i got a different solution to my problem :(adsbygoogle = window.adsbygoogle || []).push({});

f(x) continues in [a,b] interval,and differentiable (a,b) b>a>0

alpha differs 0

proove that there is b>c>a

in that formula:

http://img392.imageshack.us/my.php?image=81208753je3.gif

solution:

alpha=&

i was told that in this question i dont use "mean theorim"

but the "couchy mean theorim"

[f(b)-f(a)]/[g(b)-g(a)]=f'(c)/g'(c)

F(x)=f(x)/[x^&] G(x)=1/(x^&)

[F(b)-F(a)]/[G(b)-G(a)] = [f(b)/(b^&) - f(a)/(a^&)]/[1/(b^&) - 1/(a^&)] =

= [f(b)*(a^&) - f(a)*( b^&)]/[a^& - b^&]

F'(c) / G'(c)=[f'(c)* (1/c^&) -&*c^(-&-1)*f(c)]/[-&*c^(-&-1)]=

=[&*f(c)-c*f'(c)]/& =f(c) - c*f'(c)/&

my proffesor solved it like this after about 50 minutes when

he tried to use the mean theorim

and then he said that we need to use couchy mean theorim

i dont know how he desided that??

and the second most important problem is picking the variables.

how he picked

F(x)=f(x)/[x^&] G(x)=1/(x^&)

??

what proccess do i need to do in order to deside

that the way of solving such a problem is picking

F(x)=f(x)/[x^&] G(x)=1/(x^&)

???

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# Homework Help: Regarding the previusmean theorim question

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