I just started reading continuity, differentiability etc. So I was thinking of an example of following type:

Let $f: [a,b]\rightarrow \mathbb{R}$, where $f(x)$ is monotonically increasing continuous function and differentiable on $(a,b)$ or simply $f'(x)\geq 0$ for $x\in (a,b)$.

can we find such function for which $f'(x)$ is not continuous?

I could not find any, whatever function I take $f'(x)$ is becoming continuous. Is there such function even exists!


For every $x\in (-1,1).$ Define, $$ f(x) = 100x+ x^2\sin\frac{1}{x} \qquad\text{with $f(0)=0$}$$ which is differentiable and $$ f'(x) = 100+ 2x\sin\frac{1}{x}-\cos\frac{1}{x} \qquad\text{with $f'(0)=100$}$$ $f'$ is not continuous a $x=0$ and you might choose the interval $[a,b]$ around $x=0$ as it suite to you. Indeed

$$ f'(x)=100+2x\sin\frac{1}{x}-\cos\frac{1}{x} \ge 100+ 2x\sin\frac{1}{x}-1$$

Since $|\sin a| \le |a|$ and $-1\le-\cos a\le 1$ then $|2x\sin\frac{1}{x}|\le 2$ i.e $$2x\sin\frac{1}{x}\ge -2 $$


$$ f'(x)=100+2x\sin\frac{1}{x}-\cos\frac{1}{x} \ge 100-2-1= 97>0$$ for every $x\in \mathbb R$. so $f $ is increasing and $f'$ is not continuous at $x=0$.

  • $\begingroup$ Defined on what closed interval? $\endgroup$ – Dave Aug 17 '17 at 13:27
  • $\begingroup$ but the function is increasing in near $0$ $\endgroup$ – user467365 Aug 17 '17 at 13:28
  • 1
    $\begingroup$ Monotone?${}{}{}{}$ $\endgroup$ – David Mitra Aug 17 '17 at 13:29
  • $\begingroup$ you can choose your interval from the left of o or from the right both will give monotonicity for sure. $\endgroup$ – Guy Fsone Aug 17 '17 at 13:30
  • $\begingroup$ I don't understand, is this function monotonically increasing in $[-\epsilon,\epsilon]$, $\epsilon>0$ $\endgroup$ – user467365 Aug 17 '17 at 13:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.