y=x ³-mx ²-(m-6)x+3 đồng biến trên (0;4) 03/10/2021 Bởi Aubrey y=x ³-mx ²-(m-6)x+3 đồng biến trên (0;4)
$$\eqalign{ & y = {x^3} – m{x^2} – \left( {m – 6} \right)x + 3\,\,DB/\left( {0;4} \right) \cr & y’ = 3{x^2} – 2mx – m + 6 > 0\,\,\forall x \in \left( {0;4} \right) \cr & \Leftrightarrow 3{x^2} + 6 > m\left( {2x + 1} \right)\,\,\forall x \in \left( {0;4} \right) \cr & \Leftrightarrow g\left( x \right) = {{3{x^2} + 6} \over {2x + 1}} > m\,\,\forall x \in \left( {0;4} \right) \cr & \Rightarrow m \le \mathop {\min }\limits_{\left[ {0;4} \right]} g\left( x \right) \cr & g’\left( x \right) = {{6x\left( {2x + 1} \right) – 2\left( {3{x^2} + 6} \right)} \over {{{\left( {2x + 1} \right)}^2}}} = {{6{x^2} + 6x – 12} \over {{{\left( {2x + 1} \right)}^2}}} = 0 \cr & \Leftrightarrow \left[ \matrix{ x = 1 \hfill \cr x = – 2 \hfill \cr} \right. \cr & g\left( 0 \right) = 6;\,\,g\left( 4 \right) = 6;\,\,g\left( 1 \right) = 3 \cr & \Rightarrow \mathop {\min }\limits_{\left[ {0;4} \right]} g\left( x \right) = 3 \Rightarrow m \le 3 \cr} $$ Bình luận
$$\eqalign{
& y = {x^3} – m{x^2} – \left( {m – 6} \right)x + 3\,\,DB/\left( {0;4} \right) \cr
& y’ = 3{x^2} – 2mx – m + 6 > 0\,\,\forall x \in \left( {0;4} \right) \cr
& \Leftrightarrow 3{x^2} + 6 > m\left( {2x + 1} \right)\,\,\forall x \in \left( {0;4} \right) \cr
& \Leftrightarrow g\left( x \right) = {{3{x^2} + 6} \over {2x + 1}} > m\,\,\forall x \in \left( {0;4} \right) \cr
& \Rightarrow m \le \mathop {\min }\limits_{\left[ {0;4} \right]} g\left( x \right) \cr
& g’\left( x \right) = {{6x\left( {2x + 1} \right) – 2\left( {3{x^2} + 6} \right)} \over {{{\left( {2x + 1} \right)}^2}}} = {{6{x^2} + 6x – 12} \over {{{\left( {2x + 1} \right)}^2}}} = 0 \cr
& \Leftrightarrow \left[ \matrix{
x = 1 \hfill \cr
x = – 2 \hfill \cr} \right. \cr
& g\left( 0 \right) = 6;\,\,g\left( 4 \right) = 6;\,\,g\left( 1 \right) = 3 \cr
& \Rightarrow \mathop {\min }\limits_{\left[ {0;4} \right]} g\left( x \right) = 3 \Rightarrow m \le 3 \cr} $$