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

The extreme f(x,y)=x^2+y^2+2x^2y^2 is Minimum(0,0)

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

extreme points

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

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = 4 + 8 x^{2} + 8 y^{2} - 48 x^{2} y^{2}

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

16

f (x) = - 5 + y - 10 x

f (x, y) = - 6 x^{2} + 7 x y - 2 y^{2} + x + y

f (x) = 4 x^{3} - 12 x^{2} + 9 x - 1

\frac{1}{x ^{2} + y ^{2} - 1}

What is the extreme f(x,y)=x^2+y^2+2x^2y^2 ?
The extreme f(x,y)=x^2+y^2+2x^2y^2 is Minimum(0,0)