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

The extreme f(x)=4+x^3+y^3-3xy is Saddle(0,0),Minimum(1,1)

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extreme f(x)=4+x^3+y^3-3xy

extreme points

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

Saddle (0, 0), Minimum (1, 1)

Solution steps

Find the critical points:

(0, 0), (1, 1)

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

D (x, y) = 36 x y - 9

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(1, 1) :

Minimum

Saddle (0, 0), Minimum (1, 1)

f (x) = 3 x^{3} - 9 x + 3

y = x^{3} + x^{2} - x + 8

f (x) = 3 x 8 - x^{2}

f (x) = e^{(x^{2} - 7 x - 1)}

x^{\frac{2}{3}} - 6 x^{- \frac{1}{3}}

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