Cho mình hỏi cực trị của y=√2x+1 + √10_x bao nhiêu ạ

$\begin{array}{l} y = \sqrt {2x + 1} + \sqrt {10 - x} \\ TXD:\,\,\,D = \left[ { - \frac{1}{2};\,\,\,10} \right]\\ y' = \frac{1}{{\sqrt {2x + 1} }} - \frac{1}{{2\sqrt {10 - x} }}\\ \Rightarrow y' = 0\\ \Leftrightarrow \frac{1}{{\sqrt {2x + 1} }} - \frac{1}{{2\sqrt {10 - x} }} = 0\\ \Leftrightarrow \sqrt {2x + 1} = 2\sqrt {10 - x} \\ \Leftrightarrow 2x + 1 = 4\left( {10 - x} \right)\\ \Leftrightarrow 2x + 1 = 40 - 4x\\ \Leftrightarrow 6x = 39\\ \Leftrightarrow x = \frac{{13}}{2}\,\,\left( {tm} \right) \Rightarrow y = \frac{{3\sqrt {14} }}{2}.\\ \Rightarrow diem\,\,\,cuc\,\,\,tri\,\,\,cua\,\,\,dths\,\,\,la:\,\,\,\,\,\left( {\frac{{13}}{2};\,\,\,\frac{{3\sqrt {14} }}{2}} \right). \end{array}$