## Friday, November 18, 2016

### Unizor - Derivatives - Lagrange Theorem

Notes to a video lecture on http://www.unizor.com

Derivatives -

Lagrange Mean Value Theorem

Lagrange Mean Value Theorem

For a smooth function f(x), defined on segment [a,b] (including endpoints), there exist a point x0∈[a,b] such that

f'(x0) = [f(b)−f(a)]/(b−a)

Geometrically, this theorem states that there is a point x0 inside segment [a,b], where a tangential line is parallel to a chord connecting endpoints of function f(x) on this segment.

Proof

Proof of this theorem is based on Rolle's Theorem.

Consider a new function g(x) that is equal to a difference between f(x) and a chord connecting two endpoints of a function on segment [a,b]:

g(x) = f(x) −

−{(x−a)·[f(b)−f(a)]/(b−a)+f(a)}

This function satisfies the conditions of Rolle's Theorem:

g(a) = f(a) −

−{(a−a)·[f(b)−f(a)]/(b−a)+f(a)} = f(a) − f(a) = 0

g(b) = f(b) −

−{(b−a)·[f(b)−f(a)]/(b−a)+f(a)} = f(b) − [f(b)−f(a)+f(a)] = 0

So, g(a) = g(b) = 0

According to Rolle's Theorem, there exists point x0∈[a,b] such that

g'(x0) = 0

Let's find the derivative of function g(x) in terms of derivative of the original function f(x):

g'(x) = f I(x) − [f(b)−f(a)]/(b−a)

Now the equality to 0 of the derivative of function g(x) at point x0 in terms of original function f(x) looks like this:

f'(x0) − [f(b)−f(a)]/(b−a) = 0 from which follows

f'(x0) = [f(b)−f(a)]/(b−a)

End of proof.

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