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give me full solution and find the domain if needed

Let's solve the problem by performing the operations step-by-step and finding the domains for each case.

Given functions:
- \( f(x) = \sqrt{x - 2} \)
- \( g(x) = \frac{1}{x} \)

We are required to perform various operations and determine the domain for each.

### a. \( (f + g)(x) = f(x) + g(x) \)

\[
(f + g)(x) = \sqrt{x - 2} + \frac{1}{x}
\]

#### Domain:
- For \( \sqrt{x - 2} \), we need \( x - 2 \geq 0 \Rightarrow x \geq 2 \).
- For \( \frac{1}{x} \), \( x \neq 0 \).

Thus, the domain is \( x \geq 2 \) and \( x \neq 0 \). Therefore, the domain is \( x \in [2, \infty) \setminus \{0\} \).

### b. \( (f - g)(x) = f(x) - g(x) \)

\[
(f - g)(x) = \sqrt{x - 2} - \frac{1}{x}
\]

#### Domain:
The domain is the same as part (a), \( x \in [2, \infty) \setminus \{0\} \).

### c. \( (f \cdot g)(x) = f(x) \cdot g(x) \)

\[
(f \cdot g)(x) = \sqrt{x - 2} \cdot \frac{1}{x} = \frac{\sqrt{x - 2}}{x}
\]

#### Domain:
The same restrictions apply as before:
- \( x \geq 2 \)
- \( x \neq 0 \)

Thus, the domain is \( x \in [2, \infty) \setminus \{0\} \).

### d. \( \left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} \)

\[
\left(\frac{f}{g}\right)(x) = \frac{\sqrt{x - 2}}{\frac{1}{x}} = x \sqrt{x - 2}
\]

#### Domain:
The same restrictions apply:
- \( x \geq 2 \)
- \( x \neq 0 \)

Thus, the domain is \( x \in [2, \infty) \setminus \{0\} \).

### e. \( (f \circ g)(x) = f(g(x)) \)

\[
(f \circ g)(x) = f\left(\frac{1}{x}\right) = \sqrt{\frac{1}{x} - 2}
\]

#### Domain:
For \( f(g(x)) \) to be defined, \( \frac{1}{x} - 2 \geq 0 \).

\[
\frac{1}{x} \geq 2 \Rightarrow x \leq \frac{1}{2}
\]

Also, \( x \neq 0 \). Thus, the domain is \( x \in (-\infty, 0) \cup (0, \frac{1}{2}] \).

### f. \( (g \circ f)(x) = g(f(x)) \)

\[
(g \circ f)(x) = g\left(\sqrt{x - 2}\right) = \frac{1}{\sqrt{x - 2}}
\]

#### Domain:
For \( \sqrt{x - 2} \) to be defined, we need \( x \geq 2 \). Also, \( \sqrt{x - 2} \neq 0 \), which means \( x \neq 2 \).

Thus, the domain is \( x \in (2, \infty) \).

### g. \( (f \circ f)(x) = f(f(x)) \)

\[
(f \circ f)(x) = f\left(\sqrt{x - 2}\right) = \sqrt{\sqrt{x - 2} - 2}
\]

#### Domain:
For the expression to be valid:
- \( \sqrt{x - 2} - 2 \geq 0 \)
\[
\sqrt{x - 2} \geq 2 \Rightarrow x - 2 \ge

Perform the following fundamental operations on the given functions f(x) = √(x-2) and g(x) = 1/x. Then, determine the domain of each operation.

Learn how to perform operations on the functions f(x) = √(x-2) and g(x) = 1/x, including addition, subtraction, multiplication, division, and composition. The step-by-step solution includes finding the domain for each operation.

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