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

The extreme f(x,y)=1-x^2+2y^2 is Saddle(0,0)

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

extreme points

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

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 4

D (x, y) = - 8

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

f (x, y) = 60 x - 2 x^{2} - 3 y^{2} + 30 y

\frac{x ^{2}}{x - 4}

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

f (x, y) = ∣ x - y ∣

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

What is the extreme f(x,y)=1-x^2+2y^2 ?
The extreme f(x,y)=1-x^2+2y^2 is Saddle(0,0)