# What are the critical points of f(x) =(x^2-2x)/(x-1)^2?

Dec 3, 2015

That function has no critical points.

#### Explanation:

Note that the domain of $f$ is $\left(- \infty , 1\right) \cup \left(1 , \infty\right)$

$f ' \left(x\right) = \frac{\left(2 x - 2\right) {\left(x - 1\right)}^{2} - \left({x}^{2} - 2 x\right) \left(2 \left(x - 1\right)\right)}{x - 1} ^ 4$

$= \frac{\left(2 x - 2\right) \left(x - 1\right) - \left({x}^{2} - 2 x\right) \left(2\right)}{x - 1} ^ 3$

$= \frac{2}{x - 1} ^ 3$

$f ' \left(x\right)$ is never $0$ and

it is undefined at $x = 1$, but $1$ is not in the domain of $f$, so it is not a critical number.