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

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

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

extreme points

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

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})

sin (x) - x

f (x, y) = y + 9 - x^{2} - y^{2}

f (x) = - x^{2} + 160 x - 400

x^{2} - 4 x

x^{y}

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