Prove the following:
For every real valued function f differentiable on an interval [a,b] not containing 0 and for all pairs x1 ≠x2 in [a,b], there exists a point ξ in (x1, x2) such that
\dfrac{x_1f(x_2)-x_2f(x_1)}{x_1-x_2}=f(\xi)-\xi f'(\xi)
1sir 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 )}
1i think we have to prove
f(x_1).f(x_2)=f^2(\xi )
it seems logical for non constant function
but not getting the proof
11G(x)=\frac{f(x)}{x}.
H(x)=\frac{1}{x}.
Now from Cauchy, \frac{G(x_1)-G(x_2)}{H(x_1)-H(x_2)}=\frac{G'(\alpha)}{H'(\alpha)}.
The result follows.
1bhai soumik
" by cauchy "
wats this can u tell in detail
or is it simply dividing two equations of LMVT ???
1interesting
this theorem is called as extended mean value theorem
proof :
http://planetmath.org/encyclopedia/ExtendedMeanValueTheorem.html
1one more theorem which i came across
intermediate value theorem
http://www.calculus-help.com/the-intermediate-value-theorem/
66This 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)=tf\left(\dfrac{1}{t}\right)
in the interval [1/a, 1/b]
1sir one doubt
is this true ?
f(x_1).f(x_2)=f^2(\xi )
for some \xi →(x1,x2) ???