Toán Lớp 12: cho hàm số y=1/3mx^3-(m-1)x^2+3(m-2)x+1/2
tìm m để hàm số có cực đại cực tiểu thỏa mãn x1,x2 >0
help me ;)))
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Toán Lớp 12: cho hàm số y=1/3mx^3-(m-1)x^2+3(m-2)x+1/2
tìm m để hàm số có cực đại cực tiểu thỏa mãn x1,x2 >0
help me ;)))
Comments ( 2 )
2 < m < \dfrac{{2 + \sqrt 6 }}{2}\\
\dfrac{{2 – \sqrt 6 }}{2} < m < 0
\end{array} \right.$
y’ = 0\\
\Leftrightarrow m{x^2} – 2\left( {m – 1} \right)x + 3\left( {m – 2} \right) = 0\\
\Leftrightarrow \left\{ \begin{array}{l}
m\# 0\\
\Delta ‘ > 0\\
{x_1} + {x_2} > 0\\
{x_1}{x_2} > 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m\# 0\\
{\left( {m – 1} \right)^2} – m.3\left( {m – 2} \right) > 0\\
\dfrac{{m – 1}}{m} > 0\\
\dfrac{{m – 2}}{m} > 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m\# 0\\
{m^2} – 2m + 1 – 3{m^2} + 6m > 0\\
\left[ \begin{array}{l}
m > 2\\
m < 0
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
m > 2\\
m < 0
\end{array} \right.\\
2{m^2} – 4m – 1 < 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
m > 2\\
m < 0
\end{array} \right.\\
\dfrac{{2 – \sqrt 6 }}{2} < m < \dfrac{{2 + \sqrt 6 }}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2 < m < \dfrac{{2 + \sqrt 6 }}{2}\\
\dfrac{{2 – \sqrt 6 }}{2} < m < 0
\end{array} \right.\\
Vậy\,\left[ \begin{array}{l}
2 < m < \dfrac{{2 + \sqrt 6 }}{2}\\
\dfrac{{2 – \sqrt 6 }}{2} < m < 0
\end{array} \right.
\end{array}$