What is the extreme f(x,y)=x^3+y^3-12x-3y+15 ?

The extreme f(x,y)=x^3+y^3-12x-3y+15 is Saddle(-2,1),Maximum(-2,-1),Minimum(2,1),Saddle(2,-1)

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extreme f(x,y)=x^3+y^3-12x-3y+15

extreme points

f (x, y) = x^{3} + y^{3} - 12 x - 3 y + 15

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

Solution steps

Find the critical points:

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

\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

(- 2, 1) :

Saddle

Check critical point

(- 2, - 1) :

Maximum

Check critical point

(2, 1) :

Minimum

Check critical point

(2, - 1) :

Saddle

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

f (x) = x^{3} + 1 5^{2}

f (x) = x^{(2)} + x y + y^{(2)} - 6 x - 6 y + 5

f (x) = x^{4} - 30 x^{2} + 29

f (x, y) = e^{8 x} - x y^{3}

f (x y) = ln (4 - x - y)

What is the extreme f(x,y)=x^3+y^3-12x-3y+15 ?
The extreme f(x,y)=x^3+y^3-12x-3y+15 is Saddle(-2,1),Maximum(-2,-1),Minimum(2,1),Saddle(2,-1)