What is the equation of the line that is normal to f(x)= (2x-2)e^(2x-2) at x= 1 ?

1 Answer
Jan 10, 2018

2y+x=1

Explanation:

f(x)= (2x-2)e^(2x-2)
implies f'(x) = (4x-2)*e^(2x-2)

implies f'(1) = 2
implies (dy/dx)_"at x =1" = 2

implies (dx/dy)_"at x =1" = -1/2

Hence we get the slope of the normal: m=-1/2

Also, f(1)=0, hnce f(x) passes through (1,0)
Using point slope form of line,

(y-0)=m(x-1)
implies y=-1/2(x-1)
implies 2y+x=1