Cho mình hỏi cực trị của y=√2x+1 + √10_x bao nhiêu ạ 27/09/2021 Bởi Clara Cho mình hỏi cực trị của y=√2x+1 + √10_x bao nhiêu ạ
\[\begin{array}{l} y = \sqrt {2x + 1} + \sqrt {10 – x} \\ TXD:\,\,\,D = \left[ { – \frac{1}{2};\,\,\,10} \right]\\ y’ = \frac{1}{{\sqrt {2x + 1} }} – \frac{1}{{2\sqrt {10 – x} }}\\ \Rightarrow y’ = 0\\ \Leftrightarrow \frac{1}{{\sqrt {2x + 1} }} – \frac{1}{{2\sqrt {10 – x} }} = 0\\ \Leftrightarrow \sqrt {2x + 1} = 2\sqrt {10 – x} \\ \Leftrightarrow 2x + 1 = 4\left( {10 – x} \right)\\ \Leftrightarrow 2x + 1 = 40 – 4x\\ \Leftrightarrow 6x = 39\\ \Leftrightarrow x = \frac{{13}}{2}\,\,\left( {tm} \right) \Rightarrow y = \frac{{3\sqrt {14} }}{2}.\\ \Rightarrow diem\,\,\,cuc\,\,\,tri\,\,\,cua\,\,\,dths\,\,\,la:\,\,\,\,\,\left( {\frac{{13}}{2};\,\,\,\frac{{3\sqrt {14} }}{2}} \right). \end{array}\] Bình luận
\[\begin{array}{l}
y = \sqrt {2x + 1} + \sqrt {10 – x} \\
TXD:\,\,\,D = \left[ { – \frac{1}{2};\,\,\,10} \right]\\
y’ = \frac{1}{{\sqrt {2x + 1} }} – \frac{1}{{2\sqrt {10 – x} }}\\
\Rightarrow y’ = 0\\
\Leftrightarrow \frac{1}{{\sqrt {2x + 1} }} – \frac{1}{{2\sqrt {10 – x} }} = 0\\
\Leftrightarrow \sqrt {2x + 1} = 2\sqrt {10 – x} \\
\Leftrightarrow 2x + 1 = 4\left( {10 – x} \right)\\
\Leftrightarrow 2x + 1 = 40 – 4x\\
\Leftrightarrow 6x = 39\\
\Leftrightarrow x = \frac{{13}}{2}\,\,\left( {tm} \right) \Rightarrow y = \frac{{3\sqrt {14} }}{2}.\\
\Rightarrow diem\,\,\,cuc\,\,\,tri\,\,\,cua\,\,\,dths\,\,\,la:\,\,\,\,\,\left( {\frac{{13}}{2};\,\,\,\frac{{3\sqrt {14} }}{2}} \right).
\end{array}\]