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