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

The extreme P(y,x)=-y^3-2y^2-xy is Saddle(0,0)

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extreme P(y,x)=-y^3-2y^2-xy

extreme points

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

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (y, x) = - 1

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

x 2 - x

y = \frac{x}{x ^{2} + 1}

f (x) = - e^{- x}

f (x) = \frac{4}{3} x^{3} - 12 x^{2} + 27 x - 6

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

What is the extreme P(y,x)=-y^3-2y^2-xy ?
The extreme P(y,x)=-y^3-2y^2-xy is Saddle(0,0)