Converse of mean value theorem?

In summary: Now, consider the line 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
  • #1
xalvyn
17
0
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..
 
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  • #2
look at a picture. try two cases, convex graph, or not. lok at y = x^3 at (0,0).
 
  • #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].
 
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1. What is the converse of mean value theorem?

The converse of mean value theorem is a statement that is logically equivalent to the mean value theorem. It states that if a function is continuous on a closed interval and its derivative is never equal to zero on that interval, then the function is strictly monotonic on that interval.

2. What is the significance of the converse of mean value theorem?

The converse of mean value theorem is significant because it provides a way to check if a function is strictly monotonic on a given interval. This is useful in many real-world applications, such as in optimization problems.

3. How is the converse of mean value theorem proven?

The converse of mean value theorem can be proven using a similar approach as the mean value theorem. By using the definition of continuity and the intermediate value theorem, it can be shown that if a function is continuous and its derivative is never equal to zero on an interval, then the function must be strictly monotonic on that interval.

4. Can the converse of mean value theorem be applied to all functions?

No, the converse of mean value theorem can only be applied to continuous functions on a closed interval. Additionally, the derivative of the function must also be defined on that interval and never equal to zero.

5. Is the converse of mean value theorem an if and only if statement?

Yes, the converse of mean value theorem is an if and only if statement. This means that if the conditions of the theorem are met, then the conclusion must be true, and vice versa.

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