For #f(x) =arctan(x/3)#, what is the equation of the line tangent to #x =0 #?

Question

2149 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-11T02:59:34

#x-3y=0#

The slope of a tangent at a point of a function is derived by the value of function's derivative at that point.

As #f(x)=arctan(x/3)#, we have

#f'(x)=1/(1+(x/3)^2)xx1/3=1/(3+(x^2/3)#

#f'(0)=1/(3+(0^2/3))=1/3#

Further at #x=0# #f(x)=0#

Hence tangent at #x=0# is

#(y-0)=1/3(x-0)# or #x-3y=0#

Note that #(0,0)# is a point of inflection.
graph{(y-arctan(x/3))(3y-x)=0 [-10, 10, -5, 5]}