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

The extreme f(x)=2x^2+y^3-6xy+2 is Minimum(9/2 ,3),Saddle(0,0)

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

extreme points

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

Minimum (\frac{9}{2}, 3), Saddle (0, 0)

Solution steps

Find the critical points:

(\frac{9}{2}, 3), (0, 0)

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

D (x, y) = 24 y - 36

Check the sign of at each critical point

Check critical point

(\frac{9}{2}, 3) :

Minimum

Check critical point

(0, 0) :

Saddle

Minimum (\frac{9}{2}, 3), Saddle (0, 0)

f (x) = 4 x^{2} + 3

f (x) = (\frac{1}{12} x^{3} - 9 x + 5)

f (x) = 2 x^{2} + 18 x + 16

y = 3 x^{2} = 12 x + 11

f (x) = x^{3} - 4 x^{2} - 3 x - 6

What is the extreme f(x)=2x^2+y^3-6xy+2 ?
The extreme f(x)=2x^2+y^3-6xy+2 is Minimum(9/2 ,3),Saddle(0,0)