TITLE: show that $f^{(3)}(c) \ge 3$ for $c\in(-1,1)$
QUESTION [5 upvotes]: Let $f:I\rightarrow \Bbb{R}$, differetiable three times on the open interval $I$ which contains $[-1,1]$. Also: $f(0) = f(-1) = f'(0) = 0$ and $f(1)=1$.  
Show that there's a point $c \in (-1, 1)$ such that $f^{(3)}(c) \ge 3$
I'd be glad to get a guidace here how to start.

REPLY [9 votes]: We have from Taylor's Theorem $$f(x) = f(0) + xf'(0) + \frac{x^{2}}{2}f''(0) + \frac{x^{3}}{6}f'''(c)$$ where $c$ is some number between $0$ and $x$ and hence on putting $x = 1$ we get $$1 = f(1) = \frac{f''(0)}{2} + \frac{f'''(c_{1})}{6}\tag {1}$$ where $c_{1} \in (0, 1)$ and putting $x = -1$ we get $$0 = f(-1) = \frac{f''(0)}{2} - \frac{f'''(c_{2})}{6}\tag{2}$$ where $c_{2} \in (-1, 0)$ and subtracting equation $(2)$ from equation $(1)$ we get $$f'''(c_{1}) + f'''(c_{2}) = 6\tag{3}$$ It is now obvious that both $f'''(c_{1})$ and $f'''(c_{2})$ can't be less than $3$ (otherwise their sum will be less than $6$). Hence we must have one of these values greater than or equal to $3$.