I`m trying to generalize Lagrange's mean value theorem for $f:A \to \mathbb{R}^n$ , differentiable in $A \subset \mathbb{R}^m$.
What I have so far is that if $n=1$, we can write $g_i(t)=f(x_1,..tx_i,..,x_n$), where $x_j$ are any real numbers and $x_i$ is 1. We have $g'(t)=D_f(t)\cdot(0,..,1,...0)=\frac {\partial f} {\partial x_i}(t)$, And since $g_i(t):\mathbb{R} \to \mathbb{R}$ , we can apply the single variable mean value theorem to say that there exists a point $c_{x_i}\in [a,b] $ s.t $\frac {\partial f} {\partial x_i}(c_{x_i})=g_i'(c_{x_i})=\frac {g_i(a)-g_i(b)} {a-b}$ , for any partial derivative. However, For any partial derivative we get a different $c_{x_i}$, And I couldn`t find a way to relate $g$ back to $f$.
Also, if $n>1$, And I`d know how to handle the $n=1$ case, I know I can break down $f$ to $n$ different functions, apply the $n=1$ to each of them seperately, but then again, we get $n$ different points $c$ which aren't necessarily the same,So I got stuck here too.
I wonder if this is a good way to try and generalize the Lagrange`s Mean Value theorem for multivariable functions?Can anyone help me complete the idea, or suggest a different way of thinking?
You are stuck, because there is no solution to this problem! As you already mentioned, you will get different $c_{x_i}$'s for different $i$.
As a counterexample, choose $f: [0,2\pi] \to \mathbb R^2$ with $f(x) = \left( \cos(x), \sin(x)\right)$. Then $f(2\pi) - f(0) = \left(0, 0 \right)$, but $f'(x) = \left( - \sin(x), \cos(x)\right)$ never assumes this value, as $\sin$ and $\cos$ have no mutual roots.