Rolle theorem proof via intermediate value theorem

Question

I'm wondering if we can use the Intermediate Value Theorem to prove Rolle's Theorem.

The hypotheses of Rolle's Theorem are:

• The function should be continuous on a closed interval $[a,b]$.
• The function should be differentiable on the open interval $(a,b)$.
• $f(a)=f(b)$

The theorem then shows that there exist $c$ in $(a,b)$ such that $f'(c)=0$.

For I.V.T we need the function to be continuous and $f'(a).f'(b) \leq 0$.

If the conditions for Rolle's Theorem conditions are achieved, does it mean that $f'$ is continuous?

For the second condition we have that $$ \frac{f(x)-f(a)}{x-a} .\frac{f(x)-f(b)}{x-b}<0$$ because $$ f(x)-f(a)=f(x)-f(b)$$ and $$ 0<x-a<b-a $$ $$ a-b<x-b<0 $$

Answer 1

Here is an answer to the wrong question (using MVT to prove Rolle's), followed by an answer to the question I think you were asking.

You can almost certainly use the MVT to prove Rolle's -- indeed, Rolle's is the MVT in the special case where $f(a) = f(b)$. But usually Rolle's is used to prove the MVT, so to make this an "honest" proof, you'd need an alternative proof of the MVT.

NB Actually, having edited the question, I realize OP's asking about the INTERMEDIATE value theorem, not the MEAN value theorem.

To answer one of the questions asked: if the conditions of Rolle's theorem are achieved, does that mean that $f'$ is continuous? The answer is no. Let $$ f(x) =\begin{cases} 0 & x = 0 \\ x^2 \sin(\frac{1}{x}) & \text{else}\end{cases}. $$ Then $f$ is differentiable everywhere, has $f(-1/\pi) = f(1/\pi) = 0$, but $f'$ is not continuous at $x = 0$.

Because we cannot assume that $f'$ is continuous, your proof of Rolle via IVT doesn't seem like it's going to work, no.

Answer 2

+1 for a nice question. It is rather strange that the question's concerns were not addressed fully for such a long time and no one noticed it (perhaps due to the fact that it was marked accepted).

The basic premise of the question $f'(a)f'(b) \leq 0$ is wrong. First point is that differentiability of $f$ at $a, b$ is not given so one can't talk of $f'(a) $ and $f'(b) $. Secondly even if that is allowed by modifying the hypotheses (ie assume $f'$ exists in $[a, b] $) it does not follow that $f'(a) f'(b) \leq 0$.

The inequality in question should be $$\frac{f(x) - f(a)} {x-a} \cdot\frac{f(x) - f(b)} {x-b} \leq 0$$ but from here we can't go to $f'(a) f'(b) \leq 0$ via limiting procedure as we can't take $x\to a$ and $x\to b$ simultaneously.

The conclusion does not hold in general. However if $f$ is not constant on $[a, b] $ we can apply one of the proofs of IVT and get a subinterval $[p, q] $ such that $f(p) =f(q) =f(a) $ and $f(x) \neq f(a) $ for $x\in(p, q) $. And then one can prove easily that $f'(p) f'(q) \leq 0$.

Also another curious point is that the derivative need not be continuous (as shown in another answer here) and asker's intent is to use IVT via continuity of $f'$ to get an $f'(c) =0$. Well the derivative may not be continuous but it satisfies intermediate value property via Darboux theorem and we indeed get a point $c\in[p, q] $ with $f'(c) =0$.

The proof of Rolle's theorem as well as Darboux theorem are based on the same two ideas:

• A continuous function on a closed interval takes its minimum and maximum values.
• The sign of derivative at a point gives us information about the increasing/decreasing nature of function at a point (this is an immediate consequence of definition of derivative) so that derivative at interior extremum points must vanish.

So in essence the approach suggested in the question is a roundabout way of proving Rolle's theorem. Its best to rely on the usual proof.

Answer 3

Q. I'm wondering if we can use the Intermediate Value Theorem to prove Rolle's Theorem.

Here is a late answer from antiquity. "Yes, more than two hundred years ago Ampère did exactly that ..."

Simply use the fact that continuous functions have the IVP. There is no need to use the IVP property of the derivative, which is most commonly proved after the mean-value theorem.

Very old question but it can show up in a search. The answer is well-known among people who make it a point to be well-informed [better known as mathematical drudges].

Yes, you can prove the mean-value theorem by means of the intermediate value property. The key is to use the horizontal chord theorem which is proved pretty easily using the IVP [not the MVP which is a very different acronym].

Flett, T. M. Ampère and the horizontal chord theorem. Bull. Inst. Math. Appl.11(1975), no.1-2, 34.

From Math Review: The author points out that, in 1806, A. M. Ampère [J. École Polytechique 6 (1806), no. 13, 148–181] proved that if $f:[a,b]→R$ is continuous, $m$ is a positive integer and $δ=(b−a)/m$, then there exists $α∈[a,b−δ]$ such that $(f(α+δ)−f(α))/δ=(f(b)−f(a))/(b−a)$. A special case of this result is the positive part of the horizontal chord theorem. A somewhat sharpened version of it yields the mean value theorem.

Here are some more references for the more scholarly users of StackExchange.

REFERENCES:

[1] Pompeiu, D. (1959). Opera Matematica. Editura Academiei Republicii Populare Romˆıne. Bucharest.

[2] Samelson, H. (1979). Classroom Notes: On Rolle’s Theorem. Amer. Math. Monthly. 86(5): 486.

[3] Skau, Christian- Three pearls from elementary mathematics. Normat 48 (2000), no. 2, 56–74, 95.

