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

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

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

extreme points

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

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)

\frac{x ^{2} - 9}{x ^{2} + 9}

f (x, y) = x^{2} + x y + y^{2} + x - y + 1

f (x, y) = 3 x^{4} y - 7 x^{3} y - x^{2} + y + 1

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

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

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