What is the extreme f(x,y)=x^3+y^3-3x-12y+30 ?

The extreme f(x,y)=x^3+y^3-3x-12y+30 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+30

extreme points

f (x, y) = x^{3} + y^{3} - 3 x - 12 y + 30

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

+ 2.5 t^{2} + 6 t

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

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

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

What is the extreme f(x,y)=x^3+y^3-3x-12y+30 ?
The extreme f(x,y)=x^3+y^3-3x-12y+30 is Saddle(-1,2),Maximum(-1,-2),Minimum(1,2),Saddle(1,-2)