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 points

f (x, y) = x^{4} + 6 y^{2} - 4 x y^{3} - 1

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)

3 (x - 2)

f (x) = 2 sin (x) + 2 cos (x)

f (x, y) = y^{2} x - y x^{2} + x y

f (x) = x^{3} - 3 x + 7

f (x) = y^{4} - 4 y^{3} + 2 x^{2} + 8 x y

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)