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

The extreme f(x)=2x^3-3xy+3y^3 is Saddle(0,0),Minimum(1/(\sqrt[3]{12)}, 1/(\sqrt[3]{2)*3^{2/3}})

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

extreme points

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

Solution steps

Find the critical points:

\frac{\partial ^{2} f}{\partial x ^{2}} = 18 y

D (x, y) = 216 x y - 9

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

Minimum

y = 2 x^{3} - 9 x^{2} + 12 x + 2

f (x) = x^{4} - 50 x^{2} + 9

f (x) = - 4 x^{2} + 8 x - 2, 0 \leq x \leq 3

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

f (x, y) = x^{3} - 3 x + y^{3} + \frac{5}{2} y^{2} - 2 y

What is the extreme f(x)=2x^3-3xy+3y^3 ?
The extreme f(x)=2x^3-3xy+3y^3 is Saddle(0,0),Minimum(1/(\sqrt[3]{12)}, 1/(\sqrt[3]{2)*3^{2/3}})