Toán Hàm số y= cănx-2 + căn4-x đồng biến trên khoảng nào? 05/10/2021 By Adalyn Hàm số y= cănx-2 + căn4-x đồng biến trên khoảng nào?
\[\begin{array}{l} y = \sqrt {x – 2} + \sqrt {4 – x} \\ TXD:\,\,\,D = \left[ {2;\,\,4} \right]\\ \Rightarrow y’ = \frac{1}{{2\sqrt {x – 2} }} – \frac{1}{{2\sqrt {4 – x} }}\\ HS\,\,\,DB \Leftrightarrow y’ \ge 0\\ \Leftrightarrow \frac{1}{{2\sqrt {x – 2} }} – \frac{1}{{2\sqrt {4 – x} }} \ge 0\\ \Leftrightarrow \frac{{\sqrt {4 – x} – \sqrt {x – 2} }}{{2\sqrt {x – 2} .\sqrt {4 – x} }} \ge 0\\ \Leftrightarrow \sqrt {4 – x} – \sqrt {x – 2} \ge 0\\ \Leftrightarrow \sqrt {4 – x} \ge \sqrt {x – 2} \\ \Leftrightarrow 4 – x \ge x – 2\\ \Leftrightarrow 2x \le 6\\ \Leftrightarrow x \le 3.\\ \Rightarrow Ket\,\,hop\,\,voi\,\,\,DK:\,\,\,2 \le x \le 4\\ \Rightarrow Hs\,\,\,DB\,\,\,tren\,\,\,\left[ {2;\,\,3} \right]. \end{array}\] Trả lời
\[\begin{array}{l}
y = \sqrt {x – 2} + \sqrt {4 – x} \\
TXD:\,\,\,D = \left[ {2;\,\,4} \right]\\
\Rightarrow y’ = \frac{1}{{2\sqrt {x – 2} }} – \frac{1}{{2\sqrt {4 – x} }}\\
HS\,\,\,DB \Leftrightarrow y’ \ge 0\\
\Leftrightarrow \frac{1}{{2\sqrt {x – 2} }} – \frac{1}{{2\sqrt {4 – x} }} \ge 0\\
\Leftrightarrow \frac{{\sqrt {4 – x} – \sqrt {x – 2} }}{{2\sqrt {x – 2} .\sqrt {4 – x} }} \ge 0\\
\Leftrightarrow \sqrt {4 – x} – \sqrt {x – 2} \ge 0\\
\Leftrightarrow \sqrt {4 – x} \ge \sqrt {x – 2} \\
\Leftrightarrow 4 – x \ge x – 2\\
\Leftrightarrow 2x \le 6\\
\Leftrightarrow x \le 3.\\
\Rightarrow Ket\,\,hop\,\,voi\,\,\,DK:\,\,\,2 \le x \le 4\\
\Rightarrow Hs\,\,\,DB\,\,\,tren\,\,\,\left[ {2;\,\,3} \right].
\end{array}\]