Example 9 (2001 National High School Competition Question) The range of the function $y=x+\sqrt{x^{2}-3 x+2}$ is

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Example 9 (2001 National High School Competition Question) The range of the function $y=x+\sqrt{x^{2}-3 x+2}$ is
Solve: Fill in $\left[1, \frac{3}{2}\right) \cup[2,+\infty]$. Reason:
$$
y=x+\sqrt{x^{2}-3 x+2} \Rightarrow \sqrt{x^{2}-3 x+2}=y-x \geqslant 0 \text {. }
$$

Square both sides to get $(2 y-3) x=y^{2}-2$.
Thus, $y \neq \frac{3}{2}$ and $x=\frac{y^{2}-2}{2 y-3}$.
From $y-x=y-\frac{y^{2}-2}{2 y-3} \geqslant 0 \Rightarrow \frac{y^{2}-3 y+2}{2 y-3} \geqslant 0 \Rightarrow 1 \leqslant y<\frac{3}{2}$ or $y \geqslant 2$.
For any $y \geqslant 2$, let $x=\frac{y^{2}-2}{2 y-3}$, it is easy to see that $x \geqslant 2$. Thus, $x^{2}-3 x+2 \geqslant 0$ and $y=x+\sqrt{x^{2}-3 x+2}$.
For any $1 \leqslant y<\frac{3}{2}$, similarly let $x=\frac{y^{2}-2}{2 y-3}$, it is easy to see that $x \leqslant 1$. Thus, $x^{2}-3 x+2 \geqslant 0$ and $y=x+\sqrt{x^{2}-3 x+2}$. Therefore, the range of the function is $\left[1, \frac{3}{2}\right) \cup[2,+\infty)$.