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

The extreme f(x,y)=x^3-3x(y-2)+(y-2)^3 is Saddle(0,2),Minimum(1,3)

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

extreme points

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

Saddle (0, 2), Minimum (1, 3)

Solution steps

Find the critical points:

(0, 2), (1, 3)

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

D (x, y) = 36 x (y - 2) - 9

Check the sign of at each critical point

Check critical point

(0, 2) :

Saddle

Check critical point

(1, 3) :

Minimum

Saddle (0, 2), Minimum (1, 3)

f (x) = - x^{2} + 4 x + 9

f (x) = x^{- 2} ln (x)

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

x^{4} - 8 x^{2} + 9

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

What is the extreme f(x,y)=x^3-3x(y-2)+(y-2)^3 ?
The extreme f(x,y)=x^3-3x(y-2)+(y-2)^3 is Saddle(0,2),Minimum(1,3)