Calculus - A variation on LMVT

1

sir i am stuck after this

$\texttt{consider}\\ \\ g(x)=\frac{x}{f(x)}\\ \texttt{applying LMVT for g(x) from }x_1 \ to \ x_2 \\ \frac{g(x_1)-g(x_2)}{x_1-x_2}=g'(c)\\ \\ \frac{\frac{x_1}{f(x_1)}-\frac{x_2}{f(x_2)}}{x_1-x_2}=\frac{f(\xi )-\xi f'(\xi )}{f^2(\xi )}\\ \\ \frac{x_1f(x_2)-x_2f(x_1)}{(x_1-x_2)f(x_1)f(x_2)}=\frac{f(\xi )-\xi f'(\xi )}{f^2(\xi )}$

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student ·Apr 28 '10 at 8:10

i think we have to prove
$f (x 1 ) . f (x 2 ) = f 2 (ξ)$
it seems logical for non constant function
but not getting the proof

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11

Devil ·Apr 28 '10 at 9:47

$G (x) = x f (x) $ .

$H (x) = x 1 $ .

Now from Cauchy, $ H (x 1 ) - H (x 2 ) G (x 1 ) - G (x 2 ) = H ' (α) G ' (α) $ .

The result follows.

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student ·Apr 28 '10 at 9:53

bhai soumik

" by cauchy "

wats this can u tell in detail

or is it simply dividing two equations of LMVT ???

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student ·Apr 28 '10 at 18:37

interesting

this theorem is called as extended mean value theorem

proof :

http://planetmath.org/encyclopedia/ExtendedMeanValueTheorem.html

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student ·Apr 28 '10 at 20:05

one more theorem which i came across

intermediate value theorem

http://www.calculus-help.com/the-intermediate-value-theorem/

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kaymant ·Apr 29 '10 at 1:52

This variant of LMVT is also known as Pompeiu's mean value theorem. It could also be proved by applying LMVT to the function
$F (t) = t f ( t 1 )$
in the interval [1/a, 1/b]

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student ·Apr 29 '10 at 1:55

sir one doubt

is this true ?

$f (x 1 ) . f (x 2 ) = f 2 (ξ)$

for some $ξ$ â†’(x1,x2) ???

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