Example 11 Find the range of the function $y=x+\sqrt{x^{2}-3 x+2}$.

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From the given, we have $\sqrt{x^{2}-3 x+2}=y-x \geqslant 0$, so
$$
x^{2}-3 x+2=y^{2}-2 x y+x^{2},
$$

which simplifies to
$$
(2 y-3) x=y^{2}-2.
$$

From the above equation, we know $y \neq \frac{3}{2}$, and $x=\frac{y^{2}-2}{2 y-3}$. By $y \geqslant x=\frac{y^{2}-2}{2 y-3}$, we get
$$
\frac{y^{2}-3 y+2}{2 y-3} \geqslant 0, \frac{(y-1)(y-2)}{y-\frac{3}{2}} \geqslant 0.
$$

Therefore, $1 \leqslant y<\frac{3}{2}$ or $y \geqslant 2$.
For any $y_{0} \in[2,+\infty)$, let $x_{0}=\frac{y_{0}^{2}-2}{2 y_{0}-3}$, then
$$
x_{0}-2=\frac{y_{0}^{2}-2}{2 y_{0}-3}-2=\frac{\left(y_{0}-2\right)^{2}}{2 y_{0}-3} \geqslant 0.
$$

Thus, $x_{0} \geqslant 2$, so $x_{0}^{2}-3 x_{0}+2 \geqslant 0$, and $y_{0}=x_{0}+\sqrt{x_{0}^{2}-3 x_{0}+2}$.
For any $y_{0} \in\left[1, \frac{3}{2}\right)$, let $x_{0}=\frac{y_{0}^{2}-2}{2 y_{0}-3}$, then
$$
x_{0}-1=\frac{y_{0}^{2}-2}{2 y_{0}-3}-1=\frac{\left(y_{0}-1\right)^{2}}{2 y_{0}-3} \leqslant 0.
$$

Thus, $x_{0} \leqslant 1$, so $x_{0}^{2}-3 x_{0}+2 \geqslant 0$, and $y_{0}=x_{0}+\sqrt{x_{0}^{2}-3 x_{0}+2}$.
In summary, the range of the function is $\left[1, \frac{3}{2}\right) \cup[2,+\infty)$.
Note: We first determined the range of $y$ as $\left[1, \frac{3}{2}\right) \cup[2,+\infty)$. Is this the range of the function? The second part shows that for any $y_{0} \in \left[1, \frac{3}{2}\right) \cup[2,+\infty)$, there always exists an $x_{0}$ such that $y_{0}=x_{0}+\sqrt{x_{0}^{2}-3 x_{0}+2}$, which proves that the range of the function is $\left[1, \frac{3}{2}\right) \cup [2,+\infty)$.