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

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

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

extreme points

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

Saddle (0, 0), Minimum (- 1, - 1), Minimum (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) :

Minimum

Check critical point

(1, 1) :

Minimum

Saddle (0, 0), Minimum (- 1, - 1), Minimum (1, 1)

f (x) = 40000 - 2 x^{2}

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

f (x) = x (25 - 2 x) (\frac{48}{2} - x)

f (x) = x^{3} - y^{3} + 3 x^{2} + 3 y^{2} - 9 x + 5

f (x) = \frac{4}{3} x^{3} - \frac{25}{2} x^{2} + 6 x + 4

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