Example 8 Find the range of the function $y=x+\sqrt{x^{2}-3 x+2}$.
(2001 National High School Mathematics Competition)
Analysis: The difficulty of this problem lies in how to handle the radical in the expression.
Solution: The domain of the function is clearly $\{x \mid x \geqslant 2$ or $x \leqslant 1, x \in \mathbf{R}\}$.
From $y=x+\sqrt{x^{2}-3 x+2}$, we get: $\sqrt{x^{2}-3 x+2}=y-x \geqslant 0$. Squaring both sides and rearranging gives: $(2 y-3) x=y^{2}-2$. Thus, $y \neq \frac{3}{2}$: From $x=\frac{y^{2}-2}{2 y-3}$ and $y-x \geqslant 0$, we get: $y-\frac{y^{2}-2}{2 y-3} \geqslant 0 \Rightarrow 1 \leqslant y<\frac{3}{2}$ or $y \geqslant 2$.

For any $y \geqslant 2$, let $x=\frac{y^{2}-2}{2 y-3}$, it is easy to see that $x \geqslant 2$ belongs to the domain of the function. For any $1 \leqslant y<\frac{3}{2}$, similarly let $x=\frac{y^{2}-2}{2 y-3}$, it is easy to see that $x \leqslant 1$ belongs to the domain of the function.
Therefore, the range of the function is $\left[1, \frac{3}{2}\right) \cup[2,+\infty)$.