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

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

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

extreme points

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

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

Solution steps

Find the critical points:

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

\frac{\partial ^{2} f}{\partial x ^{2}} = 0

D (x, y) = - 36 x^{2} + 36 x - 9

Check the sign of at each critical point

Check critical point

(0, - 1) :

Saddle

Check critical point

(1, 0) :

Saddle

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

f (x, y) = x^{3} + y^{2} - 2 x y + 7 x - 8 y + 2

f (x) = (x^{2} + 11) (9 - x^{2})

4 L W

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

f (x) = \frac{1}{2 π} e^{- \frac{x ^{2}}{2}}

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