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Can you give an example of a function $f(x)$ which is differentiable but not continuously differentiable and there exists a nbd $N$ around a point $c$ such that $f'(x) >(<) 0 \forall x \in N $ and $f'(x)$ is not continuous at the point $c$?

I am basically trying to search a function which is monotone in an open interval but it is not continuously differentiable at that interval.

Can anyone please help me to find that kind of a function?

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    $\begingroup$ Duplicate of 285944 $\endgroup$
    – almagest
    Sep 25, 2019 at 6:43
  • $\begingroup$ A function $g:\Bbb R\to \Bbb R$ can be increasing and differentiable and such that $\{f'(x): |x|<r\}$ is unbounded above for every $ r>0.$ Then $f(x)=x+g(x)$ is strictly increasing and differentiable and $f'(x)=1+g'(x)\ge 1$ for all $x,$ but $f'(x)$ is discontinuous at $x=0.$ $\endgroup$ Sep 25, 2019 at 11:08

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$$f(x) = \begin{cases} 10x + x^2 \cos \left( \frac{1}{x} \right) & x \neq 0 \\ 0 & \text{otherwise} \end{cases}$$

Take $N = (-0.5 , 0.5)$ and $f'(0) =10 , f'(x) > 0$ for all $x \in N$, and $f'(x)$ is not continuous at $0$.

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Let $$g(x) = \begin{cases} x^2 \sin\left(\frac{1}{x}\right) & \text{if } x \neq 0 \\ 0 & \text{if } x = 0.\end{cases}$$ Then $$g'(x) = \begin{cases} 2x \sin\left(\frac{1}{x}\right) - \cos\left(\frac{1}{x}\right) & \text{if } x \neq 0 \\ 0 & \text{if } x = 0.\end{cases},$$ hence $g'$ is discontinuous at $x = 0$. Note that, on the interval $(-1, 1)$, we have $$|g'(x)| \le 2|x| \cdot\left|\sin\left(\frac{1}{x}\right)\right| + \left|\cos\left(\frac{1}{x}\right)\right| \le 3.$$ Hence, if we let $$f(x) = 4x + g(x),$$ then $f'(x)$ is discontinuous at $x = 0$, and on the interval $(-1, 1)$, we have $$f'(x) = 4 + g'(x) \ge 4 - |g'(x)| \ge 4 - 3 > 0.$$

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