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

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(B) The function is given by $f(x) = (x^2 - 1) |(x - 1)(x - 2)| + \cos(|x|)$.First,consider the term $|(x - 1)(x - 2)|$. This term is non-differentiab

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$2$

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(B) The function is given by $f(x) = (x^2 - 1) |(x - 1)(x - 2)| + \cos(|x|)$.First,consider the term $|(x - 1)(x - 2)|$. This term is non-differentiable at the roots of the quadratic expression inside the absolute value,which are $x = 1$ and $x = 2$.However,we must check if the factor $(x^2 - 1)$ compensates for this non-differentiability.At $x = 1$,$f(x) = (x^2 - 1) |(x - 1)(x - 2)| + \cos(|x|)$. The term $(x^2 - 1)$ has a factor of $(x - 1)$,so $(x^2 - 1)|x - 1| = (x + 1)(x - 1)|x - 1|$. Since $\lim_{x 	o 1} (x - 1)|x - 1| = 0$ and the derivative of $(x - 1)|x - 1|$ at $x = 1$ is $0$,the function is differentiable at $x = 1$.At $x = 2$,the term $|x - 2|$ is not differentiable. The expression is $f(x) = (x^2 - 1)|x - 1||x - 2| + \cos(|x|)$. Since $(x^2 - 1)|x - 1|$ is non-zero at $x = 2$ (specifically,$(2^2 - 1)|2 - 1| = 3 
eq 0$),the product $3|x - 2|$ is not differentiable at $x = 2$.Finally,consider $\cos(|x|)$. This is $\cos(x)$ for $x \ge 0$ and $\cos(-x) = \cos(x)$ for $x Therefore,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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