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

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

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

extreme points

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

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

Solution steps

Find the critical points:

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

\frac{\partial ^{2} f}{\partial x ^{2}} = 2 (x - 1)

D (x, y) = - 4 x^{2} + 12 x - 4 x y - 4 y^{2} - 9

Check the sign of at each critical point

Check critical point

(1, 1) :

Saddle

Check critical point

(2, - 1) :

Minimum

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

f (x, y) = \frac{13720}{x} + \frac{13720}{y} + 5 x y

f (x) = 4 x^{3} - 8 x

f (x) = 12 x^{5} + 45 x^{4} - 360 x^{3} + 6

f (x) = (x^{2} + 3 x - 1)^{\frac{1}{3}}

f (x) = x^{3} - 4 x^{2} - x + 1

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