What is the extreme f(x,y)=x^2y+4xy-2y^2-3 ?

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

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extreme f(x,y)=x^2y+4xy-2y^2-3

extreme points

f (x, y) = x^{2} y + 4 x y - 2 y^{2} - 3

Saddle (0, 0), Saddle (- 4, 0), Maximum (- 2, - 1)

Solution steps

Find the critical points:

(0, 0), (- 4, 0), (- 2, - 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 4

D (x, y) = - 8 y - 4 x^{2} - 16 x - 16

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(- 4, 0) :

Saddle

Check critical point

(- 2, - 1) :

Maximum

Saddle (0, 0), Saddle (- 4, 0), Maximum (- 2, - 1)

f (x) = \frac{1 - x ^{2}}{1 + x ^{2}}

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

f (x) = - \frac{3}{5} x^{5} + 9 x^{4} - 35 x^{3} + 6

f (x) = x - y

f (x, y) = x - y

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