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^&) ???