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

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

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

extreme points

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

Saddle (- 1, 2), Maximum (- 1, - 2), Minimum (1, 2), Saddle (1, - 2)

Solution steps

Find the critical points:

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

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

D (x, y) = 36 x y

Check the sign of at each critical point

Check critical point

(- 1, 2) :

Saddle

Check critical point

(- 1, - 2) :

Maximum

Check critical point

(1, 2) :

Minimum

Check critical point

(1, - 2) :

Saddle

Saddle (- 1, 2), Maximum (- 1, - 2), Minimum (1, 2), Saddle (1, - 2)

f (x) = (x - 1) (x + 1) (x + 4)

\frac{x + 1}{x ^{2} - 5 x + 6}

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

f (x) = x^{3} - 18 x^{2} + 96 x

f (x) = x^{\frac{1}{3}} (x^{2} - 9)

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