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

The extreme f(x)=x^3+2xy-2x-4y is Saddle(2,-5)

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

extreme points

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

Saddle (2, - 5)

Solution steps

Find the critical points:

(2, - 5)

\frac{\partial ^{2} f}{\partial x ^{2}} = 0

D (x, y) = - 4

Check the sign of at each critical point

Check critical point

(2, - 5) :

Saddle

Saddle (2, - 5)

f (x) = 2 x (4 - x^{2})

f (x) = x^{2} + 3 x + 1

x^{3} - 15 x y + y^{3}

f (x) = 3 x + 1 + \frac{1}{x - 1}

What is the extreme f(x)=x^3+2xy-2x-4y ?
The extreme f(x)=x^3+2xy-2x-4y is Saddle(2,-5)