Proof of a nice theorem [for young people]

[Paolo902]

Let $f$ be a differentiable function in $[a,b]$. We know that $f(a)=a$ and $f(b)=b$. Prove that there exist two different points $r,s \in (a,b)$ such that

$1/(f'(r))+1/(f'(s))=2$

Hint: use L. theorem.

[Paolo902]

Great. Congratulations, very good proof. I did the same.

[alberto.cena]

If you like to be in front of my proof, click below

Function $f$ is continuous on $[a,b]$ and $\frac{a+b}{2}$ is a value between $f(a)$ and $f(b)$ so, by Intermediate Value Theorem, there exists $c \in (a,b)$ such that $f(c) = \frac{a+b}{2}$.
We use Lagrange Theorem in the intervals $[a,c]$ and $[c,b]$. There exist $r \in (a,c)$ and $s \in (c,b)$ such that
$f'(r) = \frac{f(c) - f(a)}{c-a} = \frac{ \frac{a+b}{2} - a }{c-a} = \frac{ b - a }{2(c-a)}$
and
$f'(s) = \frac{f(b) - f(c)}{b-c} = \frac{b- \frac{a+b}{2} }{b-c} = \frac{ b - a }{2(b-c)}$.
then
$\frac{1}{f'(r)} + \frac{1}{f'(s)} = \frac{2(c-a) + 2(b-c)}{b-a}=2$

looking at the sea is better, I know it

[Paolo902]

"Fioravante Patrone":[quote="Paolo90"]
Great. Just one question: where is the sea?
[young man]

In front of me. [/quote]


You are incredible!

@5inGold:

Yes, very good hint. Go on, if you want.

[alberto.cena]

My hint

also use Intermediate Value Theorem

[Fioravante Patrone1]

"Paolo90":
Great. Just one question: where is the sea?
[young man]

In front of me.

[Paolo902]

"Fioravante Patrone":A relevant piece of information about calculus/mathematical analysis:

derivable functions do not exist, in English. Sorry

I didn't know. I beg your pardon, sir. Thanks for your correction.

"Fioravante Patrone":

[ol' man]

Great. Just one question: where is the sea?

[young man]

[Fioravante Patrone1]

A relevant piece of information about calculus/mathematical analysis:

derivable functions do not exist, in English. Sorry


[ol' man]