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

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

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

extreme points

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

Saddle (\frac{1}{3}, - 1), Minimum (1, - 1)

Solution steps

Find the critical points:

(\frac{1}{3}, - 1), (1, - 1)

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

D (x, y) = 2 (6 x - 4)

Check the sign of at each critical point

Check critical point

(\frac{1}{3}, - 1) :

Saddle

Check critical point

(1, - 1) :

Minimum

Saddle (\frac{1}{3}, - 1), Minimum (1, - 1)

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

(x + 3) (x - 6)^{2}

f (x) = x^{4} - 18 x^{2} + 1, - 6 \leq x \leq 6

f (x) = 3 x^{5} - 25 x^{3} + 60 x

x^{2} + x y + y^{2} - 7 x - 3 y

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