Let f:[-1,8] -> R be a differentiable function such that f(-1)=1,f(8)=10,
then there always exists c1, c2 and c3
Є (-1,8) such that f'(c1) + f'(c2) + f'(c3) = k, then k is equal to _______
I'm able to get this problem up to certain point, but then I doubt when adding the c's.
So by the Mean Value Theorem you have:
f'(c) = (f(b) - f(a))/(b-a)
the interval [-1,8] is your [a,b] so pluging in with your givens you should get f'(c) = 1
But after that I doubt my answer, I would suggest that c1=c2=c3, thus for k you get 3
But I will like some confirmation though.