\(y'=3x^2-2x=0\Rightarrow x=0;x=\dfrac{2}{3}\)
\(g'\left(x\right)=f'\left(\left|x+1\right|\right).\dfrac{1}{2\sqrt{\left(x+1\right)^2}}.2\left(x+1\right)=0\Leftrightarrow x=-1\)
TH2 : \(\left[{}\begin{matrix}\left|x+1\right|=0\\\left|x+1\right|=\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\left(loai\right)\\x=-\dfrac{1}{3};x=-\dfrac{5}{3}\end{matrix}\right.\)
=> hs có 2 điểm cực trị
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