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

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

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

extreme points

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

Saddle (1, - 2), Saddle (- 1, - 2)

Solution steps

Find the critical points:

(1, - 2), (- 1, - 2)

\frac{\partial ^{2} f}{\partial x ^{2}} = 0

D (x, y) = - 4 x^{2}

Check the sign of at each critical point

Check critical point

(1, - 2) :

Saddle

Check critical point

(- 1, - 2) :

Saddle

Saddle (1, - 2), Saddle (- 1, - 2)

f (x, y) = \frac{3}{4} y^{2} + \frac{1}{24} y^{3} - \frac{1}{32} y^{4} - x^{2}

f (x, y) = 2 \cdot \frac{( x ^{2} - 2 \cdot x + 3 )}{y \cdot ( y + 2 )}

f (x) = x^{y}

f (x) = 4

2 x^{3} - 3 x^{2} - 36 x + 5

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