Answer
Local minimum of $f(-2,-2)=-8$
Work Step by Step
$f_x(x,y)=2x-y+2=0$ and $f_y(x,y)=-x+2y+2=0$
Simplify the above two equations to find the critical points.
Thus, we have: $(-2,-2)$
By the second derivative test, for $(-2,-2)$
$D=f_{xx}f_{yy}-f^2_{xy}=(2)(2)-(1)^2=3 \lt 0$
Local minimum of $f(-2,-2)=-8$
