2 câu trả lời
Hàm số \(y=\sqrt{x}-x\)
ĐKXĐ: \(x\ge 0\)
Ta có: \(y'=\dfrac{1}{2\sqrt{x}}-1\)
Để hàm số đồng biến khi đó \(y'\ge 0\)
\(\Leftrightarrow \dfrac{1}{2\sqrt{x}}-1\ge 0\)
\(\dfrac{1}{2\sqrt{x}}\ge 1\)
\(\Leftrightarrow 2\sqrt{x}\le 1\)
\(\Leftrightarrow x\le \dfrac{1}{4}\)
Kết hợp với ĐKXĐ vậy hàm số đồng biến \([0; \dfrac{1}{4}]\)
\(\begin{array}{l} y = \sqrt x - x\\ D = \left[ {0; + \infty } \right)\\ \Rightarrow y' = \frac{1}{{2\sqrt x }} - 1\\ \Rightarrow y' = 0\\ \Leftrightarrow \frac{1}{{2\sqrt x }} - 1 = 0\\ \Leftrightarrow 2\sqrt x = 1\\ \Leftrightarrow \sqrt x = \frac{1}{2}\\ \Leftrightarrow x = \frac{1}{4}\\ Bang\,\,\,xet\,\,dau:\\ x\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{1}{4}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \infty \\ f'\left( x \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \,\,\,\,\\ \Rightarrow Hs\,\,\,DB\,\,tren\,\,\,\left( {0;\,\frac{1}{4}} \right)\,\,\,va\,\,ham\,\,NB\,\,\,tren\,\,\,\left( {\frac{1}{4}; + \infty } \right). \end{array}\)
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