cho hàm số $x^{3}$+(m+1) $x^{2}$ +3(m+1)x+1 xác định m để hàm số đồng biến (0 đến dương vô cùng)

Đáp án:

\(m \ge - 1\)

Giải thích các bước giải:

\(\eqalign{ & y = {x^3} + \left( {m + 1} \right){x^2} + 3\left( {m + 1} \right)x + 1 \cr & Ta\,\,co:\,\,y' = 3{x^2} + 2\left( {m + 1} \right)x + 3\left( {m + 1} \right) \cr & De\,ham\,\,so\,\,DB/\left( {0; + \infty } \right) \cr & \Rightarrow y' \ge 0\,\,\forall x \in \left( {0; + \infty } \right) \cr & \Rightarrow 3{x^2} + 2\left( {m + 1} \right)x + 3\left( {m + 1} \right) \ge 0\,\,\forall x \in \left( {0; + \infty } \right) \cr & \Leftrightarrow 3{x^2} + 2x + 3 + \left( {2x + 3} \right)m \ge 0\,\,\forall x \in \left( {0; + \infty } \right) \cr & \Leftrightarrow \left( {2x + 3} \right)m \ge - 3{x^2} - 2x - 3 \cr & \Leftrightarrow m \ge {{ - 3{x^2} - 2x - 3} \over {2x + 3}} = f\left( x \right)\,\,\,\forall x \in \left( {0; + \infty } \right)\,\,\left( {2x + 3 > 0\,\,\forall x \in \left( {0; + \infty } \right)} \right) \cr & \Rightarrow m \ge \mathop {\max }\limits_{\left[ {0; + \infty } \right)} f\left( x \right) \cr & f'\left( x \right) = {{\left( { - 6x - 2} \right)\left( {2x + 3} \right) - 2\left( { - 3{x^2} - 2x - 3} \right)} \over {{{\left( {2x + 3} \right)}^2}}} \cr & f'\left( x \right) = {{ - 12{x^2} - 18x - 4x - 6 + 6{x^2} + 4x + 6} \over {{{\left( {2x + 3} \right)}^2}}} \cr & f'\left( x \right) = {{ - 6{x^2} - 18x} \over {{{\left( {2x + 3} \right)}^2}}} = 0 \Leftrightarrow \left[ \matrix{ x = 0 \hfill \cr x = - 3 \hfill \cr} \right. \cr & Lap\,\,BBT \Rightarrow \mathop {\max }\limits_{\left[ {0; + \infty } \right)} f\left( x \right) = f\left( 0 \right) = - 1 \cr & Vay\,\,m \ge - 1 \cr} \)