What is the extreme f(x,y)=2y^2-xy+6y^2 ?

The extreme f(x,y)=2y^2-xy+6y^2 is Saddle(0,0)

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extreme f(x,y)=2y^2-xy+6y^2

extreme points

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

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 16

D (x, y) = - 1

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

- \frac{1}{80} t^{2} + \frac{9}{40} t - \frac{17}{80}

3 x^{2} - 4 x - 1

f (x) = f (x) = 3 x \frac{5}{3} - 15 x \frac{2}{3}

\frac{x + 3}{x ^{3} + 2 x ^{2} - 15 x - 36}

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

What is the extreme f(x,y)=2y^2-xy+6y^2 ?
The extreme f(x,y)=2y^2-xy+6y^2 is Saddle(0,0)