If $f:\mathbb{R}^2 \rightarrow \mathbb{R}, f(x,y) = (y-x^2)(y-2x^2)$, then $(0,0)$ is a min. point of $f$ restricted to any line through origin.

Question

$f:\mathbb{R}^2 \rightarrow \mathbb{R}, f(x,y) = (y-x^2)(y-2x^2)$. Show that the origin is a minimum point for a restriction of $f$ to any line through origin, but isn't a minimum local point to $f$.

Here's my attempt:

$\partial_x f(x,y) = -2x(y-2x²) + (y-x²)(-4x) = -2xy +4x³-4xy+4x³ = 8x³ - 6xy $

$\partial_y f(x,y) = y-2x² + y-x² = 2y-3x² $

then, partialy differentiating again, we get

$ H(f(x,y))= \begin{pmatrix} 6(4x²-y) & -6x\\ -6x & 2 \end{pmatrix} $ So $\det(H(f(0,0)) = 0 $ and then the origin is a critical degenerated point, therefore can't be a local minimum point for $f$.

Is my argument alright? And how can I think about only the restriction of f to a line through the origin? My teacher suggested analysing the signal of f. So I found:

$ f>0 $ when $(y>2x²)$ or $ (y<x²)$

$ f<0 $ when $(y>x²$ and $y<2x²)$

How can I take conclusions for the restriction now?

Thanks.

Answer 1

Let $\;y=mx\;$ be any line through the origin, then

$$f(x,mx)=(mx-x^2)(mx-2x^2)=2mx^4-3mx^3+m^2x^2\implies$$

$$f'(x)=8mx^3-9mx^2+2m^2x$$

Observe that $\;x=0\iff y=0\;$ is a critical point, and

$$f''(0)=2m^2>0\implies x=0\;\;\text{is a minimal point}$$