## Thomas' Calculus 13th Edition

Saddle point at $(0,0)$ ; Local minimum point at $f(-1,1)=-2$ and Local minimum point at $f(1,-1)=-2$
Given: $f_x(x,y)=4x^3+4y=0, f_y(x,y)=4y^3+4x=0$ Simplify the given two equations. Critical points $(0,0),(1,-1),(-1,1)$ In order to solve this problem we will have to apply Second derivative test that suggests the following conditions to calculate the local minimum, local maximum and saddle points of $f(x,y)$ or $f(x,y,z)$. 1. If $D(a,b)=f_{xx}(a,b)f_{yy}(a,b)-[f_{xy}(a,b)]^2 \gt 0$ and $f_{xx}(a,b)\gt 0$ , then $f(a,b)$ is a local minimum. 2. If $D(a,b)=f_{xx}(a,b)f_{yy}(a,b)-[f_{xy}(a,b)]^2 \gt 0$ and $f_{xx}(a,b)\lt 0$ , then $f(a,b)$ is a local maximum. 3. If $D(a,b)=f_{xx}(a,b)f_{yy}(a,b)-[f_{xy}(a,b)]^2 \lt 0$ , then $f(a,b)$ is a not a local minimum or local maximum but a saddle point. $D(0,0)=f_{xx}f_{yy}-f^2_{xy}=-16$ and $D=-16 \lt 0$ So, Saddle point at $(0,0)$ $D(1,-1)=f_{xx}f_{yy}-f^2_{xy}=128\implies D=128 \gt 0$ and $f_{xx} \gt 0$ So, Local minimum point at $f(1,-1)=-2$ For the critical point $(-1,1)$: $D=f_{xx}f_{yy}-f^2_{xy}=128 \implies D=128 \gt 0$ and $f_{xx} \gt 0$ So, Local minimum point at $f(-1,1)=-2$ Therefore, Saddle points at $(0,0)$ ; Local minimum point at $f(-1,1)=-2$ and Local minimum point at $f(1,-1)=-2$