The Mean Value Theorem states that:

Given f is a function which is continuous and differentiable on the closed interval between a and b. Then there exists a point c in (a, b) such that:

```f`(c) = f(b) - f(a)
b - a
```

That is... if you have a continuous function f between two points x=a and x=b, there will always be a point x=c on the function where the derivative f`(c) is equal to the gradient of the chord joining f(a) and f(b).

Here is my poor attempt at illustrating this:

```y

^           ,
|         ,'    ,'
|  f(c) ,'    ,'
|     .x^`. ,'
|   ,'/   .\f(b)
| ,' /  ,'  \   /
|   / ,'     `_/
f(a)/,'
|----------------------> x
|  a  c     b
```

Oh dear, that is quite poor...

The 'curve' represents our function f(x). The points where f(a) and f(b) are joined by a dotted line, and f(c) is indicated by the x. The tangent to the curve at x=c is also represented by a dotted line. The point of the theorem is that it says there will always be a point x=c in the interval between x=a and x=b where the tangent to the curve at x=c is parallel to that of the line joining the points f(a) and f(b).

This theorem is useful for Convergence Testing and Proving a function has only one root in a given interval, amongst other things. (feel fre to /msg :)

Props to buo for teaching me this shit so I could node it.

An easier way to understand this (and, in my limited experience, teach it) is to explain it with a physical metpahor:

You leave point A heading for point B. Point A is the beginning of a turnpike--a highway with no exits except the tollbooths--and Point B is your exit, where your lovely vacation and some tawdry sex await you. (Don't leave out the tawdry sex! It makes the lesson easier to remember) Needless to say, you're in a hurry. Well, there are 65 miles to cover, and you cover them in less than an hour--actually 50 minutes. The cops--if there were any--didn't pull you, so you're safe, right? Wrong.

As you hand your toll ticket to the lady in the booth, she smiles, and with your receipt, hands you a ticket for speeding: 13 miles over the limit. How can this be?

Well, dummy, here it is: Let "a" and "b" be the times you passed through points A and B, respectively. And let f(t) be your position at time t. Your average velocity for the trip is given by

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

...or, Hmmmm... 78 mph. The Mean Value Theorem demands the existence of a time "c" at which your velocity (the derivative of f(t), or f'(t))was equal to 78mph. QED. Damn you, Joseph-Louis Lagrange!

Some points you need to go over with the class, once they stop giggling:
• The function f(t) is differentiable in the open interval - I should hope so! You always have a velocity, which is the derivative of your position. Condition one, check.
• The function f(t) is continuous in the closed interval - this means that when your car is sitting at the tollbooth and you're taking the first ticket, or when you're parked at point B getting your speeding ticket, you and your car and your position must all be defined. More to the point, you won't be getting on or off of the turnpike except at places able to measure your displacement and time; hence, for the purposes of this metaphor, both are always defined. Condition two, check.
To be just a little more succinct than Jurph, the mean value theorem says (in our speeding metaphor) "You can't average a speed without going that speed for at least a moment".

Remember rise over run? That's exactly what the right side of the function is. f(a) - f(b) is the rise of the function (the difference between the top and bottom) and a - b is the run (the length of the interval of the function. So that's the slope of the straight line connecting f(a) and f(b). However, our function f(x) might have lots of curves; it can pretty much go all over the place. If we take a point on the curve of f(x), and find the slope of the curve at that point, we call that f'(x) (we say "f prime of x"). So for a point c that is between a and b, we can find a slope for that point, and we call it f'(c). In a way

In our example, a, b, and c are times - the time we lease the first toll booth, the time we arrive at the second, and some time inbetween, and f(x) is the distance we have gon at that time. The right side of the equation above is average speed, and f'(x) turns out to be the speed we're going at at that moment. So f'(c) = (f(a) - f(b))/(a - b) says that we must have been going our average speed at some point during our trip.

(P.S.: do they really do that with toll booths? That's kind of evil...)

To be precise the two hypothesis you need for the mean value theorem to hold are:
1)The function f is continuous in the closed interval a,b
2)The function f is differentiable in the open interval a,b

Here's the proof:
Let k be (f(b)-f(a))/(b-a)
Consider the function defined as:
g(x)=f(x)-kx
Then it is easy to show that:
g(a)= (bf(a)-af(b))/(b-a) = g(b)
By Rolle's Theorem the derivative of g must vanish on the open interval a,b. Thus there must exist a point c such that
g'(c)=f'(c)-k = 0
At this point c:
f'(c)=k
Thats it. Q.E.D

The mean value theorem is the most important theorem of differential calculus; it is a crucial tool in the proof of such basic results as the inverse function theorem, Taylor's theorem and the equality of mixed partial derivatives.

