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

##### 2 Answers

In slope intercept form:

#y = (-3/4)x + 29/12#

#### Explanation:

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

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

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

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

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

So:

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

Then we find:

#f(1) = 1+1-1/3 = 5/3#

#f'(1) = 1+3/3^2 = 1+1/3 = 4/3#

So the slope of the tangent at

So the normal line can be written in point slope form as:

#y - 5/3 = (-3/4)(x-1)#

From which we find:

#y = (-3/4)(x-1) + 5/3#

#=(-3/4)x+3/4 + 5/3#

#=(-3/4)x + 9/12 + 20/12#

#=(-3/4)x + 29/12#

That is:

#y = (-3/4)x + 29/12#

in slope intercept form.

First just do simple math to simplify your function

and then search a line which is tangent to your curve : this is the linear approximation :

Don't care about singularity point, this fonction is continuous everywhere. Derivate

and then apply the formula with

then you have

It's the equation of the ligne which is tangent at

the director vector is given by

here it is :

We focus on

So we have the point

Imagine a point

Then the vector

Where

I suspect there is a simpler way to do that, but this way is the most "intuitive" you will understand how thing work.