Question

Answer 1

\(g'\left(x\right)=f'\left(\left|3-2x\right|+3\right).\left(\dfrac{1}{2\sqrt{\left(3-2x\right)^2}}\right).2\left(3-2x\right)=0\Rightarrow x=\dfrac{3}{2}\)

TH2 : \(f'\left(\left|3-2x\right|+3\right)=0\Rightarrow\left[{}\begin{matrix}\left|3-2x\right|+3=\sqrt{2}\left(loại\right)\\\left|3-2x\right|+3=2\left(loại\right)\end{matrix}\right.\)

=> hs có 1 cực trị