The function f(x) = (x^2 - 1)|x^2 - 3x + 2| + | vedclass

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(D) The function $f(x) = (x^2 - 1)|(x - 1)(x - 2)| + \cos(|x|)$.First,consider the term $\cos(|x|)$. The function $|x|$ is not differentiable at $x =

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$2$

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(D) The function $f(x) = (x^2 - 1)|(x - 1)(x - 2)| + \cos(|x|)$.First,consider the term $\cos(|x|)$. The function $|x|$ is not differentiable at $x = 0$,so $\cos(|x|)$ is not differentiable at $x = 0$.Next,consider the term $(x^2 - 1)|(x - 1)(x - 2)|$. The expression $|(x - 1)(x - 2)|$ is not differentiable at the roots of $(x - 1)(x - 2) = 0$,which are $x = 1$ and $x = 2$.At $x = 1$,$f(x) = (x^2 - 1)|(x - 1)(x - 2)| + \cos(|x|)$. Since $(x^2 - 1)$ has a factor $(x - 1)$,the product $(x^2 - 1)|(x - 1)(x - 2)|$ becomes differentiable at $x = 1$ because the zero of the polynomial cancels the non-differentiability of the absolute value.At $x = 2$,the term $(x^2 - 1)$ is $3 
eq 0$. Thus,the non-differentiability of $|(x - 1)(x - 2)|$ at $x = 2$ persists.Checking $x = 2$:$Lf'(2) = \lim_{h 	o 0^-} \frac{f(2+h) - f(2)}{h} = -3 - \sin(2)$$Rf'(2) = \lim_{h 	o 0^+} \frac{f(2+h) - f(2)}{h} = 3 - \sin(2)$Since $Lf'(2) 
eq Rf'(2)$,the function is not differentiable at $x = 2$.

The function $f(x) = (x^2 - 1)|x^2 - 3x + 2| + \cos(|x|)$ is not differentiable at

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