Give an example of a function $f: (-1,1) \rightarrow \mathbb{R}$ which is continuous and monotone increasing, but which is not differentiable at 0. Explain why this does not contradict the fact that if a function is monotone increasing and differentiable at $x_0$ then $f'(x_0) \leq 0$.

My first thought was to use the function $f(x) = |x|$, however I don't believe this function is monotone increasing. Am I right?

  • $\begingroup$ $|x|$ is not monotone increasing, you are correct. $\endgroup$ – Michael Burr May 5 '16 at 15:38
  • $\begingroup$ Make your example monotone increasing by considering $$f(x)=|x|+2x.$$ $\endgroup$ – Did May 5 '16 at 21:13
  • $\begingroup$ I can't propose an edit because it's too small, but shouldn't $f'(x_0) \geq 0$, rather than the other way around? $\endgroup$ – 4D enthusiast Dec 29 '16 at 10:54

Write $f(x)=x$ if $x\geq0$ and $f(x)=2x$ if $x<0$.

To make the second conclusion, you would need $f'$ to exist throught the open interval $(-1,1)$, which is obviously not true here.

  • $\begingroup$ Does this still count even though these are two separate functions? $\endgroup$ – Ben May 5 '16 at 15:41
  • $\begingroup$ or should I let $f$ be a piece wise function consisting of the functions you mentioned? $\endgroup$ – Ben May 5 '16 at 15:54
  • $\begingroup$ I have defined $f$ piecewise. So, yes, define $f$ in this way. $\endgroup$ – Landon Carter May 5 '16 at 16:06
  • $\begingroup$ Actually it would only be necessary for f' to exist at 0, not the whole interval. $\endgroup$ – 4D enthusiast Dec 29 '16 at 11:01
  • $\begingroup$ +1 Your answer is perfectly fine and I don't know why it originally got no up votes (in my case, I mostly likely didn't see it). In fact, off-hand I can't think of a simpler example for just what was asked. $\endgroup$ – Dave L. Renfro Apr 5 at 13:34

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.