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

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

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extreme f(x,y)=3xy+x^3-y^3+1

extreme points

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

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) = 9 - 2 x + 4 y - x^{2} - 2 y^{3}

y = x (6 - 2 x)^{2}

\frac{x + 2}{x ^{3}}

f (x) = - 0.03 x^{2} + 300 x - 100000

f (x) = x^{\frac{1}{3}} (x - 4), - 125 \leq x \leq 125

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