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?
Best Answer
$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.