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

The extreme f(x,y)=4xy-2x^4-y^2+4x-2y is Saddle(0,-1),Maximum(-1,-3),Maximum(1,1)

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

extreme points

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

Saddle (0, - 1), Maximum (- 1, - 3), Maximum (1, 1)

Solution steps

Find the critical points:

(0, - 1), (- 1, - 3), (1, 1)

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

D (x, y) = 48 x^{2} - 16

Check the sign of at each critical point

Check critical point

(0, - 1) :

Saddle

Check critical point

(- 1, - 3) :

Maximum

Check critical point

(1, 1) :

Maximum

Saddle (0, - 1), Maximum (- 1, - 3), Maximum (1, 1)

f (x) = 5 (x - 3)^{2} (2 x + 1) (x + 4)^{3}

f (x) = 2 x - 1.75 x^{2} + 1. 1^{3} - 0.25 x^{4}

f (x y) = - x^{2} - y^{2} + 22 x + 18 y - 102

x^{2} + 2 x + 1

\frac{x + 4}{x ^{2} - 3 x - 28}

What is the extreme f(x,y)=4xy-2x^4-y^2+4x-2y ?
The extreme f(x,y)=4xy-2x^4-y^2+4x-2y is Saddle(0,-1),Maximum(-1,-3),Maximum(1,1)