For the real-valued functions $f(x)=x^{2}+1$ and $g(x)=\sqrt{x-4}$, find the composition $f \cdot g$ and specify its domain using interval notation.
\[
(f \cdot g)(x)=\mathbb{1}
\]

Domain of $f \cdot g: \square$
\begin{tabular}{|c|c|c|}
\hline 믐 & $\square^{\square}$ & $\sqrt{\square}$ \\
\hline 미미 & (ㅁ, ㅁ) & {$[0,0]$} \\
\hline 밈 & $(ㅁ, ㅁ]]$ & {$[\square, \square)$} \\
\hline$\varnothing$ & $\infty$ & $-\infty$ \\
\hline$x$ & & 5 \\
\hline
\end{tabular}

Step 1 ：Given the real-valued functions \(f(x)=x^{2}+1\) and \(g(x)=\sqrt{x-4}\), we are asked to find the composition \(f \cdot g\) and specify its domain using interval notation.

Step 2 ：The composition of two functions, \(f(g(x))\), means that we substitute \(g(x)\) into \(f(x)\). In this case, we need to substitute \(g(x) = \sqrt{x-4}\) into \(f(x) = x^2 + 1\).

Step 3 ：After substituting, we get \(f(g(x)) = (\sqrt{x-4})^2 + 1 = x - 4 + 1 = x - 3\). So, the composition of the functions is \((f \cdot g)(x) = x - 3\).

Step 4 ：The domain of a function is the set of all possible input values (x-values) which will produce a valid output from a particular function. For the function \(g(x) = \sqrt{x-4}\), the domain is all x-values that are greater than or equal to 4, because we can't take the square root of a negative number.

Step 5 ：Therefore, the domain of the composition will be the same as the domain of \(g(x)\), which is \([4, \infty)\).

Step 6 ：Final Answer: The composition of the functions is \((f \cdot g)(x) = \boxed{x - 3}\) and its domain is \(\boxed{[4, \infty)}\).