Given $f: I \rightarrow \mathbb{R} $ differentiable (where $I$ is an interval) and $c \in I$, are there $a,b \in I$ such that $f(b) - f(a) = f '(c) (b - a)$? Sort of like the reverse of the Mean Value Theorem. I'm not sure if it's true, but I feel like it is. Maybe with some stronger claims? Also, if it's true, can I take those two points to arbitrarily close to c? As in, given $c \in I$ and $\delta > 0$, are there $a,b \in (c - \delta, c + \delta)$ such that $f(b) - f(a) = f'(c)(b - a)$?
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I assume that you want $a\ne b$, since if $a=b$ then $f(b)-f(a)=f'(c)(b-a)$ for all $c$.
The answer is no. Let $f(x)=x^3$. Then $f'(0)=0$ and $a\ne b\implies f(b)-f(a)\ne0$.
Julián Aguirre
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Yeah, you're right... But maybe with some stronger claims it could be true? Any graph I draw I seem to be able to find a parallel line to a tangent line that crosses two different points of the graph. – violeta Feb 04 '15 at 23:12
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I strongly believe that the result will hold if $f$ is convex or concave. In my example, the point $c=0$ is an inflection point. – Julián Aguirre Feb 05 '15 at 09:59