Rolle’s theorem

Introduction

日本語 ver

This post is written Rolle’s theorem.
The mean-value theorem is proved by Rolle’s theorem.
I will write Mean-value theorem at a later.
I introduce Maximum principle because proving Rolle’s theorem need Maximum principle.

Maximum principle

It is very easy.
f is continuous function on bounded closed interval.$\implies$**
f have max value.**

Proof

This proof is difficult.
I write this proof in other posts.
Maximum Principle

Rolle’s theorem

f is continuous function on [a,b] and differentiable function on (a,b).
$$f(a) = f(b) \implies \exists ~~c ~~s.t~~ f'(c) = 0 , a<c<b$$

Proof

• f(x) is constant function
  $$\forall c \in (a,b) , f'(c) = 0$$
• else

when $\exists t ~~s.t~~f(a) < f(t)$, $\exists c ~~s.t~~ \max f(x) = f(c)$ by Maximum principle
I proof $f'(c)=0$
f is differentiable on $x = c$ and $f(c) >= f(c+h)$.
Thus
$$f'(c) = \lim_{h \rightarrow +0} \frac{f(c+h) - f(c)}{h} \leq 0$$
$$f'(c) = \lim_{h \rightarrow -0} \frac{f(c+h) - f(c)}{h} \geq 0$$
Therefore $$0 \leq \lim_{h \rightarrow -0} \frac{f(c+h) - f(c)}{h} =f'(c) = \lim_{h \rightarrow +0} \frac{f(c+h) - f(c)}{h} \leq 0$$
$$f'(c)=0$$
when$\exists t ~~s.t f(a)>f(t)$, proof is same.

Image

when $f(3)=f(5)$ , function have to turn.
This Turning point is c!!

Conclusion

Rolle’s theorem is used proof of the mean-value theorem.
I write mean-value theorem on other posts.
Mean-Value Theorem

Reference

https://mathtrain.jp/rolle