2. Find the minimum value of the function $f(x)=\sqrt{x^{2}+2 x+10}+\sqrt{x^{2}-2 x+10}$.
2. Transform the function to $f(x)=\sqrt{(x+1)^{2}+9}+\sqrt{(x-1)^{2}+9}>3+3=6$, from $\sqrt{x^{2}+2 x+10}+$ $\sqrt{x^{2}-2 x+10}=f(x)$, (1) Rationalizing the numerator gives $\frac{4 x}{\sqrt{x^{2}+2 x+10}-\sqrt{x^{2}-2 x+10}}=f(x)$, i.e., $\sqrt{x^{2}+2 x+10}-\sqrt{x^{2}-2 x+10}=\frac{f(x)}{4 x}$. (2) By (1) $\pm$ (2) we get $2 \sqrt{x^{2}+2 x+10}=f(x)+\frac{4 x}{f(x)}$, (3) $2 \sqrt{x^{2}-2 x+10}=f(x)-\frac{4 x}{f(x)}$, (4) and by (3) ${ }^{2}+$ (4) $)^{2}$, we get $x^{2}=\frac{f^{2}(x)\left[f^{2}(x)-40\right]}{4\left[f^{2}(x)-4\right]} \geqslant 0$. Thus, $f^{2}(x) \geqslant$ 40 or $f^{2}(x) \leqslant 0$ (discard), so $f(x) \geqslant 2 \sqrt{10}$, i.e., when $x=0$, the minimum value of $f(x)$ is $2 \sqrt{10}$.