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

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Answer 1 · 2017-04-08T11:49:38

#y=x#

To find an equation of a line, we need two things:

Point on the line: #(x_1,y_1)=(1,f(1))=(1,1)#
Slope of the line: #m=-1/(f'(1))=-1/(-1)=1#
(Note: Since the normal line is perpendicular to the tangent line, we take the negative reciprocal of the slope of the tangent line #f'(1)#.)

By Point-Slope Form #y-y_1=m(x-x_1)#,

#y-1=1(x-1) Rightarrow y=x#

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I hope that this was clear.

Here are more details:

#f(1)=1/(e^(2(1)-2))=1/e^0=1#

#f'(x)=(1cdote^(2x-2)-x cdot 2e^(2x-2))/(e^(2x-2))^2=((1-2x)cancel(e^(2x-2)))/(e^(2x-2))^(cancel(2))=(1-2x)/e^(2x-2)#

So,

#f'(1)=(1-2(1))/(e^(2(1)-2))=(-1)/e^0=-1#