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

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

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

extreme points

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

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

Solution steps

Find the critical points:

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

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

D (x, y) = - 36 x y - 9

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(- 1, 1) :

Minimum

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

f (x) = 0.6 x^{2} - 660 x - 6000

f (x) = \frac{x ^{2} - 1}{x ^{3}}

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

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

f (x) = 2 x^{2} + 3, - 1 \leq x \leq 2

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