a) Proof: By theorem above, there exists a ##a \in \mathbb{R}## such that for all ##x \in I## we have ##f'(x) = a##. Let ##x, y \in I##. Then, by Mean Value Theorem,
$$a = \frac{f(x) - f(y)}{x - y}$$
This can be rewritten as ##f(x) = ax - ay + f(y)##. Now, let ##g(y) = -ay + f(y)##. Then...