3. The minimum value of the function $f(x)=\sqrt{2 x^{2}-x+3}+\sqrt{x^{2}-x}$ is $\qquad$ $ـ$
3. $\sqrt{3}$.

From $\left\{\begin{array}{l}2 x^{2}-x+3 \geqslant 0, \\ x^{2}-x \geqslant 0,\end{array}\right.$ we get the domain of the function $f(x)$ as $(-\infty, 0] \cup[1,+\infty)$.
The function $f(x)$ is monotonically decreasing on $(-\infty, 0]$ and monotonically increasing on $[1,+\infty)$, hence
$$
f(x)_{\text {min }}=\min \{f(0), f(1)\}=\sqrt{3} .
$$