Find the domain and range of `f(x)=(1)/(sqrt(x-2))`

Explanation

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

### Step 1: Determine the Domain
The function \( f(x) \) involves a square root in the denominator. For the function to be defined, the expression inside the square root must be positive. Therefore, we need to solve the inequality:

\[ x - 2 > 0 \]

This simplifies to:

\[ x > 2 \]

Thus, the domain of \( f(x) \) is all values of \( x \) that are greater than 2. In interval notation, this is represented as:

\[ \text{Domain} = (2, \infty) \]

### Step 2: Determine the Range
Next, we will find the range of the function. Since \( f(x) = \frac{1}{\sqrt{x-2}} \), we note that as \( x \) approaches 2 from the right (i.e., \( x \to 2^+ \)), \( \sqrt{x-2} \) approaches 0, which means \( f(x) \) approaches infinity:

\[ f(x) \to \infty \quad \text{as} \quad x \to 2^+ \]

As \( x \) increases beyond 2, \( \sqrt{x-2} \) increases, causing \( f(x) \) to decrease. Thus, \( f(x) \) will take on all positive values but will never actually reach 0. Therefore, the range of \( f(x) \) is all positive real numbers:

\[ \text{Range} = (0, \infty) \]

### Final Answer
- **Domain**: \( (2, \infty) \)
- **Range**: \( (0, \infty) \)