Exercise about inequality - Taylor polynomial?

Consider a function defined in $$\mathbb{R}$$ such that $$f^{(3)}$$ is continuous in $$[0,1].$$ Suppose that $$f(0)=f′(0)=f′′(0)=f′(1)=f′′(1)=0$$ and $$f(1)=1$$. Prove that there exists $$c \in [0,1]$$ such that $$f^{(3)}(c)\geq 24$$.

The question was originally posted here Inequality related to Taylor polynomial

But there was some mistakes in the statement. Thus maybe it is better to write as an new question

Use Taylor’s formula to obtain $$f(\frac12)=f(0)+\frac12f’(0)+\frac{f’’(0)}{2!}(\frac12)^2+\frac{f’’’(\xi_1)}{3!}(\frac12)^3,$$ $$f(\frac12)=f(1)-\frac12f’(1)+\frac{f’’(1)}{2!}(\frac12)^2-\frac{f’’’(\xi_2)}{3!}(\frac12)^3,$$ where $$\xi_1\in[0,1/2]$$ and $$\xi_2\in[1/2,1]$$. Doing the subtraction we get $$f’’’(\xi_1)+f’’’(\xi_2)=48,$$ so one of them will be no less than $$24$$.