What is the equation of the line normal to f(x)=1/abs(x-6)  at x=1?

Jan 2, 2016

Or in general form:

$25 x + y - \frac{126}{5} = 0$

Explanation:

Unpack the definition of $f \left(x\right)$ as:

$f \left(x\right) = \left\{\begin{matrix}\frac{1}{x - 6} & \mathmr{if} x > 6 \\ - \frac{1}{x - 6} & \mathmr{if} x < 6\end{matrix}\right.$

So:

$f ' \left(x\right) = \left\{\begin{matrix}- \frac{1}{x - 6} ^ 2 & \mathmr{if} x > 6 \\ \frac{1}{x - 6} ^ 2 & \mathmr{if} x < 6\end{matrix}\right.$

In particular:

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

$f ' \left(1\right) = \frac{1}{25}$

So the slope of the tangent at $\left(1 , \frac{1}{5}\right)$ is $\frac{1}{25}$ and the perpendicular normal must have slope $- \frac{1}{\frac{1}{25}} = - 25$

The equation of the normal can be written in point slope form as:

$y - \frac{1}{5} = - 25 \left(x - 1\right)$

Hence:

$y = - 25 x + 25 + \frac{1}{5}$

That is:

$y = - 25 x + \frac{126}{5}$

in slope intercept form.

Or in general form:

$25 x + y - \frac{126}{5} = 0$