What is the extreme f(x,y)=(3x+4x^3)(y^2+2y) ?

The extreme f(x,y)=(3x+4x^3)(y^2+2y) is Saddle(0,0),Saddle(0,-2)

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extreme f(x,y)=(3x+4x^3)(y^2+2y)

extreme points

f (x, y) = (3 x + 4 x^{3}) (y^{2} + 2 y)

Saddle (0, 0), Saddle (0, - 2)

Solution steps

Find the critical points:

(0, 0), (0, - 2)

\frac{\partial ^{2} f}{\partial x ^{2}} = 2 (3 x + 4 x^{3})

D (x, y) = - 384 x^{4} y^{2} - 768 x^{4} y - 576 x^{4} - 144 x^{2} y^{2} - 288 x^{2} y - 288 x^{2} - 36 y^{2} - 72 y - 36

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(0, - 2) :

Saddle

Saddle (0, 0), Saddle (0, - 2)

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

f (x) = (740 - 18 x) x

\frac{3 x ^{2} - 11}{x ^{2} - 4}

f (x) = 2 x^{3} - 15 x^{2} + 24 x

f (x, y) = x y - \frac{1}{x} - \frac{1}{y}

What is the extreme f(x,y)=(3x+4x^3)(y^2+2y) ?
The extreme f(x,y)=(3x+4x^3)(y^2+2y) is Saddle(0,0),Saddle(0,-2)