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

1 Answer
Jan 15, 2018

derivative for slope , use 2 point form of a line

Explanation:

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 :)