What is the extreme f(x)=x^3-6xy+y^3 ?

The extreme f(x)=x^3-6xy+y^3 is Saddle(0,0),Minimum(2,2)

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extreme f(x)=x^3-6xy+y^3

extreme points

f (x) = x^{3} - 6 x y + y^{3}

Saddle (0, 0), Minimum (2, 2)

Solution steps

Find the critical points:

(0, 0), (2, 2)

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

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

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(2, 2) :

Minimum

Saddle (0, 0), Minimum (2, 2)

f (x) = 2 x^{3} - 3 x^{2} y - 12 x^{2} - 3 y^{2}

f (x) = \frac{x ^{2} + 9}{2 x}

f (x) = x - y = 10

w (x, y) = x y + \frac{e ^{y}}{y ^{2} + 1}

f (x) = x^{8} + \frac{8}{x}

What is the extreme f(x)=x^3-6xy+y^3 ?
The extreme f(x)=x^3-6xy+y^3 is Saddle(0,0),Minimum(2,2)