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