\(f\left(x\right)=\dfrac{\sqrt{3x-3}}{\sqrt{3}}+\sqrt{6-3x}\le\sqrt{\left(\dfrac{1}{3}+1\right)\left(3x-3+6-3x\right)}=2\Rightarrow max=2\Leftrightarrow x=\dfrac{5}{4}\)
\(f\left(x\right)=\sqrt{x-1}-1+\sqrt{6-3x}+1=\dfrac{x-2}{\sqrt{x-1}+1}+\sqrt{3\left(2-x\right)}+1=\left(2-x\right)\left[\sqrt{3}-\dfrac{1}{\sqrt{x-1}+1}\right]+1\ge1\Rightarrow min=1\Leftrightarrow x=2\)