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?

$f'(x)$ does change it sign at $0$ though. Calculate $f'(x)$ and try point either side of $0$ and you will see this.
