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For each of the following functions, what do the first- and the second-order optimality conditions say about wether 0 is a minimum on $\mathbb{R}$.

$f_1(x)=x^2$

$f_2(x)=x^3$

$f_3(x)=x^4$

$f_4(x)=-x^4$

So the dazzling thing is that the second derivative for all these functions gives $0$ when $0$ is input, i.e. $f_1''(0)=f_2''(0)=f_3''(0)=f_4''(0)=0$. Since the Hessian is given by the second order derivatives, and since these derivatives determine wether it is positive definite, negative definite, indefinite or singular. And since these properties determine wether a critical point is a minimum, maximum, saddle point or some other pathological situation. I get confused.

So all functions are neither a maximum, minumum or saddle point at $0$? Why would the book ask a silly question? (Im have a strong guess that I am doing things wrong here).

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1 Answer 1

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You'll need do consider higher-order derivatives, if the Hessian is indefinite. In this case but $$f_1'' (0) = 2 \neq 0$$ So you made a mistake there.

For $f_3, f_4$ you have $$f_3''(0) = f_3'''(0) = f_4''(0) = f_4'''(0) = 0$$ And $$f_3^{(4)}(0) = 24 = -f_4^{(4)}(0)$$ So you get local (global) minima / maxima there but nothing can be derived from $f_3''(0)$ and $f_4''(0)$

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  • $\begingroup$ So you can just go on looking at higher derivatives until you get possible nonzero answers to draw any conclusions?So for example, is it true that $f_2$ has a minimum and $f_4$ has a maximum at $0$? $\endgroup$
    – onimoni
    Oct 7, 2013 at 8:55
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    $\begingroup$ @user61001 Yes, that is correct. And if an odd-numbered derivative is the first non-zero one, it's a saddle point. In higher dimensions this turns out to become incredibly ugly, though. $\endgroup$
    – AlexR
    Oct 7, 2013 at 19:23

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