What is the extreme y=((1/(x-1)-2/x+z))/(2+1/x) ?

The extreme y=((1/(x-1)-2/x+z))/(2+1/x) is Saddle(0,-3)

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extreme y=((1/(x-1)-2/x+z))/(2+1/x)

extreme points

y = \frac{( \frac{1}{x - 1} - \frac{2}{x} + z )}{2 + \frac{1}{x}}

Saddle (0, - 3)

Solution steps

Find the critical points:

(0, - 3)

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

D (x, z) = - \frac{1}{( 2 x + 1 ) ^{4}}

Check the sign of at each critical point

Check critical point

(0, - 3) :

Saddle

Saddle (0, - 3)

What is the extreme y=((1/(x-1)-2/x+z))/(2+1/x) ?
The extreme y=((1/(x-1)-2/x+z))/(2+1/x) is Saddle(0,-3)