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Question Number 133191 by pticantor last updated on 19/Feb/21

$$\mathit{g}(\mathit{x},\mathit{y})={\mathit{x}}^4+{\mathit{y}}^4-2{(\mathit{x}-\mathit{y})}^2$$$$\mathit{f}\mathit{i}\mathit{n}\mathit{d}\mathit{c}\mathit{r}\mathit{i}\mathit{t}\mathit{i}\mathit{c}\mathit{a}\mathit{l}\mathit{s}\mathit{p}\mathit{o}\mathit{i}\mathit{n}\mathit{t}\mathit{s}\mathit{o}\mathit{f}\mathit{g}(\mathit{x},\mathit{y})$$$$\mathit{a}\mathit{n}\mathit{d}\mathit{h}\mathit{e}\mathit{r}\mathit{s}\mathit{n}\mathit{a}\mathit{t}\mathit{u}\mathit{r}\mathit{e}$$

Answered by Olaf last updated on 20/Feb/21

$$g(x,y)=x^4+y^4-2{(x-y)}^2$$$$▽g=\left[\begin{array}{c}\frac{\partial g}{\partial x}\\ \frac{\partial g}{\partial y}\end{array}\right]=\left[\begin{array}{c}4x^3-2(x-y)\\ 4y^3+2(x-y)\end{array}\right]$$$$▽g=\left[\begin{array}{c}0\\ 0\end{array}\right]\iff (x-y)=2x^3=-2y^3$$$$\iff (x,y)=(0,0):\mathrm{this}\mathrm{is}\mathrm{the}\mathrm{only}$$$$\mathrm{critical}\mathrm{point}.$$$$$$$$\mathrm{Hessian}\mathrm{matrix}:$$$$\left[\begin{array}{cc}\frac{{\partial }^{2}g}{\partial x^2}& \frac{{\partial }^{2}g}{\partial x\partial y}\\ \frac{{\partial }^{2}g}{\partial y\partial x}& \frac{{\partial }^{2}g}{\partial y^2}\end{array}\right]=\left[\begin{array}{cc}12x^2-2x& 2\\ 2& 12y^2-2y\end{array}\right]$$$$\mathrm{Hessian}\mathrm{matrix}\mathrm{in}(0,0):$$$$\mathrm{Hess}\left(g\right){\mid }_{(0,0)}=\left[\begin{array}{cc}0& 2\\ 2& 0\end{array}\right]$$$$\det \left(\mathrm{Hess}\left(g\right){\mid }_{(0,0)}\right)=\left|\begin{array}{cc}0& 2\\ 2& 0\end{array}\right|=-4$$$$\det \left(\mathrm{Hess}\left(g\right){\mid }_{(0,0)}\right)<0$$$$\Rightarrow (0,0)\mathrm{is}\mathrm{a}\mathrm{saddle}\mathrm{point}$$

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