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