Breanna Fisher

2022-04-10

What is the equation of the normal line of $f\left(x\right)=\frac{1}{1-2{e}^{x}}$ at x=-1?

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

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

${f}^{\prime}(-1)=\frac{2{e}^{-1}}{{(1-2{e}^{-1})}^{2}}=10.537$

We have to find slope of the normal line so the$\mathrm{\perp}$ 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)

First, we need to locate the point on x-y coordinate.

Meaning, we need f(x,y)

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

We have to find slope of the normal line so the

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

y-3.784=-0.0949(x+1)

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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