What is the extreme f(x,y)=x^4+6y^2-4xy^3-1 ?

The extreme f(x,y)=x^4+6y^2-4xy^3-1 is Saddle(1,1),Saddle(-1,-1)

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extreme f(x,y)=x^4+6y^2-4xy^3-1

extreme

Saddle (1, 1), Saddle (- 1, - 1)

Solution steps

Find the critical points:

(1, 1), (- 1, - 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 24 x y + 12

D (x, y) = 12 x^{2} (- 24 x y + 12) - 144 y^{4}

Check the sign of at each critical point

Check critical point

(1, 1) :

Saddle

Check critical point

(- 1, - 1) :

Saddle

Saddle (1, 1), Saddle (- 1, - 1)

What is the extreme f(x,y)=x^4+6y^2-4xy^3-1 ?
The extreme f(x,y)=x^4+6y^2-4xy^3-1 is Saddle(1,1),Saddle(-1,-1)