Find domain and range of following function: `y=1/(sqrt(x^(2)-x))`

Explanation

To find the domain and range of the function \( y = \frac{1}{\sqrt{x^2 - x}} \), we will follow these steps:

### Step 1: Identify the conditions for the function to be defined
The function \( y \) is defined when the denominator \( \sqrt{x^2 - x} \) is not equal to zero and is also defined for real numbers. Therefore, we need to solve the following inequalities:
1. \( x^2 - x > 0 \) (since the square root must be positive)
2. \( x^2 - x \neq 0 \) (to avoid division by zero)

### Step 2: Solve the inequality \( x^2 - x > 0 \)
We can factor the expression:
\[ x^2 - x = x(x - 1) > 0 \]
To find the intervals where this inequality holds, we will determine the critical points by setting \( x(x - 1) = 0 \):
- \( x = 0 \)
- \( x = 1 \)

### Step 3: Test intervals around the critical points
We will test the sign of \( x(x - 1) \) in the intervals:
1. \( (-\infty, 0) \)
2. \( (0, 1) \)
3. \( (1, \infty) \)

- For \( x < 0 \) (e.g., \( x = -1 \)):
\[ (-1)(-1 - 1) = (-1)(-2) = 2 > 0 \quad \text{(True)} \]

- For \( 0 < x < 1 \) (e.g., \( x = 0.5 \)):
\[ (0.5)(0.5 - 1) = (0.5)(-0.5) = -0.25 < 0 \quad \text{(False)} \]

- For \( x > 1 \) (e.g., \( x = 2 \)):
\[ (2)(2 - 1) = (2)(1) = 2 > 0 \quad \text{(True)} \]

### Step 4: Combine the results
The inequality \( x(x - 1) > 0 \) holds true in the intervals:
- \( (-\infty, 0) \)
- \( (1, \infty) \)

### Step 5: Exclude the points where the expression equals zero
Since we need \( x^2 - x \neq 0 \), we exclude \( x = 0 \) and \( x = 1 \) from our domain.

### Step 6: Write the domain
Thus, the domain of the function is:
\[ \text{Domain} = (-\infty, 0) \cup (1, \infty) \]

### Step 7: Find the range of the function
Now, we will determine the range of \( y = \frac{1}{\sqrt{x^2 - x}} \). Since \( x^2 - x > 0 \) in the domain, \( \sqrt{x^2 - x} \) will always be positive.

As \( x \) approaches \( 0 \) from the left, \( \sqrt{x^2 - x} \) approaches \( \sqrt{0 - 0} = 0 \), making \( y \) approach \( +\infty \).

As \( x \) approaches \( 1 \) from the right, \( \sqrt{x^2 - x} \) again approaches \( 0 \), making \( y \) approach \( +\infty \).

As \( x \) increases beyond \( 1 \), \( \sqrt{x^2 - x} \) increases, and thus \( y \) decreases. As \( x \) approaches \( \infty \), \( y \) approaches \( 0 \).

### Step 8: Write the range
Therefore, the range of the function is:
\[ \text{Range} = (0, \infty) \]

### Final Answer
- **Domain:** \( (-\infty, 0) \cup (1, \infty) \)
- **Range:** \( (0, \infty) \)