Proof by using mean value theorem Let $f(x)$ be a function such that it is continuous and differentiable 
Show that $(f(a)+f(b))/2 = f((a+b)/2) + ((b-a)^2 / 8)f''(c)$
I have tried many different approaches to solving this but im dead lost on how to get to the final result.
Can I please get some help on how to prove this?
 A: Edited:
Consider the 2nd degree Taylor expansion with remainders:
$\displaystyle f(a) = f\left(\frac{a+b}{2}\right) + \left(a-\frac{a+b}{2}\right)f'\left(\frac{a+b}{2}\right)+\frac{(b-a)^2}{8}f''(c_1)$, for some $c_1 \in \left(a,\frac{a+b}{2}\right)$
and similarly, $\displaystyle f(b) = f\left(\frac{a+b}{2}\right) + \left(b -\frac{a+b}{2}\right)f'\left(\frac{a+b}{2}\right)+\frac{(b-a)^2}{8}f''(c_2)$, for some $c_2 \in \left(\frac{a+b}{2},b\right)$
Thus from Darboux Theorem, we have $\exists \,c \in (c_1,c_2)$, such that $\displaystyle f''(c) = \frac{f''(c_1)+f''(c_2)}{2}$
Adding the two expressions we get the desired result.
Alternative Approach: Denote $\dfrac{a+b}{2} = c$ for brevity.
Consider the function:
$\displaystyle g(x) = f(x) - f(a)\frac{(x-b)(x-c)}{(a-b)(a-c)} - f(b)\frac{(x-a)(x-c)}{(b-a)(b-c)} - f(c)\frac{(x-a)(x-b)}{(c-a)(c-b)} \\ = f(x) - P(x)$
where, you can identify $P(x)$ to be the Lagrange Interpolating Polynomial for $f$ at the points $a,c,b$.
Since, $g(a) = g(c) = g(b) = 0$, Rolles's Theorem implies $\exists \, c_1 \in (a,c)$ and $\exists \, c_2 \in (c,b)$,
such that $g'(c_1) = 0 = g'(c_2)$
Once again apply Rolle's Theorem on $g'(x)$ in the interval $(c_1,c_2)$, then $\exists \, \xi \in (c_1,c_2)$, such that $g''(\xi) = 0$.
Now,$\displaystyle g''(x) = f''(x) - \frac{2f(a)}{(a-b)(a-c)} - \frac{2f(b)}{(b-a)(b-c)} - \frac{2f(c)}{(c-a)(c-b)}$
verify that the above identity on rearranging suitably gives the desired result.
