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

The extreme f(x,y)=x^2+3xy+y-1 is Saddle(-1/3 , 2/9)

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extreme f(x,y)=x^2+3xy+y-1

extreme points

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

Saddle (- \frac{1}{3}, \frac{2}{9})

Solution steps

Find the critical points:

(- \frac{1}{3}, \frac{2}{9})

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

D (x, y) = - 9

Check the sign of at each critical point

Check critical point

(- \frac{1}{3}, \frac{2}{9}) :

Saddle

Saddle (- \frac{1}{3}, \frac{2}{9})

f (x y) = x^{3} + y^{3} + 3 x^{2} - 18 y + 5

f (x, y) = 2 x^{3} + 2 y^{3} - 9 x^{2} + 3 y^{2} - 12 y

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

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

f (x) = 6 x^{5} - 160 x^{3} + 19

What is the extreme f(x,y)=x^2+3xy+y-1 ?
The extreme f(x,y)=x^2+3xy+y-1 is Saddle(-1/3 , 2/9)