2

So if you want to find a local minimum of a function, one way to do that would be to find an interval $[a,b)$ where the derivative is negative, and that the derivative in the interval $(b,c]$ is positive for some $c$. $f'(b)$ needs to $0$ or not defined. But my question is, is there a function where the derivative in $[a,b)$ is negative, and positive in $[b,c]$? That means there's a jump discontinuity in the derivative, so the original function would be piece-wise

Thank you!

  • 2
    $f(x) = \sqrt{x^2}$, for example. Or $f(x) = \arcsin(\sin x)$. – dfnu Dec 03 '19 at 13:10
  • 1
    It's not possible for $f'(b)$ to exist and be positive whilst $f'(b-\epsilon)$ exists and is negative for all $\epsilon\gt0$. All of the examples which others are providing create a function which is not differentiable at some point. – Peter Foreman Dec 03 '19 at 13:12
  • 3
    Derivatives must satisfy the intermediate value property (look up Darboux's theorem). So the property you are wishing for cannot happen. Derivatives, although they may be discontinuous, will never have a jump discontinuity. –  Dec 03 '19 at 13:15
  • 1
    @PeterForeman certainly. I didn't notice the inclusion of $c$ in one of the two intervals! In any case the derivative cannot be defined in $c$ – dfnu Dec 03 '19 at 13:17
  • 2
    Possible duplicate of Interpreting the significance of Darboux's Theorem – Xander Henderson Dec 03 '19 at 13:37
  • I have a question. In the context of this question, does f'(b) mean only the right-hand derivative of f at b, i.e. only $\lim_{h\to 0^+} \frac{f(b+h)-f(b)}{h}$, or does it also mean that the left-hand derivative f'(b) = $\lim_{h\to 0^-} \frac{f(b+h)-f(b)}{h}$ exists and is equal to the right-hand limit? – Adam Rubinson Dec 04 '19 at 22:19

3 Answers3

3

The answer to your question is no, since if $f: [a,b] \rightarrow \mathbb{R}$ is differentiable, the derivative $f'$ satisfies the intermediate value property.

This is called Darboux's theorem (see wikipedia for a proof).

spin
  • 11,956
1

Please look Darboux's theorem for derivatives. https://en.wikipedia.org/wiki/Darboux%27s_theorem_(analysis)

Simply states that derivatives can not have jump discontinuity. More exactly, derivatives satisfy intermediate value property.

Note: Please, feel free to edit (to enrich) this answer.

Feel_Free_To_Edit_This_Answer
  • 44
0

Is there a function where the derivative in $[a,b)$ is negative, and positive in $[b,c]$?

f is differentiable on [a,c], therefore f is continuous on [a,c]. In particular, f is continuous at b.

If the derivative in $[a,b)$ is negative, then this along with the fact that f is continuous at b means that $ \forall x < b, f(x) > f(b)$.

If the derivative at b is positive, then there exists an $s < b$ such that $f(s) < f(b)$.

Contradiction.

Adam Rubinson
  • 20,052