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