$f(x)=x^4$ 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?

  • 2
    $\begingroup$ Ah, you've just hit upon a case where the second derivative test fails. :-) $\endgroup$ – Zain Patel Jul 5 '15 at 15:28
  • $\begingroup$ Why do you think that $f^{"}(0)=0\Rightarrow f'(x)$ doesn't change its sign? $\endgroup$ – Eclipse Sun Jul 5 '15 at 15:29
  • $\begingroup$ @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? $\endgroup$ – Whyka Jul 5 '15 at 15:31
  • $\begingroup$ It is not "difference", but the "rate of difference". Your example clearly shows that $f'(x)=4x^3$ does change its sign at $x=0$. $\endgroup$ – 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 $x=0$.

  • $\begingroup$ How do I show it has a maximum without using the second-derivative test? $\endgroup$ – Whyka Jul 5 '15 at 15:36
  • 2
    $\begingroup$ "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"! $\endgroup$ – murray Jul 5 '15 at 15:56
  • $\begingroup$ @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), $a^{f(x)}>0$? $\endgroup$ – Whyka Jul 5 '15 at 16:29
  • $\begingroup$ @Whyka: Certainly if $a > 0$ and $b > 0$, then $a^b > 0$. $\endgroup$ – murray Jul 6 '15 at 16:14
  • $\begingroup$ @murray of course. The point is, b doesn't have to be larger than 0 $\endgroup$ – Whyka Jul 6 '15 at 16:45

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


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