Answer
Saddle Point of $f(0,1)=-3$
Local Minimum $f(2, 1)=-19$
Local Minimum $f(-2, 1)=-19$
Work Step by Step
We have $f_x(x,y)=4x^3-16x=0; f_y(x,y)=6y-6=0$
Solving the above two equations leads to the critical points: $(-2,1 )$ and $(0,1)$ and $(2,1)$
We will take the help of second derivative test, for $(0,1)$ as follows:
$D=f_{xx}f_{yy}-f^2_{xy}=(-16)(6)-0=-96 \lt 0$
So, Saddle Point of $f(0,1)=-3$
We will take the help of second derivative test, for $(2,1)$ as follows:
$D=f_{xx}f_{yy}-f^2_{xy}=(-16)(6)-0^2=192 \gt 0$
So, Local Minimum $f(2, 1)=-19$
We will take the help of second derivative test, for $(-2,1)$ as follows:
$D=f_{xx}f_{yy}-f^2_{xy}=(32)(6)-0^2=192 \gt 0$
So, Local Minimum $f(-2, 1)=-19$
