What is the extreme f(xy)=(3y+5x)/(xy+3) ?

The extreme f(xy)=(3y+5x)/(xy+3) is

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extreme f(xy)=(3y+5x)/(xy+3)

extreme points

f (x y) = \frac{3 y + 5 x}{x y + 3}

Saddle (\frac{3}{5}, 5), Saddle (- \frac{3}{5}, - 5), Saddle (\frac{3}{5}, 5), Saddle (- \frac{3}{5}, - 5)

Solution steps

Find the critical points:

(\frac{3}{5}, 5), (- \frac{3}{5}, - 5), (\frac{3}{5}, 5), (- \frac{3}{5}, - 5)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{2 x ( - 5 x ^{2} + 9 )}{( x y + 3 ) ^{3}}

D (x, y) = \frac{60 x ^{3} y ^{3} - 300 x ^{3} y - 900 x ^{2} - 108 x y ^{3} - 540 x y - 324 y ^{2}}{( 3 + x y ) ^{6}}

Check the sign of at each critical point

Check critical point

(\frac{3}{5}, 5) :

Saddle

Check critical point

(- \frac{3}{5}, - 5) :

Saddle

Check critical point

(\frac{3}{5}, 5) :

Saddle

Check critical point

(- \frac{3}{5}, - 5) :

Saddle

Saddle (\frac{3}{5}, 5), Saddle (- \frac{3}{5}, - 5), Saddle (\frac{3}{5}, 5), Saddle (- \frac{3}{5}, - 5)

f (x, y) = - 4 x^{2} + 8 x y - 2 y^{4} - 4

f (x) = \frac{2 ( - x + 1 )}{x ^{3}}

x^{4} + 2 x^{3} + 6 x^{2} - 9 x + 12

f (x) = ∣ 2 x - 4 ∣

y (t, x) = - 2 t + x