What is the extreme f(x,y)=x^3-6xy+3y^2+1 ?

The extreme f(x,y)=x^3-6xy+3y^2+1 is Minimum(2,2),Saddle(0,0)

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extreme f(x,y)=x^3-6xy+3y^2+1

extreme points

f (x, y) = x^{3} - 6 x y + 3 y^{2} + 1

Minimum (2, 2), Saddle (0, 0)

Solution steps

Find the critical points:

(2, 2), (0, 0)

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

D (x, y) = 36 x - 36

Check the sign of at each critical point

Check critical point

(2, 2) :

Minimum

Check critical point

(0, 0) :

Saddle

Minimum (2, 2), Saddle (0, 0)

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

f (x, y) = 4 x^{3} - 11 x y^{2} + 15 y + 12

f (x) = sin (x) + cos^{2} (x)

f (x) = cos (5 x)

\frac{x}{( x - 1 ) ^{2}}

What is the extreme f(x,y)=x^3-6xy+3y^2+1 ?
The extreme f(x,y)=x^3-6xy+3y^2+1 is Minimum(2,2),Saddle(0,0)