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

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

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

extreme points

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

Saddle (0, 0), Maximum (- 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) :

Maximum

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

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

3 x^{2} - 18 x + 15

2 x^{3} - 3 x^{2} - 12 x + 18

f (x) = e^{- x^{6}}, - 2 \leq x \leq 1

f (x) = x^{6} - 3 x^{4}

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