What is the extreme f(x)=x^3+y^3+3x^2-9y^2-1 ?

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

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extreme f(x)=x^3+y^3+3x^2-9y^2-1

extreme points

f (x) = x^{3} + y^{3} + 3 x^{2} - 9 y^{2} - 1

Minimum (0, 6), Saddle (0, 0), Saddle (- 2, 6), Maximum (- 2, 0)

Solution steps

Find the critical points:

(0, 6), (0, 0), (- 2, 6), (- 2, 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

(0, 6) :

Minimum

Check critical point

(0, 0) :

Saddle

Check critical point

(- 2, 6) :

Saddle

Check critical point

(- 2, 0) :

Maximum

Minimum (0, 6), Saddle (0, 0), Saddle (- 2, 6), Maximum (- 2, 0)

\frac{x ^{2}}{x ^{2} + 243}

f (x) = 6 sin^{2} (x) + 6 sin (x)

f (x) = 3^{2} - 6 x - 9 = 0

36 x^{5} + 540 x^{4} + 2100 x^{3} - 30

(x + 1) (x + 2) (x + 3) (x + 4)

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