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

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

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

extreme points

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

Maximum (0, 0), Saddle (0, - 1), Saddle (0, 1), Saddle (- \frac{1}{2}, 0), Minimum (- \frac{1}{2}, - 1), Minimum (- \frac{1}{2}, 1), Saddle (\frac{1}{2}, 0), Minimum (\frac{1}{2}, - 1), Minimum (\frac{1}{2}, 1)

Solution steps

Find the critical points:

(0, 0), (0, - 1), (0, 1), (- \frac{1}{2}, 0), (- \frac{1}{2}, - 1), (- \frac{1}{2}, 1), (\frac{1}{2}, 0), (\frac{1}{2}, - 1), (\frac{1}{2}, 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = 12 y^{2} - 4

D (x, y) = (24 x^{2} - 4) (12 y^{2} - 4)

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Check critical point

(0, - 1) :

Saddle

Check critical point

(0, 1) :

Saddle

Check critical point

(- \frac{1}{2}, 0) :

Saddle

Check critical point

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

Minimum

Check critical point

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

Minimum

Check critical point

(\frac{1}{2}, 0) :

Saddle

Check critical point

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

Minimum

Check critical point

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

Minimum

Maximum (0, 0), Saddle (0, - 1), Saddle (0, 1), Saddle (- \frac{1}{2}, 0), Minimum (- \frac{1}{2}, - 1), Minimum (- \frac{1}{2}, 1), Saddle (\frac{1}{2}, 0), Minimum (\frac{1}{2}, - 1), Minimum (\frac{1}{2}, 1)

f (x) = 9 - 2 x + 4 y - x^{2} - 4 y^{2}

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

\frac{5 x ^{2}}{x ^{2} + 16}

f (x) = 8 x - x^{4}, - 2 \leq x \leq 1

f (x, y) = 2 x^{2} + 4 x y^{2} - 2 x y - 10 x - 2 y + 2

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