Question

Answer 1

\(g'\left(x\right)=f'\left(\left|x-1\right|-2\right).\left(\dfrac{1}{2\sqrt{\left(x-1\right)^2}}\right).2\left(x-1\right)=0\Leftrightarrow x=1\)

TH2 : \(f'\left(\left|x-1\right|-2\right)=0\Rightarrow\left[{}\begin{matrix}\left|x-1\right|-2=0\\\left|x-1\right|-2=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3;x=-1\\x=5;x=-3\end{matrix}\right.\)

=> hs có 5 cực trị 

Answer 2

g′(x)=f′(|x−1|−2).⎛⎜ ⎜⎝12√(x−1)2⎞⎟ ⎟⎠.2(x−1)=0⇔x=1g′(x)=f′(|x−1|−2).(12(x−1)2).2(x−1)=0⇔x=1

TH2 : f′(|x−1|−2)=0⇒[|x−1|−2=0|x−1|−2=2⇔[x=3;x=−1x=5;x=−3f′(|x−1|−2)=0⇒[|x−1|−2=0|x−1|−2=2⇔[x=3;x=−1x=5;x=−3

=> HÀM SỐ CÓ 5 CỰC TRỊ