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Câu hỏi: Cho hàm số $y=x{{\left( x+1 \right)}^{2}}{{\left( x+2 \right)}^{3}}{{\left( x+3 \right)}^{4}}$. Hàm số có bao nhiêu điểm cực trị?
A. $3.$
B. $5.$
C. $1.$
D. $2.$
A. $3.$
B. $5.$
C. $1.$
D. $2.$
$\begin{aligned}
& {y}'={{\left( x+1 \right)}^{2}}{{\left( x+2 \right)}^{3}}{{\left( x+3 \right)}^{4}}+2x\left( x+1 \right){{\left( x+2 \right)}^{3}}{{\left( x+3 \right)}^{4}}+3x{{\left( x+1 \right)}^{2}}{{\left( x+2 \right)}^{2}}{{\left( x+3 \right)}^{4}}+4x{{\left( x+1 \right)}^{2}}{{\left( x+2 \right)}^{3}}{{\left( x+3 \right)}^{3}} \\
& =\left( x+1 \right){{\left( x+2 \right)}^{2}}{{\left( x+3 \right)}^{3}}\left[ \left( x+1 \right)\left( x+2 \right)\left( x+3 \right)+2x\left( x+2 \right)\left( x+3 \right)+3x\left( x+1 \right)\left( x+3 \right)+4x\left( x+1 \right)\left( x+2 \right) \right] \\
& =\left( x+1 \right){{\left( x+2 \right)}^{2}}{{\left( x+3 \right)}^{3}}\left( 10{{x}^{3}}+40{{x}^{2}}+40x+6 \right) \\
\end{aligned} $ $ {y}'=0\Leftrightarrow \left[ \begin{aligned}
& x=-1 \\
& x=-2\left( ng.k\acute{e}p \right) \\
& x=-3 \\
& x\approx -2,49 \\
& x\approx -0,18 \\
& x\approx -1,33 \\
\end{aligned} \right.$
Vậy hàm số đã cho có 5 điểm cực trị.
& {y}'={{\left( x+1 \right)}^{2}}{{\left( x+2 \right)}^{3}}{{\left( x+3 \right)}^{4}}+2x\left( x+1 \right){{\left( x+2 \right)}^{3}}{{\left( x+3 \right)}^{4}}+3x{{\left( x+1 \right)}^{2}}{{\left( x+2 \right)}^{2}}{{\left( x+3 \right)}^{4}}+4x{{\left( x+1 \right)}^{2}}{{\left( x+2 \right)}^{3}}{{\left( x+3 \right)}^{3}} \\
& =\left( x+1 \right){{\left( x+2 \right)}^{2}}{{\left( x+3 \right)}^{3}}\left[ \left( x+1 \right)\left( x+2 \right)\left( x+3 \right)+2x\left( x+2 \right)\left( x+3 \right)+3x\left( x+1 \right)\left( x+3 \right)+4x\left( x+1 \right)\left( x+2 \right) \right] \\
& =\left( x+1 \right){{\left( x+2 \right)}^{2}}{{\left( x+3 \right)}^{3}}\left( 10{{x}^{3}}+40{{x}^{2}}+40x+6 \right) \\
\end{aligned} $ $ {y}'=0\Leftrightarrow \left[ \begin{aligned}
& x=-1 \\
& x=-2\left( ng.k\acute{e}p \right) \\
& x=-3 \\
& x\approx -2,49 \\
& x\approx -0,18 \\
& x\approx -1,33 \\
\end{aligned} \right.$
Vậy hàm số đã cho có 5 điểm cực trị.
Đáp án B.