Prove that there exist $a \in [-1,1]$, such that $f'''(a)=3f(1)-3f(-1)-6f'(0)$ Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a three times differentiable function. Prove that there exists $a \in [-1,1]$ such that $$ f'''(a) = 3f(1)-3f(-1)-6f'(0)$$
Any hint/idea's how to approach this problem? Thanks!
 A: I would use Taylor's theorem with Lagrange's form of remainder.
There exists $\xi_1\in$ a closed interval containing $a$ and $x$ such that:
$$f(x)=f(a)+f'(a)(x-a)+\frac{f''(a)(x-a)^2}{2}+\frac{f'''(\xi_1)(x-a)^3}{6}$$
Now there exists an $a_1\in [0,1]$ such that 
$$f(1)=f(0)+f'(0)(1-0)+\frac{f''(0)(1-0)^2}{2}+\frac{f'''(a_1)(1-0)^3}{6} \tag{1}$$
Again there exists $a_2\in [-1,0]$ such that 
$$f(-1)=f(0)+f'(0)(-1-0)+\frac{f''(0)(-1-0)^2}{2}+\frac{f'''(a_2)(-1-0)^3}{6} \tag{2}$$
Subtract $(1)$ and $(2)$, we have $$3f(1)-3f(-1)-6f'(0)=\frac{f'''(a_1)+f'''(a_2)}{2}$$
We can prove that there exists an $a\in [a_1,a_2]$ such that $$f'''(a)=\frac{f'''(a_1)+f'''(a_2)}{2} \tag{3}$$
Rewriting we have a $a\in [a_1,a_2]$ such that $$f'''(a) = 3f(1)-3f(-1)-6f'(0)$$
(Use Darboux's property to prove $(3)$. As usual the hardest part is left to the reader! On a second thought, I thought I should write a hint.  Assume $f'''(a_1)\le f'''(a_2)$ and observe $f'''(a_1)\le \frac{f'''(a_1)+f'''(a_2)}{2} \le {f'''}(a_2)$ and switch $a_1$ and $a_2$ to complete the proof) 
