How do you find the critical points for #y = x/(x-1)^2#? Calculus Graphing with the First Derivative Identifying Stationary Points (Critical Points) for a Function 1 Answer bp 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 Answer link Related questions How do you find the stationary points of a curve? How do you find the stationary points of a function? How many stationary points can a cubic function have? How do you find the stationary points of the function #y=x^2+6x+1#? How do you find the stationary points of the function #y=cos(x)#? How do I find all the critical points of #f(x)=(x-1)^2#? Let #h(x) = e^(-x) + kx#, where #k# is any constant. For what value(s) of #k# does #h# have... How do you find the critical points for #f(x)=8x^3+2x^2-5x+3#? How do you find values of k for which there are no critical points if #h(x)=e^(-x)+kx# where k... How do you determine critical points for any polynomial? See all questions in Identifying Stationary Points (Critical Points) for a Function Impact of this question 1519 views around the world You can reuse this answer Creative Commons License