What is the extreme f(x)=x^3+3y^3+3x^2+3y^2+24 ?

The extreme f(x)=x^3+3y^3+3x^2+3y^2+24 is Minimum(0,0),Saddle(0,-2/3),Saddle(-2,0),Maximum(-2,-2/3)

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extreme f(x)=x^3+3y^3+3x^2+3y^2+24

extreme points

f (x) = x^{3} + 3 y^{3} + 3 x^{2} + 3 y^{2} + 24

Minimum (0, 0), Saddle (0, - \frac{2}{3}), Saddle (- 2, 0), Maximum (- 2, - \frac{2}{3})

Solution steps

Find the critical points:

(0, 0), (0, - \frac{2}{3}), (- 2, 0), (- 2, - \frac{2}{3})

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

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

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Check critical point

(0, - \frac{2}{3}) :

Saddle

Check critical point

(- 2, 0) :

Saddle

Check critical point

(- 2, - \frac{2}{3}) :

Maximum

Minimum (0, 0), Saddle (0, - \frac{2}{3}), Saddle (- 2, 0), Maximum (- 2, - \frac{2}{3})

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

f (x) = 9.71 x^{2} + 61.579 x + 943.59

f (x) = 2 x^{3} - 3 x^{2} + 7

x y - x - 3 y

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

What is the extreme f(x)=x^3+3y^3+3x^2+3y^2+24 ?
The extreme f(x)=x^3+3y^3+3x^2+3y^2+24 is Minimum(0,0),Saddle(0,-2/3),Saddle(-2,0),Maximum(-2,-2/3)