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

The extreme f(x,y)=x^3-y^2-3x-2y-2xy-3 is Saddle(1/3 ,-4/3),Maximum(-1,0)

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

extreme points

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

Saddle (\frac{1}{3}, - \frac{4}{3}), Maximum (- 1, 0)

Solution steps

Find the critical points:

(\frac{1}{3}, - \frac{4}{3}), (- 1, 0)

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

D (x, y) = - 12 x - 4

Check the sign of at each critical point

Check critical point

(\frac{1}{3}, - \frac{4}{3}) :

Saddle

Check critical point

(- 1, 0) :

Maximum

Saddle (\frac{1}{3}, - \frac{4}{3}), Maximum (- 1, 0)

f (x) = 2 x - \frac{32}{x}

f (x) = 2 x - \frac{4}{x}

f (x) = \frac{( 4 x - 18 )}{( x - 9 ) ^{2}}

f (x, y) = 11 x^{2} - 2 x^{3} + 2 y^{2} + 4 x y

y (x, t) = x (- 2 t + 2)

What is the extreme f(x,y)=x^3-y^2-3x-2y-2xy-3 ?
The extreme f(x,y)=x^3-y^2-3x-2y-2xy-3 is Saddle(1/3 ,-4/3),Maximum(-1,0)