An easier way to prove this Suppose $f\in C^1[a,b]$, and $f''$ exists on $(a,b)$. Show that, for any $c\in (a,b)$, there is $\xi\in(a,b)$ s.t.
$$\frac{1}{c-b}\left(\frac{f(c)-f(a)}{c-a}-\frac{f(b)-f(a)}{b-a}\right)=\frac{f''(\xi)}{2}$$
One very elegant but not trivial way is by considering the following function:
$$g(t)=\text{det}\begin{pmatrix}f(t)&t^2&t&1\\
f(c)&c^2&c&1\\
f(a)&a^2&a&1\\
f(b)&b^2&b&1\\
\end{pmatrix}$$
one can see that $g\in C^1[a,b]$ and $g''$ exists on $(a,b)$, moreover we have $g(a)=g(b)=g(c)=0$, thus by applying Rolle's theorem twice, we can get what we want.
I think this method is very elegant, but quite hard to come up with. I am trying to find an easier and more elementary approach, such as applying Taylor's theorem and mean value theorem to $f$ or a function that can be much easier to come  up with, but it seems that these approaches did not work very well. 
Does anyone have some idea about an easier way to prove this?
 A: Define the function $h(t)$ on the closed interval $[a,b]$ as
$$
h(t){}:={}\left\{\begin{array}{cl}f^{'}(a)&;\mbox{for }\,t{}={}a\\&\\\dfrac{f(t)-f(a)}{t-a}&;\mbox{for }\,t{}\in{}(a,b]\end{array}\right.
$$
Then, $h(t)$ is differentiable on $(a,b)$ and continuous on $[a,b]$. Consequently, by the mean value theorem, for $c\in(a,b)$, there exists $\theta\in(c,b)$ such that
$$
\dfrac{h(c)-h(b)}{c-b}{}={}h^{'}(\theta)\,.
$$
In addition, by Taylor's theorem, there exists $\xi\in(a,\theta)$ such that
$$
h^{'}(\theta){}={}\dfrac{(\theta-a)f^{'}(\theta)-f(\theta)+f(a)}{(\theta-a)^2}{}={}\dfrac{1}{2}f^{''}(\xi)\,.
$$

Therefore, for $c\in(a,b)$, there exists $\xi\in(a,b)$ such that
$$
\dfrac{h(c)-h(b)}{c-b}{}={}\dfrac{1}{2}f^{''}(\xi)\,.
$$ 

A: Put $\lambda:=c-a$, $\mu:=b-a$ and notice that $\lambda-\mu=c-b$. 
Define $g(x):=\frac{1}{\lambda}f(a+\lambda x)+\frac{1}{\mu}f(a+\mu x)$.
By Cauchy theorem
$$\frac{f(c)-f(a)}{c-a}-\frac{f(b)-f(a)}{b-a}=\frac{g(1)-g(0)}{1^2-0^2}
=\frac{f'(a+\lambda\zeta)-f'(a+\mu\zeta)}{2\zeta}$$
for some $\zeta\in(0,1)$ and by Lagrange theorem the last difference quotient equals
$$\frac{(\lambda\zeta-\mu\zeta)f''(\xi)}{2\zeta}=(c-b)\frac{f''(\xi)}{2}$$
for some $\xi\in(a,b)$. (We interpreted $1$ as $1^2-0^2$ instead of $1-0$ so as to kill the factor $\zeta$ coming from the numerator of the difference quotient.)
Remark: If $f\in C^2([a,b])$, one gets a very natural solution by writing the difference quotients as integrals: $\frac{f(c)-f(a)}{c-a}=\int_0^1 f'(a+\lambda t)\,dt$ and $\frac{f(b)-f(a)}{b-a}=\int_0^1 f'(a+\mu t)\,dt$, so that
$$\frac{f(c)-f(a)}{c-a}-\frac{f(b)-f(a)}{b-a}=\int_0^1 \Big(f'(a+\lambda t)-f'(a+\mu t)\Big)\,dt=\int_0^1\int_{\mu t}^{\lambda t}f''(a+s)\,ds\,dt$$
and finally
$$\frac{1}{c-b}\left(\frac{f(c)-f(a)}{c-a}-\frac{f(b)-f(a)}{b-a}\right)=\frac{1}{2}\int_0^1\int_0^1 2t f''\Big(a+\mu t+(\lambda-\mu)s\Big)\,ds\,dt$$
But $\int_0^1\int_0^1 2t\,ds\,dt=1$, so the last double integral involving $f''$ lies in the convex hull of all the values of $f''$, i.e. it belongs to the image of $f''$ (since it is an interval, thus convex).
