Evaluating

A radical function is evaluated like any other: replace x with the input. Take $f\left(x\right) = \sqrt{3x+1}-2$ and the input 5. Using the rule writes $\sqrt{3\left(5\right)+1}-2$.

Then work from the inside out. The inside is $3\left(5\right)+1$, which is 16, so the next line is $\sqrt{16}-2$. Only now take the root: the square root of 16 is 4, which leaves $4-2$, and f(5) is 2.

The root waits for the inside because it covers all of it. It does not split over a sum: $\sqrt{9+16}$ is $\sqrt{25}$, which is 5, not 3 plus 4. An input outside the domain shows up at once: f(-1) puts $\sqrt{-2}$ on the board, which has no real value.