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

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

extreme points

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

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)

x^{2} y + y^{3} - 27 y

f (x) = \frac{2 x}{x ^{2} - 4}

f (x) = 3 x^{2} - 4 x^{3}

f (x) = \frac{5 ln ( x )}{x ^{3}}

f (x) = (x^{2} - 3 x + 5) e^{- \frac{x}{3}}

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