What is the extreme f(x,y)=(x-1)^2+y^3-3y^2-9y+5 ?

The extreme f(x,y)=(x-1)^2+y^3-3y^2-9y+5 is Minimum(1,3),Saddle(1,-1)

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extreme f(x,y)=(x-1)^2+y^3-3y^2-9y+5

extreme points

f (x, y) = (x - 1)^{2} + y^{3} - 3 y^{2} - 9 y + 5

Minimum (1, 3), Saddle (1, - 1)

Solution steps

Find the critical points:

(1, 3), (1, - 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = 6 y - 6

D (x, y) = 2 (6 y - 6)

Check the sign of at each critical point

Check critical point

(1, 3) :

Minimum

Check critical point

(1, - 1) :

Saddle

Minimum (1, 3), Saddle (1, - 1)

f (x) = 3 cos (3 x) + sin (3 x)

f (x, y) = x e^{- y} + \frac{y}{x}

f (x, y) = x^{2} + 3 x y + y^{2} + 5 x - 7 y

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

f (x) = \frac{3 - 9 - x ^{2}}{x}

What is the extreme f(x,y)=(x-1)^2+y^3-3y^2-9y+5 ?
The extreme f(x,y)=(x-1)^2+y^3-3y^2-9y+5 is Minimum(1,3),Saddle(1,-1)