f(x)=≤ft(x^2-4)≤ft|x^2-5 x+6|+cos (|x|) is non-differentiable at

$f(x)=\left(x^2-4\right)|(x-2)(x-3)|+\cos x$

$f(x)=\left\{\begin{array}{c}\left(x^2-4\right)\left(x^2-5 x+6\right)+\cos x, & x \leq 2 \text { or } x \geq 3 \\ \left(4-x^2\right)\left(x^2-5 x+6\right)+\cos x, & x \in(2,3)\end{array}\right.$

only points where f(x) may be non–differentiable are x = 2 and x = 3

$f'(x)=\left\{\begin{array}{c}\left(x^2-4\right)(2 x-5)+2 x\left(x^2-5 x+6\right)-\sin x, &  x<2 \text { or } x>3 \\ \left(4-x^2\right)(2 x-5)-\left(x^2-5 x+6\right)(2 x)-\sin x, & x \in(2,3)\end{array}\right.$

f'(2 – 0) = – sin 2,

f'(2 + 0) = sin 2,

f'(3 – 0) = – 5 – sin 3

f'(3 + 0) = 5 – sin 3

Thus, f(x) is differentiable at x = 2 but not at x = 3.