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

The extreme f(x,y)=x^4+y^4-4xy+2 is Saddle(0,0),Minimum(-1,-1),Minimum(1,1)

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

extreme points

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

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) = P (x) = x^{3} + x^{2} - 5 x + 3

2 x^{3} - 3 x^{2} - 12 x + 12

f (x) = x^{3} - 3 x + 1, 0 \leq x \leq 5

f (x) = cos (7 x)

f (x) = x - 2 ln (x)

What is the extreme f(x,y)=x^4+y^4-4xy+2 ?
The extreme f(x,y)=x^4+y^4-4xy+2 is Saddle(0,0),Minimum(-1,-1),Minimum(1,1)