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Let $a,b \in \mathbb{R}$ such that $a<b$. Let $f$ differentiable on $[a,b]$ such that $f'(a) < 0 < f'(b)$.

Show that : $$\exists c\in ]a,b[, \hspace{1mm} f'(c)=0$$

My attempt :

Since the derivative changes sign, $f$ can't be strictly monotonic.

($f$ is decreasing in the neighbourhood of $a$ and increasing in the neighbourhood of $b$)

In this case there exists $x,y \in ]a,b[$ such that $f(x) = f(y)$. (Not sure about that)

We apply Rolle's theorem to conclude.

Is this valid ? How can I be more precise on the third line ? Thank you in advance.

PRORIZZLER
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    That is Darboux's theorem: derivatives have the intermediate value property. You can find a proof in https://en.wikipedia.org/wiki/Darboux%27s_theorem_(analysis) – Martin R Jul 18 '24 at 13:44
  • See also here: https://math.stackexchange.com/q/771201/42969, and here: https://math.stackexchange.com/q/1588154/42969 – Martin R Jul 18 '24 at 13:45
  • See also https://math.stackexchange.com/questions/54843/how-to-prove-that-derivatives-have-the-intermediate-value-property – Arthur Jul 18 '24 at 13:46
  • "($f$ is decreasing in the neighbourhood of $a$ and increasing in the neighbourhood of $b$)" -- this need not be true! Consider $f(x)=\begin{cases}x+x^2\sin(x^{-2})&x\ne0\0&x=0\end{cases}$. This has $f'(0)=1$, but is not increasing in any open (or half-open) interval containing $0$. – Hagen von Eitzen Jul 18 '24 at 14:23

2 Answers2

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This is a specific case of the Darboux theorem.

In your case, this is easier, since you are looking for $c$ such that $f'(c)=0$.

Since $f$ is differentiable on $[a,b]$, then $f$ is continuous on $[a,b]$.

Thus, there exists $d \in [a,b]$ such that $f(d)$ is an extremum of $f$ on $[a,b]$.

You can verify very quickly that, since $f'(a)<0$, $a$ is not an extremum of $f$. Same for $b$.

So $d\neq a$ and $d\neq b$, thus $d\in]a,b[$

By Fermat's theorem (one of his!), you have $f'(d)=0$

This $d$ is your $c$!

Martigan
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Since you want to base this on Rolle's theorem:

From $f'(a)>0$ then for sufficently small positive $h$, we have ($a<a+h<b$ and) from the limit definition of $f'(a)$, $\frac{f(a+h)-f(a)}{h}>\frac12f'(a)>0$. So with $u:=a+h\in]a,b[$, we have $f(u)>f(a)$. Using $f'(b)<0$, the same reasoning gives us $v\in]a,b[$ such that $f(v)>f(b)$.

If $f(a)<f(b)$, then by the intermediate value theorem for the continuous function $f$, there exists $w\in ]a,v[$ such that $f(w)=f(b)$, and we can apply Rolle to $[w,b]$. On the other hand, if $f(a)>f(b)$, then by the intermediate value theorem, there exists $w\in ]u,b[$ such that $f(w)=f(a)$, and we can apply Rolle to $[a,w]$. And fo course if $f(a)=f(b)$, we can apply Rolle to $[a,b]$ itself.

Hagen von Eitzen
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