What is the equation of the normal line of #f(x)=(x+5)/(e^x-1)# at #x=1#?

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Answer 1 · 2016-01-17T09:59:02

Normal Line:
#y=(e-1)^2/(5e+1)*x-(e-1)^2/(5e+1)+6/(e-1)#

Given #x=1#
solve #f(1)#:

#f(1)=6/(e-1)#
The point on the curve #(1, 6/(e-1))#

slope for the normal line #m:#

Use #y-y_1=m*(x-x_1)#

#m=(e-1)^2/(5e+1)#

#y-6/(e-1)=(e-1)^2/(5e+1)*x-(e-1)^2/(5e+1)#

#y=(e-1)^2/(5e+1)*x-(e-1)^2/(5e+1)+6/(e-1)#