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

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

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

extreme points

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

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

Solution steps

Find the critical points:

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

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

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

Check the sign of at each critical point

Check critical point

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

Minimum

Check critical point

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

Minimum

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

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

f (t) = 5 u t (t - 2)

f (x) = 3 x^{2} - 3 y^{2} + 6

x^{2} + y^{2} + 4 x - 2 y + 6

f (x) = 8 y^{3} - 12 x y + x^{3}

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