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

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

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

extreme points

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

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

Solution steps

Find the critical points:

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

Saddle

Check critical point

(0, - 3) :

Saddle

Check critical point

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

Minimum

Check critical point

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

Maximum

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

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