- #1

mathmari

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We have the function $\displaystyle{f(x,y)=y^2-3x^2y+2x^4}$ and the function $\displaystyle{g_v(t)=f(tv_1, tv_2)=t^2v_2^2-3t^3v_1^2v_2+2t^4v_1^4}$.

I have shown that $g$ has a local minimumat $t=0$

I want to show that $f$ has not a local minimum in $(0,0)$.

The gradient is \begin{equation*}\nabla f=\begin{pmatrix}-6xy+8x^3 \\ 2y-3x^2\end{pmatrix}\end{equation*} Acritical point is $(0,0)$, since $\displaystyle{\nabla f\begin{pmatrix}0 \\ 0\end{pmatrix}=\begin{pmatrix}0 \\ 0\end{pmatrix}}$.

The Hessian matrix is \begin{equation*}H_f(x,y)=\begin{pmatrix}f_{xx} & f_{xy}\\ f_{yx} & f_{yy}\end{pmatrix}=\begin{pmatrix}-6y+24x^2 & -6x\\ -6x & 2\end{pmatrix}\end{equation*}

The matrixin in $(0,0)$ is

\begin{equation*}H_f(0,0)=\begin{pmatrix}0 & 0\\ 0 & 2\end{pmatrix}\end{equation*}

The eigenvalues are, $\lambda_1=0$, $\lambda_2=2>0$. These are non-negativ, so the Hessian matrix is positiv semidefinite.

That means that $f$ has in$(0,0)$ either a minimum or a saddle point.

Is that correct so far? Now we have to show that in $(0,0)$ has a sddle point and not a minimum, right? But how? :unsure: