What is the extreme P(x,y)=(x+1)^2-(y-2)^2 ?

The extreme P(x,y)=(x+1)^2-(y-2)^2 is Saddle(-1,2)

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extreme P(x,y)=(x+1)^2-(y-2)^2

extreme points

P (x, y) = (x + 1)^{2} - (y - 2)^{2}

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) = 3 x^{2} e^{x}

f (x) = 4 y^{2} - 5 x = 2 y - 3 y^{2}

f (x, y) = x^{3} - y^{3} - 6 x y - 4

f (x) = \frac{2 ( x + 2 ) ^{2}}{x ^{2}}

f (x, y) = y^{3} x^{5} + y x

What is the extreme P(x,y)=(x+1)^2-(y-2)^2 ?
The extreme P(x,y)=(x+1)^2-(y-2)^2 is Saddle(-1,2)