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