A. \(\left( { – 4; – 3} \right)\).
B. \(\left( { – 2;0} \right).\)
C. \(\left( { – \frac{3}{2};1} \right)\)
D. \(\left( { – 3; – 2} \right).\)
Lời giải:.
Ta có \(g’\left( x \right) = \left( {2x + 3} \right).f’\left( {{x^2} + 3x + 1} \right)\)
\(g’\left( x \right) = 0 \Leftrightarrow \left[ \begin{array}{l}2x + 3 = 0\\f’\left( {{x^2} + 3x + 1} \right) = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = – \frac{3}{2}\\{x^2} + 3x + 1 = 5\\{x^2} + 3x + 1 = 1\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = – \frac{3}{2}\\x = 1\\x = – 4\\x = – 3\\x = 0\end{array} \right..\)
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