1/3x³+(m-2)x² +(m-2)x+3m+1. Tìm m để HSĐB trên (-vô cùng;1). Giúp em câu đó với ạ, em ko cô lập được m

Đáp án:

Lời giải:

\[\begin{array}{l} y = \frac{1}{3}{x^3} + \left( {m - 2} \right){x^2} + \left( {m - 2} \right)x + 3m + 1\\ \Rightarrow y' = {x^2} + 2\left( {m - 2} \right)x + m - 2 = 0\\ co\,\,\Delta ' = {\left( {m - 2} \right)^2} - m + 2 = \left( {m - 2} \right)\left( {m - 3} \right).\\ TH1:\,\,HS\,\,\,DB\,\,tren\,\,\,R\\ \Rightarrow \Delta ' \le 0\\ \Leftrightarrow \left( {m - 2} \right)\left( {m - 3} \right) \le 0\\ \Leftrightarrow 2 \le m \le 3.\\ TH2:\,\,y' = 0\,\,co\,\,2\,\,nghiem\,\,pb\,\,{x_1};\,\,{x_2}\\ \Rightarrow \Delta ' > 0\\ \Leftrightarrow \left( {m - 2} \right)\left( {m - 3} \right) > 0\\ \Leftrightarrow \left[ \begin{array}{l} m > 3\\ m < 2 \end{array} \right..\\ Ap\,\,dung\,\,\,he\,\,thuc\,\,Vi - et\,\,\,ta\,\,co:\\ \left\{ \begin{array}{l} {x_1} + {x_2} = - 2\left( {m - 2} \right)\\ {x_1}{x_2} = m - 2 \end{array} \right.\\ Bang\,\,xet\,\,dau:\\ \,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \infty \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{x_1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{x_2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \infty \\ f'\left( x \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \\ \Rightarrow Hs\,\,\,DB\,\,tren\,\,\,\left( { - \infty ;\,\,\,1} \right)\\ \Leftrightarrow 1 \le {x_1} < {x_2}\\ \Leftrightarrow \left\{ \begin{array}{l} {x_1} + {x_2} > 2\\ \left( {{x_1} - 1} \right)\left( {{x_2} - 1} \right) \ge 0 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} {x_1} + {x_2} > 2\\ {x_1}{x_2} - \left( {{x_1} + {x_2}} \right) + 1 \ge 0 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} - 2\left( {m - 2} \right) > 2\\ m - 2 + 2\left( {m - 2} \right) + 1 \ge 0 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} m - 2 < - 2\\ 3m \ge 5 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} m < 0\\ m \ge \frac{5}{3} \end{array} \right..\\ \Rightarrow k\,\,co\,\,m\,\,thoa\,\,man.\\ Vay\,\,\ 2 \leq m \leq 3\,\,\,thoa\,\,man\,\,bai\,\,toan. \end{array}\]