What is the equation of the line that is normal to #f(x)= (2x-2)/e^(2x-2) # at # x= 1 #?
1 Answer
Feb 10, 2017
Explanation:
graph{((2x-2)/(e^(2x-2))-y)(x+2y-1)=0 [-10, 10, -5, 5]}
So, the foot of the normal is P( 1, 0 ).
f'=2e^2(e^(-2x)(x-1)'+(x-1)(e^(-2x))')#
Slope of the normal
So, the equation to the normal is