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

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

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

extreme points

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

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

Solution steps

Find the critical points:

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

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

D (x, y) = 12 y - 9

Check the sign of at each critical point

Check critical point

(1, 1) :

Minimum

Check critical point

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

Saddle

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

f (x, y) = 6 x^{2} + 9 x y + 3 y^{2} + 8 x y^{2} + 5 x^{2} y

\frac{x ^{2} - 11}{x - 6}

f (x, y) = \frac{2 x y}{( 2 x + 3 y - 4 )}

f (x) = - 2 x + 3 ln (4 x), (1, 5)

f (x) = 3 x^{2} - 30 x + 72

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