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**LuisRodg****Real Member**- Registered: 2007-10-23
- Posts: 322

Hello everyone.

Today in class we were introduced to the Rolle's Theorem in which it states that if a function f(x) is continuous on [a,b] and differentiable on (a,b) then there exists a C in (a,b) in which f'(C) = 0. Now, I understand this perfectly fine but Im wondering why do I need a theorem to tell me that something exists if I can just find the derivative, set it equal to zero and I will see myself if it has a C inside of (a,b) in which f'(C)=0. I dont know but that theorem seems kind of redundant and the professor really emphasized on it as being really important.

We were also introduced to the Mean-Value Theorem which is giving me a little more trouble grasping it. The MVT states that if f a function f(x) is continuous on [a,b] and differentiable on (a,b) then there exists a C in (a,b) in which f'(C) = (f(b)-f(a))/(b-a)

Lets say that:

Z = (f(b)-f(a))/(b-a)

So I know that there exists a C in (a,b) in which f'(C) = Z. And? What does that tell me? How does that help me?

Thanks guys. Just try to clear up my head please. I love math and I usually learn it by understanding its concepts rather than memorizing some formulas so when im asked to just memorize and im not told how that helps me...It just makes it harder for me.

Thanks.

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**yttrium88****Member**- Registered: 2005-12-01
- Posts: 20

First of all, Rolle's Theorem allows for a very easy proof of the MVT. The MVT is essential to many proofs in analysis, for example, relating the second derivative definition of concave up/down with a secant definition. There are many, many instances when it is extremely difficult, if not impossible, to calculate the derivative of a differentiable function, but an application of MVT may be able to give you all you need.

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

LuisRodg wrote:

Today in class we were introduced to the Rolle's Theorem in which it states that if a function f(x) is continuous on [a,b] and differentiable on (a,b) then there exists a C in (a,b) in which f'(C) = 0.

Not quite. You left out one important condition: f(*a*) = f(*b*).

In many cases, you have a fiunction that is easy to differentiate, so you can easily find its derivative. However, there are many other functions that are not that easy to differentiate. For example, suppose you have

in the interval [0,1]. Are you sure you really want to differentiate that horrible thing to prove that the derivative is 0 somwhere in the interval (0,1)? I wouldnt. Using Rolles theorem (noting that the conditions for the theorem are all satisfied), however, we can be sure that there is a *c* ∈ (0,1) such that f′(*c*) = 0 without doing any dirty work.

*Last edited by JaneFairfax (2007-11-16 15:09:16)*

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**LuisRodg****Real Member**- Registered: 2007-10-23
- Posts: 322

Yes sorry I forgot f(a)=f(b)

Ahhh I get it. Now if the professor would had said that...

In the context that he explained it, it had no practical use. It just seemed as another formula to memorize. Now I get its purpose.

Thanks guys.

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**LuisRodg****Real Member**- Registered: 2007-10-23
- Posts: 322

I seem to have stumbled upon a problem:

Use the MVT to show that if f is differentiable on an interval I, and

if |f'(x)|<=M for all values of x in I

then

|f(x)-f(y)| <= |x-y|

No idea how to do this...

*Last edited by LuisRodg (2007-11-16 15:46:50)*

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

I think everyone is missing the big picture here. One of the beautiful aspects of mathematics is abstraction. 1 + 1 = 2 not only for poker chips, apples, or flying squires. 1 + 1 = 2 for *everything*. In the same sense, we don't just want to be able to show that x^2 + x + 3 is continuous, nor x^5 + 5x + 1. But rather, we want to show that all polynomials are continuous.

When it comes to analysis, we don't just talk about a function. We talk about all functions. We are talking about all functions that are continuous on [a, b] and differentiable on (a, b). The function Jane wrote out is a piece of cake, honestly. There are much weirder and harder things. We don't want to do a seperate proof for each of these, that would take forever (literally, forever).

It's of extreme importance that you understand the power of abstraction. It is one of mathematics greatest abilities.

Now as for your proof, it would be pretty hard without the MVT. I believe the typical approach should work, so just start out as any proof would:

Let |f'(x)| <= M for all x in I. Pick arbitrary a < b in I. Then there exists an x such that f(b)-f(a) = f'(x)(b-a) by the MVT. Taking absolute value of both sides we get that |f(b) - f(a)| = |f'(x)||b-a|.

Now you should be able to finish it off. Do you realize what this is actually proving? I mean in more or less english.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

I forgot to add in this: What the MVT is basically trying to tell you in very loose terminology is that restricting the derivative also restricts the function. In other words, the derivative is a value at a point, but it can still have great impact on values at a distance. For example, here is another simple proof which pretty much requires the use of the MVT:

Prove that if f'(x) > 0 for all x in (a, b), then f is monotone increasing on (a, b).

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**LuisRodg****Real Member**- Registered: 2007-10-23
- Posts: 322

Hey Ricky.

I wasn't able to finish it off...Could you help me with that? I never had a formal introduction with proofs so im pretty much clueless...Is there class that teaches you proofs or is that something that you learn by yourself?

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

I believe that should be a 1 rather than an M, otherwise the theorem is false.

|f(b) - f(a)| = |f'(x)||b-a|

So given that |f'(x)| <= 1, it must be that |f'(x)||b - a| <= |b -a|. Therefore, |f(b)-f(a)| <= |b-a|

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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