Problem
Assume that $f(x)$ has the $3$-order continuous derivative over $[-1,1]$, and $f(-1)=0,f(1)=1,f'(0)=0$. Prove that there exists $\xi \in (-1,1)$ such that $f'''(\xi)=3.$
Proof
According to Taylor's formula of $f(x)$ expanding at $x=0$ $$f(x)=f(0)+\frac{x^2f''(0)}{2}+\frac{x^3f'''(\xi)}{6},~~~0\lessgtr \xi \lessgtr x,$$ we obtain $$0=f(-1)=f(0)+\frac{f''(0)}{2}-\frac{f'''(\xi_1)}{6},~~~-1<\xi_1<0,$$ $$1=f(1)=f(0)+\frac{f''(0)}{2}+\frac{f'''(\xi_2)}{6},~~~0<\xi_2<1.$$ By substraction, we have $$f'''(\xi_1)+f'''(\xi_2)=6,$$ which implies, $f'''(\xi_1),f'''(\xi_2)$ are both equal to $3$，or，one of them is greater than $3$ and the other less than $3$. For the former case,the conclusion holds trivially；as for the latter one, according to the continuity of $f'''(x)$, applying the intermediate value theorem, there exists $\xi_3 \in (\xi_1,\xi_2)\subset (-1,1)$ such that$f'''(\xi_3)=3$，the conclusion holds as well.
Comment
But I suggest that, we only need the existence of $f'''(x)$, and the continuity of it is not necessary. That's because, according to Darboux's theorem, the derivative function itself has the intermediate value property.