11. (20 points) Find the range of the function
$$
f(x)=\sqrt{x^{2}-4 x+3}+\sqrt{x^{2}+4 x+5}
$$
11. It is known that the domain of the function $f(x)$ is $\{x \leqslant 1$ or $x \geqslant 3\}$.
When $x \geqslant 3$, $f(x)$ is an increasing function,
$$
f(x) \geqslant f(3)=\sqrt{26} \text {. }
$$

When $x \leqslant 1$, let $t=1-x \geqslant 0$, then
$$
\begin{array}{l}
f(x)=\sqrt{t(t+2)}+\sqrt{(3-t)^{2}+1} \\
\geqslant t+\sqrt{(3-t)^{2}+1} \\
=\sqrt{t^{2}+0^{2}}+\sqrt{(3-t)^{2}+1^{2}} \\
\geqslant \sqrt{3^{2}+1^{2}}=\sqrt{10}=f(1) .
\end{array}
$$

Therefore, the range of the function $f(x)$ is $[\sqrt{10},+\infty)$.