16. Given the function $y=3 x+\sqrt{x^{2}-2 x}$, find the range of this function.
Let $u=x-1$, then $y=3 u+3+\sqrt{u^{2}-1}$, thus $|u| \geq 1$.
Set $\sqrt{u^{2}-1}=|u|-t \geq 0$, then $0 < t \leq 1$. When $u > 0$,
$$
y=\frac{3}{2}\left(t+\frac{1}{t}\right)+3+\frac{1}{2}\left(t+\frac{1}{t}\right)-t=t+\frac{2}{t}+3
$$

Since $0 < t \leq 1$, the function is monotonically decreasing, so $y \geq 1+2+3=6 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$ (points) When $u < 0$,
$$
y=-\frac{3}{2}\left(t+\frac{1}{t}\right)+3+\frac{1}{2}\left(t+\frac{1}{t}\right)-t=-2 t+\frac{1}{t}+3 \leq 3-2 \sqrt{2}
$$
(when and only when $t=\frac{\sqrt{2}}{2}$, i.e., $x=\frac{4-3 \sqrt{2}}{4}$, the equality holds), so the range of the function is $(-\infty, 3-2 \sqrt{2}] \cup[6,+\infty) \ldots \ldots$ (15 points)