# Second partial derivative test is inconclusive

I am trying to find the critical points of the function:

$f(x,y)=2x^4-3x^2y+y^2$

and find the Max, Min and saddle points. What I've done so far is:

$f_x=8x^3-6xy=0 , f_y=-3x^2+2y=0 , f_{xx}=24x^2-6y , f_{yy}=2 , f_{xy}=-6x$

So (0,0) is the only critical point. But using the second partial derivative test:

$\Delta(0,0)=f_{xx} . f_{yy} - f^2_{xy}=0$

Which is inconclusive. Without using MATLAB or similar software and based on calculation, how can we determine whether (0,0) is Max, Min or saddle point?

Or the general question is what to do when it's inconclusive (without using software)?

Plug in $x = \epsilon$, $y = 3\epsilon^2$ and you will get that the function is greater than $0$ for all $\epsilon > 0$. Plug in $x = (3/4) \epsilon$, $y = \epsilon ^2$ and you will get that the function is less than $0$ for all $\epsilon > 0$. This implies that $(0,0)$ must be a saddle point because you take take $\epsilon > 0$ arbitrarily small.

• If $x=\dfrac{2\epsilon}{3}$ and $y=\epsilon^2$, then $f(x,y)=\dfrac{5\epsilon^4}{81}$, which is positive, not negative. – Steve Kass Apr 13 '14 at 21:43
• Your suggestion works if you change $\frac{2}{3}$ to $\frac{3}{4}$. For any $\epsilon>0$, $\,f(\dfrac{3\epsilon}{4},\epsilon^2)=-\dfrac{7\epsilon^4}{128}$. – Steve Kass Apr 13 '14 at 21:59
• How did you calculate y if $x=\dfrac{3\epsilon}{4}$? Shouldn't we use the equation $y=\dfrac{3x^2}{2}$? @SteveKass – Sam Apr 13 '14 at 22:06
• I didn’t calculate $y$. I calculated $f(x,y)=2x^4-3x^2y+y^2$ at the point $(x,y)=(\dfrac{3\epsilon}{4},\epsilon^2)$. – Steve Kass Apr 13 '14 at 22:08
• So you chose an arbitrary value ($\varepsilon^2$) for y? (I can't figure out how you got $y=\varepsilon^2$) @SteveKass – Sam Apr 13 '14 at 22:11

Hint: $$2x^4−3x^2y+y^2=(x^2-y)(2x^2-y).$$

Possibly easier hint (complete the square): $$2x^4 - 3x^2 + y^2 = (y-\frac32 x^2)^2 - \frac14x^2$$

• Still no clue, would you please elaborate? @mrf – Sam Apr 13 '14 at 20:55
• Same idea that my answer: the function can take $>0$ and $<0$ values for $(x,y)$ arbitrarily near of $(0,0)$. – Martín-Blas Pérez Pinilla Apr 13 '14 at 21:03
• I see, so can we say if the test is inconclusive, we have to examine the values in a neighborhood of the critical point? @mrf – Sam Apr 13 '14 at 21:06
• @Sam, usually is the best idea. – Martín-Blas Pérez Pinilla Apr 14 '14 at 6:44
• @mrf..can you please explain me how you complete the square?It is not like quadratic.I am in mess finding how to do in two variable case. – Believer Dec 29 '17 at 20:01

Added: Here's a way not to solve the problem (because it gives the wrong answer!)

Look at the concavity of the function at $(0,0)$ in a non-axis direction $y=ax$. $$\left.\frac{d^2}{dx^2}\right\rvert_{x=0}\left(f(x,ax)\right) = \left.\frac{d^2}{dx^2}\right\rvert_{x=0}\left(2x^4-3x^2 (a x)+(a x)^2\right)=\left. 2a^2-18ax+24x^2\right\rvert_{x=0}=2a^2.$$ This tells you that the function is concave up at $(0,0)$ in every non-axis direction, so $(0,0)$ is not a saddle point. It's not a local maximum, because $f_{yy}(0,0)>0$, so it’s a local minimum.

• Sort of. I just wanted to point out that up-concavity in every linear direction through a critical point is not sufficient evidence for a local minimum. You cannot conclude that you have a local minimum $f(0,0)=0$ just because along every linear direction through $(0,0)$, the function is either constant or concave up — that’s not a theorem. (This function is a counterexample to show that that conclusion is not valid in general.) You have here a function that is never concave down in a linear direction through $(0,0)$ but nevertheless has points arbitrarily close to zero where $f(x,y) < 0$. – Steve Kass May 9 '16 at 1:37