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: Identify the conditions for the function to be defined
The function \( f(x) \) is defined as long as the denominator is not zero and the expression under the square root is non-negative. Therefore, we need to ensure:
1. \( \sqrt{x-2} \neq 0 \)
2. \( x - 2 \geq 0 \)

### Step 2: Solve the inequalities
From the first condition:
- \( \sqrt{x-2} \neq 0 \) implies \( x - 2 \neq 0 \) or \( x \neq 2 \).

From the second condition:
- \( x - 2 \geq 0 \) implies \( x \geq 2 \).

### Step 3: Combine the conditions
Now, combining both conditions:
- From \( x \geq 2 \), we have \( x \) can be 2 or greater.
- However, from \( x \neq 2 \), we exclude 2 from the domain.

Thus, the domain of \( f(x) \) is:
\[ \text{Domain} = (2, \infty) \]

### Step 4: Determine the range of the function
Next, we will find the range of \( f(x) \). Since \( f(x) = \frac{1}{\sqrt{x-2}} \), we analyze the behavior of this function as \( x \) varies in the domain.

1. As \( x \) approaches 2 from the right (i.e., \( x \to 2^+ \)), \( \sqrt{x-2} \) approaches 0, which means \( f(x) \) approaches infinity:
\[ f(x) \to \infty \text{ as } x \to 2^+ \]

2. As \( x \) increases towards infinity, \( \sqrt{x-2} \) also increases, which means \( f(x) \) approaches 0:
\[ f(x) \to 0 \text{ as } x \to \infty \]

Since \( f(x) \) approaches infinity and tends towards 0 but never actually reaches 0, the range of \( f(x) \) is:
\[ \text{Range} = (0, \infty) \]

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