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

The extreme f(x,y)=x^3+y^3-3x^2-9y^2-9x is Saddle(-1,6),Maximum(-1,0),Minimum(3,6),Saddle(3,0)

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

extreme points

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

Saddle (- 1, 6), Maximum (- 1, 0), Minimum (3, 6), Saddle (3, 0)

Solution steps

Find the critical points:

(- 1, 6), (- 1, 0), (3, 6), (3, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 6 y - 18

D (x, y) = (6 x - 6) (6 y - 18)

Check the sign of at each critical point

Check critical point

(- 1, 6) :

Saddle

Check critical point

(- 1, 0) :

Maximum

Check critical point

(3, 6) :

Minimum

Check critical point

(3, 0) :

Saddle

Saddle (- 1, 6), Maximum (- 1, 0), Minimum (3, 6), Saddle (3, 0)

f (x, y) = \frac{1}{3} 36 - 9 x^{2} - 4 y^{2}

y + v

y = \frac{4 x}{x ^{2} + 1}

f (x) = - 5 x^{6} ln (x), (0, 4)

f (x) = 8 sin (x) + 8 cos (x)

What is the extreme f(x,y)=x^3+y^3-3x^2-9y^2-9x ?
The extreme f(x,y)=x^3+y^3-3x^2-9y^2-9x is Saddle(-1,6),Maximum(-1,0),Minimum(3,6),Saddle(3,0)