# Second Derivative Test for Multivariable Calculus Example

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?

• Probably a good idea to rewrite $$f(x,y)=(x^2)^2+2x^2(2y-4)=\Big(x^2+(2y-4)\Big)^2-(2y-4)^2$$ Commented Jan 28, 2022 at 4:20
• @CalvinKhor Again, I must be missing something...I see how that the function factors to what you said, but how is that factorization helpful? Commented Jan 28, 2022 at 18:00
• 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. Commented Jan 29, 2022 at 2:43

## 1 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.