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

The extreme f(x,y)=2x^4-2xy+y^2-2 is Saddle(0,0),Minimum(-1/2 ,-1/2),Minimum(1/2 , 1/2)

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

extreme points

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

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

Solution steps

Find the critical points:

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

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

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

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

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

Minimum

Check critical point

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

Minimum

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

f (x) = 1 - x^{3}

- 14 - x

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

f (x, y) = x^{2} + y^{2} - 6 x + 8 y + 35

f (x) = \frac{1}{3} x^{3} - 15 x^{2} + 78.2 x + 3630

What is the extreme f(x,y)=2x^4-2xy+y^2-2 ?
The extreme f(x,y)=2x^4-2xy+y^2-2 is Saddle(0,0),Minimum(-1/2 ,-1/2),Minimum(1/2 , 1/2)