What is the extreme f(x,y)=9x^2+2y^2-2xy^2 ?

The extreme f(x,y)=9x^2+2y^2-2xy^2 is Minimum(0,0),Saddle(1,3),Saddle(1,-3)

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extreme f(x,y)=9x^2+2y^2-2xy^2

extreme points

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

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

Solution steps

Find the critical points:

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

\frac{\partial ^{2} f}{\partial x ^{2}} = 4 - 4 x

D (x, y) = 18 (4 - 4 x) - 16 y^{2}

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Check critical point

(1, 3) :

Saddle

Check critical point

(1, - 3) :

Saddle

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

f (x) = 2 x^{3} - 2 x^{2} - 16 x + 1

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

f (x) = e^{x + y}

f (x) = 6 x^{3} - 15 x^{2} + 12 x - 6

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

What is the extreme f(x,y)=9x^2+2y^2-2xy^2 ?
The extreme f(x,y)=9x^2+2y^2-2xy^2 is Minimum(0,0),Saddle(1,3),Saddle(1,-3)