How can $f(x)=x^4$ have a global minimum at $x=0$ but $f''(0)=0$? $f(x)=x^4$ has a global minimum in $\Bbb R$ at the point $x=0$, but $f''(0)=0$.
This case confuses me. For every $0\neq x\in I$, $f(x)>f(0)$. So how can it be that $f''(0)=0$, following $f'(x)$ doesn't change its sign at $x=0$?
I could accept it if there was a little segment $I$ around $x=0$ fulfilling $f(x)=0$ for every $x\in I$. But I don't see why that can be the case, since, again, $x=0$ is the only $x$ fulfilling $f(x)=0$.
This contradicts my logic. Can someone help me understand how this is possible?

There is no contradiction. For a local minimum (of a sufficiently differentiable function) at $x=a$ it is a necessary condition that $f'(a)=0$ and it is sufficient that $f'(a)=0$ and $f''(a)>0$. With $f(x)=x^4$ we are in the wide range between the necessary and the sufficient.
For real fun, consider $$f(x)=\begin{cases}e^{-1/x^2}&\text{if }x\ne 0\\0&\text{if }x=0\end{cases} $$
and show that $f$ has a unique local minimum at $x=0$, whereas all derivatives exist and are equal to $0$ at $x=0$.
