Here is an exercise from a book:
Let $a<b<c$. $f$ is twice differentiable on an open interval containing $a,b,c$. Show that there exist $d\in(a,b)$ such that $$ \frac{f(c)-f(a)}{c-a}-\frac{f(b)-f(a)}{b-a}=\frac12 (c-b)f''(d). $$
Here is my proof:
Let $P(x)$ be a quadratic polynomial such that $P(x-a)=f(x)$ for $x=a,b,c$. Let $b=a+p,c=a+q$. Then the left hand side becomes $$ \frac{P(q)-P(0)}{q}-\frac{P(p)-P(0)}{p}. $$ Obviously, those two terms are identical linear polynomials in $q,p$ respectively, and we should have $$ \frac{P(q)-P(0)}{q}-\frac{P(p)-P(0)}{p}=k(q-p) $$ where $k$ is the leading coefficient of the polynomial $P$. $P''(x)=2k,\forall x$.
Now, apply Rolle's theorem twice: $\exists x_1\in(a,b),x_2\in(b,c)$, $f'(x_i)=P'(x_i-x),i=1,2$, and $\exists d\in(a,c)$ such that $f''(d)=P''(d)=2k$.
So, as a result, $$ \frac{f(c)-f(a)}{c-a}-\frac{f(b)-f(a)}{b-a}=\frac12 (c-b)f''(d). $$
It appears that the proof is complete. However, the question states that $d\in (a,b)$. I don't think this is right. I have only proven that $d\in (a,c)$.
Further, the phrase "an open interval containing $a,b,c$" seems a bit redundant. I think it is fine to just have $f$ twice differentiable on $[a,c]$.
Am I understanding it clearly?
If the claim $d\in (a,b)$ is false, could anyone give a counterexample?
$f(x)=(x-b)^{3}$ for $x >b$ and $0$ for $x \leq b$ is a counter-example to show that $d \in (a,b)$ may not be possible.