# Thread: Mean Value Theorem Proof

1. ## Mean Value Theorem Proof

Use the Mean Value Theorem to prove that for all a,b that belongs to the reel, we have :

$\displaystyle |sin(b)-sin(a)|$ $\displaystyle \leq$ $\displaystyle | b-a |$

2. Originally Posted by Larrioto
Use the Mean Value Theorem to prove that for all a,b that belongs to the reel, we have :

$\displaystyle |sin(b)-sin(a)|$ $\displaystyle \leq$ $\displaystyle | b-a |$
well, if you let $\displaystyle f(x)=\sin x,$ then by mean value theorem there exists $\displaystyle c$ between $\displaystyle a,b$ such that $\displaystyle f(b)-f(a)=(b-a)f'(c).$ that means $\displaystyle \sin b - \sin a = (b-a)\cos c.$

so $\displaystyle |\sin b - \sin a|=|b-a||\cos c| \leq |b-a|,$ because $\displaystyle |\cos c| \leq 1.$ it wan't very difficult, was it?

3. There is a second part to this problem :

Assume that f : $\displaystyle \Re \mapsto \Re$ is a differentiable function and that there exists a real number M > 0 such that |f '(x)|$\displaystyle \leq M$ for all x in the reel then show that :

$\displaystyle |f(a) - f(b)|\leq M|b-a|$for all x in the reel

4. Originally Posted by Larrioto
There is a second part to this problem :

Assume that f : $\displaystyle \Re \mapsto \Re$ is a differentiable function and that there exists a real number M > 0 such that |f '(x)|$\displaystyle \leq M$ for all x in the reel then show that :

$\displaystyle |f(a) - f(b)|\leq M|b-a|$for all x in the reel
Choose any value $\displaystyle c$, we know there exists a corresponding interval $\displaystyle [a,b]$ such that $\displaystyle |f'(c)|=\left|\frac{f(b)-f(a)}{b-a}\right|\leqslant M$ and since $\displaystyle c$ was arbitrary this must work for every interval $\displaystyle [a,b]$.

Another almost identical method would be to assume that there exists an interval $\displaystyle [a,b]$ such that $\displaystyle \left|\frac{f(b)-f(a)}{b-a}\right|>M$. But since $\displaystyle f$ is differentiable we can find a $\displaystyle c\in(a,b)$ such that $\displaystyle |f'(c)|=\left|\frac{f(b)-f(a)}{b-a}\right|\implies |f'(c)|>M$ which is a contradiction.

This is called Lipschitz's continuity by the way.

If you have any questions at all feel free to ask

5. Originally Posted by Mathstud28
Choose any value $\displaystyle c$, we know there exists a corresponding interval $\displaystyle [a,b]$ such that $\displaystyle |f'(c)|=\left|\frac{f(b)-f(a)}{b-a}\right|\leqslant M$ and since $\displaystyle c$ was arbitrary this must work for every interval $\displaystyle [a,b]$.

Another almost identical method would be to assume that there exists an interval $\displaystyle [a,b]$ such that $\displaystyle \left|\frac{f(b)-f(a)}{b-a}\right|>M$. But since $\displaystyle f$ is differentiable we can find a $\displaystyle c\in(a,b)$ such that $\displaystyle |f'(c)|=\left|\frac{f(b)-f(a)}{b-a}\right|\implies |f'(c)|>M$ which is a contradiction.

This is called Lipschitz's continuity by the way.

If you have any questions at all feel free to ask
Do you mean that whatever number we choose to be c , it's derivative will be 0 , therefore we can say that if $\displaystyle M\geq 0$ then it is $\displaystyle \geq \left|\frac{f(b)-f(a)}{b-a}\right|$ ?

6. Originally Posted by Larrioto
Do you mean that whatever number we choose to be c , it's derivative will be 0 , therefore we can say that if $\displaystyle M\geq 0$ then it is $\displaystyle \geq \left|\frac{f(b)-f(a)}{b-a}\right|$ ?
Im not quite sure what you mean by that...If the first proof doesn't make sense consider the second one. It reads better anyways.

7. ok , i get it , it's just that f(c) can be replaced by f(x), that was confusing me ...

8. Originally Posted by Mathstud28

Choose any value $\displaystyle c$, we know there exists a corresponding interval $\displaystyle [a,b]$ such that $\displaystyle |f'(c)|=\left|\frac{f(b)-f(a)}{b-a}\right|$ ...
this, although an interesting claim, but is not what mean value theorem says: you choose an interval [a,b] and then you find a < c < b that satisfies the condition. what you're doing here is

reversing things, i.e. you choose c and then you claim that there exists an interval [a,b] containing c with that condition! i'm not sure if this is always possible!

9. Originally Posted by NonCommAlg
this, although an interesting claim, but is not what mean value theorem says: you choose an interval [a,b] and then you find a < c < b that satisfies the condition. what you're doing here is

reversing things, i.e. you choose c and then you claim that there exists an interval [a,b] containing c with that condition! i'm not sure if this is always possible!
Mathworld has an "iff" in it :S...that is why I supplied a second proof that did not require that just in case. Sorry if there has been any confusion.

10. Originally Posted by Mathstud28

Mathworld has an "iff" in it :S...that is why I supplied a second proof that did not require that just in case. Sorry if there has been any confusion.
here's a simple counter-example: let $\displaystyle f(x)=x^3$ and choose $\displaystyle c=0.$ geometrically it's clear that there are no $\displaystyle a \neq b$ such that $\displaystyle f'(0)=\frac{f(b)-f(a)}{b-a}.$ you can also prove it very easily:

if there exist such $\displaystyle a,b,$ then: $\displaystyle 0=\frac{b^3-a^3}{b-a}=a^2+ab+b^2,$ which is possible iff $\displaystyle a=b=0,$ because: $\displaystyle 0=2a^2 + 2ab+2b^2=(a+b)^2+a^2+b^2.$

regarding Larrioto's question:

by mean value theorem for any real numbers $\displaystyle a,b,$ there exists $\displaystyle c$ (between $\displaystyle a,b$) such that $\displaystyle f(b)-f(a)=(b-a)f'c).$ thus: $\displaystyle |f(b)-f(a)|=|b-a||f'(c)| \leq M|b-a|.$