TITLE: Given $f(0) = f ' (0) = 0$ and $f '' (0)$ defined on $[0,h]$, show there exists $c$ in $[0,h]$ such that $f(h) = (1/2)(h^2) f '' (c)$
QUESTION [5 upvotes]: Given $f(0) = f'(0) = 0$ and $f''(x)$ defined on $[0,h]$, show there exists $c$ in $[0,h]$ such that $f(h) = (1/2)(h^2) f''(c)$.
Noting that Taylor's theorem is not available yet. The approach uses Mean Value Theorem, Intermediate Value Theorem, Rolle's theorem, etc.

REPLY [5 votes]: Define $g : [0,h] \rightarrow \mathbb{R} $ as $$ g(x) = f(x) -(x^2/h^2)f(h) $$ then you have $ g'(x) = f'(x) -(2x/h^2)f(h) $ and $ g''(x) = f''(x) -(2/h^2)f(h) $. Now observe that $ g(h) = 0 = g(0)$. Hence from Rolle's theorem there exists $ \xi \in (0,h) $ such that $ g'(\xi) = 0 $. We also have $ g'(0) = 0 $, hence Rolle's theorem again gives $ c \in [0,h] $ such that $ g''(c) = 0 $ which implies $$ f''(c) -(2/h^2)f(h) = 0. $$ So $ f(h) = (h^2/2)f''(c) $.