Question

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

Answer 1

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 `(1, 5/3)` is `4/3`, hence the slope of the normal is `-3/4`

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.

Answer 2

First just do simple math to simplify your function

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

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

`f(x) = x + 1/((x+1)/(x(x+1))+x^2/(x(x+1))`

`f(x) = x + (x(x+1))/((x^2+x+1)`

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

`f(x) = f(a)+f'(a)(x-a)`

Don't care about singularity point, this fonction is continuous everywhere. Derivate `(df)/dx` with quotient rule you obtain :

`(df)/dx = ((1 + x)^2 (2 + x^2))/(1 + x + x^2)^2`

and then apply the formula with `a = 1`

`f'(1) = 4/3`
`f(1) = 5/3`

then you have `f(x) = 5/3+4/3(x-1)`
`f(x) = y`

`4/3x-y+1/3=0`

It's the equation of the ligne which is tangent at `x = 1`

the director vector is given by `vec(u) = (-b,a)`

here it is : `vec(u) = (1,4/3)`

We focus on `x = 1` so

`4/3-y+1/3=0 => y = 5/3` when `x = 1`

So we have the point `A` `(1,5/3)` and the director vecteur `vecu = (1,4/3)`

Imagine a point `M(x,y)` (Unknown)

Then the vector `vec(AM)` is normal to `vec(u)` if and only if their scalar product is equal to `0`

`vec(AM)* vec(u) = ||vec(AM)||*||vec(u)||*cos(theta)`

Where `cos(theta)` is the angle between the two vector, if they are normal, then `theta = pi/2` and `cos(pi/2) = 0`. which implies

`vec(AM)* vec(u) = 0`

`vec(AM) = (x-1,y-5/3)`

`vec(AM)* vec(u) = x-1+4/3(y-5/3)`

`x-1+4/3(y-5/3)` must be equal `0`

`x-1+4/3(y-5/3) = 0`

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