we know that funtion |x| is not differentiable at x=0 ∴ |x2−3x+2|=|(x−1)(x−2)| Hence, it is not differentiable at x=1 and 2 Now, f(x)=(x2−1)|x2−3x+2|+cos|x| is not differentiable at x=2 For 1<x<2,f(x)=−(x2−1)(x2−3x+2)+cos‌x For 2<x<3,f(x)=(x2−1)(x2−3x+2)+cos‌x Lf′(x)=−(x2−1)(2x−3)−2x(x2−3x+2)−sin‌x L′(2)=−3‌sin‌2 Rf′(x)=(x2−1)(2x−3)+2x(x2−3x+2)−sin‌x Rf′(2)=(4−1)(4−3)+0−sin‌2=3−sin‌2 Hence, L′(2)≠Rf′(2) So, f(x) is not differentable at x=2