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

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

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

extreme points

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

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) = 6 + x^{2} + 4 x + y^{2}

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

f (x, y) = x y + \frac{50}{x} + \frac{20}{y}

f (x, y) = 49 - 9 x^{2} - y^{2}

y = X^{2} (X - 3)

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