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

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

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

extreme points

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

Saddle (0, - 3), Maximum (- 1, - 1)

Solution steps

Find the critical points:

(0, - 3), (- 1, - 1)

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

D (x, y) = - 4 x y - 4 x^{2} - 12 x - 4 y^{2} - 12 y - 9

Check the sign of at each critical point

Check critical point

(0, - 3) :

Saddle

Check critical point

(- 1, - 1) :

Maximum

Saddle (0, - 3), Maximum (- 1, - 1)

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

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

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

f (x, y) = 4 x^{2} - 12 x + y^{2} + 2 y - 10

x^{4} - 8 x^{2} + 1

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