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

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May 16, 2015

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

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