Question

The function f(x) = x2 + 2 +1 knj-bhdp-mzh​

Answer 1

Answer:

Therefore, 

∣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.

explain:

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

Lf′(2)=−3sin2

Rf′(x)=(x2−1)(2x−2)+2x(x2−3x+2)−sinx

Rf′(2)=(4−1)(4−3)+0−sin2=3−sin2

Hence, Lf′(2