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Consider the function $f:\mathbb{R}^3\rightarrow \mathbb{R}$ is given by


Does the function have a local maximum or a minimum at the origin?

My question is is there a general criteria to go about solving problems like this?. Obviously $\nabla g(0,0,0)=0$ but that does not tell you anything. Also since this is in three variables there is no "second derivative test" (using the Hessian) technique either. How does one solve this?


As an example of a problem that can be solved without using any of the "Hessian" machinery consider the following: $f(x,y)=x^2y^5+x^4y^4$. Then show that the $(0,0)$ is a saddle point of this function. Obviously the second derivative test fails here. But we have that $f(x,y)=x^2y^4(y+x^2)$ and since the second factor is both positive and negative in arbitrarily small neighborhoods of $(0,0)$ we have the desired conclusion.(The first factor of course is non negative). Thanks for the help.

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there is a second derivative test. If the Hessian is either positive definite or negative definite you have an answer. In any case, try some sample points near the origin. – Will Jagy Jan 1 '13 at 5:50
@WillJagy: Thank you. But isn't there a way to do this without using any of that machinery?. – Jack Dawkins Jan 1 '13 at 5:54
up vote 3 down vote accepted

The OP wants something other than the Hessian. Let $\epsilon>0$ be small and $k=\pm1$. Then $f(\epsilon^2,\epsilon^8,k\epsilon)=\epsilon^{16}+k\epsilon^{11}+\epsilon^{12}$, in which the dominant term is $k\epsilon^{11}$. Hence $f(\epsilon^2,\epsilon^8,k\epsilon)$ is greater than or smaller than $0$ according to whether $k>0$ or $k<0$, which implies that $(0,0,0)$ is neither a local maximum nor a local minimum of $f$.

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