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

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

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

extreme points

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

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

Solution steps

Find the critical points:

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

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

D (x, y) = - 12 y

Check the sign of at each critical point

Check critical point

(0, 1) :

Saddle

Check critical point

(0, - 1) :

Minimum

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

f (x) = \frac{x ^{3}}{3} - x^{2} - 3 x

f (x) = \frac{2 x}{x ^{2} - 9}

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

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

f (x, y) = 4 x^{2} + 2 y^{2} - 2 x y - 10 y - 2 x

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