Exact Taylor expansion of order 1

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Can we prove that for any differentiable function $f: R^n \rightarrow R$, $f(y)=f(x)+\nabla f(x)^T(z)$, for some $z$ on the line segment $[x, y]$.

Note that it is not approximation, it is exact equality. I am skeptical because this is similar to the mean value theorem which should only apply when f is on a closed interval.


We have for a closed interval [a,b] by MVT

$$f(a) = f(b) + f'(ta + (1 - t)b)(a-b)$$

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