1. ## Application of Derivatives

Let $\displaystyle f:R\to R$ be a twice differentiable function and suppose that for all $\displaystyle x\in R$,$\displaystyle f$ satisfies the following two conditions:

(i)$\displaystyle |f(x)|\leq 1$

(ii)$\displaystyle |f''(x)|\leq 1$

Prove that $\displaystyle |f'(x)|\leq 2$

2. Originally Posted by pankaj
Let $\displaystyle f:R\to R$ be a twice differentiable function and suppose that for all $\displaystyle x\in R$,$\displaystyle f$ satisfies the following two conditions:

(i)$\displaystyle |f(x)|\leq 1$

(ii)$\displaystyle |f''(x)|\leq 1$

Prove that $\displaystyle |f'(x)|\leq 2$
using Taylor's theorem, we know that for any $\displaystyle x \in \mathbb{R}$ there exists some $\displaystyle c \in \mathbb{R}$ such that $\displaystyle f(x+2)=f(x)+2f'(x)+2f''(c).$ therefore:

$\displaystyle 2|f'(x)|=|f(x+2)-f(x)-2f''(c)| \leq |f(x+2)|+|f(x)|+2|f''(c)| \leq 4,$ and hence $\displaystyle |f'(x)| \leq 2.$

a similar argument gives us this: if $\displaystyle |f(x)| \leq a$ and $\displaystyle |f''(x)| \leq b,$ for all $\displaystyle x \in \mathbb{R},$ then $\displaystyle |f'(x)| \leq 2 \sqrt{ab},$ for all $\displaystyle x \in \mathbb{R}.$

3. This is being as precise as possible