Điều kiện \(\left\{\begin{matrix} x-2\geq 0\\ 4-x\geq 0 \end{matrix}\right.\Leftrightarrow 2\leq x\leq 4\) ⇒ TXĐ: D = [2;4] \(f'(x)=\frac{1}{2\sqrt{x-1}}-\frac{1}{2\sqrt{4-x}}=\frac{\sqrt{4-x}-\sqrt{x-2}}{2\sqrt{x-2}.\sqrt{4-x}}\) \(f'(x)=0\Leftrightarrow \sqrt{4-x}=\sqrt{x-2}\Leftrightarrow x=3\in (2;4)\) \(f(2)=\sqrt{2};f(3)=2;f(4)=\sqrt{2}\) Vậy \(\underset{[2;4]}{max}f(x)=2\) khi x = 3 \(\underset{[2;4]}{min}f(x)=\sqrt{2}\) khi x = 2 hoặc x = 4