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Converse of mean value theorem?

  1. Jul 18, 2006 #1
    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..
     
  2. jcsd
  3. Jul 18, 2006 #2

    mathwonk

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    look at a picture. try two cases, convex graph, or not. lok at y = x^3 at (0,0).
     
  4. Nov 23, 2007 #3
    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 [tex]f(x) = x^2, x > 0, f(x) = 0, x \leq 0[/tex]. 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 [tex]\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)[/tex].

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

    Consider the lines [tex]L_1(t),L_2(t),L_3(t)[/tex] with slopes [tex]f'(c)<f'(a)<f'(D)[/tex] respectively, such that [tex]L_i(a-e) = f(a-e)[/tex]. (I.e., they all start at the point (a-e,f(a-e)).) To the right of the point a - e we have [tex]L_1(t) < L_2(t) < L_3(t)[/tex]. [tex]L_2(t)-f(t)[/tex] is a continuous function on [tex][a,a+h][/tex], [tex]L_2(a)-f(a) = L_2(a) - L_1(a) > 0[/tex], [tex]L_2(a+e)-f(a+e) = L_2(a+e) - L_3(a+e) < 0[/tex], so by continuity, [tex]L_2(y) = f(y)[/tex] for some y in (a,a+h). It follows that [tex]\frac{f(y)-f(a-h)}{y-(a-h)} = f'(a)[/tex].
     
    Last edited: Nov 23, 2007
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