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

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

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

extreme points

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

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

Solution steps

Find the critical points:

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

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

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

Check the sign of at each critical point

Check critical point

(0, 1) :

Saddle

Check critical point

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

Minimum

Check critical point

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

Minimum

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

f (x) = x ln (2 x)

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

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

32 x y

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

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