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

The extreme f(x)=x^2+3xy is Saddle(0,0)

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extreme f(x)=x^2+3xy

extreme points

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

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = - 9

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

f (x, y) = 72 x + 30 y - 2 x^{2} - 3 y^{2}

y = \frac{( x ^{2} + 1 )}{( x ^{2} - 4 )}

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

f (x, y) = x^{3} + y^{2} - 2 x y + 7 x - 8 y + 4

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