# What is the equation of the normal line of #f(x)= x+x/(1+x/(1+1/x))# at #x = 1#?

In slope intercept form:

#y = (-3/4)x + 29/12#

So:

Then we find:

So the normal line can be written in point slope form as:

From which we find:

That is:

in slope intercept form.

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First just do simple math to simplify your function

and then search a line which is tangent to your curve : this is the linear approximation :

I suspect there is a simpler way to do that, but this way is the most "intuitive" you will understand how thing work.

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To find the equation of the normal line at x = 1, we need to determine the slope of the tangent line at that point. First, we find the derivative of f(x) using the quotient rule. The derivative is given by f'(x) = (1 + 1/x^2) / (1 + 1/x)^2. Evaluating f'(x) at x = 1, we get f'(1) = 2.

Since the slope of the tangent line is the same as the derivative at that point, the slope of the tangent line at x = 1 is 2.

To find the equation of the normal line, we use the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is the point on the line and m is the slope. Plugging in the values x1 = 1, y1 = f(1) = 1 + 1/(1 + 1/(1/1)) = 1 + 1/(1 + 1) = 1 + 1/2 = 1.5, and m = 2, we get the equation of the normal line as y - 1.5 = 2(x - 1).

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