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

1 Answer
Jul 3, 2016

y-3.784=-0.0949(x+1)

Explanation:

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

f(-1) = 1/(1-2e^-1) = 3.784

So, f(x,y) would be f(-1, 3.784).

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

f'(x)=(2e^x)/(1-2e^x)^2
f'(-1)=(2e^-1)/(1-2e^-1)^2=10.537

We have to find slope of the normal line so the ⟂ slope would be -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(x+1)