3. Find the minimum value of the function $f(x)=\sqrt{\frac{1}{4} x^{4}-2 x^{2}-4 x+13}+\frac{1}{2} x^{2}+\frac{1}{2}$.
3. Given $f(x)=\sqrt{\left(\frac{1}{2} x^{2}-3\right)^{2}+(x-2)^{2}}+\sqrt{\left(\frac{1}{2} x^{2}-\frac{1}{2}\right)+(x-0)^{2}}$, we can let point $P\left(\frac{1}{2} x^{2}, x\right), M(3, 2), N\left(\frac{1}{2}, 0\right)$, then $f(x)=|P M|+|P N|$.

Point $P\left(\frac{1}{2} x^{2}, x\right)$ is any point on the parabola $y^{2}=2 x$, and point $M(3,2)$ is inside this parabola, while point $N\left(\frac{1}{2}, 0\right)$ is exactly the focus of the parabola. The directrix of the parabola is $l: x=-\frac{1}{2}$. By the definition of a parabola, $|P N|=|P Q|$, where $|P Q|$ is the distance from point $P$ to the directrix $l$. Therefore, $f(x)=|P M|+|P N|=|P M|+|P Q| \geqslant\left|M Q_{0}\right|$, where $Q_{0}$ is the foot of the perpendicular from $M$ to $l$. Suppose $M Q_{0}$ intersects the parabola $y^{2}=2 x$ at point $P_{0}$, then $Q_{0}\left(-\frac{1}{2}, 2\right), P_{0}(2,2)$. Thus, $f(x)$ has its minimum value when and only when $P$ is $(2, 2)$, i.e., $x=2, f(x)$ has the minimum value of $M Q_{0}$. And $\left|M Q_{0}\right|=\frac{7}{2}$, so the minimum value of $f(x)$ is $\frac{7}{2}$.