Find $(f \circ g)(x)$ and write the domain in interval notation. \[ f(x)=\frac{1}{x-2}, g(x)=\sqrt{x-2} \] The domain of $f \circ g$ is \begin{tabular}{|ccc} $\infty$ & $-\infty$ & $(\square, \square)$ \\ {$[\square, \square]$} & $(\square, \square]$ & {$[\square, \square)$} \\ $\square \cup \square$ & \\ $\times x$ & 5 \\ \hline \end{tabular}

Step 1 ：\( (f \circ g)(x) = f(g(x)) = f(\sqrt{x-2}) \)

Step 2 ：\( (f \circ g)(x) = \frac{1}{\sqrt{x-2}-2} \)

Step 3 ：Since the denominator cannot be zero, we need to find the values of x that make \(\sqrt{x-2}-2 \neq 0\)

Step 4 ：Solving the equation \(\sqrt{x-2}-2 \neq 0\) gives us \(x \neq 6\)

Step 5 ：Also, since the square root function is only defined for non-negative values, we have \(x-2 \geq 0\) or \(x \geq 2\)

Step 6 ：So, the domain of \(f \circ g\) is \([2, 6) \cup (6, \infty)\)

Step 7 ：\(\boxed{(f \circ g)(x) = \frac{1}{\sqrt{x-2}-2}}\)

Step 8 ：\(\boxed{domain \ f \circ g(x) = [2, 6) \cup (6, \infty)}\)