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

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

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

extreme points

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

D (x, y) = 36 x^{2} - 36 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)

f (x) = 2 x^{3} + 3 x^{2} - 12 x - 18

f (x, y) = 2 x y + \frac{1}{x} + \frac{4}{y}

f (x) = 4 x^{3} e^{- x}

f (x) = x^{3} + y^{3} - 36 x y

f (x, y) = 400 - 16 x^{2} - 36 y^{2}

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