Question 86: Planar vectors $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$ satisfy $|\vec{a}|=3,|\vec{b}|=4,|\vec{a}-\vec{b}|=2 \sqrt{3}$, then the minimum value of the function $\mathrm{f}(\mathrm{x})=\left|\mathrm{x} \overrightarrow{\mathrm{a}}+\frac{1}{\mathrm{x}} \overrightarrow{\mathrm{b}}\right|$ $(x \neq 0)$ is $\qquad$ -
Hence, the minimum value of $f(x)$ is $\sqrt{37}$ when $x=\frac{2 \sqrt{3}}{3}$, equality holds.