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

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

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

extreme points

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

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 0

D (x, y) = - 9

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

f (x, y) = x^{2} y^{3} - (x - 2 y)^{2} + 6 x^{4} y = 3 x y

f (x) = 2 x^{4} - 20 x^{2} + 18

- x^{2} + 4 x + 6

f (x, y) = x^{2} y - 15 x y^{2} + 12 y

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

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