TITLE: Given $f: I \rightarrow \mathbb{R} $ differentiable and $c \in I$, are there $a,b \in I$ such that $f(b) - f(a) = f '(c) (b - a)$?
QUESTION [5 upvotes]: 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 $f(b) - f(a) = f'(c)(b - a)$?

REPLY [6 votes]: 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 $a\ne b\implies f(b)-f(a)\ne0$.