Solution: ∂f ∂x (x,y) = ex(x2 +2x−y2) = 0 ∂f ∂y (x,y) = −2yex = 0 ∴y = 0 ⇒ x = 0,−2 the critical points are (0,0) and (−2,0) (3 points) ∂2f ∂x2 ...
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