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

The extreme f(x,y)=-x^4-y^4+4xy is Saddle(0,0),Maximum(-1,-1),Maximum(1,1)

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

extreme points

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

Saddle (0, 0), Maximum (- 1, - 1), Maximum (1, 1)

Solution steps

Find the critical points:

(0, 0), (- 1, - 1), (1, 1)

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

D (x, y) = 144 x^{2} y^{2} - 16

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(- 1, - 1) :

Maximum

Check critical point

(1, 1) :

Maximum

Saddle (0, 0), Maximum (- 1, - 1), Maximum (1, 1)

x^{2} - 4

f (x) = 3 x^{3} - 2 x^{2} + 6

K (a, b) = 4 a + 9 ab

e^{- 2.5 x^{2}}

f (x) = - 2 x^{3} + 18 x^{2} - 48 x + 19

What is the extreme f(x,y)=-x^4-y^4+4xy ?
The extreme f(x,y)=-x^4-y^4+4xy is Saddle(0,0),Maximum(-1,-1),Maximum(1,1)