What is the extreme f(x,y)=x^2-2xy+1/3 y^3-3y ?

The extreme f(x,y)=x^2-2xy+1/3 y^3-3y is Minimum(3,3),Saddle(-1,-1)

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extreme f(x,y)=x^2-2xy+1/3 y^3-3y

extreme points

f (x, y) = x^{2} - 2 x y + \frac{1}{3} y^{3} - 3 y

Minimum (3, 3), Saddle (- 1, - 1)

Solution steps

Find the critical points:

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

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

D (x, y) = 4 y - 4

Check the sign of at each critical point

Check critical point

(3, 3) :

Minimum

Check critical point

(- 1, - 1) :

Saddle

Minimum (3, 3), Saddle (- 1, - 1)

f (x, y) = 3 x - 4 y + 7

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

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

f (x) = x^{4} - 8 x^{2} + 6

f (x, y) = 4 x^{2} + y^{2} - 8 x + 6 y + 8

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