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\} $

BOOK'S Answer is $\left\{ 0\right\} $For that $\lim_{x\rightarrow1}$$2(x-1)\cos\left\{ \frac{1}{x-1}\right\} -1+\sin\frac{1}{x-1}$ must be equal to zero .I am unable to prove it

  • $\begingroup$ It is differentiable at $x = 1$. Check your derivative. $\endgroup$ – jonsno Nov 1 '17 at 10:53
  • $\begingroup$ @samjoe That was just a typing mistake,i have corrected it $\endgroup$ – Kislay Tripathi Nov 1 '17 at 11:00
  • $\begingroup$ @GuyFsone Ok then please answer it with correct one $\endgroup$ – Kislay Tripathi Nov 1 '17 at 11:20
  • $\begingroup$ There is a problem on how you define the derivative at $x=1$. and the limit as proposed from your book is not correct unless you missed something. See the answer below $\endgroup$ – Guy Fsone Nov 1 '17 at 11:48

YOUR ERROR you use the derivative function to find the derivate at the fake singularity $x= 1$

You should proceed as follows. Since $x=1$ is a fake singularity (or fake discontinuity) of your function By definition of derivative we have,

\begin{split}f'(1)&=& \lim_{x\to 1}\frac{ f(x)-f(1)}{x-1} \\&=& \lim_{x\to 1}\frac{ (x-1)^{2}\cos\left\{ \frac{1}{x-1}\right\} -|x| +1}{x-1} \\&=& \lim_{x\to 1}(x-1)^{}\cos\left\{ \frac{1}{x-1}\right\} + \frac{ -|x| +1}{x-1} = -1\end{split}

Since we have, $$\lim_{x\to 1}| (x-1)^{}\cos\left\{ \frac{1}{x-1}\right\} |\le\lim_{x\to 1 }|x-1| = 0 $$ and $\text{for } ~~~|x-1|\le \frac12\Longleftrightarrow 0< \frac12\le x\le \frac 32$ we have,

$$ \frac{ -|x| +1}{x-1} =\frac{ -x +1}{x-1} = -1~~$$

Conlcusion $$f'(x)={\begin{cases} 2(x-1)\cos\left\{ \frac{1}{x-1}\right\} +1-\sin\frac{1}{x-1} & x\in(-\infty,0)\\ 2(x-1)\cos\left\{ \frac{1}{x-1}\right\} -1-\sin\frac{1}{x-1} & 0\leq x<1\\ -1 & x=1 \end{cases}}$$ Hence is not differentiable ONLY at $x= 0$

Warning!! Do not confused the following: $f'$ is not continuous at $x=0$ and at $x=-1$

  • $\begingroup$ Brother but i have seen many questions in examples that are solved on the basis of continuity of it's derivative $\endgroup$ – Kislay Tripathi Nov 1 '17 at 12:46
  • $\begingroup$ Yes but here the context is different . $x=1$ is a fake singularity. we cannot do that here. for the other point the regularity freely occur so you can apply that rule for every point except at $x=1,$or$ x=0$ $\endgroup$ – Guy Fsone Nov 1 '17 at 12:58

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