> 
> let $f$ be a nonnegative and differentiable twice in the interval $[-1,1]$
> 
> 
> Prove: if $f(0)=0$ and $f'(0)=0$ then $f''(0)\geq 0$
> 
> 
> 1. Are all the assumptions on $f$ necessary for the result to hold ?
> 2. what can be said if $f''(0)= 0$ ?
> 
> 
> 

Looking at the taylor polynomial and lagrange remainder we get:

$$f(x)=f(0)+f'(0)x+\frac{f''(c)x^2}{2}$$

$$f(x)=\frac{f''(c)x^2}{2}$$

Because the function is nonnegative and $\frac{x^2}{2}\geq 0$ so $f''(c)\geq 0$

For 1., all the data is needed but I can not find a valid reason.

For 2., can we conclude that the function the null function?