I trying to solve this problem in Advanced Calc by Buck, sec 3.6 problem 9:

Let $f(x,y)=(y-x^{2})(y-2x^{2})$. Show that the origin is a critical point for $f$ which is a saddle point, although on any line through the origin, $f$ has a local minimum at $(0,0)$.


We have \begin{align*} f(x,y) & =(y-x^{2})(y-2x^{2})\\ & =y^{2}-3x^{2}y+2x^{4} \end{align*}

so that \begin{align*} f_{1}(x,y) & =-6xy+8x^{3} & & (1)\\ f_{2}(x,y) & =2y-3x^{2} & & (2) \end{align*} set (1) and (2) equal to zero and solve:

from (2) $2y-3x^{2}=0\,\Longrightarrow y=\frac{3}{2}x^{2}$

in (1) $-6xy+8x^{3}=0\,\Longrightarrow-9x^{3}+8x^{3}=0\,\Longrightarrow x=0,\Longrightarrow y=0$ hence (0,0) is a critical point of $f$.

(0,0) is a saddle point: By Theorem 19 page 157: If the determinant of the hessian (2nd order partial derivatives matrix) is negative at $p_{0}$, then $p_{0}$is a saddle point.

hence $H=\left[\begin{array}{cc} f_{11} & f_{12}\\ f_{21} & f_{22} \end{array}\right]=\left[\begin{array}{cc} -6y+24x^{2} & -6x\\ -6x & 2 \end{array}\right]$


at (0,0) $\Delta=0$, therefore the second derivative test is inconclusive.

However since $f(x,y)=(y-x^{2})(y-2x^{2})$, then if $x^{2}<y<2x^{2}$ we have $f(x,y)<0$, also if ($y<x^{2}$ or $y>2x^{2}$) we have $f(x,y)>0$, and we have $f(x,y)=0$ if $x=y=0$. Therefore the critical point (0,0) is a saddle point.

I don't feel confident about the proof of (0,0) is a saddle point. Also I am not sure how to do the last part (any line through the origin, $f$ has a local minimum at $(0,0)$) any help appreciated.


Yout proof that $(0,0)$ is a saddle point is correct. Lines should be understood as straight lines, hence, if $y=cx$, where $c\neq0$, $$ f(x,cx)=(cx-x^2)(cx-2x^2)=x^2(c-x)(c-2x)>0 $$ for $x\neq0$, but near 0. The cases $c=0$ and $x\equiv0$ are obvious.

  • $\begingroup$ @Izak $c$ is a constant, hence if $c\neq0$, for small $x$ such are also $c-x$ and $c-2x$. And we are interesting in a local minimum. $\endgroup$ – Przemysław Scherwentke Nov 16 '14 at 6:18
  • $\begingroup$ Got it, thank you @Przemysław Scherwentke $\endgroup$ – Izak Nov 16 '14 at 6:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.