Homework Help: Regarding the previusmean theorim question

1. Nov 26, 2008

transgalactic

i got a different solution to my problem :

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^&)

???