What is the extreme f(z)=x^2-y^2-2x+4y+6 ?

The extreme f(z)=x^2-y^2-2x+4y+6 is Saddle(1,2)

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extreme f(z)=x^2-y^2-2x+4y+6

extreme points

f (z) = x^{2} - y^{2} - 2 x + 4 y + 6

Saddle (1, 2)

Solution steps

Find the critical points:

(1, 2)

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

D (x, y) = - 4

Check the sign of at each critical point

Check critical point

(1, 2) :

Saddle

Saddle (1, 2)

f (x) = \frac{x ^{3}}{3} - x^{2} - 3 x - 1, 2 \leq x \leq 7

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

f (x) = 416 - x x

f (x) = x^{\frac{4}{5}} - 4

f (x) = (\frac{5 x ^{4}}{ln ( x )})

What is the extreme f(z)=x^2-y^2-2x+4y+6 ?
The extreme f(z)=x^2-y^2-2x+4y+6 is Saddle(1,2)