A question regarding Rolle's theorem

Question

Suppose that $f(x)$ is continuous in the closed interval $[a,b]$ and that $f(a)=f(b)=0$. In addition, suppose that its derivative exists for all $c \in(a,b)$ except for finite number of points in $(a,b)$, and that in those points $$ \lim_{x\to c}f'(x)=+\infty. $$ Can I still deduce that there exists $c\in(a,b)$ such that $f'(c)=0$?

My attempt: By the assumptions $f$ has a tangent line at every point in $(a,b)$ - including vertical tangent lines (like $x=0$ in $x\mapsto \sqrt[3]{x}$ ). The points where there is a vertical tangent can not be local maximum/minimum points. Since maximum/minimum point must exists in this case, it must be a point with zero derivative. Is this argument true?

"The points where there is a vertical tangent can not be local maximum/minimum points." Why? Can you fully justify this statement? — David K, Commented Jul 16 at 13:19
When the derivative of a function doesn't exist, you can't assume the function leaves a point in the same direction it enters the point. — David K, Commented Jul 16 at 13:39
If $\lim \ldots = \infty$ means the limit is $+\infty$, your argument is correct. (Would also hold if the limit is $-\infty$ at some points.) But if it only means $\lim_{x \to c} \lvert f'(x)\rvert = +\infty$, it's not. — Dermot Craddock, Commented Jul 16 at 13:56

Ryszard Szwarc · Accepted Answer · 2025-07-17 09:27:20Z

6

The answer is yes. Moreover the assumptions can be substantially weakened. Namely, it suffices to assume that each interval $(a,x_1),(x_1,x_2),\ldots,(x_n,b)$ contains a point, where the derivative is positive. In particular no limit assumptions at the points $x_k$ are needed.

Let $a<x_1<\ldots <x_n<b,$ where $f$ is not differentiable at $x_k,$ $1\le k\le n.$ Assume the conclusion does not hold, i.e. $f'(x)\neq 0$ for $x\neq a,x_1,\ldots, x_n,b.$ Then, basing on the Rolle theorem, the function $f$ is strictly monotone on every interval $[a,x_1],[x_1,x_2],\ldots, [x_n,b]$ (see the spoiler). Therefore the derivative is negative or positive on each interval $(a,x_1),(x_1,x_2),\ldots,(x_n,b).$ As each interval contains a point, where the derivative is positive, then the derivative is positive on each interval. Hence $f$ is strictly increasing on each interval $[a,x_1],[x_1,x_2],\ldots, [x_n,b].$ Thus $f$ is strictly increasing, which leads to $f(a)<f(b),$ a contradiction.

The following well known fact has been used in the answer. Assume $f$ is continuous on $[c,d]$ and $f'$ does not vanish on $(c,d).$ Then $f$ is strictly monotone. Indeed assume by contradiction, the opposite. Then the function $g(x,y)=f(y)-f(x)$ defined on $a\le x<y\le b$ admits nonnegative and nonpositive values. By IMV there are $x<y$ such that $g(x,y)=0,$ i.e. $f(x)=f(y). $ By the Rolle theorem $f'$ vanishes at some $z,$ $x<z<y,$ a contradiction.

edited Jul 17 at 9:27

answered Jul 16 at 13:49

Ryszard Szwarc

44.9k3 gold badges19 silver badges62 bronze badges

$\begingroup$ Thanks @Ryszard Szwarc. What is Darboux condition ? $\endgroup$
– boaz
Commented Jul 16 at 13:55
$\begingroup$ @boaz Intermediate value theorem for derivatives. $\endgroup$
– Dermot Craddock
Commented Jul 16 at 13:56
$\begingroup$ The same as intermediate value theorem. $\endgroup$
– Ryszard Szwarc
Commented Jul 16 at 13:56
$\begingroup$ I have modified the reasoning so that the Darboux property is not needed. $\endgroup$
– Ryszard Szwarc
Commented Jul 16 at 14:42
$\begingroup$ I see now how you invoke Rolle to say that if a differentiable function is not strictly monotonic in an interval, there is some point where the derivative is zero. This is nice. You could perhaps include a proof ot this fact in the answer. $\endgroup$
– Gribouillis
Commented Jul 16 at 15:05

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Gribouillis · Accepted Answer · 2025-07-16 14:23:05Z

3

With your assumptions, if $c\in (a,b)$ satisfies $\lim_{x\to c}f'(x)=+\infty$ then $f$ does not reach its maximum or its minimum at $c$.

