3. The function $f(x)=2 \sqrt{x-1}+\sqrt{6-2 x}$ achieves its maximum value when $x=$
Solution: $f(x)=\sqrt{2} \sqrt{2 x-2}+\sqrt{6-2 x} \leq \sqrt{3} \sqrt{4}=2 \sqrt{3}$, when $\frac{2 x-2}{2}=\frac{6-2 x}{1}, x=\frac{7}{3}$, the equality holds.