Mathematics - Continuity - Differentiability Question with Solution | TestHub

$f(x)=|x^2-3x+2|+\cos |x|=|(x-1)(x-2)|+\cos |x|$

$$f(x)=\{\begin{array}{ll}{x}^2-3x+2+\cos x,& x<0\\ x^2-3x+2+\cos x,& 0\le x<1\\ -(x^2-3x+2)+\cos x,& 1\le x<2\\ x^2-3x+2+\cos x,& x>2\end{array}$$

$$f^{\prime }(x)=\{\begin{array}{ll}2x-3-\sin x,& x<0\\ 2x-3-\sin x,& 0\le x<1\\ -2x+3-\sin x,& 1\le x<2\\ 2x-3-\sin x,& x>2\end{array}$$

It is clear that $f(x)$ is not differentiable at $x=1$.

Thus, $f^{\prime }(1^{-})=-1-\sin 1$

and $f^{\prime }(1^{+})=1-\sin 1$.