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

May 30, 2016

$y = - 4 x + \frac{11}{2}$

#### Explanation:

$f \left(x\right) = 1 + \frac{1}{1 + \frac{1}{x}}$

$= 1 + \frac{x}{x + 1}$

$= 1 + \frac{x + 1 - 1}{x + 1}$

$= 2 - \frac{1}{x + 1}$

$= 2 - {\left(x + 1\right)}^{- 1}$

So:

$f ' \left(x\right) = {\left(x + 1\right)}^{- 2} = \frac{1}{x + 1} ^ 2$

Then:

$f \left(1\right) = \frac{3}{2}$

$f ' \left(1\right) = \frac{1}{4}$

Any line perpendicular to a line of slope $m$ will have slope $- \frac{1}{m}$.

The slope of the tangent line is $\frac{1}{4}$, so the slope of the normal is $- 4$.

So we want the equation of a line of slope $- 4$ through the point $\left(1 , \frac{3}{2}\right)$

In point slope form it can be expressed:

$y - \frac{3}{2} = - 4 \left(x - 1\right)$

Adding $\frac{3}{2}$ to both sides we get:

$y = - 4 x + 4 + \frac{3}{2} = - 4 x + \frac{11}{2}$

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

$y = - 4 x + \frac{11}{2}$

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