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I found the following problem: Find all relative min/max and saddle points for the function $f(x,y)=x^4 +4x^2y-8x^2$. All the derivatives work out easily enough:

$f_{x}=4x^3+8xy-16x$

$f_{y}=4x^2$

$f_{xx}=12x^2+8y-16$

$f_{yy}=0$

$f_{xy}=8x$

So then to find the critical points, $f_x=f_y=0$ happens whenever $x=0$ and $y=$anything. So you calculate $f_{xx}f_{yy}-(f_{xy})^2=0$ at for any $(0,y)$.

So the Second Derivative Test is inconclusive, so I don't know which points could possibly be max or min points or saddle points. I can look at the graph in geogebra and visually verify, but how would I know with pencil and paper only which spots along the y axis are min/max/saddle? I know I could do something like $f(0,5)$ and compare it to $f(0.01,5)$ and $f(-0.01,5)$ and that shows me it is a local min. But I wouldn't want to do that for every point down the axis, obviously.

Is there something simple I'm missing?

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  • $\begingroup$ Probably a good idea to rewrite $$f(x,y)=(x^2)^2+2x^2(2y-4)=\Big(x^2+(2y-4)\Big)^2-(2y-4)^2$$ $\endgroup$ Commented Jan 28, 2022 at 4:20
  • $\begingroup$ @CalvinKhor Again, I must be missing something...I see how that the function factors to what you said, but how is that factorization helpful? $\endgroup$
    – ruferd
    Commented Jan 28, 2022 at 18:00
  • $\begingroup$ well I just have a hunch it’s useful but I have not sat down and finished the problem. It makes it easy to systematically check that the y axis is full of minimum points (and I didn’t have to see the graph for that). It hints that you can probably find a saddle point by comparing with $x^2-y^2$. I don’t know. Hence why I said ‘probably’. Good luck. $\endgroup$ Commented Jan 29, 2022 at 2:43

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$f(x,y)=x^2(x^2+4y-8) ; f(0,y)=0 $ for all $y$

let's choose the point $(\epsilon_1,y+\epsilon_2)$ in a neighborhood of $(0,y)$, $\epsilon_1$ and $\epsilon_2$ are very small positive or negative numbers : $f(\epsilon_1,y+\epsilon_2)=\epsilon_1^2(\epsilon_1^2+4y+4\epsilon_2-8)$

if $y=2$, $f(\epsilon_1,2+\epsilon_2)=\epsilon_1^2(\epsilon_1^2+4\epsilon_2)$ : we can get either positive or negative values for $f(\epsilon_1,y+\epsilon_2)$ for some $(\epsilon_1,\epsilon_2)$ in every neighborhood of $(0,0)$ therefore $(0,2)$ is a saddle point.

if $y>2$, $4y-8>0$ and $f(\epsilon_1,y+\epsilon_2)=\epsilon_1^2(\epsilon_1^2+4\epsilon_2+4y-8)>0$ for every $(\epsilon_1,\epsilon_2)$ in some neighborhood of $(0,0)$ therefore $(0,y)$ is a minimum point.

if $y<2$ , $4y-8<0$ and $f(\epsilon_1,y+\epsilon_2)=\epsilon_1^2(\epsilon_1^2+4\epsilon_2+4y-8)<0$ for every $(\epsilon_1,\epsilon_2)$ in some neighborhood of $(0,0)$ therefore $(0,y)$ is a maximum point.

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