1. Find the range of the function $f(x)=2 x+3+\sqrt{-2 x^{2}+12 x-14}$.
1. Let $f(x)=m$, then $-2 x+m-3=\sqrt{-2 x^{2}+12 x-14}$.

Let $\Gamma_{1}: y=-2 x+m-3, \Gamma_{2}: y=\sqrt{-2 x^{2}}+12-14$, i.e., $\frac{y^{2}}{4}+\frac{(x-3)^{2}}{2}=1(y \geqslant 0)$.
Draw the graph (omitted), $\Gamma_{1}$ is a family of lines with a slope of -2, $\Gamma_{2}$ is the upper part of an ellipse centered at $(3,0)$ with the major axis parallel to the $y$-axis. From the graph, $\Gamma_{1}$ passes through the point $A(3-\sqrt{2}, 0)$, the minimum value of $m$ is $9-2 \sqrt{2}$, and $\Gamma_{1}$ is tangent to $\Gamma_{2}$ in the first quadrant when $m$ is maximum. By $-2 x+m-3=\sqrt{-2 x^{2}+12 x-14}$, we have a repeated root, i.e., $6 x^{2}-4 m x+m^{2}-6 m+23=0$ has a repeated root, then $\Delta=0$, i.e., $m^{2}-18 m+69=0$, solving for $m$ gives $m=9+2 \sqrt{3}$ (discard the negative value). Therefore, the range of $f(x)$ is $[9-2 \sqrt{2}, 9+2 \sqrt{3}]$.