12. Let $x \in \mathbf{R}$, then the minimum value of the function $f(x)=|2 x-1|+|3 x-2|+|4 x-3|+|5 x-4|$ is
$f(x)=2\left|x-\frac{1}{2}\right|+3\left|x-\frac{2}{3}\right|+4\left|x-\frac{3}{4}\right|+5\left|x-\frac{4}{5}\right|$ consists of 15 terms, the middle term is $\left|x-\frac{3}{4}\right|$, so the minimum value of $f(x)$ is $f\left(\frac{3}{4}\right)=\frac{3}{2}-1+\frac{9}{4}-2+4-\frac{15}{4}=1$.