For the real-valued functions $f(x)=x^{2}+1$ and $g(x)=\sqrt{x-6}$, find the composition $f \circ g$ and specify its domain using interval notation. \[ (f \circ g)(x)=\mathbb{1} \] Dornain of $f \circ \mathrm{g}: \square$ \begin{tabular}{|c|c|c|} \hline$\frac{\text { 믐 }}{}$ & $\square^{\square}$ & $\sqrt{\square}$ \\ \hline 미미 & $(\square, \square)$ & {$[\square, \square]$} \\ \hline DUD & $(\square, \square]$ & {$[\square, \square)$} \\ \hline$\varnothing$ & $\infty$ & $-\infty$ \\ \hline$x$ & & 5 \\ \hline \end{tabular}

Step 1 ：The composition of two functions, \(f(g(x))\), means that we substitute \(g(x)\) into \(f(x)\). In this case, we substitute \(g(x) = \sqrt{x-6}\) into \(f(x) = x^2 + 1\).

Step 2 ：The domain of the composition function is the set of all real numbers that \(g(x)\) can take such that \(f(g(x))\) is defined. Since \(g(x) = \sqrt{x-6}\), the domain of \(g(x)\) is \(x \geq 6\). This is because the square root of a negative number is not a real number.

Step 3 ：Therefore, the domain of \(f \circ g\) is also \(x \geq 6\).

Step 4 ：The composition \(f \circ g\) is \(x - 5\).

Step 5 ：\(\boxed{(f \circ g)(x)=x - 5}\)

Step 6 ：The domain of \(f \circ g\) is \([6, \infty)\).

Step 7 ：\(\boxed{\text{Domain of } f \circ g: [6, \infty)}\)