Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If a function is differentiable and monotone on the interval $(a, b)$, then its derivative is also monotone on $(a, b)$.

How do you prove this statement is wrong?

Can you please provide an example?

share|cite|improve this question

This thing


share|cite|improve this answer
+1 just because it's so funny. – Git Gud Jan 13 '13 at 20:23
It's 2-valued at one interval. Need eraser. – alancalvitti Jan 13 '13 at 20:24
We need more pictures on MSE :) – StuartHa Jan 13 '13 at 21:21

Consider $f(x)=x^3$ on $[-1,1]$. Then it is clearly monotone, but $f'(x)=3x^2$ which is clearly not monotone.

share|cite|improve this answer
Google search has improved! search for x^3, 3*x^2 – user13107 Jan 14 '13 at 1:16

No, it shouldn't.

Explanation of this fact can be based on the convexity of the function, which depends on the second derivative of the function.

In other words: if the function is monotone on some interval, as well as its derivative, the function doesn't have point(s) of inflection. In case the function is monotone, but its derivative is not, the function has point(s) of inflection on the interval.

When function is monotone, but its derivative is not, function changes type of its convexity still being monotone. Of course, this happens in the case if the necessary convexity condition is true.

Several people have already posted $x^3$ as example. Here are some plots: the function and its first and second derivatives.


Another example:

random function

share|cite|improve this answer
+1 for a general explanation instead of just a counter-example – Kevin Jan 14 '13 at 7:31

$f(x)=x^3$, with $-1<x< 1$.

share|cite|improve this answer

Just for another example $f(x)=x^3+x$ has non-zero derivative $3x^2+1$.

The only thing you need for $f(x)$ to be increasing is for the derivative to be non-negative (and if you want strictly increasing you need zeros of the derivative to be isolated). Take your favourite wiggly non-negative function (not too pathological) and integrate it - eg $\sin^2 x$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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