For #f(x) =x/(x^2+1)#, what is the equation of the line tangent to #x =3 # and at what point is the tangent line horizontal?

1 Answer
Nov 26, 2015
  1. Tangent line is: #y=-0.08x+0.54#

  2. The tangent line is horizontal at: #x=-1# and #x=1#

Explanation:

To find tangent lines we first have to calculate #f'(x)#

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

#f'(x)=(1*(x^2+1)-2x*x)/(x^2+1)^2#

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

  1. To calculate tangent at #x_0=3# we have to calculate #f'(x_0)#

#f'(3)=(1-3^2)/(3^2+1)=-8/10--0.8#

#f(3)=3/(9+1)=3/10=0.3#

Now we have to find #b# for which #y=-0.8x+b# passes through #(3;0.3)#

#0.3=-0.08*3+b#

#b=0.3+0.24#

#b=0.54#

Answer 1: The tangent is #y=-0.8x+0.54#

To solve the second part we have to find points for which #f'(x)=0#

#(1-x^2)/(x^2+1)^2=0#

The fraction is zero if and only if its numerator is zero.

#1-x^2=0#

#x^2=1#

#x=-1 vv x=1#

Answer 2: Tangent lines of #f(x)# are horizontal for #x=-1# and #x=1#