined by the critical points:
1. \( (-\infty, 1) \)
2. \( (1, 2) \)
3. \( (2, \infty) \)

- **Interval \( (-\infty, 1) \)**: Choose \( x = 0 \):
\[ (0 - 1)(0 - 2) = (-1)(-2) = 2 > 0 \quad \text{(True)} \]

- **Interval \( (1, 2) \)**: Choose \( x = 1.5 \):
\[ (1.5 - 1)(1.5 - 2) = (0.5)(-0.5) = -0.25 < 0 \quad \text{(False)} \]

- **Interval \( (2, \infty) \)**: Choose \( x = 3 \):
\[ (3 - 1)(3 - 2) = (2)(1) = 2 > 0 \quad \text{(True)} \]

### Step 5: Combine the results
From the tests, we find that the expression \( (x - 1)(x - 2) > 0 \) is true in the intervals:
\[ (-\infty, 1) \quad \text{and} \quad (2, \infty) \]

### Step 6: Consider the first condition for the numerator
Since the numerator \( x \) must also be defined, we require \( x \geq 0 \). Therefore, we need to intersect this condition with our previous results.

### Step 7: Find the intersection of intervals
The intersection of \( [0, 1) \) and \( (-\infty, 1) \) gives us:
\[ [0, 1) \]
The intersection of \( [0, 1) \) and \( (2, \infty) \) gives us:
\[ (2, \infty) \]

### Final Domain
Thus, the domain of the function is:
\[ [0, 1) \cup (2, \infty) \]