What is the extreme f(x,y)=x^3-y^3-2xy+7 ?

The extreme f(x,y)=x^3-y^3-2xy+7 is Saddle(0,0),Maximum(-2/3 , 2/3)

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extreme f(x,y)=x^3-y^3-2xy+7

extreme points

f (x, y) = x^{3} - y^{3} - 2 x y + 7

Saddle (0, 0), Maximum (- \frac{2}{3}, \frac{2}{3})

Solution steps

Find the critical points:

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

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

D (x, y) = - 36 x y - 4

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

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

Maximum

Saddle (0, 0), Maximum (- \frac{2}{3}, \frac{2}{3})

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

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

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

f (x, y) = 2 x^{2} + 3 y^{2} - 7

f (x) = \frac{x ^{2} + 4}{x}

What is the extreme f(x,y)=x^3-y^3-2xy+7 ?
The extreme f(x,y)=x^3-y^3-2xy+7 is Saddle(0,0),Maximum(-2/3 , 2/3)