Is there any reason that this generalization of the mean value theorem would fail?
Let A be a subset of Rn that is differentiably connected, and let f : A –> R be continuously differentiable at every point in A. If x and y are two points in A then there exists a point c in A such that
f(y)-f(x) = (grad(f(c)))*(y-x)
It seems like this would work to me, however every generalization I've run into specifies that A must be open, so I assume if it failed it would be on account of that, however I can't figure out why.
Thanks
Best Answer
A hint: Consider the auxiliary function $\phi(t):=f(x+t(y-x))$ for $0\leq t\leq 1$. In using this function you make an essential assumption about the two points $x,y\in A$. This assumption is the missing link; it has nothing to do with $A$ being open or what.