Thread: Can Someone Explain the Mean Value Theorem to me?

Originally Posted by drain

Let f be continous on [a,b] and differenciable on (a,b).

There shall exists a point x in (a,b) such that,

f'(x) = MEAN VALUE.

The mean value is the average, that is,

(f(b)-f(a))/(b-a)

Hence,

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

lol I get that. I mean why is it useful? Like some sorts of applications of it?

Originally Posted by drain

I mean why is it useful? Like some sorts of applications of it?

What do you mean?

It is not useful in applications. It is useful in proofs. It is perhaps, the most important theorem in calculus. So many proofs are derived from it.

Originally Posted by ThePerfectHacker

What do you mean?

Is this a math joke?

I guess my question is like, what sorts of problems will we be asked to solve with it? (In a first semester calculus course)

Originally Posted by drain

Is this a math joke?

Yes. I have a talent for coming up with math jokes (though not necessarily funny).

For example, look at my signature. "Elementary Complex Analysis". There is a topic in math called Complex analysis, thus if you call your book like that it sounds like it is both elementary and complicated. I found that funny.

I guess my question is like, what sorts of problems will we be asked to solve with it? (In a first semester calculus course)

You be asked to find what c makes the mean value.

Example,

Given f(x)=x^2 on [-1,1]
Find a number c in (-1,1) such that,
f'(c)=Mean Value

Thanks man. BTW, are you from the past? I didn't know the USSR was still around~!

Originally Posted by drain

Thanks man. BTW, are you from the past? I didn't know the USSR was still around~!

I came from the Union of Soviet Socialist Republic. But its own power destroyed it (like a blackhole) so today it no longer exists. But I can still say I came from there.

Well they have good boxers from that area.