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

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Answer 1 · 2016-07-03T02:45:32

#y-3.784=-0.0949(x+1)#

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)#