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

The extreme y=((x+3)(x^2-2r))/(x^3) is Saddle(-3, 9/2)

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extreme y=((x+3)(x^2-2r))/(x^3)

extreme points

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

Saddle (- 3, \frac{9}{2})

Solution steps

Find the critical points:

(- 3, \frac{9}{2})

\frac{\partial ^{2} f}{\partial x ^{2}} = 0

D (x, r) = - \frac{( 4 x + 18 ) ^{2}}{x ^{8}}

Check the sign of at each critical point

Check critical point

(- 3, \frac{9}{2}) :

Saddle

Saddle (- 3, \frac{9}{2})

14 x^{2} + \frac{3500}{x}

f (x, y) = x^{3} + y^{3} - 12 x - 3 y + 15

f (x) = x^{3} + 1 5^{2}

f (x) = x^{(2)} + x y + y^{(2)} - 6 x - 6 y + 5

f (x) = x^{4} - 30 x^{2} + 29

What is the extreme y=((x+3)(x^2-2r))/(x^3) ?
The extreme y=((x+3)(x^2-2r))/(x^3) is Saddle(-3, 9/2)