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

The extreme f(x,y)=3x^4-x^2+3y^2 is Saddle(0,0),Minimum(-1/(sqrt(6)),0),Minimum(1/(sqrt(6)),0)

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

extreme

Saddle (0, 0), Minimum (- \frac{1}{6}, 0), Minimum (\frac{1}{6}, 0)

Solution steps

Find the critical points:

(0, 0), (- \frac{1}{6}, 0), (\frac{1}{6}, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 6

D (x, y) = 6 (36 x^{2} - 2)

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(- \frac{1}{6}, 0) :

Minimum

Check critical point

(\frac{1}{6}, 0) :

Minimum

Saddle (0, 0), Minimum (- \frac{1}{6}, 0), Minimum (\frac{1}{6}, 0)

What is the extreme f(x,y)=3x^4-x^2+3y^2 ?
The extreme f(x,y)=3x^4-x^2+3y^2 is Saddle(0,0),Minimum(-1/(sqrt(6)),0),Minimum(1/(sqrt(6)),0)