# What is the equation of the normal line of f(x)=(x+5)/(e^x-1) at x=1?

Normal Line:
$y = {\left(e - 1\right)}^{2} / \left(5e+1\right) \cdot x - {\left(e - 1\right)}^{2} / \left(5e+1\right) + \frac{6}{e - 1}$

#### Explanation:

Given $x = 1$
solve $f \left(1\right)$:

$f \left(1\right) = \frac{6}{e - 1}$
The point on the curve $\left(1 , \frac{6}{e - 1}\right)$

slope for the normal line $m :$

Use $y - {y}_{1} = m \cdot \left(x - {x}_{1}\right)$

$m = {\left(e - 1\right)}^{2} / \left(5e+1\right)$

$y - \frac{6}{e - 1} = {\left(e - 1\right)}^{2} / \left(5e+1\right) \cdot x - {\left(e - 1\right)}^{2} / \left(5e+1\right)$

$y = {\left(e - 1\right)}^{2} / \left(5e+1\right) \cdot x - {\left(e - 1\right)}^{2} / \left(5e+1\right) + \frac{6}{e - 1}$