# `f(x) = x/(x - 1)` Find all relative extrema, use the second derivative test where applicable.

First derivative can be found using the quotient rule.

`f'(x) = ((x-1)*1 - x*x)/(x-1)^2`

`f'(x) = (-x^2+x-1)/(x-1)^2`

At critical points, f'(x) = 0

-x^2+x-1=0

Roots can be found by the using the standard equation for quadratic roots.

`x = (-1+-sqrt(1^2-4*(-1)*(-1)))/(2*(-1))`

`x = (-1+-sqrt(-7))/-2`

There are no real roots, therefore, f(x)...

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First derivative can be found using the quotient rule.

`f'(x) = ((x-1)*1 - x*x)/(x-1)^2`

`f'(x) = (-x^2+x-1)/(x-1)^2`

At critical points, f'(x) = 0

-x^2+x-1=0

Roots can be found by the using the standard equation for quadratic roots.

`x = (-1+-sqrt(1^2-4*(-1)*(-1)))/(2*(-1))`

`x = (-1+-sqrt(-7))/-2`

There are no real roots, therefore, f(x) doesn't have any critical points.

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