What is the extreme f(x)=2x^3-6xy+3y^2 ?

The extreme f(x)=2x^3-6xy+3y^2 is Minimum(1,1),Saddle(0,0)

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extreme f(x)=2x^3-6xy+3y^2

extreme points

f (x) = 2 x^{3} - 6 x y + 3 y^{2}

Minimum (1, 1), Saddle (0, 0)

Solution steps

Find the critical points:

(1, 1), (0, 0)

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

D (x, y) = 72 x - 36

Check the sign of at each critical point

Check critical point

(1, 1) :

Minimum

Check critical point

(0, 0) :

Saddle

Minimum (1, 1), Saddle (0, 0)

f (x) = 168000 x - 100 x^{2}

f (x) = x^{2} (6 - x)^{3}

f (x) = x^{2} + x y + y^{2} - 31 y + 320

f (x, y) = - 2 x^{3} - 2 y^{3} - 6 x y + 1

f (x) = 3 e^{x} - x e^{x}

What is the extreme f(x)=2x^3-6xy+3y^2 ?
The extreme f(x)=2x^3-6xy+3y^2 is Minimum(1,1),Saddle(0,0)