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Say I have the following vector valued function $f: \mathbb{R} \rightarrow \mathbb{R}^2$ defined by $f(x) = ({x^2},{x^3})$. Clearly the derivative $f'(x) = ({2x},{3x^2})$. But, does there exist a $0 < b < 1$ such that $f'(b) = f(1) - f(0)$? Obviously simple arithmetic says no, but doesn't this contradict the mean value theorem?

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    $\begingroup$ The theorem is only for $\mathbb R \to \mathbb R$. $\endgroup$ – Karolis Juodelė Feb 13 '14 at 18:45
  • $\begingroup$ There is no exact analog of the mean value theorem for vector-valued functions $\endgroup$ – Omnomnomnom Feb 13 '14 at 18:46
  • $\begingroup$ $\exists\ c_{1}$ and $c_{2}$ such that $2 = 2$ and $3 = 6c_{2}$ where $0 < c_{\rm i} < 1$. $i = 1, 2$. $\endgroup$ – Felix Marin Feb 13 '14 at 19:11

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