IMC 2005 · Day 2 · P10

Let $$g(x) = -\frac{f(-1)}{2}\, x^2 (x-1) - f(0)(x^2 - 1) + \frac{f(1)}{2}\, x^2 (x+1) - f'(0)\, x (x-1)(x+1).$$ It is easy to check that $g(\pm 1) = f(\pm 1)$, $g(0) = f(0)$ and $g'(0) = f'(0)$.

Apply Rolle's theorem for the function $h(x) = f(x) - g(x)$ and its derivatives. Since $h(-1) = h(0) = h(1) = 0$, there exist $\eta \in (-1, 0)$ and $\vartheta \in (0, 1)$ such that $h'(\eta) = h'(\vartheta) = 0$. We also have $h'(0) = 0$, so there exist $\varrho \in (\eta, 0)$ and $\sigma \in (0, \vartheta)$ such that $h''(\varrho) = h''(\sigma) = 0$. Finally, there exists a $\xi \in (\varrho, \sigma) \subset (-1, 1)$ where $h'''(\xi) = 0$. Then $$f'''(\xi) = g'''(\xi) = -\frac{f(-1)}{2} \cdot 6 - f(0) \cdot 0 + \frac{f(1)}{2} \cdot 6 - f'(0) \cdot 6 = \frac{f(1) - f(-1)}{2} - f'(0).$$

The expression $\dfrac{f(1) - f(-1)}{2} - f'(0)$ is the divided difference $f[-1, 0, 0, 1]$ and there exists a number $\xi \in (-1, 1)$ such that $f[-1, 0, 0, 1] = \dfrac{f'''(\xi)}{3!}$.