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

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

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

extreme points

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

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

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

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

\frac{3 x ^{2}}{x ^{2} - 16}

f (x) = \frac{x}{x ^{2} + 25}

f (x, y) = 4 - \frac{1}{9} x^{2} - \frac{1}{16} y^{2}

f (x) = ln (\frac{1}{x + y})

f (x) = x^{2} e^{x} [- 3.1]

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