2
$\begingroup$

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!

$\endgroup$
  • 1
    $\begingroup$ $f(x) = \sqrt{x^2}$, for example. Or $f(x) = \arcsin(\sin x)$. $\endgroup$ – dfnu Dec 3 '19 at 13:10
  • 1
    $\begingroup$ 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. $\endgroup$ – Peter Foreman Dec 3 '19 at 13:12
  • 3
    $\begingroup$ 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. $\endgroup$ – Robert Wolfe Dec 3 '19 at 13:15
  • 1
    $\begingroup$ @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 $c$ $\endgroup$ – dfnu Dec 3 '19 at 13:17
  • 2
    $\begingroup$ Possible duplicate of Interpreting the significance of Darboux's Theorem $\endgroup$ – Xander Henderson Dec 3 '19 at 13:37
2
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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).

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

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.