Finding the critical points of $f(x,y)=(y-x^2)(y-2x^2)$

Question

Finding the critical points of $f(x,y)=(y-x^2)(y-2x^2)$

I know that $(a,b)$ is a critical point $\iff \nabla f(a,b)=(0,0)$

So $\nabla f(x,y)= (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y})$

• $\frac{\partial f}{\partial x}[(y-x^2)(y-2x^2)]=8x^3-6xy$
• $\frac{\partial f}{\partial y}[(y-x^2)(y-2x^2)]=-3x^2+2y$

$\nabla f(x,y)=(8x^3-6xy, -3x^2+2y)=(0,0) \iff (x,y)=(0,0)$

So finding the Hessian matrix at $(0,0)$

• $\frac{\partial ^2f}{\partial x^2}[(y-x^2)(y-2x^2)]=\frac{\partial f}{\partial x}[8x^3-6xy]=24x^2-6y$
• $\frac{\partial ^2f}{\partial y^2}[(y-x^2)(y-2x^2)]=\frac{\partial f}{\partial y}[-3x^2+2y]=2$
• $\frac{\partial ^2f}{\partial x \partial y}[(y-x^2)(y-2x^2)]=\frac{\partial f}{\partial x}[8x^3-6xy]=-6x$

$H(0,0)=\begin{pmatrix} 0 & 0 \\ 0 & 2 \end{pmatrix}$

Finding the eigenvalues for $H(0,0)$:

$\det(H(0,0)-\lambda I)= \begin{vmatrix} -\lambda & 0 \\ 0 & 2-\lambda \end{vmatrix} = \lambda^2-2\lambda$

$\lambda^2-2\lambda=0 \iff \begin{cases} \lambda_1 = 0 \\ \lambda_2 = 2 \end{cases}$

So I can't really tell if $f(x,y)$ has any relative maximum nor minimum or a saddle point, and I don't know how to continue...

Edit: I know that this is a duplicate post, but I don't understand the answer, and the question is 4 years old.

Answer 1

Interpreting the linked OP's solution.

For $x^2<\color{red}{y=1.5x^2}<2x^2$: $$f(x,y)=(y-x^2)(y-2x^2)=(1.5x^2-x^2)(1.5x^2-2x^2)<0.$$ For $\color{red}{y=0.5x^2}<x^2$: $$f(x,y)=(y-x^2)(y-2x^2)=(0.5x^2-x^2)(0.5x^2-2x^2)>0.$$ Hence, it is a saddle point.

Answer 2

We have that

$$f(x,y)=(y-x^2)(y-2x^2)=y^2-3x^2y+2x^4=\left(y-\frac32x^2\right)^2-\frac14x^4$$

then it is a saddle point, indeed

• $y-\frac32x^2=0 \implies f(x,y)\le 0$

• $x=0 \implies f(x,y)\ge 0$