3. Find the minimum value of the function $f(x)=\sqrt{x^{2}+6 x+10}+\sqrt{x^{2}-6 x+10}$.
3. Let $f(x)=\sqrt{(x+3)^{2}+(1-0)^{2}}+\sqrt{(x-3)^{2}+(1-0)^{2}}$, and let points $P(x, 1), F_{1}(3,0), F_{2}(-3, 0)$ form a triangle, then $f(x)=\left|P F_{1}\right|+\left|P F_{2}\right|>\left|F_{1} F_{2}\right|=6$. Let $b=1$, then we have $\sqrt{(x+3)^{2}+(b-0)^{2}}+\sqrt{(x-3)^{2}+(b-n)^{2}}=f(x)$, this equation represents the locus of points $(x, b)$ whose sum of distances to two fixed points $F_{1}(3,0)$, $F_{2}(-3,0)$ is $f(x)\left(>\mid F_{1} F_{2}\right.$). This represents an ellipse, whose equation is $\frac{x^{2}}{\frac{f^{2}(x)}{4}}+$ $\frac{b^{2}}{\frac{f^{2}(x)}{4}-9}=1$. Substituting $b=1$ yields $\frac{4 x^{2}}{f^{2}(x)}+\frac{4}{f^{2}(x)-36}=1$, then $x^{2}=\frac{f^{2}(x)\left[f^{2}(x)-40\right]}{4\left[f^{2}(x)-36\right]} \geqslant 0$. Therefore, $f^{2}(x) \geqslant 40$ (discard $f^{2}(x) \leqslant 36$), hence when $x=0$, $f(x)_{\text {min }}=2 \sqrt{10}$.