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

The extreme f(x,y)=x^2-3xy+x/y is

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

extreme points

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

Saddle (0, \frac{1}{3}), Saddle (0, - \frac{1}{3}), Saddle (0, \frac{1}{3}), Saddle (0, - \frac{1}{3})

Solution steps

Find the critical points:

(0, \frac{1}{3}), (0, - \frac{1}{3}), (0, \frac{1}{3}), (0, - \frac{1}{3})

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{2 x}{y ^{3}}

D (x, y) = \frac{4 x}{y ^{3}} - \frac{1}{y ^{4}} - \frac{6}{y ^{2}} - 9

Check the sign of at each critical point

Check critical point

(0, \frac{1}{3}) :

Saddle

Check critical point

(0, - \frac{1}{3}) :

Saddle

Check critical point

(0, \frac{1}{3}) :

Saddle

Check critical point

(0, - \frac{1}{3}) :

Saddle

Saddle (0, \frac{1}{3}), Saddle (0, - \frac{1}{3}), Saddle (0, \frac{1}{3}), Saddle (0, - \frac{1}{3})

f (x) = 9 x - (5 x) ln (x), (0, \infty)

f (x, y) = x^{3} + y^{3} - 27 x y

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

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

f (x) = 3 - 4 x - \frac{4}{x}