What is the extreme f(x)=x^3-y^3+3x^2+3y^2-9x+5 ?

The extreme f(x)=x^3-y^3+3x^2+3y^2-9x+5 is Minimum(1,0),Saddle(1,2),Saddle(-3,0),Maximum(-3,2)

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extreme f(x)=x^3-y^3+3x^2+3y^2-9x+5

extreme points

f (x) = x^{3} - y^{3} + 3 x^{2} + 3 y^{2} - 9 x + 5

Minimum (1, 0), Saddle (1, 2), Saddle (- 3, 0), Maximum (- 3, 2)

Solution steps

Find the critical points:

(1, 0), (1, 2), (- 3, 0), (- 3, 2)

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

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

Check the sign of at each critical point

Check critical point

(1, 0) :

Minimum

Check critical point

(1, 2) :

Saddle

Check critical point

(- 3, 0) :

Saddle

Check critical point

(- 3, 2) :

Maximum

Minimum (1, 0), Saddle (1, 2), Saddle (- 3, 0), Maximum (- 3, 2)

f (x) = \frac{4}{3} x^{3} - \frac{25}{2} x^{2} + 6 x + 4

f (x y) = x^{2} y - 15 x y^{2} + 12 y

f (x) = 12 x^{2} - 18 x

- 6 x^{5} + 180 x^{3} - 1350 x + 14

ln (1 - 3 x)

What is the extreme f(x)=x^3-y^3+3x^2+3y^2-9x+5 ?
The extreme f(x)=x^3-y^3+3x^2+3y^2-9x+5 is Minimum(1,0),Saddle(1,2),Saddle(-3,0),Maximum(-3,2)