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

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

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

extreme points

f (x, y) = 3 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) = - 36 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{4 x}{4 + x ^{2} + y ^{2}}

y = cos (x) - 3 x

f (x) = x^{2} - 15 x + 26

y = x^{4} - x^{2}

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

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