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

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Answer 1 · 2016-05-30T13:59:45

#y = -4x+11/2#

#f(x) = 1 + 1/(1+1/x)#

#= 1 + x/(x+1)#

#= 1+(x+1-1)/(x+1)#

#= 2-1/(x+1)#

#= 2-(x+1)^(-1)#

So:

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

Then:

#f(1) = 3/2#

#f'(1) = 1/4#

Any line perpendicular to a line of slope #m# will have slope #-1/m#.

The slope of the tangent line is #1/4#, so the slope of the normal is #-4#.

So we want the equation of a line of slope #-4# through the point #(1, 3/2)#

In point slope form it can be expressed:

#y-3/2 = -4(x-1)#

Adding #3/2# to both sides we get:

#y = -4x+4+3/2 = -4x+11/2#

So the equation of the normal in point intercept form is:

#y = -4x+11/2#

graph{(y - (1 + 1/(1+1/x)))(y+4x-11/2) = 0 [-10.17, 9.83, -2.4, 7.6]}