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

The extreme f(x,y)=xy^2+3xy-x+2 is Saddle(0,(-3+sqrt(13))/2),Saddle(0,(-3-sqrt(13))/2)

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

extreme

Saddle (0, \frac{- 3 + 13}{2}), Saddle (0, \frac{- 3 - 13}{2})

Solution steps

Find the critical points:

(0, \frac{- 3 + 13}{2}), (0, \frac{- 3 - 13}{2})

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

D (x, y) = - 4 y^{2} - 12 y - 9

Check the sign of at each critical point

Check critical point

(0, \frac{- 3 + 13}{2}) :

Saddle

Check critical point

(0, \frac{- 3 - 13}{2}) :

Saddle

Saddle (0, \frac{- 3 + 13}{2}), Saddle (0, \frac{- 3 - 13}{2})

What is the extreme f(x,y)=xy^2+3xy-x+2 ?
The extreme f(x,y)=xy^2+3xy-x+2 is Saddle(0,(-3+sqrt(13))/2),Saddle(0,(-3-sqrt(13))/2)