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

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

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

extreme points

f (x) = 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) = x^{\frac{1}{3}} (x + 6)^{\frac{2}{3}}

f (x) = 2 x^{3} + 3 x^{2} - 36 x + 46

f (x) = \frac{1 + x ^{2}}{4 - x ^{2}}

y = x^{3} + 4 x^{2} + 5 x - 2

\frac{x - 4}{x ^{2}}

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