For the functions $f(x)=\sqrt{x-1}$ and $g(x)=3 x$ find the following. (a) $(f \circ g)(x)$ and its domain (b) $(g \circ f)(x)$ and its domain

Step 1 ：Let's start with the given functions: \(f(x) = \sqrt{x - 1}\) and \(g(x) = 3x\).

Step 2 ：For part (a), we need to find the composition of \(f\) and \(g\), denoted as \((f \circ g)(x)\). This means we substitute \(g(x)\) into \(f(x)\).

Step 3 ：So, \((f \circ g)(x) = f(g(x)) = \sqrt{3x - 1}\).

Step 4 ：The domain of this function is the set of all x-values for which this function is defined. In this case, the square root function is defined for all values greater than or equal to 1. So, the domain of \((f \circ g)(x)\) is \(x \geq \frac{1}{3}\).

Step 5 ：For part (b), we need to find the composition of \(g\) and \(f\), denoted as \((g \circ f)(x)\). This means we substitute \(f(x)\) into \(g(x)\).

Step 6 ：So, \((g \circ f)(x) = g(f(x)) = 3\sqrt{x - 1}\).

Step 7 ：The domain of this function is the set of all x-values for which this function is defined. In this case, the square root function is defined for all values greater than or equal to 1. So, the domain of \((g \circ f)(x)\) is \(x \geq 1\).

Step 8 ：Final Answer: \(\boxed{(a) (f \circ g)(x) = \sqrt{3x - 1}, \text{ with domain } x \geq \frac{1}{3}. (b) (g \circ f)(x) = 3\sqrt{x - 1}, \text{ with domain } x \geq 1.}\)