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In the general situation of $f:S\to R^m$ where $S\subset R^n$. There is a form of the mean value theorem: $a\cdot (f(y)-f(x))=a\cdot (f'(z)(y-x))$ which requires a vector $a$ and dot products.

Is it possible to create a generalization of the mean value theorem that doesn't involve dot products. Something like $$f(y)-f(x) = cf'(z)(y-x)$$ where $c$ is some real number. The idea being that the path $f(t(y-x)+ x)$ where $t\in [0,1]$ should have a tangent $f'(z)(y-x)$ that is parallel to $f(y)-f(x)$. Now there is no reason that they should have the same length though and so a constant scaling factor $c$ is needed. And to avoid some pathological cases maybe some conditions are needed such as $f(y)\neq f(x)$.

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"The idea being that the path $f(t(y-x)+ x)$ where $t\in [0,1]$ should have a tangent $f'(z)(y-x)$ that is parallel to $f(y)-f(x)$."

Although this sounds like something which could hold, it is false in general. Consider a mapping $f:[0,1]\to\mathbb{R}^3$ defined by $f(x) = (t,\sin(2\pi t),\cos(2\pi t))$. Then $f(0) = (0,0,1)$ and $f(1) = (1,0,1)$ and hence $f(1)-f(0) = (1,0,0)$. However, there is no point $t$ such that $f'(t) = (1,0,0)$.

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