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

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

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

extreme points

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

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 6

D (x, y) = - 4

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

y = x^{3} - 3 x + 2

y = x^{3} - 3 x + 1

f (x) = - 2 x^{4} + 8 x^{3} - 1

f (x, y) = - 3 \cdot x^{2} - 2 \cdot y^{2} + 2 \cdot x \cdot y + 1167 \cdot x + 1479 \cdot y

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

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