The form of the mean value theorem discussed in the other writeups in this node is standard in first-year calculus books, but it does not generalize to higher dimensions. To see this, think about paths in the plane, that is functions γ: [0, 1] → R2. If the mean value theorem were true in the same form for functions into R2, it would follow that whenever γ(0) = γ(1), there is a point 0 ≤ t ≤ 1 with γ′(t) = 0. That is, any differentiable closed path has a critical point. Of course, this is not true. Just think of a circle traversed at constant speed. The problem is that when you have more than one range dimension, you can come back to your starting point without passing through zero: the intermediate value theorem is no longer true.

What is true is that you cannot travel more distance than your maximum derivative permits; Jurph's tollbooth speeding ticket example still works, with a caveat. The toll collector can say "you were going at least 78 miles per hour", but not specify the exact value of your derivative, that is, she can't say "at some point you were going exactly 78 miles per hour in the direction between your endpoints". For remember, in two dimensions, a derivative is a velocity vector.

Thus, to establish the general mean value theorem, we just have to remove the appeal to the intermediate value theorem at the end, and prove an inequality rather than an existence theorem:

Mean value theorem. Suppose V is a (not necessarily finite dimensional) Banach space, and f: [0, 1] → V is a differentiable function. For any points x ≠ y ∈ [0, 1], put xt = (1 − t) x + t y. Then

||f(y) − f(x)||V ≤ |y − x| sup {||f′(xt)||V | 0 ≤ t ≤ 1}.

This isn't the "linear transformation" derivative, which lives in V* = L(R; V), but the "first-year calculus" derivative, limh→0 (f(x+h) − f(x)) / h, which is a point of V. It turns out not to matter. If you don't understand this remark, ignore it and use your calculus intuition.

The words are bigger, but all this says is that if f is differentiable, its change between two points (the left side) is limited by the maximum speed of that change between the points (the right side).

Proof of the theorem. The mean value theorem is a fundamental theorem of analysis which depends crucially on the completeness of the real numbers, and the proof follows a common pattern for such theorems. A one-sentence summary is, "Suppose it were false, then by the least upper bound axiom there must be a highest number for which it is still true; but then it must extend a bit past this point." Formally, choose any

M > sup {||f′(xt)||V | 0 ≤ t ≤ 1};

that is, M exceeds our maximum speed. Let E be the set of all t for which f doesn't break the higher speed limit given by M:

E = {t ∈ [0, 1] | ||f(xt) − f(x)||V ≤ M t |y − x|}.

We want to show that 1 ∈ E, because this implies that

||f(y) − f(x)||V ≤ M |y − x|

and since this is true independent of M, ||f(y) − f(x)||V is actually bounded by the infimum of all the M we could have chosen; that is

||f(y) − f(x)||V ≤ |y − x| sup {||f′(xt)||V | 0 ≤ t ≤ 1},

which is the desired conclusion. Hence let s be the largest element of E, and let us show that s = 1. Suppose not, and choose t > s, t ∈ [0, 1]. By the triangle inequality,

||f(xt) − f(x)||V ≤ ||f(xt) − f(xs)||V + ||f(xs) − f(x)||V.

The second term here is no problem: since we supposed s ∈ E, it must be that

||f(xs) − f(x)||V ≤ M s |y − x|.

As for the first, remember that M exceeds the derivative of f at every point of [0, 1], in particular at xs: M > ||f′(xs)||V. Thus, if t is sufficiently close to s,

||f(xt) − f(xs)||V < M |xt − xs| = M (t − s) |y − x|,

since ||f(xt) − f(xs) − f′(xs)(xt − xs)||V = o(|xt − xs|) = o(t − s) as t approaches s. (This is the definition of the derivative.) Hence

||f(xt) − f(x)||V ≤ M (t − s) |y − x| + M s |y − x| = M t |y − x|,

that is t ∈ E, which contradicts the choice of s as the largest element of E. Hence s = 1 and we are done.     ///

This proof comes in essence from volume 1 of Lars Hörmander's masterpiece The analysis of linear partial differential operators. A more accessible reference for general differential calculus is Foundations of modern analysis by Jean Dieudonné.

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