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