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

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

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

extreme points

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

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 8

D (x, y) = - 16

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

f (x) = 3 x^{4} + 9 x^{3} + 11

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

f (x) = 2 x^{2} + 5 x - 1

f (x) = x^{4} - 10 x^{2} + 9

f (x) = x^{2} y + y^{3} - 48 y

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