What is the extreme f(x,y)=2x^2-3xy+4 ?

The extreme f(x,y)=2x^2-3xy+4 is Saddle(0,0)

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extreme f(x,y)=2x^2-3xy+4

extreme points

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

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^{4} - 18 x^{2} + 11

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

f (x) = - x^{3} + 3 x^{2} + 9 x - 6

f (x, y) = \frac{x ^{2}}{8} + \frac{y ^{2}}{2} - 1

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

What is the extreme f(x,y)=2x^2-3xy+4 ?
The extreme f(x,y)=2x^2-3xy+4 is Saddle(0,0)