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

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

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

extreme points

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

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}} = - 4

D (x, y) = 48 x^{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)

f (x) = x 32 - x^{2}

x \cdot ln (x)

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

f (x, y) = 3 + x y - x - 2 y

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

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