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

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

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

extreme points

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

Saddle (0, 0), Minimum (- \frac{1}{2}, - \frac{1}{4}), Minimum (\frac{1}{2}, \frac{1}{4})

Solution steps

Find the critical points:

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

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

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}{4}) :

Minimum

Check critical point

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

Minimum

Saddle (0, 0), Minimum (- \frac{1}{2}, - \frac{1}{4}), Minimum (\frac{1}{2}, \frac{1}{4})

3 + 4 x^{2} - x^{4}

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

f (x, y) = \frac{x ^{4} + 2 x ^{3} y ^{2} + 2}{3 ( x - 1 ) y}

f (x) = (x - 1)^{2} + 4 - 4 x^{2}

x^{2} - 5 x y + 4 y^{2} - 2 x + 5 y

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