What is the extreme f(x,y)=x^3-y^2-6xy-4 ?

The extreme f(x,y)=x^3-y^2-6xy-4 is Maximum(-6,18),Saddle(0,0)

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extreme f(x,y)=x^3-y^2-6xy-4

extreme points

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

Maximum (- 6, 18), Saddle (0, 0)

Solution steps

Find the critical points:

(- 6, 18), (0, 0)

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

D (x, y) = - 12 x - 36

Check the sign of at each critical point

Check critical point

(- 6, 18) :

Maximum

Check critical point

(0, 0) :

Saddle

Maximum (- 6, 18), Saddle (0, 0)

f (x) = x^{2} - 6 x + 9

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

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

f (x) = - 4 cos (3 x) - 4, (0, 2 π)

f (t) = 5 - 6 u (t - 2) + 3 u (t - 5)

What is the extreme f(x,y)=x^3-y^2-6xy-4 ?
The extreme f(x,y)=x^3-y^2-6xy-4 is Maximum(-6,18),Saddle(0,0)