## University Calculus: Early Transcendentals (3rd Edition)

Saddle point at $(0,0)$ Local minimum point at $f(-1,1)=-2$ and Local minimum point at $f(1,-1)=-2$
Since, we have $f_x(x,y)=4x^3+4y=0, f_y(x,y)=4y^3+4x=0$ After solving the above two equations, we get The critical point is: $(0,0),(1,-1),(-1,1)$ As per the second derivative test, for $(0,0)$ we have $D=f_{xx}f_{yy}-f^2_{xy}=-16$ and $D=-16 \lt 0$ Thus, we have: Saddle points at $(0,0)$ As per the second derivative test, for $(1,-1)$ we have $D=f_{xx}f_{yy}-f^2_{xy}=128$ and $D=128 \gt 0$ and $f_{xx} \gt 0$ Thus, we have: Local minimum point at $f(1,-1)=-2$ As per the second derivative test, for $(-1,1)$ we have $D=f_{xx}f_{yy}-f^2_{xy}=128$ and $D=128 \gt 0$ and $f_{xx} \gt 0$ Thus, we have: Local minimum point at $f(-1,1)=-2$ Hence, we have: Saddle point at $(0,0)$ Local minimum point at $f(-1,1)=-2$ and Local minimum point at $f(1,-1)=-2$