2exthe $\star \star$ Given the function $f(x)=\frac{\sin (\pi x)-\cos (\pi x)+2}{\sqrt{x}}\left(\frac{1}{4} \leqslant x \leqslant \frac{5}{4}\right)$, then the minimum value of $f(x)$ is $\qquad$ L.
Since on $\left[\frac{1}{4}, \frac{5}{4}\right]$, $\sin \pi\left(x-\frac{1}{4}\right) \geqslant 0$, we have
$$
f(x)=\frac{\sqrt{2} \sin \left(\pi x-\frac{\pi}{4}\right)+2}{\sqrt{x}} \geqslant \frac{2}{\sqrt{x}} \geqslant \frac{2}{\sqrt{\frac{5}{4}}}=\frac{4 \sqrt{5}}{5} \text {. }
$$

At $x=\frac{5}{4}$, $f(x)$ attains its minimum value $\frac{4 \sqrt{5}}{5}$.