What is the inverse function of #f(x)=1/(1+x)#. How inverse functions appear in graph?

Question

Charles Arocha · Accepted Answer

Please see below.

We have #y=f(x)=1/(1+x)# i.e. #y(1+x)=1# or #1+x=1/y# or #x=1/y-1=(1-y)/y# Hence inverse of #f(x)=1/(1+x)# is #g(x)=(1-x)/x=1/x-1# and The graph of #f(x)=1/(1+x)# appears as graph{1/(1+x) [-10, 10, -5, 5]} The graph of #g(x)=(1-x)/x# appears as graph{(1+x)/x [-10, 10, -5, 5]} In case of inverse functions, they are reflection of each other in the line #y=x#. As is seen from the graph below graph{(y-x)(y-1/(1+x))(y+1-1/x)=0 [-10, 10, -5, 5]} As #f(x)=1/(1+x)#, #f'(x)=-1/(1+x)^2# and #f'(g(x))=-1/(1+(1-x)/x)^2=-1/(1/x^2)=-x^2# and #g'(x)=-1/x^2# As such #g'(x)=1/(f'(g(x)))#

Luke Alvarez · Answer

See below

#f(x)=1/(1+x)# #f^(-1)(x)# #y=1/(1+x)# #1/y=1+x=>x=1/y-1=(1-y)/y# #:.# #x=(1-y)/y# Substituting #y=x# #y=(1-x)/x#, #color(white)(8888)color(blue)(f^(-1)(x)=(1-x)/x)# Derivative of #g(x)# #g(x)=(1-x)/x# We can express this as #(1-x)x^(-1)=x^(-1)-x*x^(-1)=x^(-1)-1# #d/dx(x^(-1)-1)=-x^(-2)=-1/x^2# #color(blue)(g'(x)=-1/x^2)# To show #g'(x)=1/(f'(g(x))# we need to first find #f'(x)# #1/(1+x)=(1+x)^-1# #d/dx(1+x)^(-1)=-(1+x)^(-2)*1=-1/(1+x)^2# #color(blue)(f'(x)=-1/(1+x)^2)# #f'(g(x))=-1/(1+g(x))^2=-1/(1+(1-x)/x)^2# #->=-1/((x+1-x)/x)^2=-1/(1/x^2)=-x^2# So: #1/(f'(g(x)))=1/(-x^2)=-1/x^2=g'(x)=-1/x^2# This probably looks a bit intimidating, because it includes a lot of simplifying to make the differentiation easier. Hope it helps though.