converse of mean value theorem?

[xalvyn]

hi...

was wondering, does the converse of the mean value theorem hold?

that is, given any function f(x), and a tangent to the graph of y = f(x) at any point, can we always construct two points on the graph (with the tangent lying between) such that the line joining them is parallel to the tangent?

thanks to anyone who can share some insight..

[mathwonk]

look at a picture. try two cases, convex graph, or not. lok at y = x^3 at (0,0).

[rudinreader]

One sufficient condition for a MVT converse to hold at a is that f'(x) is strictly increasing in a neighborhood of a. "Monotonic" increasing is not enough, as in the example of f(x) = x^2, x > 0, f(x) = 0, x \leq 0. f'(0) = 0, but for any x < 0 < y, we have f(y)-f(x) = f(y) > 0 = f'(0)(y-x).

Theorem: Fix a number a. If f' is strictly increasing in some neighborhood (a-e,a+e) of a, then there exists x < a < y such that (f(y)-f(x))/(y-x) = f'(a).

proof:

By the mean value theorem, there exists c in (a-e,a), d in (a,a+e), D in (a-e,a+e) such that \frac{f(a)-f(a-e)}{e} = f'(c) < f'(a) < f'(d) = \frac{f(a+e)-f(a)}{e}, \ \frac{f(a+e)-f(a-e)}{2e} = f'(D).

If D = a, we are done. Otherwise, assume a < D < a + e. (a-e<D<a case done similarly.)

Consider the lines L_1(t),L_2(t),L_3(t) with slopes f'(c)<f'(a)<f'(D) respectively, such that L_i(a-e) = f(a-e). (I.e., they all start at the point (a-e,f(a-e)).) To the right of the point a - e we have L_1(t) < L_2(t) < L_3(t). L_2(t)-f(t) is a continuous function on [a,a+h], L_2(a)-f(a) = L_2(a) - L_1(a) > 0, L_2(a+e)-f(a+e) = L_2(a+e) - L_3(a+e) < 0, so by continuity, L_2(y) = f(y) for some y in (a,a+h). It follows that \frac{f(y)-f(a-h)}{y-(a-h)} = f'(a).