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Use the second derivative test?

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?

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ƒxy is the derivative of ƒx in terms of y

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Tamara J. | Math Tutoring - Algebra and Calculus (all levels)Math Tutoring - Algebra and Calculus (al...
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   ƒ(x, y) = x2 + y2 +xy - x + 1

   ƒx = 2x + y - 1  ,   ƒy = 2y + x + 1

   critical point when ƒxy=0:     (1, -1)

   ƒxx = 2  ,   ƒyy = 2  ,   ƒxy = 1 

The second derivative test requires you to find ƒxx and ƒxx ƒyy - ƒ2xy at the critical point:

   at (1, -1):    ƒxx = 2   ,     ƒxx ƒyy - ƒ2xy = (2)(2) - (1)2 = 4 - 1 = 3

Since ƒxx>0 and ƒxx ƒyy - ƒ2xy>0 at (1, -1), then (1, -1) is a local minimum.