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

2 Answers
Dec 27, 2015

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 #(1, 5/3)# is #4/3#, hence the slope of the normal is #-3/4#

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.

Dec 27, 2015

First just do simple math to simplify your function

#f(x) = x + x/(1+x/(1+1/x))#

#f(x) = x + 1/(1/x+x/(x+1)#

#f(x) = x + 1/((x+1)/(x(x+1))+x^2/(x(x+1))#

#f(x) = x + (x(x+1))/((x^2+x+1)#

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

#f(x) = f(a)+f'(a)(x-a)#

Don't care about singularity point, this fonction is continuous everywhere. Derivate #(df)/dx# with quotient rule you obtain :

#(df)/dx = ((1 + x)^2 (2 + x^2))/(1 + x + x^2)^2 #

and then apply the formula with #a = 1#

#f'(1) = 4/3#
#f(1) = 5/3#

then you have #f(x) = 5/3+4/3(x-1)#
#f(x) = y#

#4/3x-y+1/3=0#

It's the equation of the ligne which is tangent at #x = 1#

the director vector is given by #vec(u) = (-b,a)#

here it is : #vec(u) = (1,4/3)#

We focus on #x = 1# so

#4/3-y+1/3=0 => y = 5/3# when #x = 1#

So we have the point #A# #(1,5/3)# and the director vecteur #vecu = (1,4/3)#

Imagine a point #M(x,y)# (Unknown)

Then the vector #vec(AM)# is normal to #vec(u)# if and only if their scalar product is equal to #0#

#vec(AM)* vec(u) = ||vec(AM)||*||vec(u)||*cos(theta)#

Where #cos(theta)# is the angle between the two vector, if they are normal, then #theta = pi/2# and #cos(pi/2) = 0#. which implies

#vec(AM)* vec(u) = 0#

#vec(AM) = (x-1,y-5/3)#

#vec(AM)* vec(u) = x-1+4/3(y-5/3)#

# x-1+4/3(y-5/3)# must be equal #0#

# x-1+4/3(y-5/3) = 0#

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