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

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

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

extreme points

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

Saddle (\frac{1.03461 E 15}{2.0 E 15}, \frac{1.03461 E 15}{4.0 E 15}), Saddle (- \frac{1.03461 E 15}{2.0 E 15}, \frac{1.03461 E 15}{4.0 E 15}), Minimum (\frac{6.33518 E 15}{2.5 E 15}, \frac{6.33518 E 15}{5.0 E 15}), Minimum (- \frac{6.33518 E 15}{2.5 E 15}, \frac{6.33518 E 15}{5.0 E 15})

Solution steps

Find the critical points:

(\frac{1.03461 E 15}{2.0 E 15}, \frac{1.03461 E 15}{4.0 E 15}), (- \frac{1.03461 E 15}{2.0 E 15}, \frac{1.03461 E 15}{4.0 E 15}), (\frac{6.33518 E 15}{2.5 E 15}, \frac{6.33518 E 15}{5.0 E 15}), (- \frac{6.33518 E 15}{2.5 E 15}, \frac{6.33518 E 15}{5.0 E 15})

\frac{\partial ^{2} f}{\partial x ^{2}} = 12 y^{2}

D (x, y) = 12 y^{2} (12 x^{2} - 8 y) - 64 x^{2}

Check the sign of at each critical point

Check critical point

(\frac{1.03461 E 15}{2.0 E 15}, \frac{1.03461 E 15}{4.0 E 15}) :

Saddle

Check critical point

(- \frac{1.03461 E 15}{2.0 E 15}, \frac{1.03461 E 15}{4.0 E 15}) :

Saddle

Check critical point

(\frac{6.33518 E 15}{2.5 E 15}, \frac{6.33518 E 15}{5.0 E 15}) :

Minimum

Check critical point

(- \frac{6.33518 E 15}{2.5 E 15}, \frac{6.33518 E 15}{5.0 E 15}) :

Minimum

Saddle (\frac{1.03461 E 15}{2.0 E 15}, \frac{1.03461 E 15}{4.0 E 15}), Saddle (- \frac{1.03461 E 15}{2.0 E 15}, \frac{1.03461 E 15}{4.0 E 15}), Minimum (\frac{6.33518 E 15}{2.5 E 15}, \frac{6.33518 E 15}{5.0 E 15}), Minimum (- \frac{6.33518 E 15}{2.5 E 15}, \frac{6.33518 E 15}{5.0 E 15})

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