\(g'\left(x\right)=f'\left(\left|2x+1\right|-4\right)\left(\dfrac{1}{2\sqrt{\left(2x+1\right)^2}}\right).2\left(2x+1\right)=0\Rightarrow x=-\dfrac{1}{2}\)
TH2 : \(f'\left(\left|2x+1\right|-4\right)=0\Rightarrow\left[{}\begin{matrix}\left|2x+1\right|-4=-2\\\left|2x+1\right|-4=2\\\left|2x+1\right|-4=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2};x=x=-\dfrac{3}{2}\\x=\dfrac{5}{2};x=-\dfrac{7}{2}\\x=3;x=0\end{matrix}\right.\)
=> hs có 7 cực trị
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