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

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

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

extreme points

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

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

Solution steps

Find the critical points:

(0, 3), (0, 1), (2, 3), (2, 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = 6 y - 12

D (x, y) = (6 x - 6) (6 y - 12)

Check the sign of at each critical point

Check critical point

(0, 3) :

Saddle

Check critical point

(0, 1) :

Maximum

Check critical point

(2, 3) :

Minimum

Check critical point

(2, 1) :

Saddle

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

f (x) = - 5 x + 2 ln (2 x)

\frac{x + 3}{x ^{2} - 2 x - 15}

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

f (x) = 2 x^{3} - 36 x^{2} + 192 x, 3 \leq x \leq 9

f (x) = 2 x^{2} - 3 x - 5

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