Question

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

Answer 1

`y = -4x+11/2`

Explanation:

`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]}