What is the extreme f(x,y)=sqrt(16+(x-2)^2-4(y+1)^2) ?

The extreme f(x,y)=sqrt(16+(x-2)^2-4(y+1)^2) is Saddle(2,-1)

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extreme f(x,y)=sqrt(16+(x-2)^2-4(y+1)^2)

extreme points

f (x, y) = 16 + (x - 2)^{2} - 4 (y + 1)^{2}

Saddle (2, - 1)

Solution steps

Find the critical points:

(2, - 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{4 ( x ^{2} - 4 x + 20 )}{( x ^{2} - 4 x + 16 - 4 y ^{2} - 8 y ) - 4 y ^{2} - 8 y + x ^{2} + 16 - 4 x}

D (x, y) = - \frac{64}{( x ^{2} - 4 x + 16 - 4 y ^{2} - 8 y ) ^{2}}

Check the sign of at each critical point

Check critical point

(2, - 1) :

Saddle

Saddle (2, - 1)

f (x) = 3 x^{4} - 4 x^{3} - 8, - 1 \leq x \leq 2

4 x^{4} + 4 y^{4} - 2 x y

K (y, z) = 3 yz

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

f (x) = \frac{x}{x}

What is the extreme f(x,y)=sqrt(16+(x-2)^2-4(y+1)^2) ?
The extreme f(x,y)=sqrt(16+(x-2)^2-4(y+1)^2) is Saddle(2,-1)