# Can a function's derivative switch signs without crossing zero?

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!

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

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.

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).

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 $$f(s) < f(b)$$.