Is this proof of Rolle's Theorem a valid one?

Question

Let $f:\Bbb [a,b]\to \Bbb R$ satisfy the following :

(i) $f(x)$ is continuous in $[a,b]$

(ii) $f(x)$ is derivable in $(a,b)$

(iii) $f(a)=f(b)$

then, $\exists c\in (a,b)$ such that $f'(c)= 0.$

I tried to prove this theorem as follows:

Since, $f(x)$ is continuous in $[a,b]$ so by Maximum-Minimum Theorem $f$ attains a maximum and a minimum value in $[a,b]$ and we denote them by $M$ amd $m$ respectively.

If $M=m$ then $f(x)=m=M$ for all $x\in [a,b]$ and hence, $f'(x)=0,\forall x\in (a,b)\subset [a,b]$ and so, the theorem is true.

If $m\neq M$ we assume $f(a)=f(b)=M$ then $\exists c\in (a,b)$ such that, $f(c)=m.$

We know that,

Let $c$ be an interior point of the interval $I$ at which $f:I\to \Bbb R$ has a relative extremum. If the derivative of $f$ at $c$ exists, then $f'(c) = 0.$

Since, $f$ is derivable in $(a,b)$ so, $f'(c)=0.$

Similarly, if $f(a)=f(b)\neq M$ then $\exists c\in (a,b)$ such that, $f(c)=M.$

We know that,

Let $c$ be an interior point of the interval $I$ at which $f:I\to \Bbb R$ has a relative extremum. If the derivative of $f$ at $c$ exists, then $f'(c) = 0.$

Since, $f$ is derivable in $(a,b)$ so, $f'(c)=0.$

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Till now, all the books I read, never used this proof. It appeared strange. I am curious to know if the above proof is a valid one or not as I am unable to find any mistake.

The proofs which I found in different books are:

Book 1:

Book 2:

(This one uses a different version of Rolle's Theorem)

Book 3:

I know that some users doesn't like the use of images but in here, I am really afraid to type these massive proofs in the books.

Answer 1

Your proof is correct and complete. The idea is to use compactness of the domain $[a, b]$ to deduce that $f$ must attain both a minimum $m$ and maximum $M$. There is a trivial case where $m = M$, in which case $f$ must be the constant function mapping each element of the domain to $M$, in which case any $x \in (a, b)$ will have $f' (x) = 0$. Otherwise, $m < M$, and you've considered two sub-cases:

(1) $f (a) (= f (b)) = M$. In this case, $m < M$ implies there exists $c \in (a, b)$ such that $f (c) = m$. Then claim $f' (c) = 0$.

(2) $f (a) (= f (b)) < M$. In this case, there exists $d \in (a, b)$ such that $f (d) = M$. Then claim $f' (d) = 0$.

The only thing I would change in your proof is the wording:

If $m \ne M$ we assume $f (a) = f (b) = M$...

(emphasis is mine). This, at least to me, was a bit confusing because I didn't see the complementary case later on. You can say something like

If $m \ne M$, we consider two cases: either $f (a) = f (b) = M$, or $f (a) = f (b) \ne M$...

(kind of how I discussed it above) and proceed as you have, analyzing the two possibilities. Regardless, this is not a problem with the logic of the proof itself, which is sound.

Answer 2

I think, in text books, Rolle's theorem comes before the local extremum concept like the limit concept comes before the derivative concept.

I remember I tried to prove the Rolle's theorem as follows:

Theorem (Rolle's Theorem): Let $f:\Bbb [a,b]\to \Bbb R$ be a function, continuous on $[a,b]$, differentiable on $(a,b)$ and such that $f(a)=f(b)$. Then, $\exists c\in (a,b)$ such that $f'(c)= 0.$

Proof: Since $f(x)$ is continuous on the closed interval $[a,b]$, by the Maximum-Minimum Theorem, the maximum and the minumum values of $f(x)$ are attained in $[a,b].$ Since, $f(a)=f(b)$, either the maximum or the minumum must be attained in the open interval $(a,b)$. Suppose that the minumum is attained. The maximum case can be proved similarly. Now, $\exists c\in (a,b)$ such that $f(x)\geq f(c)$ for all $x\in [a,b].$ Then, we have the following observations, since $f'(c)$ exists:

1. $\frac{f(x)-f(c)}{x-c}\geq 0$ for all $x>c$ which implies that $f'(c)\geq 0,$
2. $\frac{f(x)-f(c)}{x-c}\leq 0$ for all $x<c$ which implies that $f'(c)\leq 0.$

Hence $f'(c)=0.$