The mean value theorem

As soon as one draws the picture for Rolle's Theorem one ought to wonder what would happen in $f(a)\neq f(b)$. It sure looks like some tangent line should have a slope equal to the slope of the line segment joining (a,f(a)) and (b,f(b)).

Theorem 7957 (The Mean Value Theorem)

Suppose a < b, $f:[a,b]\rightarrow (-\infty,\infty)$ is continuous, f'(x) exists for every $x\in(a,b)$. Then for some $c\in(a,b)$ we have f'(c)(b-a) = f(b) - f(a).

The geometric interpretation follows by dividing through by b-a. The proof is a trick, plain and simple. A rabbit out of a hat if you will. Define $R:[a,b]\rightarrow(-\infty,\infty)$ by R(x) = (f(x)-f(a))(b-a) - (f(b)-f(a))(x-a). R (for rabbit, or for Rolle, take your pick) satisfies all the hypotheses of Rolle's Theorem, so for some $c\in(a,b)$ we have R'(c) = 0. We can calculate

$$\begin{array} {rcl} R'(x) &=& (f'(x) - 0)(b-a) - (f(b)-f(a))(1-0)\\ & = & f'(x)(b-a) - (f(b)-f(a))\end{array}$$

and so

0 = R'(c) = f'(c)(b-a)- (f(b) - f(a))

This is the same as f'(c)(b-a) = f(b) - f(a). QED

Let me repeat, no one cares what the value of c is. It is enough to know it is there.

Usually, the Mean Value Theorem is interpreted in terms of slopes if we look at the graph of y = f(x):

Just as importantly, it can be interpreted in terms of area if we look a the graph of y = f'(x):