What is the extreme f(x,y)=x^3+y^3+6x^2-9y^2-4 ?

The extreme f(x,y)=x^3+y^3+6x^2-9y^2-4 is Minimum(0,6),Saddle(0,0),Saddle(-4,6),Maximum(-4,0)

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extreme f(x,y)=x^3+y^3+6x^2-9y^2-4

extreme points

f (x, y) = x^{3} + y^{3} + 6 x^{2} - 9 y^{2} - 4

Minimum (0, 6), Saddle (0, 0), Saddle (- 4, 6), Maximum (- 4, 0)

Solution steps

Find the critical points:

(0, 6), (0, 0), (- 4, 6), (- 4, 0)

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

D (x, y) = (6 x + 12) (6 y - 18)

Check the sign of at each critical point

Check critical point

(0, 6) :

Minimum

Check critical point

(0, 0) :

Saddle

Check critical point

(- 4, 6) :

Saddle

Check critical point

(- 4, 0) :

Maximum

Minimum (0, 6), Saddle (0, 0), Saddle (- 4, 6), Maximum (- 4, 0)

f (x) = \frac{1}{9} (x^{4} - 4 x^{3})

f (x) = x^{3} + 2 x^{2} - x - 10

f (x) = \frac{x ^{3} + 1}{x}

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

cos (x) - sin (x)

What is the extreme f(x,y)=x^3+y^3+6x^2-9y^2-4 ?
The extreme f(x,y)=x^3+y^3+6x^2-9y^2-4 is Minimum(0,6),Saddle(0,0),Saddle(-4,6),Maximum(-4,0)