# Determining whether a critical point is max,min or saddle

Consider $f(x,y,z)=x^2+7y^2-2xy+6yz+2y^3+3z^2$. Show the origin is a critical point and classify the critical point as local min, local max or a saddle point.

I think I've shown that the origin is a critcal point by finding the $\nabla f(x,y,z)=(2x-2y,14y-2x+6z+6y^2,6y+6z)=\vec{0}$ gives the 3 equations:

$2x-2y=0\implies x=y$

$14y-2x+6z+6y^2=0$

and

$6y+6z=0\implies y=-z$

But I'm not too sure how to continue from this point.

EDIT:

I tried to find the eigenvalues of the Hessian, but I am having trouble finding the eigenvalues. I got the characteristic polynomial as:

$144-\lambda^3+30\lambda^2-124\lambda-12\lambda^2 y-96\lambda y+ 144y$

which I have no idea how to even begin to factor to get my eigenvalues. Is there some sort of shortcut I'm not seeing?

• I don't think I understand why your $f$ takes $w$ as an argument and then doesn't use it. Doing so makes this problem a bit weird because $(0,0,0,w)$ will be a critical point for all $w$. – Ian Nov 2 '15 at 14:41
• Sorry, that's a typo. There shouldn't be a $w$ there. Thanks for pointing it out. – Gretchen Nov 2 '15 at 14:42
• You also have a $w$ in $\nabla f$. Anyway, now you should use the second derivative test. Do you know about eigenvalues? (There is a way to answer the question without eigenvalues, but it is not so obvious why it works.) – Ian Nov 2 '15 at 14:43
• By Second Derivative Test, you mean checking the determinant of the Hessian of $f$? Yes, I know about eigenvalues. – Gretchen Nov 2 '15 at 14:45
• OK. The point is a local max if all eigenvalues of the Hessian are negative, a local min if all eigenvalues of the Hessian are positive, and a saddle point if some eigenvalues of the Hessian are positive and others are negative. If all are nonnegative but some are zero or all are nonpositive but some are zero, then the second derivative test doesn't help. – Ian Nov 2 '15 at 14:49