# How do you find the critical points for y = x/(x-1)^2?

##### 1 Answer
May 16, 2015

Critical points would be those at which either $\frac{\mathrm{dy}}{\mathrm{dx}} = 0$ or it does not exist.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{\left(x - 1\right)}^{2} - x .2 \left(x - 1\right)}{x - 1} ^ 4$

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

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

$\frac{\mathrm{dy}}{\mathrm{dx}}$ would be equal to zero at x= $- 1$ and it does not exist at x=1. The critical points are thus x=$- 1$ and 1