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

The extreme f(x,y)=3x^2+x^3y+4y^2 is Saddle(2,-1),Saddle(-2,1)

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

extreme points

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

Saddle (2, - 1), Saddle (- 2, 1)

Solution steps

Find the critical points:

(2, - 1), (- 2, 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = 8

D (x, y) = 8 (6 x y + 6) - 9 x^{4}

Check the sign of at each critical point

Check critical point

(2, - 1) :

Saddle

Check critical point

(- 2, 1) :

Saddle

Saddle (2, - 1), Saddle (- 2, 1)

f (- \infty, \infty) = (x - 2)^{2} + 1

f (x) = x^{3} + 3 x^{2} - 9 x + 2

f (x) = 2 x^{3} + 6 x^{2} - 5

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

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

What is the extreme f(x,y)=3x^2+x^3y+4y^2 ?
The extreme f(x,y)=3x^2+x^3y+4y^2 is Saddle(2,-1),Saddle(-2,1)