What is the extreme f(x,y)=(sqrt(x^2+y^2-16))/(x-3) ?

The extreme f(x,y)=(sqrt(x^2+y^2-16))/(x-3) is Saddle(16/3 ,0)

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extreme f(x,y)=(sqrt(x^2+y^2-16))/(x-3)

extreme points

f (x, y) = \frac{x ^{2} + y ^{2} - 16}{x - 3}

Saddle (\frac{16}{3}, 0)

Solution steps

Find the critical points:

(\frac{16}{3}, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{x ^{2} - 16}{( x - 3 ) ( x ^{2} + y ^{2} - 16 ) x ^{2} + y ^{2} - 16}

D (x, y) = \frac{6 x ^{6} - x ^{5} y ^{2} - 66 x ^{5} + 15 x ^{4} y ^{2} + 48 x ^{4} - 2 x ^{3} y ^{4} - 84 x ^{3} y ^{2} + 1424 x ^{3} + 12 x ^{2} y ^{4} + 48 x ^{2} y ^{2} - 3408 x ^{2} - x y ^{6} - 18 x y ^{4} + 912 x y ^{2} - 5888 x + 3 y ^{6} - 1872 y ^{2} + 17664}{( x - 3 ) ^{5} ( x ^{2} + y ^{2} - 16 ) ^{3}}

Check the sign of at each critical point

Check critical point

(\frac{16}{3}, 0) :

Saddle

Saddle (\frac{16}{3}, 0)

f (x) = x^{2} + x y + y^{2} - 12 x + 1

f (x) = x^{2} - 8 x + 10

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

f (x) = x^{2} + 9 x - 5

f (x) = \frac{ln ( x )}{x}, 1 \leq x \leq 3

What is the extreme f(x,y)=(sqrt(x^2+y^2-16))/(x-3) ?
The extreme f(x,y)=(sqrt(x^2+y^2-16))/(x-3) is Saddle(16/3 ,0)