* Rolle's Theorem and the Mean Value Theorem

Rolle's theorem is an important basic result about differentiable functions. Like many basic results in the calculus it seems very obvious. It just says that between any two points where the graph of the differentiable function f (x) cuts the x-axis there must be a point where f'(x) = 0. The following picture illustrates the theorem.

Figure 1.9: Rolle: somewhere between a and b, the graph must be flat.

Like most important theorems, Rolle's theorem has to be stated rather carefully in order to make sure that it is true. I will now state a simple, but precise, version of the theorem:

Theorem 1.40 (Rolle's Theorem)   Suppose that f (x) is a differentiable function whose derivative is a continuous function. Suppose that f (a) = 0 and f (b) = 0 (with a < b, let's say). Then there must be at least one point c between a and b (a < c < b) at which f'(c) = 0.

(The precise behaviour of f (x) outside the interval a$\le$x$\le$b is not really relevant and the theorem can be stated in a more general form.)

Suppose we accept this result as being intuitively obvious (it can be proved). Now consider the following extension.

As before, let f (x) be a function that is differentiable and whose derivative is continuous. Choose any points a < b and define a new function F(x) by the formula

F(x) = f (x) - f (a) - $${\frac{x-a}{b-a}}$$(f (b) - f (a))

F is also a differentiable function with a continuous derivative. The derivative is easy to calculate--it is

F'(x) = f'(x) - $${\frac{f(b)-f(a)}{b-a}}$$.

Now notice that F(a) = F(b) = 0 (check this). So we can apply Rolle's theorem to F(x) and say that there must be a point c between a and b at which F'(c) = 0. But this says that there is a point c between a and b at which

0 = f'(c) - $${\frac{f(b)-f(a)}{b-a}}$$

Rearranging this equation we get the following important theorem

Theorem 1.41 (Mean Value Theorem)   If f (x) is a differentiable function with a continuous derivative and if a < b are any two points then there is a point c between a and b at which

f'(c) = $${\frac{f(b)-f(a)}{b-a}}$$

This result is also `geometrically obvious' once you draw the right picture.

Figure 1.10: Mean Value Theorem: somewhere between a and b, the graph must be parallel to the chord AB.

The expression (f (b) - f (a))/(b - a) is just the slope of the chord AB. So the theorem is saying that there must be a point c between a and b at which the slope of the graph is equal to the slope of the chord AB.

This may seem very simple--it is very simple--but it is also very important. In most of its applications we think of the result in the following way, which is easily derived.

f (b) = f (a) + (b - a)f'(c)

(Note that the theorem does not tell us what c is--just that it must be there.)

Here's a simple application of this theorem. Suppose that f'(x) > 0 between a and b (a < b). Then (b - a)f'(c) must be positive (whatever c is). So f (b) must be bigger than f (a). So, in an obvious sense, the function must be increasing.

Here's another typical application. Let f (x) = tan(x). Then f'(x) = 1/cos²(x) > 1. So we must have

tan(b) - tan(a) > b - a

This result is rubbish. For example, tan($\pi$) = tan(0) = 0. But 0 - 0 is not bigger than $\pi$ - 0. What's gone wrong?

The problem here is that tan(x) is not a differentiable function between 0 and $\pi$--it blows up at $\pi$/2. When using these theorems, no matter how `obvious' they are, you really must check that the conditions of the theorem are precisely satisfied.