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I'm doing a problem that asks me to find these things for $f(x,y)=y^3+3x^2y-6x^2-6y^2+2$.

I found $f_x=6xy-12x$ and $f_y=3y^2+3x^2-12y$

And found two of the zeros/points listed in the answer key, $(0,0), (0,4)$.

The book says that points $(1,1), (-1,-1)$ are saddle points but I don't see how they got those x and y values as zeros of the second derivatives. In the case of x=1 and y=1, $f_x=6(1)(1)-12(1)=6-12=-6.$

Can someone please explain this?

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    $\begingroup$ Have you found the hessian matrix with respect to this function? What are its eigenvalues when evaluated at $(0,0)$ and $(0,4)$? $\endgroup$ – Git Gud Oct 15 '16 at 6:59
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    $\begingroup$ How were you taught to determine if the critical points are maxima, minima or saddle points? $\endgroup$ – Git Gud Oct 15 '16 at 7:03
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    $\begingroup$ Oh, I see, It's exactly the same thing, only your instructor chewed your food and fed it directly into your mouth. OK, and now I realize I misread your question, sorry. Give me a minute. $\endgroup$ – Git Gud Oct 15 '16 at 7:09
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    $\begingroup$ MSE is an amazing place to learn. And the book is wrong. Saddle points are critical points and $(1,1)$ and $(-1,-1)$ are not critical points. $\endgroup$ – Git Gud Oct 15 '16 at 7:17
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    $\begingroup$ You did miss a couple of points, though. And those might be saddle points. Try $(-2,2)$ and $(2,2)$. $\endgroup$ – Git Gud Oct 15 '16 at 7:32
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Finding the Max (downhill in every direction)/Min (uphill in every direction)/Saddle points (down hill along one axis, uphill along the other) for $f(x,y)=y^3+3x^2y-6x^2-6y^2+2$.

Equation to test the points you find to see if they are one of the types you are looking for:

$D$(hessian matrix)$=f_{xx}*f_{yy}-(f_{xy})^2$

-> If D > 0 then the point plugged into its equation is an extrema point. If D < 0 then the point is a saddle point. In the case that D is zero, the test is inconclusive.

-> Need $f_x$ an $f_y$ as well to find the critical points. -> Will use $f+{xx}$ to determine if an extrema point is a maximum point or a minimum point.

Needed partial derivatives:

$f_x=6xy-12x$, $f_y=3y^2+3x^2-12y$

$f_{xx}=6y-12$, $f_{yy}=6y-12$

$f_{xy}=6x$

$D=(6y-12)*(6y-12)-(6x)^2$

Final equation:

$D=36y^2-144y+144$

Finding the critical points (those which are candidates for being a max, min, or saddle point):

->Set $f_x=f_y=0$

$6xy-12x=0$ -----and------ $3y^2+3x^2-12y=0$

-> There is an x or y in every term of both equations. -> Thus setting both x and y to zero sets both equations to zero.

-> First critical point: (0,0)

Now,

-> Use left hand side to find a equation for x.

$6xy-12x=0$ -> $6xy=12x$ -> $\frac{12x}{x}=6y$ -> $12=6y$

-> Thus: y=2. -> Plug y value into right hand side equation.

$3(2^2)+3x^2-12(2)=0$

$12+3x^2-24=0$

$3x^2=12$

-> Thus x = -2 || 2

-> This yields the following critical points: (-2, 2) and (2, 2).

Finally,

-> Notice that setting x to zero, with any y value makes the left hand equation equal to zero. If x equals zero on the right hand side:

$3y^2+3(0^2)-12y$

$(3y^2-12y)=0$

$3(y^2-4y)=0$ -> $y^2-4y=0$

-> Thus y= 4, yielding the following critical point: (0, 4).

-> Therefore the critical points are: (0, 0), (0, 4), (2, 2), (-2, 2).

Testing the points using the formula for D:

$D(0, 0)=36(0)^2-144(0)+144=144$

-> D is positive so this is an extrema point. Plug the point into $f_{xx}$.

$f_{xx}(0,0)=6(0)-12=-12$

-> $f_{xx}$ is negative, the slope in all directions from this point goes downwards, so this is a local max.

$D(0, 4)=36(4)^2-144(4)+144=144$

-> D is positive so this is an extrema point. Plug the point into $f_{xx}$.

$f_{xx}(0,0)=6(4)-12=18$

-> $f_{xx}$ is positive, the slope in all directions from this point goes upwards, so this is a local min.

$D(2, 2)=36(2)^2-144(2)+144=-288$

-> D is negative meaning this is a saddle point.

$D(-2, 2)=36(2)^2-144(2)+144=-288$

-> D is negative meaning this is a saddle point as well.

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