What is the extreme f(x,y)=sqrt(x^2-4y^2+16) ?

The extreme f(x,y)=sqrt(x^2-4y^2+16) is Saddle(0,0)

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extreme f(x,y)=sqrt(x^2-4y^2+16)

extreme points

f (x, y) = x^{2} - 4 y^{2} + 16

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{4 ( x ^{2} + 16 )}{( x ^{2} - 4 y ^{2} + 16 ) x ^{2} - 4 y ^{2} + 16}

D (x, y) = - \frac{64}{( x ^{2} + 16 - 4 y ^{2} ) ^{2}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

f (x) = x^{3} + 3 x^{2} + 3 x + 1

16 (x - 1)^{4} - 4 (x - 1)^{2} + 1

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

f (x y) = x^{3} + y^{3} + 3 x^{2} - 18 y + 5

f (x, y) = 2 x^{3} + 2 y^{3} - 9 x^{2} + 3 y^{2} - 12 y

What is the extreme f(x,y)=sqrt(x^2-4y^2+16) ?
The extreme f(x,y)=sqrt(x^2-4y^2+16) is Saddle(0,0)