Indeed, let $x_n >c$ be a sequence of points such that $f$ is differentiable everywhere in $(c, x_n)$ and $x_n\to c$. For $n$ large enough one has $f'(x)>0$ on $(c, x_n)$. By the mean value theorem, $\exists y_n\in (c, x_n)$ such that \begin{equation} f(x_n) - f(c) = f'(y_n)(x_n - c)> 0 \end{equation} hence $f(c)< f(x_n)$ and $f(c)$ is not a maximum value.

The same reasoning applied on the left of $c$ shows that $f(c)$ is not a minimum value.

It follows that if $f$ is not constant, there is a point in $(a, b)$ where $f$ has a global maximum or a global minimum and at that point $f$ is differentiable, hence its derivative is zero.

Rolle's theorem is therefore valid with these assumptions.

answered Jul 16 at 14:23

Gribouillis

16.9k1 gold badge17 silver badges31 bronze badges

$\begingroup$ You can argue even simpler (IMO), similarly as in the proof of the usual Rolle's theorem. $f$ must have a minimum or maximum in the interior, at $c \in (a, b)$. $f(c) = + \infty$ is impossible at a maximum is impossible because $f(c+h) \le f(c)$ for $h > 0$, and also impossible at a minimum. $\endgroup$
– Martin R
Commented Jul 16 at 14:29
$\begingroup$ @MartinR I think there is a priori a difference between $f'(c) = +\infty$ and $\lim_{x\to c} f'(x) = +\infty$. This needs some argument. $\endgroup$
– Gribouillis
Commented Jul 16 at 14:33
$\begingroup$ I see what you mean. $\lim_{x\to c} f'(x) = +\infty$ implies $f'(c) = +\infty$ if $f$ is continuous at $c$. – Btw, the statement is true even if we allow $f'(c) = +\infty$ or $f'(c) = -\infty$ at the exceptional points. $\endgroup$
– Martin R
Commented Jul 16 at 14:36
$\begingroup$ @MartinR One could also probably relax the finiteness condition. $\endgroup$
– Gribouillis
Commented Jul 16 at 14:41

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Martin R · Accepted Answer · 2025-07-17 09:04:34Z

This follows from the generalization of Rolle's theorem:

Consider a real-valued, continuous function $f$ on a closed interval $[a, b]$ with $f(a) = f(b)$. If for every $x$ in the open interval $(a, b)$ the right-hand limit $$ f'(x^{+}):=\lim _{h\to 0^{+}}{\frac {f(x+h)-f(x)}{h}} $$ and the left-hand limit $$ f'(x^{-}):=\lim _{h\to 0^{-}}{\frac {f(x+h)-f(x)}{h}} $$ exist in the extended real line $[−\infty, \infty]$, then there is some number $c$ in the open interval $(a, b)$ such that one of the two limits $f'(c^+)$ and $f'(c^-)$ is $\ge 0$ and the other one is $\le 0$ (in the extended real line).

The proof is identical to the proof of the “usual” Rolle theorem: $f$ must attain its maximum or its minimum at some interior point $c \in (a, b)$. In the case of a maximum is $f'(c^+) \le 0$ and $f'(c^-) \ge 0$, in the case of a minimum it is the other way around.

If $f'(c)$ happens to exist (in the extended line) then necessarily $f'(c) = 0$, so the following is an immediate consequence:

If $f: [a, b] \to \Bbb R$ is continuous with $f(a) = f(b)$, and for every $x \in (a, b)$ exists $$ f'(x) =\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}} \in [-\infty, \infty] $$ in the extended real line, then there is some $c \in (a, b)$ with $f'(c) = 0$.

These assumptions are satisfied in your case, because $ \lim_{x\to c} f'(x)=+\infty$ implies that $f'(c)=+\infty$. The statement is even true if the conditions are relaxed to $$ \lim_{x\to c} f'(x)=+\infty \text{ or } \lim_{x\to c} f'(x)=-\infty $$ at the exceptional points.

Strangely, Wikipedia does not contain a generalization of Fermat's interior extremum theorem to derivatives in the extended real line, which would suffice here. — Gribouillis, Commented Jul 16 at 15:50
@Gribouillis: That is perhaps because the most general form (the first part of what I quoted) allows that the right and left derivative are different. In that case the derivative might not exist at an interior extremum. — Martin R, Commented Jul 16 at 15:55

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