Let f(x, y)=x^2+y^2+xy-x+y. Use the second derivative test to show that x=1, y=-1 is a local mini-mum (and thus an absolute minimum it is the only critical point) of f. fxx=2 fyy=2 What to do next?

ƒ(x, y) = x^{2} + y^{2} +xy - x + 1

ƒ_{x} = 2x + y - 1 , ƒ_{y} = 2y + x + 1

critical point when ƒ_{x}=ƒ_{y}=0: (1, -1)

ƒ_{xx} = 2 , ƒ_{yy} = 2 , ƒ_{xy} = 1

The second derivative test requires you to find ƒ_{xx} and ƒ_{xx} ƒ_{yy} - ƒ^{2}_{xy} at the critical point:

at (1, -1): ƒ_{xx} = 2 , ƒ_{xx} ƒ_{yy} - ƒ^{2}_{xy} = (2)(2) - (1)^{2} = 4 - 1 = 3

Since ƒ_{xx}>0 and ƒ_{xx} ƒ_{yy} - ƒ^{2}_{xy}>0 at (1, -1), then (1, -1) is a local minimum.

## Comments

How did you get fxy?

ƒ

_{xy}is the derivative of ƒ_{x}in terms of y