# How do you find the inflection points for the function f(x)=x/(x-1)?

$f ' \left(x\right) = \frac{1 \cdot \left(x - 1\right) - x \cdot 1}{{\left(x - 1\right)}^{2}} = \frac{- 1}{{\left(x - 1\right)}^{2}} = - {\left(x - 1\right)}^{- 2}$
$f ' ' \left(x\right) = 2 {\left(x - 1\right)}^{- 3} = \frac{2}{{\left(x - 1\right)}^{3}}$
Since $f ' ' \left(x\right) < 0$ when $x < 1$ and $f ' ' \left(x\right) > 0$ when $x > 1$, there is a concavity change only at $x = 1$; however, the original function $f \left(x\right)$ is undefined at $x = 1$, so it cannot have an inflection point there. Hence, there is no inflection point.