The bonus question:
Let $f(x)$ be defined for $a \leq x \leq b$. Assuming appropriate properties of continuity and derivability, prove for $a < x < b$ that $ \frac{\frac{f(x)-f(a)}{x-a} - \frac{f(b)-f(a)}{b-a}}{x-b} = \tfrac{1}{2}f^{\prime \prime}(\beta) $ where $\beta$ is some number between $a$ and $b$.
I am thinking that its just the Mean Value Theorem.
$f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}$
I have rewritten the problem as this:
$\frac{f^{\prime}(c)-f^{\prime}(d)}{c-d}$ for $c$ and $d$ within $(a,b)$.
Then I get lost, I don't see where the contant $1/2$ comes into play and how to have the denomiator be $x-b$.