What is the extreme f(x)=x^3+y^3-3x^2-3y^2-9x ?

The extreme f(x)=x^3+y^3-3x^2-3y^2-9x is Saddle(-1,2),Maximum(-1,0),Minimum(3,2),Saddle(3,0)

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extreme f(x)=x^3+y^3-3x^2-3y^2-9x

extreme points

f (x) = x^{3} + y^{3} - 3 x^{2} - 3 y^{2} - 9 x

Saddle (- 1, 2), Maximum (- 1, 0), Minimum (3, 2), Saddle (3, 0)

Solution steps

Find the critical points:

(- 1, 2), (- 1, 0), (3, 2), (3, 0)

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

D (x, y) = (6 x - 6) (6 y - 6)

Check the sign of at each critical point

Check critical point

(- 1, 2) :

Saddle

Check critical point

(- 1, 0) :

Maximum

Check critical point

(3, 2) :

Minimum

Check critical point

(3, 0) :

Saddle

Saddle (- 1, 2), Maximum (- 1, 0), Minimum (3, 2), Saddle (3, 0)

f (x) = 2 x^{3} - 36 x^{2} + 210 x - 15

f (x) = - x^{3} + 3 x^{2} + 9 x - 3

\frac{x}{x ^{2} + 16}

f (x) = - 4 cos (3 x) - 4

f (x, y) = 200 y^{2} + x^{2} - x^{2} y

What is the extreme f(x)=x^3+y^3-3x^2-3y^2-9x ?
The extreme f(x)=x^3+y^3-3x^2-3y^2-9x is Saddle(-1,2),Maximum(-1,0),Minimum(3,2),Saddle(3,0)