What is the extreme f(x,y)=x^3+xy^2+6xy ?

The extreme f(x,y)=x^3+xy^2+6xy is Saddle(0,0),Saddle(0,-6),Minimum(sqrt(3),-3),Maximum(-sqrt(3),-3)

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extreme f(x,y)=x^3+xy^2+6xy

extreme points

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

Saddle (0, 0), Saddle (0, - 6), Minimum (3, - 3), Maximum (- 3, - 3)

Solution steps

Find the critical points:

(0, 0), (0, - 6), (3, - 3), (- 3, - 3)

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

D (x, y) = 12 x^{2} - 4 y^{2} - 24 y - 36

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(0, - 6) :

Saddle

Check critical point

(3, - 3) :

Minimum

Check critical point

(- 3, - 3) :

Maximum

Saddle (0, 0), Saddle (0, - 6), Minimum (3, - 3), Maximum (- 3, - 3)

f (x) = 5 cos (2 x) + 5

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

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

f (x) = 5 (x - 3)^{2} (2 x + 1) (x + 4)^{3}

f (x) = 2 x - 1.75 x^{2} + 1. 1^{3} - 0.25 x^{4}

What is the extreme f(x,y)=x^3+xy^2+6xy ?
The extreme f(x,y)=x^3+xy^2+6xy is Saddle(0,0),Saddle(0,-6),Minimum(sqrt(3),-3),Maximum(-sqrt(3),-3)