# Find The set of points where function is not differentiable

Question Let f(x) = $\begin{cases} (x-1)^{2}\cos\left\{ \frac{1}{x-1}\right\} -|x| & x\neq1\\ -1 & x=1 \end{cases}$.Find the set of points where f(x) is not differentiable.

MY APPROACH f(x) =$\begin{cases} (x-1)^{2}\cos\left\{ \frac{1}{x-1}\right\} +x & x\in(-\infty,0)\\ (x-1)^{2}\cos\left\{ \frac{1}{x-1}\right\} -x & 0\leq x<1\\ -1 & x=1 \end{cases}$

This shows function is continuous at every point.

$f'(x)={\begin{cases} 2(x-1)\cos\left\{ \frac{1}{x-1}\right\} +1+\sin\frac{1}{x-1} & x\in(-\infty,0)\\ 2(x-1)\cos\left\{ \frac{1}{x-1}\right\} -1+\sin\frac{1}{x-1} & 0\leq x<1\\ 0 & x=1 \end{cases}}$

My Answer is $\left\{ 0,1\right\}$