What is the extreme f(x)=x^3+y^3+9x^2-3y^2-8 ?

The extreme f(x)=x^3+y^3+9x^2-3y^2-8 is Minimum(0,2),Saddle(0,0),Saddle(-6,2),Maximum(-6,0)

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extreme f(x)=x^3+y^3+9x^2-3y^2-8

extreme points

f (x) = x^{3} + y^{3} + 9 x^{2} - 3 y^{2} - 8

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

Solution steps

Find the critical points:

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

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

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

Check the sign of at each critical point

Check critical point

(0, 2) :

Minimum

Check critical point

(0, 0) :

Saddle

Check critical point

(- 6, 2) :

Saddle

Check critical point

(- 6, 0) :

Maximum

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

f (x, y) = 6 x - x^{2} + 2 y^{4} - 16 y^{2} + 5

f (x) = x^{\frac{2}{3}} (20 - x)

y = x^{2} + 6 z + z^{2}

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

f (x) = 5 x^{2} y - 3 x y^{2} - 4 x + 3 y

What is the extreme f(x)=x^3+y^3+9x^2-3y^2-8 ?
The extreme f(x)=x^3+y^3+9x^2-3y^2-8 is Minimum(0,2),Saddle(0,0),Saddle(-6,2),Maximum(-6,0)