What is the extreme f(x,y)=4+x^3-3xy+y^3 ?

The extreme f(x,y)=4+x^3-3xy+y^3 is Saddle(0,0),Minimum(1,1)

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extreme f(x,y)=4+x^3-3xy+y^3

extreme points

f (x, y) = 4 + x^{3} - 3 x y + 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) = x^{4} (1 - x)^{8}

f (x) = - x e^{- \frac{x}{4}}

f (x) = 2 x^{3} + 6 x^{2} - 18 x + 1, - 6 \leq x \leq 2

f (x, y) = x^{2} + y^{2} - 12 x + 4 y - 11

f (x) = 6 x + 7 x^{(\frac{6}{7})}

What is the extreme f(x,y)=4+x^3-3xy+y^3 ?
The extreme f(x,y)=4+x^3-3xy+y^3 is Saddle(0,0),Minimum(1,1)