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

The extreme f(x,y)=x^3y+3x^2-y is Saddle(1,-2)

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

extreme points

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

Saddle (1, - 2)

Solution steps

Find the critical points:

(1, - 2)

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

D (x, y) = - 9 x^{4}

Check the sign of at each critical point

Check critical point

(1, - 2) :

Saddle

Saddle (1, - 2)

f (x) = 4 x^{3} + 3 x^{2} - 6 x + 1

f (x) = x^{4} - 2 x^{2} + 1

f (x, y) = x y + \frac{1}{x} + \frac{1}{y}

f (x) = \frac{x ^{2} x - 1}{1 - x}

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

What is the extreme f(x,y)=x^3y+3x^2-y ?
The extreme f(x,y)=x^3y+3x^2-y is Saddle(1,-2)