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

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

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

extreme points

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

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

f (x) = x^{3} - 12 x^{2} - 27 x - 26

f (x) = \frac{6}{x ^{2} + 3}

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