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

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Answer 1 · 2017-02-10T05:21:16

#x+2y-1=0#. See normal-inclusive Socratic graph.

graph{((2x-2)/(e^(2x-2))-y)(x+2y-1)=0 [-10, 10, -5, 5]}

#f =2e^2e^(-2x)(x-1)=0#, at x = 1.

So, the foot of the normal is P( 1, 0 ).

f'=2e^2(e^(-2x)(x-1)'+(x-1)(e^(-2x))')#

#=2e^2(e^(-2x)-2e^(-2x)(x-1))=2#, at x = 1.

Slope of the normal #= -1/(f')=-1/2#.

So, the equation to the normal is

#y-0=-1/2(x-1)#, giving

#x+2y-1=0#