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

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

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extreme f(x,y)= 1/4 x^4-1/2 x^2+y^2

extreme

Saddle (0, 0), Minimum (- 1, 0), Minimum (1, 0)

Solution steps

Find the critical points:

(0, 0), (- 1, 0), (1, 0)

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

D (x, y) = 2 (3 x^{2} - 1)

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(- 1, 0) :

Minimum

Check critical point

(1, 0) :

Minimum

Saddle (0, 0), Minimum (- 1, 0), Minimum (1, 0)

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