What is the equation of the normal line of f(x)=11−2ex at x=-1?

Explanation:First, we need to locate the point on x-y coordinate.Meaning, we need f(x,y)f(−1)=11−2e−1=3.784So, f(x,y) would be f(-1,3.784).Now, we can take the derivative of f(x) to find the slope at x=-1f′(x)=2ex(1−2ex)2f′(−1)=2e−1(1−2e−1)2=10.537We have to find slope of the normal line so the ⊥ slope would be−110.537=−0.0949Finally, we can plugin all the numbers in a point-slope equation. That will give us following:y-3.784=-0.0949(x+1)

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High School
Calculus and Analysis
Calculus 1
Derivatives

Breanna Fisher

Answered question

2022-04-10

What is the equation of the normal line of

f (x) = \frac{1}{1 - 2 e^{x}}

at x=-1?

Answer & Explanation

kinoNoidae1wj

Beginner2022-04-11Added 7 answers

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

f (- 1) = \frac{1}{1 - 2 e^{- 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) = \frac{2 e^{x}}{{(1 - 2 e^{x})}^{2}}

f^{'} (- 1) = \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(x+1)

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