What is the extreme f(x)=x^2-y^2-5x-4y ?

The extreme f(x)=x^2-y^2-5x-4y is Saddle(5/2 ,-2)

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extreme f(x)=x^2-y^2-5x-4y

extreme points

f (x) = x^{2} - y^{2} - 5 x - 4 y

Saddle (\frac{5}{2}, - 2)

Solution steps

Find the critical points:

(\frac{5}{2}, - 2)

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

D (x, y) = - 4

Check the sign of at each critical point

Check critical point

(\frac{5}{2}, - 2) :

Saddle

Saddle (\frac{5}{2}, - 2)

20 q - 2 q^{2}

f (x) = \frac{1}{x - 2}, 0 \leq x \leq 1

f (r) = (r - 8)^{3}

f (x) = - x^{3} + 4 x^{2}

y = - (x + 4)^{2} + 8

What is the extreme f(x)=x^2-y^2-5x-4y ?
The extreme f(x)=x^2-y^2-5x-4y is Saddle(5/2 ,-2)