Example 7 Find the minimum value of the function $f(x)=\sqrt{2 x^{2}-3 x+4}+\sqrt{x^{2}-2 x}$.
Solve: First, consider the domain of $f(x)$. From $2 x^{2}-3 x+4 \geqslant 0, x^{2}-2 x \geqslant 0$, we have $x \in(-\infty, 0] \cup$ $[2,+\infty)$

It is easy to see that $y=2 x^{2}-3 x+4$ and $y=x^{2}-2 x$ are both decreasing on $(-\infty, 0]$ and increasing on $[2,+\infty)$. Therefore, $f(x)$ is decreasing on $(-\infty, 0]$ and increasing on $[2,+\infty)$. Hence,
$$
f_{\text {min }}(x)=\min \{f(0), f(2)\}=\min \{2, \sqrt{6}\}=2
$$

Thus, the minimum value of $f(x)$ is 2.