Consider the functions $f(x)=2 x-8$ and $g(x)=\sqrt{x+6}$. Determine each of the following. \[ f \circ g(x)= \] Give the domain of $f \circ g(x)$. \[ g \circ f(x)= \] Give the domain of $g \circ f(x)$.

Step 1 ：\(f \circ g(x) = f(g(x)) = 2g(x) - 8\)

Step 2 ：Substitute \(g(x)\) into the equation: \(f \circ g(x) = 2\sqrt{x+6} - 8\)

Step 3 ：Find the domain of \(f \circ g(x)\), the value inside the square root must be greater than or equal to zero: \(x+6 \geq 0\)

Step 4 ：Solving for x gives \(x \geq -6\). So, the domain of \(f \circ g(x)\) is \([-6, \infty)\)

Step 5 ：\(g \circ f(x) = g(f(x)) = \sqrt{f(x)+6}\)

Step 6 ：Substitute \(f(x)\) into the equation: \(g \circ f(x) = \sqrt{2x-8+6} = \sqrt{2x-2}\)

Step 7 ：Find the domain of \(g \circ f(x)\), the value inside the square root must be greater than or equal to zero: \(2x-2 \geq 0\)

Step 8 ：Solving for x gives \(x \geq 1\). So, the domain of \(g \circ f(x)\) is \([1, \infty)\)

Step 9 ：\(\boxed{f \circ g(x) = 2\sqrt{x+6} - 8}\) with domain \(\boxed{[-6, \infty)}\)

Step 10 ：\(\boxed{g \circ f(x) = \sqrt{2x-2}}\) with domain \(\boxed{[1, \infty)}\)