Use the second derivative test to identify any critical points and determine whether each critical point is a maximum, minimum, saddle point, or none of these. f(x, y) = 5x2 - 4xy + y² – 6y (х, у, 2) 3D (| 10, — 4,2 minimum

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Q: 3x4 – 4x3 – 12x² + 18 f(x, y) = 12(1+ 4y²)

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A: Critical points using second derivative test

Q: f (x, y) = xª – 4.x³ + 2x² + 8xy + 1.

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