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

The extreme F(x,y)=x^3-3xy+3y^2+1 is Minimum(1/2 , 1/4),Saddle(0,0)

Popular >

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

extreme points

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

Minimum (\frac{1}{2}, \frac{1}{4}), Saddle (0, 0)

Solution steps

Find the critical points:

(\frac{1}{2}, \frac{1}{4}), (0, 0)

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

D (x, y) = 36 x - 9

Check the sign of at each critical point

Check critical point

(\frac{1}{2}, \frac{1}{4}) :

Minimum

Check critical point

(0, 0) :

Saddle

Minimum (\frac{1}{2}, \frac{1}{4}), Saddle (0, 0)

f (x) = x y + 5 x - 5

g (x) = \frac{x ^{5} + I n ( x )}{x ^{2} - x}

e^{x} (24 - x^{2})

x^{3} + 3 x y^{2} - 15 x - 12 y

f (x) = x + \frac{4}{x + 1}

What is the extreme F(x,y)=x^3-3xy+3y^2+1 ?
The extreme F(x,y)=x^3-3xy+3y^2+1 is Minimum(1/2 , 1/4),Saddle(0,0)