What is the extreme f(x,y)=x^3-3xy+y^2+y-5 ?

The extreme f(x,y)=x^3-3xy+y^2+y-5 is Minimum(1,1),Saddle(1/2 , 1/4)

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extreme f(x,y)=x^3-3xy+y^2+y-5

extreme points

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

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

Solution steps

Find the critical points:

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

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

D (x, y) = 12 x - 9

Check the sign of at each critical point

Check critical point

(1, 1) :

Minimum

Check critical point

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

Saddle

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

f (x) = \frac{( 8 x ^{3} - x ^{2} - 108 )}{4 x ^{2}}

f (x) = 3 x - 4

- (x + 1)^{2} - 5 (y - 3)^{2}

f (x) = ln (x + y + 1)

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

What is the extreme f(x,y)=x^3-3xy+y^2+y-5 ?
The extreme f(x,y)=x^3-3xy+y^2+y-5 is Minimum(1,1),Saddle(1/2 , 1/4)