What is the extreme f(x)=18x^2-32y^2-36x-128y-110 ?

The extreme f(x)=18x^2-32y^2-36x-128y-110 is Saddle(1,-2)

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extreme f(x)=18x^2-32y^2-36x-128y-110

extreme points

f (x) = 18 x^{2} - 32 y^{2} - 36 x - 128 y - 110

Saddle (1, - 2)

Solution steps

Find the critical points:

(1, - 2)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 64

D (x, y) = - 2304

Check the sign of at each critical point

Check critical point

(1, - 2) :

Saddle

Saddle (1, - 2)

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

f (x) = 16 - x^{2}, - 4 \leq x \leq 4

f (x) = 6 + 3 x - 3 x^{2}

f (x) = x (16 - 47 + 2 x) (\frac{47}{2} - x)

f (x, y) = 3 x - 4 y + 5

What is the extreme f(x)=18x^2-32y^2-36x-128y-110 ?
The extreme f(x)=18x^2-32y^2-36x-128y-110 is Saddle(1,-2)