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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 $(0,0)$.

Solution:

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

$\Delta=-12y+48x^{2}-36x^{2}=12(x^{2}-y)$

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.

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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.

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  • $\begingroup$ @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

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