Đáp án:
\[m > 0\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
f\left( x \right) = a{x^2} + bx + c \ge 0,\,\,\,\forall x \in R \Leftrightarrow \left\{ \begin{array}{l}
a > 0\\
\Delta ' \le 0
\end{array} \right.\\
y = \frac{{m{x^3}}}{3} - \frac{{m{x^2}}}{2} + \left( {m + 1} \right)x - 15\\
\Rightarrow y' = \frac{{3m{x^2}}}{3} - \frac{{2mx}}{2} + \left( {m + 1} \right) = m{x^2} - mx + \left( {m + 1} \right)\\
y' \ge 0,\,\,\,\forall x \in R\\
\Leftrightarrow m{x^2} - mx + \left( {m + 1} \right) \ge 0,\,\,\,\forall x \in R\\
\Leftrightarrow \left\{ \begin{array}{l}
m > 0\\
\Delta \le 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m > 0\\
{m^2} - 4m.\left( {m + 1} \right) \le 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m > 0\\
- 3{m^2} - 4m \le 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m > 0\\
3{m^2} + 4m \ge 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m > 0\\
\left[ \begin{array}{l}
m \ge 0\\
m \le - \frac{4}{3}
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow m > 0
\end{array}\)
Vậy \(m > 0\)
