Đáp án:
\[m \le - 1\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
y = \frac{1}{3}{x^3} + \left( {m - 1} \right){x^2} + \left( {2m + 1} \right)x + m\\
\Rightarrow y' = {x^2} + 2\left( {m - 1} \right){x^2} + \left( {2m + 1} \right)
\end{array}\)
Hàm số đã cho nghịch biến trên khoảng (0;3) khi và chỉ khi
\(\begin{array}{l}
y' \le 0,\,\,\,\,\forall x \in \left( {0;3} \right)\\
\Leftrightarrow {x^2} + 2\left( {m - 1} \right)x + 2m + 1 \le 0,\,\,\,\,\forall x \in \left( {0;3} \right)\\
\Leftrightarrow \left( {{x^2} - 2x + 1} \right) + 2m\left( {x + 1} \right) \le 0,\,\,\,\,\forall x \in \left( {0;3} \right)\\
\Leftrightarrow 2m \le \frac{{ - {x^2} + 2x - 1}}{{x + 1}},\,\,\,\,\forall x \in \left( {0;3} \right)\\
\Leftrightarrow 2m \le \mathop {\min }\limits_{\left( {0;3} \right)} f\left( x \right) = \frac{{ - {x^2} + 2x - 1}}{{x + 1}}\\
f\left( x \right) = \frac{{ - {x^2} + 2x - 1}}{{x + 1}} \Rightarrow f'\left( x \right) = \frac{{\left( { - 2x + 2} \right)\left( {x + 1} \right) - \left( { - {x^2} + 2x - 1} \right)}}{{{{\left( {x + 1} \right)}^2}}}\\
\Rightarrow f'\left( x \right) = \frac{{ - 2{x^2} + 2 + {x^2} - 2x + 1}}{{{{\left( {x + 1} \right)}^2}}} = \frac{{ - {x^2} - 2x + 3}}{{{{\left( {x + 1} \right)}^2}}}\\
f'\left( x \right) = 0 \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = - 3
\end{array} \right.\\
f\left( 0 \right) = - 1;\,\,\,\,\,\,\,f\left( 1 \right) = 0;\,\,\,\,\,\,f\left( 3 \right) = - 1\\
\Rightarrow \mathop {\min }\limits_{\left[ {0;3} \right]} = f\left( 0 \right) = f\left( 3 \right) = - 1\\
\Rightarrow m \le - 1
\end{array}\)
