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

Dec 27, 2015

In slope intercept form:

$y = \left(- \frac{3}{4}\right) x + \frac{29}{12}$

#### Explanation:

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

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

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

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

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

So:

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

Then we find:

$f \left(1\right) = 1 + 1 - \frac{1}{3} = \frac{5}{3}$

$f ' \left(1\right) = 1 + \frac{3}{3} ^ 2 = 1 + \frac{1}{3} = \frac{4}{3}$

So the slope of the tangent at $\left(1 , \frac{5}{3}\right)$ is $\frac{4}{3}$, hence the slope of the normal is $- \frac{3}{4}$

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

$y - \frac{5}{3} = \left(- \frac{3}{4}\right) \left(x - 1\right)$

From which we find:

$y = \left(- \frac{3}{4}\right) \left(x - 1\right) + \frac{5}{3}$

$= \left(- \frac{3}{4}\right) x + \frac{3}{4} + \frac{5}{3}$

$= \left(- \frac{3}{4}\right) x + \frac{9}{12} + \frac{20}{12}$

$= \left(- \frac{3}{4}\right) x + \frac{29}{12}$

That is:

$y = \left(- \frac{3}{4}\right) x + \frac{29}{12}$

in slope intercept form.

Dec 27, 2015

First just do simple math to simplify your function

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

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 :

$f \left(x\right) = f \left(a\right) + f ' \left(a\right) \left(x - a\right)$

Don't care about singularity point, this fonction is continuous everywhere. Derivate $\frac{\mathrm{df}}{\mathrm{dx}}$ with quotient rule you obtain :

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{{\left(1 + x\right)}^{2} \left(2 + {x}^{2}\right)}{1 + x + {x}^{2}} ^ 2$

and then apply the formula with $a = 1$

$f ' \left(1\right) = \frac{4}{3}$
$f \left(1\right) = \frac{5}{3}$

then you have $f \left(x\right) = \frac{5}{3} + \frac{4}{3} \left(x - 1\right)$
$f \left(x\right) = y$

$\frac{4}{3} x - y + \frac{1}{3} = 0$

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

the director vector is given by $\vec{u} = \left(- b , a\right)$

here it is : $\vec{u} = \left(1 , \frac{4}{3}\right)$

We focus on $x = 1$ so

$\frac{4}{3} - y + \frac{1}{3} = 0 \implies y = \frac{5}{3}$ when $x = 1$

So we have the point $A$ $\left(1 , \frac{5}{3}\right)$ and the director vecteur $\vec{u} = \left(1 , \frac{4}{3}\right)$

Imagine a point $M \left(x , y\right)$ (Unknown)

Then the vector $\vec{A M}$ is normal to $\vec{u}$ if and only if their scalar product is equal to $0$

$\vec{A M} \cdot \vec{u} = | | \vec{A M} | | \cdot | | \vec{u} | | \cdot \cos \left(\theta\right)$

Where $\cos \left(\theta\right)$ is the angle between the two vector, if they are normal, then $\theta = \frac{\pi}{2}$ and $\cos \left(\frac{\pi}{2}\right) = 0$. which implies

$\vec{A M} \cdot \vec{u} = 0$

$\vec{A M} = \left(x - 1 , y - \frac{5}{3}\right)$

$\vec{A M} \cdot \vec{u} = x - 1 + \frac{4}{3} \left(y - \frac{5}{3}\right)$

$x - 1 + \frac{4}{3} \left(y - \frac{5}{3}\right)$ must be equal $0$

$x - 1 + \frac{4}{3} \left(y - \frac{5}{3}\right) = 0$

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