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

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

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

extreme points

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

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

Solution steps

Find the critical points:

(0, 1), (0, - 1), (1, 0), (- 1, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 18 x

D (x, y) = 324 x^{2} - 324 y^{2}

Check the sign of at each critical point

Check critical point

(0, 1) :

Saddle

Check critical point

(0, - 1) :

Saddle

Check critical point

(1, 0) :

Minimum

Check critical point

(- 1, 0) :

Maximum

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

ln (x + 2) (x - 1)^{2}

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

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

y = \frac{x}{x ^{2} + 9}

x^{3} + 3 x^{2} - 2

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