# 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?

• Ah, you've just hit upon a case where the second derivative test fails. :-) – Zain Patel Jul 5 '15 at 15:28
• Why do you think that $f^{"}(0)=0\Rightarrow f'(x)$ doesn't change its sign? – Eclipse Sun Jul 5 '15 at 15:29
• @EclipseSun because $f''(x)$ shows the difference in $f'(x)$ at the same point, but 0 difference means no change in $f'(x)$... Or am I wrong? – Whyka Jul 5 '15 at 15:31
• It is not "difference", but the "rate of difference". Your example clearly shows that $f'(x)=4x^3$ does change its sign at $x=0$. – Eclipse Sun Jul 5 '15 at 15:33

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 $x=0$.
• "maximum" should be "minimum" in the answer at hand. How show this function has a minimum at $x=0$? simple: use the definition of "minimum"! – murray Jul 5 '15 at 15:56
• @murray yes, pardon me. I forgot my definitions. And yes, it should be minimum. For every $0\neq x\in \Bbb R$, $f(x)>f(0)$. Just to make sure - is it true that for every a>0 and function f(x), $a^{f(x)}>0$? – Whyka Jul 5 '15 at 16:29
• @Whyka: Certainly if $a > 0$ and $b > 0$, then $a^b > 0$. – murray Jul 6 '15 at 16:14
$f'(x)$ does change it sign at $0$ though. Calculate $f'(x)$ and try point either side of $0$ and you will see this.