Suppose that f is twice differentiable on $[a, b]$ and $f'(a)=f'(b)=0$ show $\exists\xi\in(a,b)$ such that $|f''(\xi)|\geq\frac{4}{(b-a)^2}|f(b)-f(a)|$
There's an answer available on this site but it uses integration which I cannot use here
I tried applying Taylor's theorem about a and b but this only got me $\exists\xi_1,\xi_2\in(a,b)$ such that $|f(b)-f(a)|=|f''(\xi_i)|\frac{(b-a)^2}{2}$ for i=1,2
This seems close to the answer but not quite correct, anyone have any ideas? Thanks
Consider the value $f\left(\frac{a+b}2\right)$ using Taylor series expansions at $a$ and at $b$:
$$f\left(\frac{a+b}2\right)=f(a)+f’(a)\left(\frac{a+b}2-a\right)+\frac 12\cdot f’’(\xi_1)\left(\frac{a+b}2-a\right)^2$$ for some $\xi_1\in \left(a,\frac{a+b}2\right)$ and $$f\left(\frac{a+b}2\right)=f(b)+f’(b)\left(\frac{a+b}2-b\right)+\frac 12\cdot f’’(\xi_2)\left(\frac{a+b}2-b\right)^2$$ for some $\xi_2\in \left(\frac{a+b}2,b\right).$ Subtracting and applying triangle inequality (noting that $f’(a)=f’(b)=0$) yield $$|f(b)-f(a)|\leq \frac 12(|f’’(\xi_2)|+|f’’(\xi_1)|)\cdot \frac{(b-a)^2}4\leq |f’’(\xi)|\cdot \frac {(b-a)^2}4, $$ where $\xi$ is chosen among $\xi_1$ and $\xi_2$ to yield $\max(|f’’(\xi_1)|,|f’’(\xi_2)|)$. The result follows. QED
