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

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Answer 1 · 2015-05-16T07:42:39

Critical points would be those at which either $dy/dx=0$ or it does not exist.

$dy/dx=( (x-1)^2 -x.2(x-1))/(x-1)^4$

= $((x-1) -2x)/(x-1)^3$

= $-(x+1)/(x-1)^3$

$dy/dx$ 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

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