Friday, November 18, 2016
Unizor Derivatives - Cauchy Theorem
Notes to a video lecture on http://www.unizor.com
Derivatives -
Caushy Mean Value Theorem
Cauchy Mean Value Theorem
For two smooth functions f(x) and g(x), defined on segment [a,b] (including endpoints), there exist a point x0∈[a,b] such that the following is true:
f'(x0)/g'(x0) = [f(b)−f(a)]/[g(b)−g(a)]
(with obvious restrictions on denominators not equal to zero).
Proof
Proof of this theorem is based on Rolle's Theorem.
Consider a new function h(x) defined as:
h(x) = f(x) − g(x)·[f(b)−f(a)]/[g(b)−g(a)]
This function satisfies the conditions of Rolle's Theorem:
h(a) = f(a) −g(a)·[f(b)−f(a)]/[g(b)−g(a)] =
= [f(a)·g(b)−f(b)·(g(a)]/[g(b)−g(a)]
h(b) = f(b) − g(b)·[f(b)−f(a)]/[g(b)−g(a)] =
= [f(a)·g(b)−f(b)·(g(a)]/[g(b)−g(a)]
So, h(a) = h(b)
According to Rolle's Theorem, there exists point x0∈[a,b] such that
h'(x0) = 0
Let's find the derivative of function h(x) in terms of derivatives of the original functions f(x) and g(x):
h'(x) = f'(x) − g'(x)·[f(b)−f(a)]/[g(b)−g(a)]
Now the equality to 0 of the derivative of function h(x) at point x0 in terms of original functions f(x) and g(x) looks like this:
0 = f'(x0) − g'(x0)·[f(b)−f(a)]/[g(b)−g(a)]
from which follows
f'(x0)/g'(x0) = [f(b)−f(a)]/[g(b)−g(a)]
End of proof.
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