# Proof of an Inversion of the Mean Value Theorem

• Nov 28th 2011, 08:22 PM
limddavid
Proof of an Inversion of the Mean Value Theorem
Question:

Let f:[a,b]-> R be a continuous function on the closed interval [a,b] and differentiable on the open interval (a,b). Show that for any c in (a,b) that is not a point of maximum of minimum for f', there exist $\displaystyle x_1,x_2$ in (a,b) such that:

$\displaystyle f'(c)=\frac{f(x_2)-f(x_1)}{x_2-x_1}$

This seems so matter-of-fact to me, and I know it's simply a sort of inversion of the Mean value Theorem, but I can't see how I would approach this. It looked really easy at first sight, but I'm lost..

All I know is that if there's an interval (x1,x2), there is at least one point such that the derivative at that point is equal to $\displaystyle \frac{f(x_2)-f(x_1)}{x_2-x_1}$. (Mean Value Theorem). But I feel like it's not applicable in this proof..

• Nov 30th 2011, 12:16 AM
Opalg
Re: Proof of an Inversion of the Mean Value Theorem
Quote:

Originally Posted by limddavid
Question:

Let f:[a,b]-> R be a continuous function on the closed interval [a,b] and differentiable on the open interval (a,b). Show that for any c in (a,b) that is not a point of maximum of minimum for f', there exist $\displaystyle x_1,x_2$ in (a,b) such that:

$\displaystyle f'(c)=\frac{f(x_2)-f(x_1)}{x_2-x_1}$

This seems so matter-of-fact to me, and I know it's simply a sort of inversion of the Mean value Theorem, but I can't see how I would approach this. It looked really easy at first sight, but I'm lost..

You're right, this seems to be harder than it looks. Here's an outline solution.

Define a new function g on the interval [a,b] by $\displaystyle g(x) = f(x) - f'(c)(x-c).$ Then g is also continuous on [a,b] and differentiable on (a,b). If there are two distinct points $\displaystyle x_1,x_2$ in [a,b] with $\displaystyle g(x_1) = g(x_2)$ then $\displaystyle \frac{f(x_2)-f(x_1)}{x_2-x_1} = f'(c)$, as required.

Suppose that there do not exist two such points, so that g takes each of its values only once. In other words, g is an injective function on the interval [a,b]. Use the intermediate value theorem to deduce that g must be a monotone function on the interval. It then follows that g'(x) must always have the same sign (either $\displaystyle g'(x)\geqslant0$ for all x in [a,b], or $\displaystyle g'(x)\leqslant0$ for all x in [a,b]). But $\displaystyle g'(c) = 0$, which means that g' has either a maximum or a minimum at c, and hence so does f' (because g' and f' only differ by a constant).

Taking the contrapositive, it follows that if f' does not have a max or a min at c, then points $\displaystyle x_1,x_2$ with the desired property must exist.