[4] Oxtoby, J. C., Horizontal chord theorems Amer. Math. Monthly 79 (1972), 468–475.

Answer 4

Yes, we can use the IVT to prove Rolle's theorem, but not in the way you expect.

The idea is to construct a sequence of nested intervals, each half as wide as the previous, where the value of $f$ is the same at both endpoints. These will converge to a point where $f'(x)=0$.

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Set $a_0=a$ and $b_0=b$. The $n$th interval will be $[a_n,b_n]\subset[a_{n-1},b_{n-1}]$, with length $b_n-a_n=(b_{n-1}-a_{n-1})/2=(b_0-a_0)/2^n$.

(We'll start with $n=0$ and use induction, of course.) Consider the function

$$g(x)=f(x)-f\left(x+\tfrac{b_n-a_n}{2}\right)$$

defined for $a_n\leq x\leq\tfrac{a_n+b_n}{2}$, so $\tfrac{a_n+b_n}{2}\leq x+\tfrac{b_n-a_n}{2}\leq b_n$. Note that $g$ is continuous on this interval, because $f$ is continuous. Also, since $f(a_n)=f(b_n)$,

$$g(a_n)=f(a_n)-f\left(\tfrac{a_n+b_n}{2}\right)=-g\left(\tfrac{a_n+b_n}{2}\right).$$

If this value is $g(a_n)=0$, then $f(a_n)=f\left(\tfrac{a_n+b_n}{2}\right)=f(b_n)$, and we can simply bisect the interval and proceed with either half: $[a_{n+1},b_{n+1}]=\left[a_n,\tfrac{a_n+b_n}{2}\right]$ or $\left[\tfrac{a_n+b_n}{2},b_n\right]$. (If this happens for both $n=0$ and $n=1$, then we should take the interval not containing $a_0$ nor $b_0$, so we don't get convergence to an endpoint. That is, $[a_2,b_2]=\left[\tfrac{3a+b}{4},\tfrac{a+b}{2}\right]$ or $\left[\tfrac{a+b}{2},\tfrac{a+3b}{4}\right]$, instead of $\left[a,\tfrac{3a+b}{4}\right]$ or $\left[\tfrac{a+3b}{4},b\right]$.)

If $g(a_n)\neq0$, then $g$ has opposite signs at the endpoints of its domain, so by IVT, there is some $x\in\left(a_n,\tfrac{a_n+b_n}{2}\right)$ such that $g(x)=0$. Set $[a_{n+1},b_{n+1}]=\left[x,x+\tfrac{b_n-a_n}{2}\right]$, so $f(a_{n+1})=f(b_{n+1})$.

Now we have the desired sequence of nested intervals; they converge to some point $c\in(a,b)$.

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We assumed $f$ is differentiable on $(a,b)$, so $f'(c)$ exists.

If $c=a_m$ for some $m\in\mathbb N$, then $a_n=c$ for all $n>m$ (since $c=a_m\leq a_n\leq c$ by the nesting). This implies

$$0=\frac{f(b_n)-f(a_n)}{b_n-a_n}=\frac{f(b_n)-f(c)}{b_n-c}$$

$$0=\lim_{n\to\infty}\frac{f(b_n)-f(c)}{b_n-c}=f'(c).$$

Similarly if $c=b_m$ for some $m$ then $f'(c)=0$. So assume $c\in(a_m,b_m)$ for all $m$. (Thus the following denominators are non-zero.)

$$0=\frac{f(b_n)-f(a_n)}{b_n-a_n}=\frac{f(b_n)-f(c)+f(c)-f(a_n)}{b_n-a_n}$$

$$=\frac{b_n-c}{b_n-a_n}\cdot\frac{f(b_n)-f(c)}{b_n-c}+\frac{c-a_n}{b_n-a_n}\cdot\frac{f(c)-f(a_n)}{c-a_n}.$$

From the definition of the derivative, for any $\varepsilon>0$, there exists $m\in\mathbb N$ such that

$$-\varepsilon<\frac{f(b_n)-f(c)}{b_n-c}-f'(c)<\varepsilon$$

for all $n>m$, and there also exists $l\in\mathbb N$ such that

$$-\varepsilon<\frac{f(a_n)-f(c)}{a_n-c}-f'(c)<\varepsilon$$

for all $n>l$. Therefore, for all $n>\max\{l,m\}$,

$$0=\frac{b_n-c}{b_n-a_n}\cdot\frac{f(b_n)-f(c)}{b_n-c}+\frac{c-a_n}{b_n-a_n}\cdot\frac{f(c)-f(a_n)}{c-a_n}$$

$$<\frac{b_n-c}{b_n-a_n}\cdot\Big(f'(c)+\varepsilon\Big)+\frac{c-a_n}{b_n-a_n}\cdot\Big(f'(c)+\varepsilon\Big)$$

$$=(1)\big(f'(c)+\varepsilon\big)$$

$$0-\varepsilon<f'(c)$$

and similarly

$$0=\frac{b_n-c}{b_n-a_n}\cdot\frac{f(b_n)-f(c)}{b_n-c}+\frac{c-a_n}{b_n-a_n}\cdot\frac{f(c)-f(a_n)}{c-a_n}$$

$$>\frac{b_n-c}{b_n-a_n}\cdot\Big(f'(c)-\varepsilon\Big)+\frac{c-a_n}{b_n-a_n}\cdot\Big(f'(c)-\varepsilon\Big)$$

$$=(1)\big(f'(c)-\varepsilon\big)$$

$$0+\varepsilon>f'(c)$$

so we have $-\varepsilon<f'(c)<\varepsilon$. Since $\varepsilon$ was arbitrary, we conclude that $f'(c)=0$.