What is the extreme f(x,y)=(-x-ln(x^2+y^2-81))/(666) ?

The extreme f(x,y)=(-x-ln(x^2+y^2-81))/(666) is Saddle(-1-sqrt(82),0),Minimum(sqrt(82)-1,0)

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extreme f(x,y)=(-x-ln(x^2+y^2-81))/(666)

extreme points

f (x, y) = \frac{- x - ln ( x ^{2} + y ^{2} - 81 )}{666}

Saddle (- 1 - 82, 0), Minimum (82 - 1, 0)

Solution steps

Find the critical points:

(- 1 - 82, 0), (82 - 1, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{- y ^{2} + x ^{2} - 81}{333 ( x ^{2} + y ^{2} - 81 ) ^{2}}

D (x, y) = - \frac{x ^{2} + y ^{2} + 81}{110889 ( x ^{2} + y ^{2} - 81 ) ^{3}}

Check the sign of at each critical point

Check critical point

(- 1 - 82, 0) :

Saddle

Check critical point

(82 - 1, 0) :

Minimum

Saddle (- 1 - 82, 0), Minimum (82 - 1, 0)

f (x) = sin (2 x) - x, - \frac{π}{2} \leq x \leq \frac{π}{2}

f (x) = 5 x^{4} - x^{5} + 8

f (x) = 5 x^{4} - x^{5} + 7

f (x) = 5 x^{4} - x^{5} + 4

17 r^{2} - r^{3}

What is the extreme f(x,y)=(-x-ln(x^2+y^2-81))/(666) ?
The extreme f(x,y)=(-x-ln(x^2+y^2-81))/(666) is Saddle(-1-sqrt(82),0),Minimum(sqrt(82)-1,0)