What is the equation of the normal line of #f(x)=x/(x-1) # at #x=4 #?

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Answer 1 · 2018-01-15T17:39:35

derivative for slope , use 2 point form of a line

#f^'(x) = ((x-1) - (x))/(x-1)^2# = #-1/(x-1)^2#

hence slope of normal is

#(x-1)^2# by #m1.m2= -1# ( tangent perpendicular to normal )

find value of #y ( or f(x))# @ #x=4#
#y = 4/3#
and value of slope is #9#

by 2 point form

#y - 4/3 = 9 (x - 4)#

hence normal equation is:
#3y -27x = -104 # or #27x - 3y = 104 #

hope u find it helpful :)