What is the extreme f(x)=x^3-y^3+6xy-10 ?

The extreme f(x)=x^3-y^3+6xy-10 is Saddle(0,0),Minimum(2,-2)

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extreme f(x)=x^3-y^3+6xy-10

extreme points

f (x) = x^{3} - y^{3} + 6 x y - 10

Saddle (0, 0), Minimum (2, - 2)

Solution steps

Find the critical points:

(0, 0), (2, - 2)

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

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

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(2, - 2) :

Minimum

Saddle (0, 0), Minimum (2, - 2)

f (x) = \frac{4860}{x} + 11 x + 710272

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

f (x) = 6 + 5 x - 5 x^{2}

f (x, y, z) = ln (4 - x - y)

\frac{0.5 x ^{2} + 4 x - 10}{x - 6}

What is the extreme f(x)=x^3-y^3+6xy-10 ?
The extreme f(x)=x^3-y^3+6xy-10 is Saddle(0,0),Minimum(2,-2)