\(y'=-3x^2+2\left(m+3\right)x-\left(m^2+2m\right)\)
\(y''=-6x+2\left(m+3\right)\)
Hàm đạt cực đại tại x=2 khi:
\(\left\{{}\begin{matrix}y'\left(2\right)=0\\y''\left(2\right)< 0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}-12+4\left(m+3\right)-\left(m^2+2m\right)=0\\-12+2\left(m+3\right)< 0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}m=0\\m=2\end{matrix}\right.\\m< 3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}m=0\\m=2\end{matrix}\right.\)
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