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

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

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

extreme points

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

Saddle (0, 2), Saddle (0, 0), Minimum (\frac{1}{3}, 1), Maximum (- \frac{1}{3}, 1)

Solution steps

Find the critical points:

(0, 2), (0, 0), (\frac{1}{3}, 1), (- \frac{1}{3}, 1)

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

D (x, y) = 36 x^{2} - 4 y^{2} + 8 y - 4

Check the sign of at each critical point

Check critical point

(0, 2) :

Saddle

Check critical point

(0, 0) :

Saddle

Check critical point

(\frac{1}{3}, 1) :

Minimum

Check critical point

(- \frac{1}{3}, 1) :

Maximum

Saddle (0, 2), Saddle (0, 0), Minimum (\frac{1}{3}, 1), Maximum (- \frac{1}{3}, 1)

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