# A note to Lagrange mean value theorem.

[section]1. Introduction

Lagrange mean value theorem is one of most important theorems in calculus. It is an important tool to study the property of function. It sets up a "bridge" between function and derivable function. But Lagrange mean value theorem only affirms existence of "intermediate point" [xi], not states its other properties. Recently, some people have studied asymptotic qualities of [xi] and obtained good results. In this paper, on the base of summarizing related results, we study the monotonicity, continuity and derivable property of [xi].

[section]2. Several lemmas

Lemma 1. (Lagrange mean value theorem) Assume that f(x) is continuous on the closed interval [a, b] and derivable on the open interval (a, b). Then for [for all]x [member of] (a, b], there at least exists a point [xi] [member of] (a, x), such that

f'([xi]) = [f(x) - f(a)]/[x - a], (1)

In reference [1-4], they obtain some important results, among them the most typical result is as follows.

Lemma 2. Assume that f(x) is a first-order continuous and derivable function on interval [a, b], and f'(x) - f'(a) is a-order infinitesimal of x - a, where a > 0. Then [xi] with (1) has asymptotic estimator

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[section]3. Main results and their proofs

Assume that f(x) satisfies the conditions of lemma 1 when x [member of] [a, b], then for [for all]x [member of] (a, b], when a is fixed, the "intermediate point" [xi] changes with x and [xi] has important properties as follows.

Theorem. Assume that f(x) is continuous on the closed interval [a, b] and derivable on the open interval (a, b), and f(x) is second-order continuous and derivable in interval (a, b), f'(x) is strictly monotone in interval (a, b), f"(x) is always positive or always negative in interval (a, b), then we have

(i) The point [xi] with (1) is a uniform function about x, notes [xi] = [xi](x);

(ii) [xi] = [xi](x) is a monotone increasing function about x;

(iii) [xi] = [xi](x) is a continuous function about x;

(iv) [xi] = [xi](x) is a derivable function about x and

[xi]'(x) = [f'(x) - f'([xi](x))]/[(x - a)f"([xi](x)). (3)

Proof. (i) Because f'(x) is strictly monotone in interval (a, b), we can easily prove that

(i) holds.

(ii) Assume that f'(x) is monotone increasing in interval (a, b), for [for all][x.sub.1], [x.sub.2] [member of] (a, b) and [x.sub.1] < [x.sub.2], by given condition and lemma 1 we have

f([x.sub.2] - f(a) = f'([xi]([x.sub.2]))([x.sub.2] - a), f([x.sub.1] - f(a) = f'([xi]([x.sub.1]))([x.sub.1] - a).

So f([x.sub.2]) - f([x.sub.1]) = [f'([xi]([x.sub.2])) - f'([xi]([x.sub.1])) ([x.sub.1] - a) + f'([xi]([x.sub.2]))([x.sub.2] - [x.sub.1]). Also by f([x.sub.2]) - f([x.sub.1]) = f'([eta])([x.sub.2] - [x.sub.1]), hence

where [x.sub.1] < [eta] [x.sub.2], a < [xi]([x.sub.1]) < [x.sub.1], a < [xi]([x.sub.2]) < [x.sub.2], a < [x.sub.1] < [x.sub.2] < b. Because f'(x) is monotone increasing, we have f'([eta]) > f'([xi]([x.sub.1])), and

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But f'([eta])([x.sub.2] - [x.sub.1]) = f([x.sub.2]) - f([x.sub.1]), f'([xi]([x.sub.1]))([x.sub.1] - a) = f([x.sub.1]) - f(a), f'([xi]([x.sub.2]))([x.sub.2] - a) = f([x.sub.2]) - f(a),

So f'([eta]) - f'([xi]([x.sub.2]))]([x.sub.2] - a) > 0, f'([eta]) - f'([xi]([x.sub.2])) > 0, f'([xi])([x.sub.2])) - f'([xi]([x.sub.1])) > 0.

By monotone increasing f'(x), we have

[xi]([x.sub.2]) > [xi]([x.sub.1]).

When f'(x) is monotone decreasing in interval (a, b), the proof is same to above proof. (iii) By given conditions and Lemma 1, we have

f'([xi](x)) = [f(x) - f(a)]/[x - a],

and

f'([xi](x + h)) = [f(x + h) - f(a)]/[x + h - a]

so

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By (ii), we know that [xi] = [xi](x) is a monotone function about x. When h [not equal to] 0, also by Lemma 1, we have

f'([eta](x + h)) - f(xi](x)) = f"([eta])[[xi](x + h) - [xi](x)],

where [eta] lies between [xi](x + h) and [eta](x), so

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That is to say, [xi] = [xi](x) is continuous in interval (a, b).

(iv) By definition of derivative, we have

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Thus, the proof is complete.

Finally, it deserves to mention that, in the course of teaching higher mathematics, some people consider the application of this theorem as an incorrect proof method. In fact, this thinking is not right. we give an example.

Example. Assume that f(x) is quadratic differentiable on the closed interval [a, b] and f"(x) > 0, we try to prove that the function

g(x) = [f(x) - f(a)]/[x - a]

is a monotone increasing function in the open interval (a, b). Proof By Lemma 1, we have

f(x) - f(a) = f'([xi](x))(x - a), a < [xi](x) < x [less than or equal to] b.

So

g(x) = [f(x) - f(a)]/[x - a] = f'([xi](x)).

By above theorem we know that [xi] = [xi](x) is a monotone increasing and derivable function in the open interval (a, b), also by derivative rules of compound function, we have

g'(x) = f"([xi](x))[xi]'(x),

but f'(x) > 0 and [xi]'(x) > 0, so

g'(x) = f"([xi](x))[xi]'(x) > 0.

Thus, the proof is complete.

References

 Wang Zewen, etc, the Inverse Problem to Mean Value Theorem of Differentials and Its Asymptotic Property, Journal of East China Geological Institute, 26(2003), No.2, 126-128.

 Li Wenrong, Asymptotic Property of intermediate point to Mean Value Theorem of Differentials, Mathematics in Practice and Theory, 15(1985), No.2, 46-48.

 Zhang Guangfan, A Note on Mean Value Theorem of Differentials, Mathematics in Practice and Theory, 18(1988), No.2, 87-89.

 Dai Lihui, etc, the Trend of Change of to Mean Value Theorem of Differentials, Chinese Journal of Engineering Mathematics, 10(1994), No.4, 178-181.

 Wu Liangsen, etc, Mathematics Analysis, East China Normal University Publisher, 8(2001), No.3, 119-123.

Dewang Cui, Wansheng He and Hongming Xia

Department of Mathematics, Tianshui Normal University,

Tianshui, Gansu 741001, P.R.China

(1) This work is supported by the Gansu Provincial Education Department Foundation 0608-04.
Author: Printer friendly Cite/link Email Feedback Cui, Dewang; He, Wansheng; Xia, Hongming Scientia Magna Report 9CHIN Jan 1, 2009 1135 On fuzzy number valued Lebesgue outer measure. Some properties of ([alpha], [beta])-fuzzy BG-algebras. Lagrange equations Mean value theorems (Calculus)