Prove: if $f(0)=0$ and $f'(0)=0$ then $f''(0)\geq 0$ 
let $f$ be a nonnegative and differentiable twice in the interval $[-1,1]$
Prove: if $f(0)=0$ and $f'(0)=0$ then $f''(0)\geq 0$

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*Are all the assumptions on $f$ necessary for the result to hold ?


*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?

For 2) you could have $f(x)=x^4$ which satisfies all properties and conclusions including $f''(0)=0$, yet $f(x)$ is not identically zero.
For 1) relaxing the condition that $f(0)=0$, we could look at $f(x)=\cos(x)+3$ (which has instead $f(0)=4$).  It satisfies the property that $f(x)$ is non-negative, is twice differentiable on $[-1,1]$ and that $f'(0)=0$.  However, $f''(0)=-\cos(0)=-1$ is not non-negative.
Relaxing the condition that $f(x)$ be non-negative on the interval, letting $f(x)=-x^2$ one has $f(0)=0, f'(0)=0$, yet $f''(0)=-2$ is not non-negative
Relaxing the condition that $f$ be twice differentiable makes no sense since then $f''(0)$ can't be talked about.
Try searching yourself as well for a function $f(x)$ which violates only the hypothesis that $f'(0)=0$ which has $f''(0)<0$.
With these counterexamples, we conclude that in fact every hypothesis was necessary.
