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

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

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

extreme points

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

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

Solution steps

Find the critical points:

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

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

D (x, y) = 6 (24 x^{2} - 2)

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

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

Minimum

Check critical point

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

Minimum

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

5 x^{2} + 3, - 1 \leq x \leq 2

f (x) = x^{4} - 6 x^{\frac{3}{2}}, x > 0

f (0.1, 0) = (2 x - x^{2}) (2 y - y^{2})

f (x) = 600 t - 15 t^{2}

f (x, y) = x + y + 5

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