# What is the equation of the normal line of f(x)=1/(1-2e^(x) at x=-1?

Jul 3, 2016

$y - 3.784 = - 0.0949 \left(x + 1\right)$

#### Explanation:

First, we need to locate the point on x-y coordinate.
Meaning, we need $f \left(x , y\right)$

$f \left(- 1\right) = \frac{1}{1 - 2 {e}^{-} 1} = 3.784$

So, $f \left(x , y\right)$ would be $f \left(- 1 , 3.784\right)$.

Now, we can take the derivative of $f \left(x\right)$ to find the slope at $x = - 1$.

$f ' \left(x\right) = \frac{2 {e}^{x}}{1 - 2 {e}^{x}} ^ 2$
$f ' \left(- 1\right) = \frac{2 {e}^{-} 1}{1 - 2 {e}^{-} 1} ^ 2 = 10.537$

We have to find slope of the normal line so the ⟂ slope would be $- \frac{1}{10.537} = - 0.0949$

Finally, we can plugin all the numbers in a point-slope equation. That will give us following:

$y - 3.784 = - 0.0949 \left(x + 1\right)$