Bài tập 5.23 trang 203 SBT Toán 11

a) Ta có:

\(f'\left( x \right) = m{x^2} - 6x + m\)

Để \(f'\left( x \right) > 0,\forall x\) thì:

\(\left\{ \begin{array}{l}
m > 0\\
\Delta \prime  < 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
m > 0\\
9 - {m^2} < 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
m > 0\\
\left[ \begin{array}{l}
m <  - 3\\
m > 3
\end{array} \right.
\end{array} \right. \Leftrightarrow m > 3\)

b) Ta có:

\(g'\left( x \right) = m{x^2} - mx + m + 1\)

Để \(g'\left( x \right) < 0,\forall x\) thì

\(\begin{array}{l}
\left\{ \begin{array}{l}
m < 0\\
\Delta  < 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
m < 0\\
{m^2} - 4m\left( {m + 1} \right) < 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
m < 0\\
 - 3m2 - 4m < 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
m < 0\\
m <  - \frac{4}{3}\\
m > 0
\end{array} \right. \Leftrightarrow m <  - \frac{4}{3}
\end{array}\)

-- Mod Toán 11 HỌC247