Answer
Saddle Point of $f(0,-1)=2$
Work Step by Step
We have $f_x(x,y)=10x+4y+4=0, f_y(x,y)=4x-4y-4=0$
Solving the above equations leads to the critical point:
$(0,-1)$
We will take the help of second derivative test, for $(0,-1)$
$D=f_{xx}f_{yy}-f^2_{xy}=(10)(-4)-(4)^2=-56 \lt 0$
Thus, we have: saddle Point of $f(0,-1)=2$
