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

The extreme f(x,y)=-x^4+y^4+x^2-y^2 is

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extreme f(x,y)=-x^4+y^4+x^2-y^2

extreme points

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

Saddle (0, 0), Minimum (0, - \frac{1}{2}), Minimum (0, \frac{1}{2}), Maximum (- \frac{1}{2}, 0), Saddle (- \frac{1}{2}, - \frac{1}{2}), Saddle (- \frac{1}{2}, \frac{1}{2}), Maximum (\frac{1}{2}, 0), Saddle (\frac{1}{2}, - \frac{1}{2}), Saddle (\frac{1}{2}, \frac{1}{2})

Solution steps

Find the critical points:

(0, 0), (0, - \frac{1}{2}), (0, \frac{1}{2}), (- \frac{1}{2}, 0), (- \frac{1}{2}, - \frac{1}{2}), (- \frac{1}{2}, \frac{1}{2}), (\frac{1}{2}, 0), (\frac{1}{2}, - \frac{1}{2}), (\frac{1}{2}, \frac{1}{2})

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

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

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(0, - \frac{1}{2}) :

Minimum

Check critical point

(0, \frac{1}{2}) :

Minimum

Check critical point

(- \frac{1}{2}, 0) :

Maximum

Check critical point

(- \frac{1}{2}, - \frac{1}{2}) :

Saddle

Check critical point

(- \frac{1}{2}, \frac{1}{2}) :

Saddle

Check critical point

(\frac{1}{2}, 0) :

Maximum

Check critical point

(\frac{1}{2}, - \frac{1}{2}) :

Saddle

Check critical point

(\frac{1}{2}, \frac{1}{2}) :

Saddle

Saddle (0, 0), Minimum (0, - \frac{1}{2}), Minimum (0, \frac{1}{2}), Maximum (- \frac{1}{2}, 0), Saddle (- \frac{1}{2}, - \frac{1}{2}), Saddle (- \frac{1}{2}, \frac{1}{2}), Maximum (\frac{1}{2}, 0), Saddle (\frac{1}{2}, - \frac{1}{2}), Saddle (\frac{1}{2}, \frac{1}{2})

f (x) = 4 x^{3} + 3 x^{2} - 60 x - 12

f (x) = (x + 5)^{3} + 7

f (x) = \frac{4 x}{x ^{2} - 4}

f (x) = 4 x e^{- x}

f (x) = x^{3} + 3 x y^{2} - 15 x + y^{3} - 15 y

What is the extreme f(x,y)=-x^4+y^4+x^2-y^2 ?
The extreme f(x,y)=-x^4+y^4+x^2-y^2 is