What is the extreme f(x)=x^3-3xy^2-3y^2+12y ?

The extreme f(x)=x^3-3xy^2-3y^2+12y is Saddle(1,1),Saddle(-2,-2)

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extreme f(x)=x^3-3xy^2-3y^2+12y

extreme points

f (x) = x^{3} - 3 x y^{2} - 3 y^{2} + 12 y

Saddle (1, 1), Saddle (- 2, - 2)

Solution steps

Find the critical points:

(1, 1), (- 2, - 2)

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

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

Check the sign of at each critical point

Check critical point

(1, 1) :

Saddle

Check critical point

(- 2, - 2) :

Saddle

Saddle (1, 1), Saddle (- 2, - 2)

x^{2} y - 6 y^{2} - 3 x^{2}

f (x) = - x^{2} + 7 x - 7

f (x) = x^{3} + x^{2} - 46 x + 80

f (x) = 3 x^{4} + 8 x^{3} - 6 x^{2} + 24 x

3 x (x - 3)^{3}

What is the extreme f(x)=x^3-3xy^2-3y^2+12y ?
The extreme f(x)=x^3-3xy^2-3y^2+12y is Saddle(1,1),Saddle(-2,-2)