Mean value theorem

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For any function that is continuous on [a, b] and differentiable on (a, b) there exists some c in the interval (a, b) such that the secant joining the endpoints of the interval [a, b] is parallel to the tangent at c.

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.

This theorem was first developed by Lagrange (1736 – 1813). It is one of the most important results in differential calculus, as well as one of the most important theorems in mathematical analysis, and is essential in proving the fundamental theorem of calculus. The mean value theorem can be used to prove Taylor's theorem, of which it is a special case.

1 Formal statement
2 Proof
3 Cauchy's mean value theorem
- 3.1 Proof of Cauchy's mean value theorem
4 Mean value theorems for integration
5 See also
6 External links

[edit] Formal statement

Topics in calculus
Fundamental theorem Limits of functions Continuity Vector calculus Tensor calculus Mean value theorem
Differentiation
Product rule Quotient rule Chain rule Implicit differentiation Taylor's theorem Related rates Table of derivatives
Integration
Lists of integrals Improper integrals Integration by: parts, disks, cylindrical shells, substitution, trigonometric substitution

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

$f ' (c) = \frac{f(b) - f(a)}{b - a}.$

The mean value theorem is a generalization of Rolle's theorem, which assumes f(a) = f(b), so that the right-hand side above is zero.

The mean value theorem is still valid in a slightly more general setting, one only needs to assume that f : [a, b] → R is continuous on [a, b], and that for every x in (a, b) the limit

$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

exists as a finite number or equals ±∞.

[edit] Proof

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:

$y = \frac{f(b) - f(a)}{b - a}(x - a) + f(a)$

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

$g(a) = g(b) \qquad \Rightarrow \qquad f(a) + ra = f(b) + rb$

$\Rightarrow \qquad r = - \frac{ f(b) - f(a) }{ b - a}$

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

$f ' (c) = g ' (c) - r = 0 - r = \frac{ f(b) - f(a) }{ b - a}$

as required.

[edit] Cauchy's mean value theorem

Cauchy's mean value theorem, also known as the extended mean value theorem, is the more general form of the mean value theorem. It states: If functions $f (t)$ and $g (t)$ are both continuous on the closed interval $[a, b]$ , differentiable on the open interval $(a, b)$ , and $g'(t)$ is not zero on that open interval, then there exists some $c$ in $(a, b)$ , such that

$\frac {f'(c)} {g'(c)} = \frac {f(b) - f(a)} {g(b) - g(a)}.$

Cauchy's mean value theorem can be used to prove l'Hopital's rule. The mean value theorem is the special case of Cauchy's mean value when $g (t) = t$ (or more generally when $g (t)$ is affine and not constant, meaning $g (t) = p t + q$ where $p$ and $q$ are constants and $p \ne 0$ ).

[edit] Proof of Cauchy's mean value theorem

The proof of Cauchy's mean value theorem is based on the same idea as the proof of the mean value theorem. First we define a new function h(t) and then we aim to transform this function so that it satisfies the conditions of Rolle's theorem.

$h(t) = f(t) - m g(t)\,$

where m is a constant. We choose m so that

$h(a) = h(b) \qquad \Rightarrow \qquad m = \frac {f(b) - f(a)} {g(b) - g(a)}$

Since h is continuous and h(a) = h(b), by Rolle's theorem, there exists some c in (a, b) such that h′(c) = 0, i.e.

$h'(c) = 0 \ = \ f'(c) - \frac {f(b) - f(a)} {g(b) -g(a)} g'(c)$

$\Rightarrow \qquad \frac {f'(c)} {g'(c)}\ = \ \frac {f(b) - f(a)} {g(b) - g(a)}$

as required.

[edit] Mean value theorems for integration

The first mean value theorem for integration states

If G : [a, b] → R is a continuous function and φ : [a, b] → R is an integrable positive function, then there exists a number x in (a, b) such that

$\int_a^b G(t)\varphi (t) \, dt=G(x) \int_a^b \varphi (t) \, dt.$

In particular for φ(t) = 1, there exists x in (a, b) such that

$\int_a^b G(t) \, dt=\ G(x)(b - a).\,$

The second mean value theorem for integration is stated as follows

If G : [a, b] → R is a monotonically decreasing (not necessarily positive) function and φ : [a, b] → R is an integrable function, then there exists a number x in (a, b) such that

$\int_a^b G(t)\varphi(t)\,dt = G(a+0) \int_a^x \varphi(t)\,dt + G(b-0) \int_x^b \varphi(t)\,dt.$