What is the equation of the normal line of #f(x)= x+x/(1+x/(1+1/x))# at #x = 1#?
2 Answers
In slope intercept form:
#y = (-3/4)x + 29/12#
Explanation:
#f(x) = x + x/(1+x/(1+1/x))#
#=x + x/(1+x^2/(x+1))#
#=x + (x(x+1))/(x^2+x+1)#
#=x+(x^2+x+1-1)/(x^2+x+1)#
#=x+1-1/(x^2+x+1)#
So:
#f'(x) = 1+(2x+1)/(x^2+x+1)^2#
Then we find:
#f(1) = 1+1-1/3 = 5/3#
#f'(1) = 1+3/3^2 = 1+1/3 = 4/3#
So the slope of the tangent at
So the normal line can be written in point slope form as:
#y - 5/3 = (-3/4)(x-1)#
From which we find:
#y = (-3/4)(x-1) + 5/3#
#=(-3/4)x+3/4 + 5/3#
#=(-3/4)x + 9/12 + 20/12#
#=(-3/4)x + 29/12#
That is:
#y = (-3/4)x + 29/12#
in slope intercept form.
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 :
Don't care about singularity point, this fonction is continuous everywhere. Derivate
and then apply the formula with
then you have
It's the equation of the ligne which is tangent at
the director vector is given by
here it is :
We focus on
So we have the point
Imagine a point
Then the vector
Where
I suspect there is a simpler way to do that, but this way is the most "intuitive" you will understand how thing work.