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Differential Calculus

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5 Sep 2007 17:05:38 IST

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what is mean value theorum?give the provement also

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what is mean value theorum?give the provement also

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khyati gupta

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5 Sep 2007 22:14:46 IST

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IT IS GIVEN IN THE 12TH CLASS NCERT , SO REFER THAT 4 UR PROBLEM....

Avinash Sharma

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12 Sep 2007 02:59:55 IST

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Mean Value Theorem and it?s proof

In calculus, the mean value theorem states, roughly, that given a section of a smooth curve, there is a point on that section at which the derivative (slope) of the curve is equal to the "average" derivative of the section. It is used to prove theorems that make global conclusions about a function on an interval starting from local hypotheses about derivatives at points of the interval.

This theorem can be understood concretely by applying it to motion: if a car travels one hundred miles in one hour, so that its average speed during that time was 100 miles per hour, then at some time its instantaneous speed must have been exactly 100 miles per hour.

Let f : [a, b] ® R be a continuous function on the closed interval [a, b], and differentiable on the open interval (a, b). Then, there exists some c in (a, b) such that

f'(c) = { f(b) - f(a)} / (b - a )

An understanding of this and the point-slope formula will make it clear that the equation of a secant (which intersects (a, f(a)) and (b, f(b)) ) is:

y = [{f(b) - f(a)} / (b - a )] (x - a) + f(a)

The formula ( f(b) - f(a) ) / (b - a) gives the slope of the line joining the points (a, f(a)) and (b, f(b)), which we call a chord of the curve, while f ' (x) gives the slope of the tangent to the curve at the point (x, f(x) ). Thus the Mean value theorem says that given any chord of a smooth curve, we can find a point lying between the end-points of the chord such that the tangent at that point is parallel to the chord. The following proof illustrates this idea.

Define g(x) = f(x) + rx, where r is a constant. Since f is continuous on [a, b] and differentiable on (a, b), the same is true of g. We choose r so that g satisfies the conditions of Rolle's theorem, which means

g(a) = g(b) Þ f(a) + r a = f(b) + r b

Þ r = -{f(b) - f(a) }/ (b - a)

By Rolle's theorem, since g is continuous and g(a) = g(b), there is some c in (a, b) for which g '(c) = 0, and it follows from g(x) = f(x) + rx that,

f' (c) = g' (c) - r = 0 - r = {f(b) - f(a) }/ (b - a)

as required.

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