Proof of an Inversion of the Mean Value Theorem
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 in (a,b) such that:
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 . (Mean Value Theorem). But I feel like it's not applicable in this proof..
Suggestions please?(Doh) (Headbang) Thank you!
Re: Proof of an Inversion of the Mean Value Theorem
You're right, this seems to be harder than it looks. Here's an outline solution.
Originally Posted by limddavid
Define a new function g on the interval [a,b] by Then g is also continuous on [a,b] and differentiable on (a,b). If there are two distinct points in [a,b] with then , 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 for all x in [a,b], or for all x in [a,b]). But , 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 with the desired property must exist.