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

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

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

extreme points

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

Saddle (\frac{1}{2}, - \frac{1}{2}), Saddle (- \frac{1}{2}, - \frac{1}{2}), Minimum (\frac{3}{4 2}, - \frac{1}{4}), Minimum (- \frac{3}{4 2}, - \frac{1}{4})

Solution steps

Find the critical points:

(\frac{1}{2}, - \frac{1}{2}), (- \frac{1}{2}, - \frac{1}{2}), (\frac{3}{4 2}, - \frac{1}{4}), (- \frac{3}{4 2}, - \frac{1}{4})

\frac{\partial ^{2} f}{\partial x ^{2}} = - 4 x^{2} - 6 y

D (x, y) = - 96 x^{4} - 48 x^{2} y^{2} - 120 x^{2} y - 4 x^{2} + 24 y^{3} - 12 y^{2}

Check the sign of at each critical point

Check critical point

(\frac{1}{2}, - \frac{1}{2}) :

Saddle

Check critical point

(- \frac{1}{2}, - \frac{1}{2}) :

Saddle

Check critical point

(\frac{3}{4 2}, - \frac{1}{4}) :

Minimum

Check critical point

(- \frac{3}{4 2}, - \frac{1}{4}) :

Minimum

Saddle (\frac{1}{2}, - \frac{1}{2}), Saddle (- \frac{1}{2}, - \frac{1}{2}), Minimum (\frac{3}{4 2}, - \frac{1}{4}), Minimum (- \frac{3}{4 2}, - \frac{1}{4})

f (x) = e^{7 x^{2} + y^{2} + 1}

P (X, y) = 36 X^{2} - 9 y^{2}

\frac{5}{- 5 e x ^{2} - 5 y ^{2} + 5}

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

f (x) = 6 + x^{2} + 4 x + y^{2}

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